Post by: Saladin on May 04, 2010, 12:42:04 pm
Post by: Saladin on May 04, 2010, 12:42:52 pm
Please post your doubts here.
Post by: Khey [Rainbow] on May 04, 2010, 05:56:16 pm
Post by: Saladin on May 04, 2010, 05:58:37 pm
Post by: Khey [Rainbow] on May 04, 2010, 06:04:04 pm
(a) Show the at the start of the year, after the annual deposit has been made, the amount in the account was $1343.29
(b) Find the amount at the start of the12th year, after annual deposit has been made. Give your answer to 2.d.p
Post by: Khey [Rainbow] on May 04, 2010, 06:09:04 pm
(i) AB is perpendicular to BC
(ii) BM = CM, where M is the midpoint of AC
(b) Find also an equation of circle passing through A, B & C
Post by: Saladin on May 04, 2010, 06:11:50 pm
Therefore we can use the
Therefore, we can notice similarities
Also
n is positive, so
Post by: Saladin on May 04, 2010, 06:12:20 pm
Post by: Khey [Rainbow] on May 04, 2010, 06:15:24 pm
Sorry 4 da trouble
By the way where do u live? n which skool do u go to?
Post by: Ghost Of Highbury on May 04, 2010, 06:16:10 pm
(a) For a triangle ABC, show that
(i) AB is perpendicular to BC
(ii) BM = CM, where M is the midpoint of AC
(b) Find also an equation of circle passing through A, B & C
okay, i can help in a)i) part
for AB to be perpendicular to BC, the product of their gradients shud be -1
lets try, 'm' of AB = (9-3)/(5-1) = 6/4 = 3/2
'm' of BC = (7-9)(8-5) = -2/3
3/2 * -2/3 = -1
hence proved.
a)ii) first find the co-ordinates of M i.e ((8+1)/2, (7+3)/2) = (4.5,5)
length of BM =
length of CM=
b) Eqn of a circle = (x-h)^2 + (y-k)^2 = r^2
h,k = co-ordinates of the center of the circle, here of M
h = 4.5
k = 5
r = 4.03
(x-4.5)^2 + (y-5)^2 = 16.25
Hence proved
@Dude - SOrry i cudnt help much, i used my add. math knowledge to solve this for him. U have a lotta Qs to solve.. :)
Post by: Saladin on May 04, 2010, 06:18:32 pm
A bank account pays 11.5% per annum compound interest, Shan deposites $400 in this account at the start of the year.
(a) Show the at the start of the year, after the annual deposit has been made, the amount in the account was $1343.29
(b) Find the amount at the start of the12th year, after annual deposit has been made. Give your answer to 2.d.p
Are you missing a part of the question or anything?
no yikes, i havnt seen any1 wid so many doubts (no offence)
Post by: Khey [Rainbow] on May 04, 2010, 06:20:38 pm
but u say its 90
Post by: Saladin on May 04, 2010, 06:22:31 pm
Q1) isnt 2! times 40 = 80??
but u say its 90
i corrected my post.
Post by: Khey [Rainbow] on May 04, 2010, 06:23:19 pm
nothing else is missing..
n sorry 4 da trouble
Post by: Saladin on May 04, 2010, 06:36:38 pm
A bank account pays 11.5% per annum compound interest, Shan deposites $400 in this account at the start of the year.
(a) Show the at the start of the year, after the annual deposit has been made, the amount in the account was $1343.29
(b) Find the amount at the start of the12th year, after annual deposit has been made. Give your answer to 2.d.p
Please tell me where u got this question from, if it is a ppq, please post it.
(a)
(b)
The answer for b may be wrong, please give me the q from which u got it from.
Post by: MasterMath on May 04, 2010, 06:37:57 pm
so you'll get n= 10 as you said but p=-3
:)
I can help any1 if needed :)
Post by: Khey [Rainbow] on May 04, 2010, 06:40:51 pm
its alryt if u cant solve it
Post by: Ghost Of Highbury on May 04, 2010, 06:42:33 pm
okay, i can help in a)i) part
for AB to be perpendicular to BC, the product of their gradients shud be -1
lets try, 'm' of AB = (9-3)/(5-1) = 6/4 = 3/2
'm' of BC = (7-9)(8-5) = -2/3
3/2 * -2/3 = -1
hence proved.
a)ii) first find the co-ordinates of M i.e ((8+1)/2, (7+3)/2) = (4.5,5)
length of BM ==
=
= 4.03
length of CM==
=
b) Eqn of a circle = (x-h)^2 + (y-k)^2 = r^2
h,k = co-ordinates of the center of the circle, here of M
h = 4.5
k = 5
r = 4.03
(x-4.5)^2 + (y-5)^2 = 16.25
Hence proved
@Dude - SOrry i cudnt help much, i used my add. math knowledge to solve this for him. U have a lotta Qs to solve.. :)
@Khey, MasterMath - Did u check this answer? If any errors, plz tell me, i tried solving it.
Post by: Khey [Rainbow] on May 04, 2010, 06:44:49 pm
Post by: Saladin on May 04, 2010, 06:47:20 pm
Post by: Saladin on May 04, 2010, 06:51:17 pm
yah its right!
can u chek if mine is rite???
Post by: MasterMath on May 04, 2010, 06:54:27 pm
The G.P Sqnce is : P + P(1+r%) + P(1+r%)^2 ...P(1+r%)^n-1
Thats the sqnce for n years ...
We know that the sum for any G.P is a1((1-(r)^n)/1-r) Where r is the common ratio or w.e
so it becomes 400((1-(1+r%)^12)/1-(1+r%))
Giving you 9364.563977
For the triangle question
Part a was explained correctly .. you do the slope product = -1
For part b
You use distance fromula to find distance between BM and MC .. Given that M is the midpt so you use midpt formula which is Ax+Cx /2 same for y .. Gives you that point M is 4.5,5
So the distance between B and M is Radical 65 / 2 .. radical for 65 only
you show you get the same answer for applying it for between BM and MC ..
And now part C ..
They want the circle that passes through all of them .. It makes sense that M should be the center cause BM = MC = AM .. (remember M is the midpt of AC and you just proved BM = MC )
So the formula of the circle becomes : (x-4.5)2+(y-5)2= Part b answer ^2 ( equal to 65/4)
Simply done :D
Post by: Saladin on May 04, 2010, 07:01:49 pm
Post by: MasterMath on May 04, 2010, 07:08:05 pm
Post by: Khey [Rainbow] on May 04, 2010, 07:13:43 pm
nothin else..=D
Post by: MasterMath on May 04, 2010, 07:19:29 pm
Post by: halosh92 on May 06, 2010, 10:31:13 am
-q2
for paper K these are the questions:
q4
q5a
q7
thxx aloot
Post by: halosh92 on May 06, 2010, 07:07:13 pm
The circle C, with centre (a, b) and radius 5, touches the x-axis at (4, 0)
(a) Write down the value of a and the value of b
Post by: T.Q on May 07, 2010, 12:18:45 am
could someone explain this:
The circle C, with centre (a, b) and radius 5, touches the x-axis at (4, 0)
(a) Write down the value of a and the value of b
solve simultaneously
first equ. distance between the two points = radius
second equ. equation of the circle
then substitute
Post by: halosh92 on May 07, 2010, 08:59:10 am
solve simultaneously
first equ. distance between the two points = radius
second equ. equation of the circle
then substitute
could u just type the equations for me
am rili sorry :)
Post by: abuelzouz on May 07, 2010, 09:55:31 am
Post by: T.Q on May 07, 2010, 10:17:31 am
could u just type the equations for me
am rili sorry :)
srry , u dont solve simultaneously
the solution is attached
Post by: halosh92 on May 07, 2010, 11:18:09 am
srry , u dont solve simultaneously
the solution is attached
got it !! thx alot!! ;D ;D ;D ;D
Post by: T.Q on May 07, 2010, 12:51:58 pm
got it !! thx alot!! ;D ;D ;D ;D
u r welcome
in my drawing i made some mistakes b=5 , a=4
Post by: abuelzouz on May 07, 2010, 01:47:14 pm
i still dont get the trapezium rule!
!! ??
Post by: T.Q on May 07, 2010, 01:50:00 pm
!! ??
check this http://www.astarmathsandphysics.com/a_level_maths_notes/C2/a_level_maths_notes_c2_trapezium_rule.html (http://www.astarmathsandphysics.com/a_level_maths_notes/C2/a_level_maths_notes_c2_trapezium_rule.html)
Post by: halosh92 on May 07, 2010, 03:36:58 pm
(a) Sketch C. (2 marks)
(b) Write down the exact coordinates of the points at which C meets the coordinate axes.
(3 marks)
(c) Solve, for x in the interval 0 <x < 2pie,
cos ( x + pie/4) = 0.5,
giving your answers in terms of p.
plzzz plzz could anyone solve this...im having a brain breakdown god dammiiitttttt
Post by: Saladin on May 07, 2010, 03:40:07 pm
Post by: halosh92 on May 07, 2010, 03:53:00 pm
Post by: Saladin on May 07, 2010, 04:07:29 pm
(b) You know the x-axis intercepts. They are
(c)
There are two possible values of
Now you also are told to find all intercepts, so also y axis. Simple make x=0 and calculate the value of y through the equation to get
They are:
And you will get
Hope that helps.
Post by: halosh92 on May 07, 2010, 04:35:06 pm
Ok, I answered question 1. Just in degrees. It is simple thegraph moved 45 degrees back.
(b) You know the x-axis intercepts. They areand
. They are simple the places where the original
those values.
(c)
There are two possible values of
Now you also are told to find all intercepts, so also y axis. Simple make x=0 and calculate the value of y through the equation to get
They are:and
And you will getand
.
Hope that helps.
am rili sorry i got everything except part c
i mean first of all im equating x=0 in the equation but the answer is 0.9999.....
and where did u get the pie/3 from?
Post by: Saladin on May 07, 2010, 06:10:10 pm
am rili sorry i got everything except part c
i mean first of all im equating x=0 in the equation but the answer is 0.9999.....
and where did u get the pie/3 from?
No, X does not equal 0. You have to find out the two values of
Post by: halosh92 on May 07, 2010, 06:20:49 pm
No, X does not equal 0. You have to find out the two values of
ok i found them as pie/3 and 5pie/3 ??
Post by: Saladin on May 07, 2010, 06:29:21 pm
Post by: halosh92 on May 07, 2010, 06:38:29 pm
Here u go.
thankyou soooooooo muchhhhhhhhhhh
ure a life savor and sorry for the trouble :-[
Post by: Summer :] on May 09, 2010, 05:11:58 pm
(a) For a triangle ABC, show that
(i) AB is perpendicular to BC
(ii) BM = CM, where M is the midpoint of AC
(b) Find also an equation of circle passing through A, B & C
(b)
the equation of the circle is (x-h)^2 + (y-b)^2 = r^2
where (h,x) is the midpoint of the circle and r = radius of the circle
substitute the values of A (1,3) in the place of (x,y) in the equation..
like this
(1-h)^2 + (3-b)^2 = r^2
do the same with points B and C
with B : (5-h)^2 +(9-b)^2 = r^2
with C : (8-h)^2 + (7-b)^2 - r^2
the radius of the circle is the same
therefore all the equations above should be equal
(1-h)^2 + (3-b)^2 = (5-h)^2 + (9-b)^2 = (8-h)^2 + (7-b)^2
rearrange them and you'll get one of the values of either h or B
then subsitute this value to get the other values
you'll get the midpoint of the circle
all you need is the radius now
so use one of the points example A which is (1,3)
(1-h)^2 +(3-b)^2 = r^2
to get it since you have the values of the midpoint (h,b) which u would've have obtained previously ::)
Post by: Monica on May 14, 2010, 05:55:03 am
You can make it sticky the day b4 the exam, other than that its been there for 10 days and no one posted.
Post by: Blizz_rb93 on May 14, 2010, 09:03:40 am
trust me ppl will start posting soon :D
Post by: wakemeup on May 14, 2010, 04:45:12 pm
http://www.srepapmaxeeeerf.org/A%20Level/Maths/Edexcel/C1/C1%202008-01.pdf
Q7 (d)
Q10 (a)
Post by: vanibharutham on May 14, 2010, 05:52:14 pm
u have to find x4 to get the sequence
x1 = 1
x2 = -0.5
x3 = 1
x4 = -0.5
see the sequence?
all evens are -0.5
all odds are 1
therefore x2008 = -0.5
Its a repeating sequence
10a)
hmm
easy way to do this:
you know that the curve will cross the x-axis when y = 0
therefore, x = -3 and x = 1
You should also note that it is a cubic function
and that when x = 0 , y = 3
you know have three points
(0,3) (-3,0) and (1,0)
make sure it goes through those points and that it is cubic
here is a sketch:
http://www.wolframalpha.com/input/?i=plot+x%C2%B3+%2B+x%C2%B2+-5x+%2B+3 (http://www.wolframalpha.com/input/?i=plot+x%C2%B3+%2B+x%C2%B2+-5x+%2B+3)
Post by: Monica on May 14, 2010, 09:09:01 pm
No it's good to be sticky before 10 days! Cause people don't study a day before the exam so once they have questions they start posting!! I'm preparing some questions for C1 but please don't close the forum or move it or smthn :p
trust me ppl will start posting soon :D
Well there are many other subjects too, we can't make all subjects sticky, can we?
Members must learn how to search a bit more. =]
Post by: Blizz_rb93 on May 14, 2010, 10:03:11 pm
Well there are many other subjects too, we can't make all subjects sticky, can we?Well everyone is trying to study all the subjects preparing for everything also :P so we need to give everyone a chance and when it's a sticky it's more easier than people posting threads on their own on specific topics, just go to the C1 or C2 or whatever Sticky threat and post your questions on it, much easier =]
Members must learn how to search a bit more. =]
Post by: Blizz_rb93 on May 14, 2010, 10:35:29 pm
Now sometimes they ask u to find the square root or cubic root of a big number
what's an easy way to finding it out??
like for example they'd give us a large number say 8096 and they want the cubic root of it or the square root
how would we find it?
Post by: Meticulous on May 14, 2010, 10:56:42 pm
Post by: Blizz_rb93 on May 15, 2010, 08:24:29 am
Trial and error. I did that and ended up with a 100.My point was for any big number and not that number in specific :p
like i dont get how is a fast way to calculate a sq. or cubic root for a large value??
Post by: vanibharutham on May 15, 2010, 11:22:36 am
they wont give u something so hard in the exam :)
Post by: Blizz_rb93 on May 15, 2010, 11:47:20 am
Post by: Blizz_rb93 on May 15, 2010, 12:27:41 pm
I got this question regarding Trigonometry equation solving..
Here's the question:
Solve for 0 < x < 180, the equation:
a) Sin(x+10) = (Square root of 3) / 2
b) cos2x = -0.9, giving your answers to 1 decimal place
Thing is i dont get how do u find questions when they got parts like that where they want sin(x+10 degrees ) :s
Post by: Saladin on May 15, 2010, 12:29:46 pm
Hi
I got this question regarding Trigonometry equation solving..
Here's the question:
Solve for 0 < x < 180, the equation:
a) Sin(x+10) = (Square root of 3) / 2
b) cos2x = -0.9, giving your answers to 1 decimal place
Thing is i dont get how do u find questions when they got parts like that where they want sin(x+10 degrees ) :s
Which paper did you take them from. That way, I can check my answers.
Post by: immortal on May 15, 2010, 03:04:29 pm
Hia) sin(x+10)=(Square root of 3) / 2
I got this question regarding Trigonometry equation solving..
Here's the question:
Solve for 0 < x < 180, the equation:
a) Sin(x+10) = (Square root of 3) / 2
b) cos2x = -0.9, giving your answers to 1 decimal place
Thing is i dont get how do u find questions when they got parts like that where they want sin(x+10 degrees ) :s
(x+10)=sin-1(Square root of 3) / 2
(x+10)=60
x=50
2nd quadrant (x+10)=120
x=110
b)cos2x = -0.9
2x=cos-10.9
2ndquad 2x=180-25.84 3rdquad 2x=180+25.84
x=77.1 x=102.9
Post by: Blizz_rb93 on May 15, 2010, 06:42:32 pm
Post by: astarmathsandphysics on May 15, 2010, 06:50:56 pm
Post by: wakemeup on May 15, 2010, 07:05:26 pm
10a)
hmm
easy way to do this:
you know that the curve will cross the x-axis when y = 0
therefore, x = -3 and x = 1
You should also note that it is a cubic function
and that when x = 0 , y = 3
you know have three points
(0,3) (-3,0) and (1,0)
make sure it goes through those points and that it is cubic
here is a sketch:
http://www.wolframalpha.com/input/?i=plot+x%C2%B3+%2B+x%C2%B2+-5x+%2B+3 (http://www.wolframalpha.com/input/?i=plot+x%C2%B3+%2B+x%C2%B2+-5x+%2B+3)
Ok, I got 7(d) but how exactly do you know whether the curve in 10(a) goes up then down or down then up.
Post by: astarmathsandphysics on May 15, 2010, 07:16:20 pm
Post by: halosh92 on May 20, 2010, 02:18:51 pm
thx :)
Post by: Saladin on May 20, 2010, 06:43:18 pm
Q2
thx :)
I am sure you can figure it out now.
Post by: halosh92 on May 20, 2010, 06:52:09 pm
I am sure you can figure it out now.
there is my doubt exactly ;D
why did u subtract "9" from 30 ???
thx alot ;)
Post by: Saladin on May 20, 2010, 06:53:54 pm
there is my doubt exactly ;D
why did u subtract "9" from 30 ???
thx alot ;)
It is the sum of terms 10 to 30. Inclusive of the 10th term.
Thus, you have to subtract the sum of the first 9 terms to get the answer.
I hope that answers your question.
Post by: halosh92 on May 20, 2010, 07:17:31 pm
It is the sum of terms 10 to 30. Inclusive of the 10th term.
Thus, you have to subtract the sum of the first 9 terms to get the answer.
I hope that answers your question.
so we always exclude!
thx a bunch! ;)
Post by: Blizz_rb93 on May 22, 2010, 09:13:28 pm
Thank you.
Post by: astarmathsandphysics on May 22, 2010, 10:22:10 pm
Post by: Blizz_rb93 on May 22, 2010, 10:41:12 pm
jan 2010 q4 and 6Thank you very much!!
Post by: 7ooD on May 23, 2010, 01:17:19 am
Post by: zabady on May 23, 2010, 01:21:10 am
Post by: Saladin on May 23, 2010, 05:54:41 am
i have a doubt at jan 2008 6 c
Because what you have to do is simple give a function, that will map the current graph to have the maximum point in the y-axis.
This means moving the graph 2 units horizontally to the rite.
Thus, the graph is
Post by: Mr.Lonely on May 23, 2010, 07:40:15 am
Post by: astarmathsandphysics on May 23, 2010, 09:13:21 am
Post by: Blizz_rb93 on May 23, 2010, 01:42:45 pm
GUYS I NEED C1 2010 JAN MARK SCHEME ASAP ... PLZ HELP ME :)
Post by: coatseville on May 23, 2010, 01:57:23 pm
Thankyou :)
Post by: wakemeup on May 23, 2010, 02:27:15 pm
Post by: coatseville on May 23, 2010, 02:37:03 pm
Post by: Blizz_rb93 on May 23, 2010, 04:44:29 pm
how do u find this? o.o
Post by: Saladin on May 23, 2010, 04:56:32 pm
(3 [surd 7])^2
how do u find this? o.o
That basically means
So it basically means
Post by: Blizz_rb93 on May 23, 2010, 05:02:45 pm
That basically means
So it basically means
ohh okay lol, that's easy!!
thanks :D
Post by: raimunts on May 23, 2010, 05:38:23 pm
Post by: astarmathsandphysics on May 23, 2010, 05:50:51 pm
Any sigma notation question in C1 uses a standard formula I think so just use that
Post by: zabady on May 23, 2010, 06:10:06 pm
Post by: astarmathsandphysics on May 23, 2010, 06:43:15 pm
Post by: Saladin on May 23, 2010, 06:45:56 pm
is this edexcel c1
yes.
Post by: zabady on May 23, 2010, 07:20:48 pm
Post by: Saladin on May 23, 2010, 07:27:02 pm
plz someone answer my question upstairs the exam is tommorow
on it now.
Post by: Saladin on May 23, 2010, 07:42:08 pm
Find the area of the trapezium.
Then take the remaining triangle away.
Then
Post by: SGVaibhav on May 23, 2010, 08:01:15 pm
You have to use the t****zium rule.
Find the area of the t****zium.
Then take the remaining triangle away.
Then
hahaha loool its censored auto :P
Post by: Saladin on May 23, 2010, 08:16:29 pm
i know!!!
Post by: zabady on May 23, 2010, 08:56:14 pm
Post by: Blizz_rb93 on May 23, 2010, 09:10:59 pm
Post by: raimunts on May 23, 2010, 09:17:41 pm
What's the formula for finding the length of a line on a graph??
You mean when co-ordinates are given for line AB where A is (x1,y1) B is (x2,y2)
Then length of line = ?[(x1-x2)²+(y1-y2)²]
I hope this helps.
Post by: raimunts on May 23, 2010, 09:21:51 pm
You mean when co-ordinates are given for line AB where A is (x1,y1) B is (x2,y2)
Then length of line = ?[(x1-x2)²+(y1-y2)²]
I hope this helps.
Sorry there...instead of the question mark (?) it should be square root
Post by: Blizz_rb93 on May 23, 2010, 09:34:36 pm
thanks anyways though =]
Post by: coatseville on May 24, 2010, 11:34:59 am
Could you please explain how to do part b?
Thankyou SO MUCH :)
Post by: coatseville on May 24, 2010, 03:28:02 pm
How did you find it like just say easy or hard because we cant discuss until 24 hours has passed!
Post by: Ukhti-R on May 24, 2010, 05:19:21 pm
Nvm i figured it out anyways...
How did you find it like just say easy or hard because we cant discuss until 24 hours has passed!
Even things as small as that cannot be discussed.
Please refrain from discussing the exam paper until 24 hours have passed.
Thank you :)
Post by: Blizz_rb93 on May 24, 2010, 06:16:54 pm
Even things as small as that cannot be discussed.we cant say if it was hard or easy?? :/
Please refrain from discussing the exam paper until 24 hours have passed.
Thank you :)
Post by: Ukhti-R on May 24, 2010, 06:36:33 pm
we cant say if it was hard or easy?? :/
Nahh mate. :)
Post by: T.Q on May 25, 2010, 09:34:40 am
Post by: astarmathsandphysics on May 25, 2010, 11:34:18 am
Post by: halosh92 on May 25, 2010, 01:03:04 pm
Post by: coatseville on May 25, 2010, 03:06:00 pm
Yah i think i did well also :)
But what do you think the Grade threshold will be?
I think it mite be lower than other years... like maybe 60/75 or so what do you think?
Post by: Ukhti-R on May 25, 2010, 09:09:33 pm
Post by: halosh92 on May 26, 2010, 03:47:12 pm
anyone has 2009 june ms !!!!!!!!!!! rili urgent!
Post by: Saladin on May 26, 2010, 05:11:54 pm
hey!
anyone has 2009 june ms !!!!!!!!!!! rili urgent!
here
Post by: halosh92 on May 26, 2010, 05:43:36 pm
here
THANKU ;)
Post by: halosh92 on May 26, 2010, 08:57:34 pm
Q9a
plzzzz help and explain in details.
anyone? :)
Post by: Saladin on May 27, 2010, 04:36:39 pm
edexcel 2009 june
Q9a
plzzzz help and explain in details.
anyone? :)
on it now.
Post by: Saladin on May 27, 2010, 04:52:25 pm
Therefore, the volume is equal to,
Now you need to find what makes the surface area.
2 sectors + 2 rectangles + curved face
So,
I hope that explains everything.
Post by: halosh92 on May 27, 2010, 05:25:17 pm
We need to findin terms of r.
Therefore, the volume is equal to,
Now you need to find what makes the surface area.
2 sectors + 2 rectangles + curved face
So,
I hope that explains everything.
gr8!!!!!!!!!!!!!! THANKYOU ;)
Post by: Saladin on May 27, 2010, 05:31:22 pm
gr8!!!!!!!!!!!!!! THANKYOU ;)
Engraved signing off. 8)
Post by: halosh92 on May 28, 2010, 08:14:16 pm
Q9a
someone plzz explain why were not starting fomr zero :s
Post by: Saladin on May 28, 2010, 08:15:52 pm
may/june 2007 C2 edexcel
Q9a
someone plzz explain why were not starting fomr zero :s
Your question does not make sense.
Post by: halosh92 on May 28, 2010, 08:26:04 pm
Your question does not make sense.
if u look at the marking scheme the curve is staring from 0.5 not from 0
Post by: Saladin on May 28, 2010, 09:04:05 pm
if u look at the marking scheme the curve is staring from 0.5 not from 0
exactly. I didn't get it either....
Post by: Darkstar3000 on May 30, 2010, 10:19:30 am
F'(x)=x^2 - 2 + 1/x^2
a) Find the value of f''(x) at x=4
b) Given that f(3) = 0, find f(x)
c) Prove that f is an increasing function
my problem is part c i can do the first two, this is from Mixed exercise 9D Q3 in the c2 text book
Post by: Saladin on May 30, 2010, 10:29:53 am
the function f, defined for x is an element of R, x>0, is such that
F'(x)=x^2 - 2 + 1/x^2
a) Find the value of f''(x) at x=4
b) Given that f(3) = 0, find f(x)
c) Prove that f is an increasing function
my problem is part c i can do the first two, this is from Mixed exercise 9D Q3 in the c2 text book
On the case!
Enjoy.
Post by: Darkstar3000 on May 30, 2010, 12:52:58 pm
Post by: Saladin on May 30, 2010, 12:59:26 pm
thx but where is (x - 1/x)^2 coming from
That is the factorized version of f(x).
Post by: Darkstar3000 on May 30, 2010, 01:03:43 pm
That is the factorized version of f(x).
thank you
Post by: Saladin on May 30, 2010, 01:16:34 pm
thank you
Here to help.
Post by: halosh92 on May 30, 2010, 04:16:19 pm
hey
june 2005
Q10b
i got the gradient but i cant seem to do whats next , could someone do it ?
thx :D
Post by: cooldude on May 30, 2010, 04:51:51 pm
edexcel
hey
june 2005
Q10b
i got the gradient but i cant seem to do whats next , could someone do it ?
thx :D
y'=2-(16/x^3)
the minimum point is-->
2-(16/x^3)=0
8=x^3
x=2
we can show it is minimum by finding the second derivate-->
y''=48/x^2
f''(2)=12
therefore since the second derivate is positive it is a minima
therefore x>2
Post by: halosh92 on May 30, 2010, 05:41:04 pm
y'=2-(16/x^3)
the minimum point is-->
2-(16/x^3)=0
8=x^3
x=2
we can show it is minimum by finding the second derivate-->
y''=48/x^2
f''(2)=12
therefore since the second derivate is positive it is a minima
therefore x>2
thxxx :)
Post by: halosh92 on May 31, 2010, 10:22:35 am
Q10 c
Post by: cooldude on May 31, 2010, 10:29:09 am
jan 2005 edexcel
Q10 c
erm there isnt ne 10th q for the paper uve specified ???
Post by: Saladin on May 31, 2010, 10:29:26 am
jan 2005 edexcel
Q10 c
R u sure?
There is no question 10.
Post by: halosh92 on May 31, 2010, 10:38:34 am
R u sure?
There is no question 10.
oh am so sorry ;D
i meant question 9c ;D
Post by: Saladin on May 31, 2010, 10:41:59 am
oh am so sorry ;D
i meant question 9c ;D
Find the stationary point at which
This is the maximum point. Now simply equate the x value found into the equation.
Post by: cooldude on May 31, 2010, 10:47:18 am
oh am so sorry ;D
i meant question 9c ;D
A=80x-2x^2-pix^2/2
dA/dx=80-4x-pix
d^2A/dx^2=-4-pi
and therefore since (-4-pi) this is negative, thus the value in part b is a maxima
Post by: halosh92 on May 31, 2010, 10:57:11 am
this question also:
Q3
Post by: Saladin on May 31, 2010, 01:43:05 pm
thx everyone
this question also:
Q3
2 hours.
Post by: Saladin on May 31, 2010, 02:28:09 pm
Post by: Saladin on May 31, 2010, 02:57:14 pm
First you need to find the angle of the sector of the circle. The angle is 60o because the semi-circle is divided into three equal parts as two other quarter circles combine to for two separate circles. Its like this, you need to view the other two quartiles as triangles. Thus the sector of B and C is one of the three. And half a circle is 180o. Thus dividing it by three will give you
So, now you need to understand that the gap is formed when the two segments of the circle is taken away from sector.
So you simply do this.
Find the area of the sector!
Now the area of the section!
Now the area of the shaded region is the area of [Area of section - 2 X Area of Sector].
This will lead to the image attached.
Post by: halosh92 on May 31, 2010, 03:02:44 pm
Alrite here it comes.
First you neejavascript:void(0);d to find the angle of the sector of the circle. The angle is 60o because the semi-circle is divided into three equal parts as two other quarter circles combine to for two separate circles. Its like this, you need to view the other two quartiles as triangles. Thus the sector of B and C is one of the three. And half a circle is 180o. Thus dividing it by three will give you
So, now you need to understand that the gap is formed when the two segments of the circle is taken away from sector.
So you simply do this.
Find the area of the sector!
Now the area of the section!
Now the area of the shaded region is the area of [Area of section - 2 X Area of Sector].
This will lead to the image attached.
THX ALOT DUDE gr8 :)
Post by: halosh92 on June 02, 2010, 08:56:18 pm
i got the cordinates but didnt get this one
(0, 1/sqrt2)
Post by: saadmannan on June 03, 2010, 01:30:17 am
i have doubt in Jan 2005 Q9a i am not able to solve the equation
the sign in the question that is given is positive while im getting negative
i checked the mark scheme if anything was wrong but my method was exactly same as in MS but the answer is different
i have posted the question below
Post by: Saladin on June 03, 2010, 06:09:42 am
hi
i have doubt in Jan 2005 Q9a i am not able to solve the equation
the sign in the question that is given is positive while im getting negative
i checked the mark scheme if anything was wrong but my method was exactly same as in MS but the answer is different
i have posted the question below
In 2 hours.
Post by: cooldude on June 03, 2010, 08:04:11 am
Q5b
i got the cordinates but didnt get this one
(0, 1/sqrt2)
y=cosx+cos(pi/4)
y=cosx+(1/root 2)
x=0
y=1+(1/root 2) (as cos0=1)
therefore intersection with the y-axis-->
(0,1+(1/root 2))
Post by: cooldude on June 03, 2010, 08:13:55 am
hi
i have doubt in Jan 2005 Q9a i am not able to solve the equation
the sign in the question that is given is positive while im getting negative
i checked the mark scheme if anything was wrong but my method was exactly same as in MS but the answer is different
i have posted the question below
2x+2y+(pi*x)=80
2y=80-pi*x-2x
ar=2xy+((pi*(x^2))/2)
ar=x(80-pi*x-2x)+((pi*(x^2))/2)
ar=80x-(pi*(x^2))-2(x^2)+((pi*(x^2))/2)
ar=80x-((pi*(x^2))/2)-2(x^2)
ar=80x-(x^2)(pi/2+2)
hence proved
Post by: halosh92 on June 03, 2010, 09:09:06 am
y=cosx+cos(pi/4)
y=cosx+(1/root 2)
x=0
y=1+(1/root 2) (as cos0=1)
therefore intersection with the y-axis-->
(0,1+(1/root 2))
for the second step
y=cosx+(1/root 2)
am getting the 1/root 2 as 0.707 as decimals???? how did u get it as fraction form?
otherwise i get the rest
thx :D
Post by: cooldude on June 03, 2010, 09:19:30 am
for the second step
y=cosx+(1/root 2)
am getting the 1/root 2 as 0.707 as decimals???? how did u get it as fraction form?
otherwise i get the rest
thx :D
yeah its the same thing 0.707 is the decimal form of 1/root 2, its actually a known fact the values of sin, cos, tan of 0,30,45,60,90 are all given in our cie book so i guess thats how i knw it, ill just give u a list of the values so u can learn them for future qs-->
sin0=0, sin30=1/2, sin45=1/root 2, sin60=(root 3)/2, sin 90=1, cos0=1, cos 30=(root 3)/2, cos 45=1/root 2, cos 60=1/2, cos 90=0, tan0=0, tan 30=1/root 3, tan45=1, tan 60=root 3, tan 90=1/0 (undefined)
Post by: Sue T on June 03, 2010, 09:40:00 am
the formula isnt in th data booklet so do we need 2 learn all those volume n area formulas 4 all th shapes????
Post by: studz on June 03, 2010, 09:52:19 am
in jan 09 we had to use the formula for surface area of a cylinder 2 slove ques 10 a
the formula isnt in th data booklet so do we need 2 learn all those volume n area formulas 4 all th shapes????
yeah Sue we do..but not for all the shapes...m assuming only for cylinder sphere.. and some times for a cuboid too..but yeah thats it i guess..
Post by: wakemeup on June 03, 2010, 05:01:58 pm
http://www.xtremepapers.net/Edexcel/Mathematics/2008%20Jan/6664_01_que_20080109.pdf
I get some of the answers but how exactly do I get all 6? I'm still kinda confused in trigonometric functions. So if anyone could give some notes about how to find other angles after finding the first one through the equation, that would be really helpful.
Thanks
Post by: Ghost Of Highbury on June 03, 2010, 05:10:15 pm
Could someone please explain how to do 4b in this paperi can get 3 values..cud u provide the ms link?
http://www.xtremepapers.net/Edexcel/Mathematics/2008%20Jan/6664_01_que_20080109.pdf
I get some of the answers but how exactly do I get all 6? I'm still kinda confused in trigonometric functions. So if anyone could give some notes about how to find other angles after finding the first one through the equation, that would be really helpful.
Thanks
Post by: wakemeup on June 03, 2010, 05:25:49 pm
http://www.xtremepapers.net/Edexcel/Mathematics/2008%20Jan/6664_01_rms_20080306.pdf
Post by: wakemeup on June 03, 2010, 05:31:06 pm
http://www.xtremepapers.net/Edexcel/Mathematics/2008%20Jun/6664-01%20Core%20Mathematics%20C2.pdf
MS:
http://www.xtremepapers.net/Edexcel/Mathematics/2008%20Jun/6664_01_msc_20080612.pdf
Post by: Ghost Of Highbury on June 03, 2010, 05:34:43 pm
sin x = +/- sqrt(0.6) = =/- 0.7745
x = sin(-1) 0.7745 = 50.7
180-50.7 = 129.3
x = sin(-1) -0.7745 = -50.7
180 - (-50.7 ) = 230.7
360 - 50.7 = 309.7
answers : 50.7, 129.3, 230.7, 309.7
Post by: wakemeup on June 03, 2010, 05:41:25 pm
sin^2 x = 0.6
sin x = +/- sqrt(0.6) = =/- 0.7745
x = sin(-1) 0.7745 = 50.7
180-50.7 = 129.3
x = sin(-1) -0.7745 = -50.7
180 - (-50.7 ) = 230.7
360 - 50.7 = 309.7
answers : 50.7, 129.3, 230.7, 309.7
Thanks. Could you please do the other one too?
Post by: Ghost Of Highbury on June 03, 2010, 05:43:48 pm
Post by: Ghost Of Highbury on June 03, 2010, 05:48:00 pm
y - 3x
cos y = -0.5
y = cos-1 -0.5 = 120
360+120 = 480. 480/3 = 160
360 - 120 = 240 . 240/3 = 80
x = 80
x = 160
3x = 120
x = 40
x = 360-40 = 320
idk how they got 200 and 280 :-/
Post by: Majdovic14 on June 05, 2010, 11:48:15 am
1)cuboid
2)Prism
3)cylinder
4)semi circle
5)Rectangular box
Post by: Ghost Of Highbury on June 05, 2010, 12:09:41 pm
Volume - Lenght*Breadth*Height
Surface area = 2(lb + bh + hl)
2) Prism
Volume = Area of floor*height
Surface area = 2 * area of base + perimeter of base * H
3) Cylinder - Pir^2h (Vol.)
Surface area --> 2pir^2 + 2pirh
4) Semi-circle - > Area - (pi*r^2)/2
5) Box - if u mean cuboid? answered, if u mean cube - VOlume - l*B*H
Surface area = 4(LB)
Post by: Ibiza279 on June 05, 2010, 09:41:54 pm
Post by: astarmathsandphysics on June 05, 2010, 10:00:32 pm
Post by: astarmathsandphysics on June 06, 2010, 07:24:46 am
y coordinate of A
(x-2)^2+(y+1)^1=169/4
(-4-2)^2+(y+1)^1=169/4
36+(y+1)^1=169/4
(y+1)^2=6.25 so y=+-sqrt(6.25)-1=-3.5 or 1.5 but y<0 so -3.5
the x coordinates of A and B are 2-12/2=6=-4 aand 2+12/2=8
y coordinate of B
(x-2)^2+(y+1)^1=169/4
(8-2)^2+(y+1)^1=169/4 so 36+(y+1)^1=169/4
(y+1)^2=6.25 so y=+-sqrt(6.25)-1=-3.5 or 1.5 but y<0 so -3.5
A=(-4,-3.5) B=(8,-3.5)
Post by: Ibiza279 on June 06, 2010, 10:56:01 am
Post by: astarmathsandphysics on June 06, 2010, 12:18:38 pm
Post by: Saladin on June 06, 2010, 12:24:52 pm
im havin difficulty in math c2 jan 2010 question 8 part c
In 30 mins.
Post by: Saladin on June 06, 2010, 12:46:26 pm
Find the area underneath the curve between (4,0) and (5,0)
Because this area in combination with the region are will make a triangle.
The area of the triangle is
So, now simply take away the excess region from the triangle.
Post by: Ibiza279 on June 06, 2010, 05:16:24 pm
January 2010 Question 7 (d)
Find the area underneath the curve between (4,0) and (5,0)
Because this area in combination with the region are will make a triangle.
The area of the triangle is
So, now simply take away the excess region from the triangle.
aaah now i get it, Thanks alot dude ! :)
Post by: halosh92 on June 07, 2010, 03:51:55 pm
if cos thita= -2/3 and cos thita = 1
in the first answer we get 132 and 228
but for the seocnd part we got 0
dont we also get 90 and 270 ?
Post by: Saladin on June 07, 2010, 04:00:56 pm
in trignometrical identity.
if cos thita= -2/3 and cos thita = 1
in the first answer we get 132 and 228
but for the seocnd part we got 0
dont we also get 90 and 270 ?
Please post the exact question. And post as many as you can today, because for today, I am focused on C2.
Post by: Alpha on June 07, 2010, 04:05:47 pm
in trignometrical identity.
if cos thita= -2/3 and cos thita = 1
in the first answer we get 132 and 228
but for the seocnd part we got 0
dont we also get 90 and 270 ?
cos theta = 1 is 0 and 360, if 360 is inclusive.
90 and 270 are values for cos theta = 0.
You are confusing with sin theta = 1 and -1.
Post by: Saladin on June 07, 2010, 05:17:47 pm
cos theta = 1 is 0 and 360, if 360 is inclusive.
90 and 270 are values for cos theta = 0.
You are confusing with sin theta = 1 and -1.
Its good to see you helping Alpha, very much appreciated.
Post by: Alpha on June 07, 2010, 05:22:32 pm
Its good to see you helping Alpha, very much appreciated.
Thank you. :)
Love trigonometry.
I'm just wondering...
Stupid question: what is C2?
Heard about M1, M2, S1, S2... Never C2 ???
Post by: halosh92 on June 07, 2010, 05:24:22 pm
one more question
Q7b
Post by: Saladin on June 07, 2010, 05:24:47 pm
Thank you. :)
I'm just wondering...
Stupid question: what is C2?
Heard about M1, M2, S1, S2... Never C2 ???
It essentially means, Core Mathematics 2. It is a part of the AS Edexcel Mathematics course. You have to do C1 & C2 and another Application Unit such as Mechanics, Statistics or Decision mathematics.
What Qual are you doing? International Baccalaureate perhaps?
Post by: Saladin on June 07, 2010, 05:26:01 pm
ok thx alpha.
one more question
Q7b
On it now.
Post by: Alpha on June 07, 2010, 05:27:31 pm
It essentially means, Core Mathematics 2. It is a part of the AS Edexcel Mathematics course. You have to do C1 & C2 and another Application Unit such as Mechanics, Statistics or Decision mathematics.
What Qual are you doing? International Baccalaureate perhaps?
Ah simply Core... ;D
I'm doing Core and Statistics, A Level. But we always refer to it as main papers here. Thank you. ;D
Halosh, I can't open the file?
Post by: Alpha on June 07, 2010, 05:42:16 pm
ok thx alpha.
one more question
Q7b
V = 4x(x^2 – 45x + 500).
V = 4x(x - 25)(x - 20).
Volume cannot be negative; x > 25.
Post by: halosh92 on June 07, 2010, 05:44:40 pm
its paper A7
Post by: Saladin on June 07, 2010, 05:46:15 pm
ok thx alpha.
one more question
Q7b
The range of the values is
Post by: Alpha on June 07, 2010, 05:49:35 pm
Ya, I got you. Inequalities.
Post by: halosh92 on June 07, 2010, 05:55:02 pm
V > 0.
Ya, I got you. Inequalities.
but the ms says:
0<x<20
Post by: Alpha on June 07, 2010, 06:17:21 pm
but the ms says:
0<x<20
Okay, I guess Dude eliminated the 4x.
4x(x - 25)(x - 20) > 0
3 critical values:
x = 0, x = 20, x = 25.
V = 4x(x - 25)(x - 20)
As x tends to infinity; V tends to infinity.
As x tends to -ve infinity; V tends to -ve infinity.
See the curve in the attachment.
V > 0; upper part lying above the line.
Therefore, 0 < x < 20, and x > 25.
Now, why is x > 25 eliminated...
Post by: Alpha on June 07, 2010, 06:19:56 pm
Post by: halosh92 on June 07, 2010, 06:25:31 pm
Okay, I guess Dude eliminated the 4x.
4x(x - 25)(x - 20) > 0
3 critical values:
x = 0, x = 20, x = 25.
V = 4x(x - 25)(x - 20)
As x tends to infinity; V tends to infinity.
As x tends to -ve infinity; V tends to -ve infinity.
See the curve in the attachment.
V > 0; upper part lying above the line.
Therefore, 0 < x < 20, and x > 25.
Now, why is x > 25 eliminated...
ure amazing thankyou so much :D
maybe the ms is wrong those are just practice papers.. ;)
and thx to engraved also :D
Post by: Alpha on June 07, 2010, 06:26:29 pm
I'm getting tired today...
Length is 50. x is being cut from both sides. Logically, x cannot be greater than 25: you cannot cut longer than the length. ;)
Post by: Alpha on June 07, 2010, 06:28:01 pm
ure amazing thankyou so much :D
maybe the ms is wrong those are just practice papers.. ;)
and thx to engraved also :D
Welcome that much. ;)
Thank you Dude too, much appreciated. He helps a lot here.
Nah, MS is right. ;)
Post by: Saladin on June 07, 2010, 06:32:28 pm
Welcome that much. ;)
Thank you Dude too, much appreciated. He helps a lot here.
Nah, MS is right. ;)
Thank you for your appreciaton, which Qual do you do, Edexcel or CIE?
Post by: Alpha on June 07, 2010, 06:35:09 pm
How about you?
Post by: halosh92 on June 07, 2010, 07:41:50 pm
paper A6
Q7
thx
Post by: Saladin on June 07, 2010, 07:44:24 pm
http://freeexampapers.com/Dndex.php?d=QSBMZXZlbC9NYXRocy9FZGV4Y2VsL1Jlc291cmNlcw==
paper A6
Q7
thx
[/quote
On it now.
Post by: Ghost Of Highbury on June 07, 2010, 07:51:04 pm
(b) can be done by lookin at the graph itself..check for the highest and lowest points, and rite em down,
(c) f(x) = 2.5 means
y = 2.5
5sin 3x = 2.5
sin 3x = 1/2 ..say (a = 3x)
sin a = 0.5
a = (sin)-1 0.5 = 30
a = 3x = 30
x = 10
3x = 180-30 = 150
x = 50
3x = 360+30 = 390
x = 130
---b = 3x-----
b = 150
b= 360+150 = 510
3x = 510
x = 170
Post by: halosh92 on June 07, 2010, 07:58:39 pm
U just gotta plot the graph correctly.
(b) can be done by lookin at the graph itself..check for the highest and lowest points, and rite em down,
(c) f(x) = 2.5 means
y = 2.5
5sin 3x = 2.5
sin 3x = 1/2 ..say (a = 3x)
sin a = 0.5
a = (sin)-1 0.5 = 30
a = 3x = 30
x = 10
3x = 180-30 = 150
x = 50
3x = 360+30 = 390
x = 130
---b = 3x-----
b = 150
b= 360+150 = 510
3x = 510
x = 170
thankyou very much :D
Post by: halosh92 on June 07, 2010, 08:32:11 pm
Post by: Saladin on June 07, 2010, 08:43:36 pm
same paper Q8iia
On it now.
Post by: Saladin on June 07, 2010, 08:54:57 pm
Post by: halosh92 on June 07, 2010, 10:16:32 pm
just want to check
Post by: astarmathsandphysics on June 07, 2010, 10:21:38 pm
Post by: MasterMath on June 07, 2010, 10:26:43 pm
Post by: Alpha on June 08, 2010, 01:15:22 am
ok thx alpha.
one more question
Q7b
For this one, it's a question of 1 mark only... So, you won't have to do all that long method.
Besides, it's asking you to state the answer.
Simple logic again, look at the width of the rectangle. Its 40 cm. So maximum of x cannot be more than half of the width, 0 < x < 20.
Post by: halosh92 on June 08, 2010, 06:28:54 am
2x^3+x^(1/2)+1+2x^(-1) differentiates to 6x^2+1/2x^(-1/2)-2x^(-2)thanx
Post by: Saladin on June 08, 2010, 06:30:08 am
thanx
Any more?
Post by: halosh92 on June 08, 2010, 06:51:08 am
Any more?
am doing pastpapers now if i have more ill post
Post by: halosh92 on June 08, 2010, 07:00:11 am
Q4a
Post by: MasterMath on June 08, 2010, 07:02:44 am
practice paper A5Link it here please ? and ill answer you
Q4a
Post by: halosh92 on June 08, 2010, 07:05:37 am
Link it here please ? and ill answer you
http://www.freeexampapers.com/Dndex.php?d=QSBMZXZlbC9NYXRocy9FZGV4Y2VsL0MyL1ByYWN0aWNlIFBhcGVycw==
and not only part a the whole question ;D
Post by: MasterMath on June 08, 2010, 07:08:30 am
http://www.freeexampapers.com/Dndex.php?d=QSBMZXZlbC9NYXRocy9FZGV4Y2VsL0MyL1ByYWN0aWNlIFBhcGVycw==Okay give me a sec opening it :D
and not only part a the whole question ;D
Post by: MasterMath on June 08, 2010, 07:09:18 am
http://www.freeexampapers.com/Dndex.php?d=QSBMZXZlbC9NYXRocy9FZGV4Y2VsL0MyL1ByYWN0aWNlIFBhcGVycw==Doesnt work for some reason :S .. can you copy paste the question , sorry for that =/!
and not only part a the whole question ;D
Post by: Saladin on June 08, 2010, 07:11:45 am
Post by: halosh92 on June 08, 2010, 07:13:18 am
Doesnt work for some reason :S .. can you copy paste the question , sorry for that =/!
sure np
(a) Write down formulae for sin (A + B) and sin (A - B).
Using X = A + B and Y = A- B, prove that
Sin X + sin Y = 2 sin X+Y/2 cos X-Y/2
b) hence solve for 0<=x<360
sin 4 thita + sin 2 thita = 0
Post by: MasterMath on June 08, 2010, 07:18:59 am
Woking on it now.It opened but why is the syllabus number 6663 ? mines 6664 written on the paper >_<
Post by: halosh92 on June 08, 2010, 07:22:49 am
It opened but why is the syllabus number 6663 ? mines 6664 written on the paper >_<
those are practice papers their not pastpapers for edexcel
Post by: MasterMath on June 08, 2010, 07:30:32 am
those are practice papers their not pastpapers for edexcelOkay you expand Sin(X) which is equal to Sin(A+B)= SinACosB + SinBCosA and expand Sin(Y) which is equal to Sin(A-B)=SinACosB -SinBCosA .. then add them cause they said Sin(X) + Sin (Y) which gives 2SinACosB .. then im stuck >_<
Post by: MasterMath on June 08, 2010, 07:31:48 am
Okay you expand Sin(X) which is equal to Sin(A+B)= SinACosB + SinBCosA and expand Sin(Y) which is equal to Sin(A-B)=SinACosB -SinBCosA .. then add them cause they said Sin(X) + Sin (Y) which gives 2SinACosB .. then im stuck >_<oh i got it ! .. X+Y / 2 .. is A .. check it from the equation .. and X-Y/2 is B ! .. so just replace and here you go!
Post by: MasterMath on June 08, 2010, 07:32:29 am
oh i got it ! .. X+Y / 2 .. is A .. check it from the equation .. and X-Y/2 is B ! .. so just replace and here you go!Cause X = A+B .. and Y= A-B .. so X + Y = 2A .. but we want it A so divide by 2.. same for X-Y :D
Post by: MasterMath on June 08, 2010, 07:34:51 am
Post by: halosh92 on June 08, 2010, 07:41:35 am
And for b) .. i dont see any variable in the equation O_O . what are we supposed to solve for >_<
for the a) part where do they get the cos from ??? :S
and part b) the zero is the thita its a typing error
Post by: MasterMath on June 08, 2010, 07:46:16 am
for the a) part where do they get the cos from ??? :Sokay ill slowly explain part a) .. They said X=A+B .. and Y=A-B .. and then they said sin(X) + sin(Y) = ignore this part now..
and part b) the zero is the thita its a typing error
sin(X) = sin(A+B) substitution.
sin(Y) = sin(A- B) substitution.
Now lets work with the expansion of sin(A+B) = sinAcosB +sinBcosA
and sin(A-B)= sinAcosB -sinBcosA..
okay now lets get back to the formula..
sin(X) + sin(Y) = sin(A+B) + sin(A-B) = sum of expansions= sinAcosB + sinBcosA + sinAcosB -sinBcosA = sinAcosB + sinAcosB=2sinAcosB..
okay all good till now?
Post by: MasterMath on June 08, 2010, 07:49:05 am
So A must be equal to X+Y/2 .. and B must be equal to X-Y/2 .. to verify this lets try adding X and Y and dividing by 2 ..we should get A ..
X= A+B Y = A-B .. X+Y =A+B+A-B=2A .. then divide by 2 .. = A!!!
So our equation is correct .. so we can simply replace A by X+Y/2 and B by X-Y/2 .. so we reached the final equation they wanted!
Got it?
Post by: halosh92 on June 08, 2010, 07:51:44 am
okay ill slowly explain part a) .. They said X=A+B .. and Y=A-B .. and then they said sin(X) + sin(Y) = ignore this part now..
sin(X) = sin(A+B) substitution.
sin(Y) = sin(A- B) substitution.
Now lets work with the expansion of sin(A+B) = sinAcosB +sinBcosA
and sin(A-B)= sinAcosB -sinBcosA..
okay now lets get back to the formula..
sin(X) + sin(Y) = sin(A+B) + sin(A-B) = sum of expansions= sinAcosB + sinBcosA + sinAcosB -sinBcosA = sinAcosB + sinAcosB=2sinAcosB..
okay all good till now?
ok i got the expansion and all the other things but one tiny thing :
"Now lets work with the expansion of sin(A+B) = sinAcosB +sinBcosA"
shouldnt the expansion be sinA + SinB
where does the cos come from??? or is there some kind of rule?
Post by: MasterMath on June 08, 2010, 07:57:15 am
ok i got the expansion and all the other things but one tiny thing :Yes there is a rule for sin(X+Y)= sinXcosY +sinYcosX for any values of X and Y.. and also sin(X-Y)=sinXcosY -sinYcosX ... you got to memorize those .. and here are some more
"Now lets work with the expansion of sin(A+B) = sinAcosB +sinBcosA"
shouldnt the expansion be sinA + SinB
where does the cos come from??? or is there some kind of rule?
cos(X+Y)=cosXcosY -sinXsinY cos(X-Y)= cosXcosY +sinXsinY
tan(X+Y)=(tanX+ tanY)/1-tanXtanY tan(X-Y)=(tanX-tanY)/1+tanXtanY
and more more..
Sin2x= 2sinxcosx
Cos2x= Cos2x-sin2x which is also equal to = 1-2sin2x = 2cos2x-1
and tan2x= 2tanx/1-tan2x
Post by: MasterMath on June 08, 2010, 08:00:25 am
Yes there is a rule for sin(X+Y)= sinXcosY +sinYcosX for any values of X and Y.. and also sin(X-Y)=sinXcosY -sinYcosX ... you got to memorize those .. and here are some moreI have more , here are some .. dunno if needed for C2 maybe C3 ..
cos(X+Y)=cosXcosY -sinXsinY cos(X-Y)= cosXcosY +sinXsinY
tan(X+Y)=(tanX+ tanY)/1-tanXtanY tan(X-Y)=(tanX-tanY)/1+tanXtanY
and more more..
Sin2x= 2sinxcosx
Cos2x= Cos2x-sin2x which is also equal to = 1-2sin2x = 2cos2x-1
and tan2x= 2tanx/1-tan2x
X= (P+Q)/2 .. Y=(P-Q)/2
sinp + sinq = 2sinXcosY
sinp - sinq= 2cosXsinY
cosp + cosq =2cosXcosY
cosP - cosq= -2sinXsinY
Post by: halosh92 on June 08, 2010, 08:04:32 am
Yes there is a rule for sin(X+Y)= sinXcosY +sinYcosX for any values of X and Y.. and also sin(X-Y)=sinXcosY -sinYcosX ... you got to memorize those .. and here are some more
cos(X+Y)=cosXcosY -sinXsinY cos(X-Y)= cosXcosY +sinXsinY
tan(X+Y)=(tanX+ tanY)/1-tanXtanY tan(X-Y)=(tanX-tanY)/1+tanXtanY
and more more..
Sin2x= 2sinxcosx
Cos2x= Cos2x-sin2x which is also equal to = 1-2sin2x = 2cos2x-1
and tan2x= 2tanx/1-tan2x
OMG THANKYOU SO MUCH
:'( but i never took any of this r u sure this is AS ??? ???
am rili sorry for the trouble thx again ;)
Post by: MasterMath on June 08, 2010, 08:06:29 am
OMG THANKYOU SO MUCHyep all those are needed , they'll help you atleast instead of proving them in the exam!
:'( but i never took any of this r u sure this is AS ??? ???
am rili sorry for the trouble thx again ;)
Post by: Saladin on June 08, 2010, 11:53:44 am
Post by: halosh92 on June 08, 2010, 11:54:19 am
This is from the old syllabus, this is now in C3. You do not need to know it.
;D ;D ;D ;D ;D ;D ;D ;D ;D ;D
Post by: halosh92 on June 08, 2010, 01:23:52 pm
PAPER A4
Q2b
Post by: Saladin on June 08, 2010, 03:14:33 pm
http://www.freeexampapers.com/Dndex.php?d=QSBMZXZlbC9NYXRocy9FZGV4Y2VsL0MyL1ByYWN0aWNlIFBhcGVycw==
PAPER A4
Q2b
On it now.
Post by: Saladin on June 08, 2010, 03:19:04 pm
You can just use your calculator to solve this. If you have an fx-991 MS.
Post by: halosh92 on June 08, 2010, 03:22:47 pm
Here are the steps
You can just use your calculator to solve this. If you have an fx-991 MS.
how did u get this exactly
(n^3+4n+3)
and my calculator is fx-115MS
Post by: Saladin on June 08, 2010, 03:26:44 pm
how did u get this exactly
(n^3+4n+3)
and my calculator is fx-115MS
You have to use long division.
Post by: Saladin on June 08, 2010, 03:27:51 pm
how did u get this exactly
(n^3+4n+3)
and my calculator is fx-115MS
I would suggest you get the calculator that I have, because it can solve quadratic equations, and it is allowed.
Post by: halosh92 on June 08, 2010, 03:36:06 pm
I would suggest you get the calculator that I have, because it can solve quadratic equations, and it is allowed.
i dont think its possible but can u explain it though
Post by: Saladin on June 08, 2010, 03:37:03 pm
i dont think its possible but can u explain it though
Its hard to show long division here.
I will try to scan. Alrite?
Post by: Saladin on June 08, 2010, 03:39:47 pm
Post by: GodOfWar on June 08, 2010, 09:03:48 pm
Here u go!
Or u could use synthetic division.
Post by: halosh92 on June 08, 2010, 09:16:08 pm
could someone gimmi the volume and total surface area of a prism, sphere, cube , cylinder
thx
URGENTTTTTTTTTTTTTTTT
Post by: Summer :] on June 09, 2010, 07:00:25 am
thx engraved.
could someone gimmi the volume and total surface area of a prism, sphere, cube , cylinder
thx
URGENTTTTTTTTTTTTTTTT
Surface area of a cylinder : 2 X pi X r X h
T.surface area of a closed cylinder: 2 x (pi) x r (r +h)
T. suface area of an open cylinder: 2 x pi x r x h + pi(r)^2
Volume of cylinder: pi x (r)^2 x h
Surface area of a sphere: 4 pi r 2
Volume: 4 x pi x (r)^3 / 3
Surface area of a cube: 6(a)^2
where a is the side
Volume of the cube: a x a x a = a^3
http://www.teacherschoice.com.au/maths_library/area%20and%20sa/area_9.htm
and that about prisms..because there are many types of prisms..
Post by: Blizz_rb93 on August 30, 2010, 10:27:22 pm
its in the C1 Textbook in Chapter 1 Mixed Exercise
Question 6
(8/27) to the power of 2/3
how do i solve this? i dont get questions when they come like that
and is the subject doubts threads all active? seems like no one goes on anymore cause its still summer?
Post by: S.M.A.T on August 30, 2010, 11:04:42 pm
hey can anyone help me with this question
its in the C1 Textbook in Chapter 1 Mixed Exercise
Question 6
(8/27) to the power of 2/3
how do i solve this? i dont get questions when they come like that
and is the subject doubts threads all active? seems like no one goes on anymore cause its still summer?
download the page below
Post by: Blizz_rb93 on September 07, 2010, 10:39:39 am
i also would like to ask another small question
now when we have an equation y=6-10x-4x^2 and we move the values to the normal way like putting 4x^2 in the beginning then followed by -10x then by -6, does moving the values affect the signs?? if so, how?
Post by: astarmathsandphysics on September 07, 2010, 02:15:15 pm
Post by: Blizz_rb93 on September 07, 2010, 03:15:23 pm
reflection in x axis. All the y vlues change signso it becomes 4x^2 + 10x + 6 ?
Post by: S.M.A.T on September 08, 2010, 03:53:21 am
thank you!
i also would like to ask another small question
now when we have an equation y=6-10x-4x^2 and we move the values to the normal way like putting 4x^2 in the beginning then followed by -10x then by -6, does moving the values affect the signs?? if so, how?
I am showing you with an example:
Suppose the question asked to solve the equation 6-10x-4x^2=0
Since before solving we move the values to the normal way like putting 4x^2 in the beginning,so multiply -1 on both side of the '= 'sign i.e (6-10x-4x^2)*-1=0*-1
so the equation become 4x^2+10x-6=0
Not 4x^2 + 10x + 6
Post by: The Golden Girl =D on September 25, 2010, 05:53:27 pm
Anyways can some one exaplin to me Completing the Square :'(
http://www.themathpage.com/alg/complete-the-square.htm#complete
cuz i opened this amazing page but i don't get the following :( ;
in letter c) y did we do + 4 !?!?
e) y did we do 25/4 .... i don't get it ?
f) y did we do 9/4
Plz someone help
Post by: The Golden Girl =D on September 25, 2010, 06:01:40 pm
I got a doubt but i know it's a very minute thing but for me it's not cuz idk how much time i've been trying to read books search sites to get to Understand this the ...and the thing that pisses me off is that when my teacher exapleined it he said didn't ya take it in IGCSE Maths ...Man i never EVER saw this thing in my life .
Anyways can some one exaplin to me Completing the Square :'(
http://www.themathpage.com/alg/complete-the-square.htm#complete
cuz i opened this amazing page but i don't get the following :( ;
in letter c) y did we do + 4 !?!?
e) y did we do 25/4 .... i don't get it ?
f) y did we do 9/4
Plz someone help
Nvm i get it :-[
Post by: elemis on September 25, 2010, 06:56:44 pm
Nvm i get it :-[
I have a better explanation if you want ?
Post by: astarmathsandphysics on September 25, 2010, 10:34:35 pm
http://www.astarmathsandphysics.com/a_level_maths_notes/C1/a_level_maths_notes_c1_completing_the_square_and_solving_quadratic_equations.html
Post by: The Golden Girl =D on September 26, 2010, 04:25:54 pm
ya sure :)
Post by: astarmathsandphysics on September 28, 2010, 10:04:48 pm
a) Find the equation of the diagonal BD. ( I got x + y = 1)
b) Given that the side BC has gradient 5/3, find the coordinates of B and hence of D.
I don't get part b) at all
Post by: Deadly_king on September 29, 2010, 05:31:02 am
The points A (-3,-4) and C (5,4) are the ends of the diagonal of a rhombus ABCD.
a) Find the equation of the diagonal BD. ( I got x + y = 1)
b) Given that the side BC has gradient 5/3, find the coordinates of B and hence of D.
I don't get part b) at all
You got it almost right........except for the part BO.
Anyway let me get this clear for you :
You did obtain equation of line BC ----> 3y = 5x - 13
Now solve simultaneously with x + y = 1 since line BC meets line BD at B.
I'll do it using the substitution method :
From equation 2 --> y = 1 - x
Replace in equation 1 :
3(1 - x) = 5x - 13
3 - 3x = 5x - 13
-8x = -16
x = 2
When x = 2 ----> y = 1 - (2) = -1
Hence coordinates of B (2, -1)
The midpoint of BD = midpoint of AC = (1,0). You calculated it in the first part when finding equation of diagonal BD.
Let coordinates of D be (x,y)
Hence (1,0) = ( (x+2)/2 , (y+-1)/2 )
Therefore (x+2)/2 = 1 ----> x = 0 and (y-1)/2 = 0 ----> y = 1
Coordinates of D (0,1)
Hope it helps :)
Post by: Ivo on October 08, 2010, 12:01:55 am
Q1) The equation of a curve is y=ax^2 - 2bx + c, where a, b and c are constants with a > 0.
a) Find, in terms of a, b and c, the coordinates of the vertex of the curve.
b) Given that the vertex of the curve lies on the line y=x, find an expression for c in terms of a and b. Show that in this case, whatever the value of b, c >= -1/4a (>= means greater than or equal to).
For this question, I found answers to part a), but am stuck on b). I know part a) answer is in co-ordinates (b/a , c-(b^2/a)). I also know some of part b), c is [b(b+1)/a], so I basically don't get why the show that thing is that.
Q2) a) Express 9x^2 +12x + 7 in the form (ax + b)^2 + c where a, b and c are constants whose values are to be found.
b) Find the set of values taken by: 1/(9x^2 +12x + 7) for real values of x.
For this question, I could get a) with a = 3, b=2, c=3. But have no idea what "real values of x" mean and please provide reasons why.
Q3) a) Express 9x^2 - 36x + 52 in the form (Ax-B)^2 + C, where A, B and C are integers. Hence, or otherwise, find the set of values taken by 9x^2 - 36x + 52 for real x. Again I don't know what "real x" means, nor how to work out the answer.
Many thanks in advance, this is too tricky for me ;(
Post by: elemis on October 08, 2010, 04:46:25 am
Post by: elemis on October 08, 2010, 05:22:28 am
Q2) a) Express 9x^2 +12x + 7 in the form (ax + b)^2 + c where a, b and c are constants whose values are to be found.
b) Find the set of values taken by: 1/(9x^2 +12x + 7) for real values of x.
For this question, I could get a) with a = 3, b=2, c=3. But have no idea what "real values of x" mean and please provide reasons why.
COULD SOMEONE PLEASE CHECK THIS EXPLANATION !
Real numbers are any positive or negative numbers which might be whole numbers or numbers like pi or e.
Now you should notice that
Inputing a very large positive or negative number into x we see that it tends towards zero on both cases.
But, I then differentiate the equation to find the maximum points (which is when x = -2/3)
Hence,
Post by: Ivo on October 08, 2010, 05:22:36 pm
COULD SOMEONE PLEASE CHECK THIS EXPLANATION !
Real numbers are any positive or negative numbers which might be whole numbers or numbers like pi or e.
Now you should notice thatis the equation of some asymptote. Hence, the value it should be tending to some value, but will never reach it.
Inputing a very large positive or negative number into x we see that it tends towards zero on both cases.
But, I then differentiate the equation to find the maximum points (which is when x = -2/3)
Hence,
OK, thanks for your response. Your answer/range is right, however I didn't understand why from your last sentence. Is it because 1/0 doesn't exist, but why 1/3? What's the general 'rule'?
Also, hardest question is Q1), any ideas, I've thought about it for a long time now. :(
Thanks for your help so far!
Post by: Deadly_king on October 08, 2010, 06:20:22 pm
OK, thanks for your response. Your answer/range is right, however I didn't understand why from your last sentence. Is it because 1/0 doesn't exist, but why 1/3? What's the general 'rule'?
Also, hardest question is Q1), any ideas, I've thought about it for a long time now. :(
Thanks for your help so far!
Sorry for late reply guys :(
Yeah....Ari is right there :)
Hmm.........from part(i) of the question, you can find the turning point to be (-2/3 , 3)
So you should replace the value of x in the equation to get the range of value of y. Since (3x + 2)2 is a perfect square......y will in no way be less than zero.
Post by: Ivo on October 08, 2010, 09:10:23 pm
The equation of a curve is
a) Find, in terms of a, b and c, the coordinates of the vertex of the curve.
b) Given that the vertex of the curve lies on the line
I've partially answered the question (I've done parts a) and a bit of b), but don't know how to show that the statement in b) is correct).
Here are my answers so far:
a)
b)
Many many thanks in advance!
Post by: elemis on October 09, 2010, 05:01:47 am
OK, thanks for your response. Your answer/range is right, however I didn't understand why from your last sentence. Is it because 1/0 doesn't exist, but why 1/3? What's the general 'rule'?
Since it is some form of an asymptote the Y value will go closer and closer to zero but NEVER reach zero.
Remember I found the max point to be -2/3 ? That was the x coordinate.
Inputting -2/3 into the original equation gives a Y VALUE of 1/3
Since this is the max point the curve's maximum value that it can attain is 1/3
Post by: **RoRo** on October 09, 2010, 02:17:14 pm
Solve the simultaneous equations:
2x + 2y = 7
x2 - 4y2 = 8
The answers according to the textbook are: x=3, y=1/2 OR x=6 1/3, y= -2 5/6
Thanks a lot in advance! :)
Post by: Dibss on October 09, 2010, 02:32:06 pm
Solve the simultaneous equations:
2x + 2y = 7
x2 - 4y2 = 8
First eq rearranged.
x= (7-2y)/2
Substitute the x in the second eq.
[(7-2y)2]/22 - 4y2 = 8
(49-28y+4y2)/4 -4y2= 8
49 - 28y + 4y2 -16y2 = 32
Now just solve like a normal quadratic.
Did this in a hurry, will check up the answer and get back to you in a few (:
Post by: Dibss on October 09, 2010, 02:38:08 pm
First eq rearranged.Okay so you get the eq
x= (7-2y)/2
Substitute the x in the second eq.
[(7-2y)2]/22 - 4y2 = 8
(49-28y+4y2)/4 -4y2= 8
49 - 28y + 4y2 -16y2 = 32
Now just solve like a normal quadratic.
Did this in a hurry, will check up the answer and get back to you in a few (:
12y2 + 28y -17 = 0
Solve it
y = 0.5 or y = 2.83
Substitute values into the first eq and get x values.
Makes sense or you have any questions? (:
Post by: Dibss on October 09, 2010, 02:43:41 pm
Post by: Dibss on October 09, 2010, 03:02:19 pm
Sorry, can't be of help - I'm as lost as you :/
Post by: Ivo on October 09, 2010, 03:28:19 pm
Okay being the blondie that I am, solved the first two parts in a minute then got stuck on the proving for eternity. Decided to post up at least the parts I managed and read that you already have the same answers xP
Sorry, can't be of help - I'm as lost as you :/
Well at least I have company! :) Anyway, thanks for your attempt. Now I know I'm not the only one!
Post by: Freaked12 on October 09, 2010, 05:16:38 pm
(i) y = ax^2 - 2bx + c, we differentiate to get:
dy/dx = 2ax - 2b.
We know dy/dx=0 at the vertex, we get the coordinates you asked for
(ii) Now given that the vertex is on y=x, this tells as that:
b/a = c - (b^2/a) => c= (b(b+1))/a
This equation relates 'c' with 'a' and 'b'.. Also the factor of 4 and => sign might make you think of a discriminant, why?
The equation for 'c' is in fact a quadratic in 'b':
b^2 + b - ac = 0.
We require D => 0 for real roots to exist, and since we are dealing with a real cartesian system this has to be the only case, therefore:
(1)^2 - 4*(1)*(-ac) >= 0 => 1 + 4ac >=0
and so we get c >= -1/(4a), since we know a>0/
Post by: Dibss on October 09, 2010, 05:28:20 pm
We require D >= 0 for real roots to exist, and since we are dealing with a real cartesian system this has to be the only case, therefore:
(1)^2 - 4*(1)*(-ac) >= 0 => 1 + 4ac >=0
Can't believe I didn't figure that out :/
+Rep (:
Post by: Freaked12 on October 09, 2010, 05:33:40 pm
Post by: Ivo on October 09, 2010, 06:56:43 pm
Perfect answer, and your method would make sense in context with the question. This question came out of a chapter in the textbook dealing with quadratics - more specifically completing the square and about discriminant
Thank you for your (and that of your tutor)'s efforts - greatly appreciated and now I understand why! I will + rep you, definitely!!! 8)
Post by: Deadly_king on October 10, 2010, 08:14:13 am
Yo tell me if this solution is okay
(i) y = ax^2 - 2bx + c, we differentiate to get:
dy/dx = 2ax - 2b.
We know dy/dx=0 at the vertex, we get the coordinates you asked for
(ii) Now given that the vertex is on y=x, this tells as that:
b/a = c - (b^2/a) => c= (b(b+1))/a
This equation relates 'c' with 'a' and 'b'.. Also the factor of 4 and => sign might make you think of a discriminant, why?
The equation for 'c' is in fact a quadratic in 'b':
b^2 + b - ac = 0.
We require D => 0 for real roots to exist, and since we are dealing with a real cartesian system this has to be the only case, therefore:
(1)^2 - 4*(1)*(-ac) >= 0 => 1 + 4ac >=0
and so we get c >= -1/(4a), since we know a>0/
Good job pal :)
I got stuck at the same place as Ivo........I thought about the discriminant part but could not quite link it :D
+ rep btu i need to spread the love first ;)
Post by: astarmathsandphysics on October 12, 2010, 08:49:23 am
Post by: Ivo on October 26, 2010, 03:30:42 pm
Simplify...
(http://img716.imageshack.us/img716/1497/simplify.jpg)
Many thanks!
Post by: Deadly_king on October 26, 2010, 04:02:10 pm
I've tried it so many times, can't seem to get it!
Simplify...
(http://img716.imageshack.us/img716/1497/simplify.jpg)
Many thanks!
Is the answer 13/3 + (40,/3)/9 ???
,/ represents square root
Post by: Ivo on October 26, 2010, 05:04:48 pm
Is the answer 13/3 + (40,/3)/9 ???
,/ represents square root
Indeed it its! Could you please show how you got it?
I took out a factor of
Post by: Deadly_king on October 26, 2010, 06:18:19 pm
Indeed it its! Could you please show how you got it?
I took out a factor of, but could never get the answer!
OK....i'll elaborate.
I'll simplify the numbers first.
(,/3)-3 + (,/3)-2 + (,/3)-1 + (,/3)0 + (,/3)1 + (,/3)2 + (,/3)3
Now we can simplify : Anything to the power of zero is 1 and a square root to the power of 2 results in the number being squared
(,/3)-3 + 3-1 + (,/3) + 1 + (,/3) + 3 + (,/3)3
Now we add the numbers only.
1 + 3 + 1/3 + (,/3)-3 + (,/3)-1 + (,/3)1 + (,/3)3
Anything to the power of negative factor can be written as 1 over the value
Example : x-1 ----> 1/x
Hence => 13/3 + 1/(,/3)3 + 1/(,/3) + (,/3) + (,/3)3
Now you need to rationalise, that is multiply by (,/3)/(,/3) so that you obtained all the roots in the numerator. Moreover (,/3)2 = 3 ---> (,/3)3 = 3(,/3)
13/3 + (,/3)/9 + (,/3)/3 + (,/3) + 3(,/3)
Now forget about the (,/3) and add its coefficients to get the required solution ;)
13/3 + 40(,/3)/9
Hope it helps :)
Post by: Ivo on October 27, 2010, 12:47:15 am
OK....i'll elaborate.
I'll simplify the numbers first.
(,/3)-3 + (,/3)-2 + (,/3)-1 + (,/3)0 + (,/3)1 + (,/3)2 + (,/3)3
Now we can simplify : Anything to the power of zero is 1 and a square root to the power of 2 results in the number being squared
(,/3)-3 + 3-1 + (,/3) + 1 + (,/3) + 3 + (,/3)3
Now we add the numbers only.
1 + 3 + 1/3 + (,/3)-3 + (,/3)-1 + (,/3)1 + (,/3)3
Anything to the power of negative factor can be written as 1 over the value
Example : x-1 ----> 1/x
Hence => 13/3 + 1/(,/3)3 + 1/(,/3) + (,/3) + (,/3)3
Now you need to rationalise, that is multiply by (,/3)/(,/3) so that you obtained all the roots in the numerator. Moreover (,/3)2 = 3 ---> (,/3)3 = 3(,/3)
13/3 + (,/3)/9 + (,/3)/3 + (,/3) + 3(,/3)
Now forget about the (,/3) and add its coefficients to get the required solution ;)
13/3 + 40(,/3)/9
Hope it helps :)
OK thanks a lot for your detailed explanation! I just didn't understand the last bit, I somehow got
Post by: Deadly_king on October 29, 2010, 05:49:30 am
OK thanks a lot for your detailed explanation! I just didn't understand the last bit, I somehow got, instead of
. Where could I have gone wrong?
I guess you forgot 1/9. ;)
To the extent I had reached in my explanation, you just needed to add 1/9 + 1/3 + 1 + 3 = 40/9
These are all coefficients of (,/3), so it's similar to adding x/9 + x/3 + x + 3x = 40x/9
You just need to replace x by (,/3) to obtain 40(,/3)/9 :D
Post by: Ivo on October 29, 2010, 11:51:04 am
I guess you forgot 1/9. ;)
To the extent I had reached in my explanation, you just needed to add 1/9 + 1/3 + 1 + 3 = 40/9
These are all coefficients of (,/3), so it's similar to adding x/9 + x/3 + x + 3x = 40x/9
You just need to replace x by (,/3) to obtain 40(,/3)/9 :D
Thanks for your explanations! I made a schoolboy error - I thought
Post by: Deadly_king on October 29, 2010, 02:45:22 pm
Thanks for your explanations! I made a schoolboy error - I thought. Silly me! + rep for you anyway, thanks!
No problem buddy ;)
Try to be more careful next time ;D
I guess you just +rep someone, so you need to wait for 2hrs to +rep another person.
Don't worry about it ;)
Post by: Ivo on October 29, 2010, 03:11:04 pm
Let
The polynomial
Calculate the two possible values of
Very hard for me, here's my working so far. My answer's wrong though.
I found the polynomial to be:
Then I said that because it's a perfect square:
Many thanks in advance!
Post by: Tyserius on October 30, 2010, 03:32:32 am
Post by: Ivo on October 30, 2010, 12:34:42 pm
Your mistake is that it should be (a+3)/(1+a) instead of (a-3)/(1+a) as x + (-a-3)/(1+a) = x - (a+3)/(1+a). I tried many ways to solve but I don't seem to get the 2nd value for a. The first value should be a = 3?
Thanks for your hint there Tyserius! I've solved it now:
The polynomial is:
Therefore,
So quadratic:
Giving solutions of
Many thanks once again for pointing out that error, as a result an + rep comes your way! :)
Post by: Tyserius on November 03, 2010, 01:09:02 am
Post by: astarmathsandphysics on November 11, 2010, 09:13:18 pm
Post by: Ivo on November 13, 2010, 12:38:39 pm
Find the coordinates of the stationary points on the curve with equation
Find the set of real values of
For the 1st part, I got the stationary points as:
Many thanks!
Post by: elemis on November 13, 2010, 01:17:27 pm
Post by: elemis on November 13, 2010, 01:23:31 pm
Post by: Ivo on November 13, 2010, 05:56:19 pm
http://answers.yahoo.com/question/index?qid=20101106111046AAZVRlB (http://answers.yahoo.com/question/index?qid=20101106111046AAZVRlB)
Thanks for the link, but I still don't really understand what they are trying to say. Hope you can help.
Post by: astarmathsandphysics on November 14, 2010, 11:23:55 am
Post by: **RoRo** on November 18, 2010, 09:06:49 pm
1. Can someone please explain to me how do you come to a conclusion about Question 9 (e) in June 2005 paper?
2. The answer to question 7 (d) in the marking scheme is 26 years old? How is that? [January 2006]
Thanks in advance!! :)
Post by: Alpha on November 19, 2010, 04:38:34 am
I have 2 questions:
1. Can someone please explain to me how do you come to a conclusion about Question 9 (e) in June 2005 paper?
2. The answer to question 7 (d) in the marking scheme is 26 years old? How is that? [January 2006]
Thanks in advance!! :)
Can you post the link to those papers please? Or upload them?
I think it's Edexcel, right? Don't find the papers on Xtreme. :-\
Post by: elemis on November 19, 2010, 04:50:45 am
3000 = n/2 (2*500 + (n-1)200)
64000 = 800n +200n2
2n2+8n-640=0
n2 +4n -320 = 0
(n+20)(n-16)= 0
n= -20 or 16
Disallow -20 and consider only 16.
The above formula INCLUDES her 11th birthday. So we must add 16 to 10 to give 26.
Post by: Alpha on November 19, 2010, 04:53:29 am
For part 2 :
3000 = n/2 (2*500 + (n-1)200)
64000 = 800n +200n2
2n2+8n-640=0
n2 +4n -320 = 0
(n+20)(n-16)= 0
n= -20 or 16
Disallow -20 and consider only 16.
The above formula INCLUDES her 11th birthday. So we must add 16 to 10 to give 26.
Where did you find the papers? I'm still looking for them. :-\
And thank you Lion. :)
Post by: elemis on November 19, 2010, 04:57:55 am
If you input 100 into the equation a + (n-1)d to find the amount he would be paying in tthe 100 th month you will see that you get an answer of -49.
However, a person cant pay $-49 to a bank hence making it a nonsensical answer.
Post by: elemis on November 19, 2010, 04:59:49 am
Where did you find the papers? I'm still looking for them. :-\
And thank you Lion. :)
I downloaded them off the net. However, the site I got it off is now dead.
Post by: Alpha on November 19, 2010, 05:05:08 am
Thank you. ;)
Post by: **RoRo** on November 19, 2010, 07:36:12 am
Can you post the link to those papers please? Or upload them?
I think it's Edexcel, right? Don't find the papers on Xtreme. :-\
I have the past papers in a booklet here!
I tried to find the link on www.freeexampapers.com but it just won't open! :/
Check this link, it has some of the past papers: https://eiewebvip.edexcel.org.uk/PastPapers/Default.aspx
Post by: elemis on November 19, 2010, 07:37:08 am
I have the past papers in a booklet here!
I tried to find the link on www.freeexampapers.com but it just won't open! :/
Check this link, it has some of the past papers: https://eiewebvip.edexcel.org.uk/PastPapers/Default.aspx
I've answered your doubts already.
Post by: **RoRo** on November 19, 2010, 08:08:26 am
For the 1st part :
If you input 100 into the equation a + (n-1)d to find the amount he would be paying in tthe 100 th month you will see that you get an answer of -49.
However, a person cant pay $-49 to a bank hence making it a nonsensical answer.
Thank you very much!
& I apologize for the late reply, my Internet got dc for a while!
Post by: elemis on November 19, 2010, 08:09:47 am
Thank you very much!
& I apologize for the late reply, my Internet got dc for a while!
No worries. The explanation for the 2nd question is here :
For part 2 :
3000 = n/2 (2*500 + (n-1)200)
64000 = 800n +200n2
2n2+8n-640=0
n2 +4n -320 = 0
(n+20)(n-16)= 0
n= -20 or 16
Disallow -20 and consider only 16.
The above formula INCLUDES her 11th birthday. So we must add 16 to 10 to give 26.
Post by: **RoRo** on November 19, 2010, 02:14:03 pm
No worries. The explanation for the 2nd question is here :
I read your explanation, but see, in part (c) we found that the total of the allowances up to and including her 18th birthday was 39600...how come her allowance would stop at the age of 26 at 32000? It's supposed to increase, right?
& one more question please:
In the specimen paper, [it has no particular date info, so I'll type the question here]:
The curve C has equation y=f(x) and the point P (3,5) lies on C.
Given that f'(x) = 3x2-8x+6
(a) Find f(x)
The answer is f(x)=x3-4x2+6x-4
(b) Not needed now
The point Q also lies on C, and the tangent to C at Q is parallel to the tangent to C at P.
(c) Find the x-coordinate of Q.
I found the gradient to be 9 [using the formula for f'(x)]
& it's correct according to the marking scheme.
And then I equated the formula:
3x2-8x+6=9
I got 2 solutions: x=3, x=-1/3
Which one would be the correct one? The MS says it's -1/3 but I don't understand why!
Post by: elemis on November 19, 2010, 05:36:39 pm
I read your explanation, but see, in part (c) we found that the total of the allowances up to and including her 18th birthday was 39600...how come her allowance would stop at the age of 26 at 32000? It's supposed to increase, right?
& one more question please:
In the specimen paper, [it has no particular date info, so I'll type the question here]:
The curve C has equation y=f(x) and the point P (3,5) lies on C.
Given that f'(x) = 3x2-8x+6
(a) Find f(x)
The answer is f(x)=x3-4x2+6x-4
(b) Not needed now
The point Q also lies on C, and the tangent to C at Q is parallel to the tangent to C at P.
(c) Find the x-coordinate of Q.
I found the gradient to be 9 [using the formula for f'(x)]
& it's correct according to the marking scheme.
And then I equated the formula:
3x2-8x+6=9
I got 2 solutions: x=3, x=-1/3
Which one would be the correct one? The MS says it's -1/3 but I don't understand why!
Its -1/3 since you already know the coordinates of P i.e. it is 3,5
Post by: **RoRo** on November 19, 2010, 06:15:35 pm
Its -1/3 since you already know the coordinates of P i.e. it is 3,5
OHHHHHH!!
OMG! That's so stupid of me!
& I've spent like half an hour trying to figure out why did the Mark Scheme say so!!
Thanks a lot!! ;D
Post by: Helium on December 06, 2010, 05:30:19 pm
Anyone mind,
Q)Perry left school driving towards the lake one hour before jaidee.
jaidee drove in the opposite direction going 6mph slower then perry for one hour after which time they
were 174 min apart. What was perry's speed.
I'm stuck.
Post by: astarmathsandphysics on December 06, 2010, 10:04:04 pm
Post by: **RoRo** on December 08, 2010, 08:21:59 pm
In the marking scheme, the answer is -1/4<k<3 and they've got some really weird explanation, check this link: http://www.scribd.com/doc/29694732/Edexcel-GCE-January-2010-Core-Mathematics-1-C1-Marking-Scheme
Can someone please explain this to me?
Thanks!
Post by: Freaked12 on December 08, 2010, 09:02:05 pm
you have to use b^2-4ac<0
b^2 - 4ac is less than zero, then there are no solutions. This means that there are no values of x giving a value of y of zero, hence the graph of the curve will not cross the x-axis.
b^2 - 4ac = 0 then there are two equal values for x (that is, the
line of the graph touches the x-axis once)
your result would match the mark scheme if you had used the correct formula.
Hope this helps
Post by: **RoRo** on December 08, 2010, 09:07:07 pm
I didn't understand what you wrote in bold! :/
Post by: Freaked12 on December 08, 2010, 09:10:57 pm
you have to write b^2-4ac<0 to show there are no real roots of K.
This will ultimately answer your questioin
Post by: Freaked12 on December 08, 2010, 09:14:51 pm
Post by: **RoRo** on December 08, 2010, 09:17:25 pm
I was just saying by equating b^2-4ac to zero, you are showing that the equation has one solution but the question is saying find all the values of k where where roots are not real so this is wrong.
you have to write b^2-4ac<0 to show there are no real roots of K.
This will ultimately answer your questioin
Ohhh...yes, yes, yes!
Thank you for the clarification!
Post by: HUSH1994 on December 10, 2010, 06:31:07 am
given that f:x- 2x^2 + 8x-18
find the least value of k for which f is one-one,how do u do it?
Post by: blackberry on December 21, 2010, 07:34:42 pm
Given that for all values of x: 3x2+12x+5=p(x+q)2 +r
a)find the values of p,q and r
b)solve the equation 3x2+12x+5=0
---------------------------------------------------------------------------------------
a)i got this one!
b)i tried using the quadratic formula to solve it and i got -2+(1/3)(surd 21) however the answer is -2+(surd7/surd3). The answer was obtained by completing the square. When i solved both the answers using a calculator the answers were the same.
This question is from a past paper but i dont know which one, and i need to know if my answer is acceptable as the textbook doesnt mention it, so can anyone tell me which past paper this question if from??
thanxx!
Post by: The Golden Girl =D on December 22, 2010, 07:03:17 am
heyy guys i have a doubt, its from the c1 textbook, chapter 2, mixed exercise 2G, Q.5
Given that for all values of x: 3x2+12x+5=p(x+q)2 +r
a)find the values of p,q and r
b)solve the equation 3x2+12x+5=0
---------------------------------------------------------------------------------------
a)i got this one!
b)i tried using the quadratic formula to solve it and i got -2+(1/3)(surd 21) however the answer is -2+(surd7/surd3). The answer was obtained by completing the square. When i solved both the answers using a calculator the answers were the same.
This question is from a past paper but i dont know which one, and i need to know if my answer is acceptable as the textbook doesnt mention it, so can anyone tell me which past paper this question if from??
thanxx!
I did it by Completing the Square . you can do it by Quadratic formula or completing the Square ,both are correct ;)
x2+ 4x + 5/3
x2 + 4x + (2)2 - (2)2 + 5/3 =0
[x+2]2- 4 + 5/3
[x+2]2= 7/3
x+2 = positive negative surd 7/3
x= -2 positive or negative surd 7/3
Therefore ;
x = -2+ surd 7/3 OR x= -2-surd 7/3
Let me know if you don't get it :)
Post by: blackberry on December 22, 2010, 11:17:44 am
quadratic formula answer : -2+(1/3)(surd21)= -0.472
square completion answer: -2+(surd7/surd3)= -0.472
they want us to express the answer in surd form, so my answer is the quadratic formula one and the one they mentioned in the textbook was the square completion one. Even though they both give the same answer i.e. -0.472, i wanted to know if my answer in surd form is acceptable by the examiners so i wanted to know which paper this q is from??
thanxxx for the help!!! <3
Post by: The Golden Girl =D on December 22, 2010, 01:31:50 pm
Post by: elemis on December 22, 2010, 01:43:38 pm
well i know how to do it by completing the square but what i meant was:
quadratic formula answer : -2+(1/3)(surd21)= -0.472
square completion answer: -2+(surd7/surd3)= -0.472
they want us to express the answer in surd form, so my answer is the quadratic formula one and the one they mentioned in the textbook was the square completion one. Even though they both give the same answer i.e. -0.472, i wanted to know if my answer in surd form is acceptable by the examiners so i wanted to know which paper this q is from??
thanxxx for the help!!! <3
Surd form is acceptable.
Post by: blackberry on December 22, 2010, 03:42:41 pm
if anyone can just please tell me which paper this is from i'd be glad!!
Post by: The Golden Girl =D on December 27, 2010, 11:25:47 am
= surd 245
therefore ; 7surd5 ----> this is what I can NOT seem to get I got the surd245 but I can't get this one :-\
Hmm in Q 7 , I don't get how to get X2008 ?! .....
All the Q went well for me Except for Part c) ???
Can Someone Please do it for me in DETAIL .
Thank You In an Advance :)
Post by: Dibss on December 27, 2010, 11:37:35 am
In January 2008 Q 4 , I can't seem to get the Answer Ends like this :
= surd 245
therefore ; 7surd5 ----> this is what I can NOT seem to get I got the surd245 but I can't get this one :-\
I'll use the word root instead of surd just because I'm used to it. :P
245 = 49 x 5 right?
so root 245 = root 49 x root 5
= 7 x root 5
Understand?
Post by: Dibss on December 27, 2010, 11:44:58 am
c) solve 2p^2 + 3p + 1 =1
to get p = 0 (not possible) and p = -3/2
Let me know if I need to elaborate.
d) This one was tricky and I remember being confused by it.
You're meant to substitute your value of p in x2, x3, x4 etc.
Then you'll notice that
x1, x2, x3, x4 = 1, -1/2, 1, -1/2
See the pattern?
So for x2008 (a positve postion) the value will be -1/2
(:
Post by: The Golden Girl =D on December 27, 2010, 11:47:48 am
Just checked your second question
c) solve 2p^2 + 3p + 1 =1
to get p = 0 (not possible) and p = -3/2
Let me know if I need to elaborate.
d) This one was tricky and I remember being confused by it.
You're meant to substitute your value of p in x2, x3, x4 etc.
Then you'll notice that
x1, x2, x3, x4 = 1, -1/2, 1, -1/2
See the pattern?
So for x2008 (a positve postion) the value will be -1/2
(:
Thanks I get My Mistake , Thanks Again Sis =]
Post by: Dibss on December 27, 2010, 11:49:44 am
Post by: The Golden Girl =D on December 27, 2010, 06:26:05 pm
MS :http://www.scribd.com/doc/15813129/Edexcel-Mathematics-c1-GCE-Core-1-As-Level-January-2009-Mark-scheme-666301
QP: Attached iA =]
Post by: Dibss on December 27, 2010, 07:21:38 pm
So your answer should be like the red line in the figure below:
(http://www.bbc.co.uk/schools/gcsebitesize/maths/images/graph_18.gif)
(:
Post by: The Golden Girl =D on December 27, 2010, 07:53:56 pm
You draw the graph just like you would draw y=1/x but farther from the asymptotes.
So your answer should be like the red line in the figure below:
(http://www.bbc.co.uk/schools/gcsebitesize/maths/images/graph_18.gif)
(:
Thanks (Jazaki Allaha Kair) =]
Post by: Dibss on December 27, 2010, 08:12:15 pm
Post by: elemis on December 28, 2010, 04:36:13 am
In January 2009 Q8 , Can Someone Please tell me how to draw the reciprocal (y = 2/x ) => I don't know how to , I just drew random lines :S
MS :http://www.scribd.com/doc/15813129/Edexcel-Mathematics-c1-GCE-Core-1-As-Level-January-2009-Mark-scheme-666301
QP: Attached iA =]
This is all in the C1 textbook. There are a few days till the actual exam. What have you been doing ?
Post by: elemis on December 28, 2010, 06:06:27 am
Post by: The Golden Girl =D on December 28, 2010, 06:11:44 am
Post by: elemis on December 28, 2010, 06:16:03 am
Yes . Can you tell me how to do it in the Right Way.
Yeah, like I said. Check the C1 math textbook.
Whisper/Dibbs has already answered your query on the previous page.
Post by: The Golden Girl =D on January 02, 2011, 08:28:34 am
January 2010 ;
Q7]c) My Answer is 25 Where as is the MS it says 205 , Can Someone Please do it for me :$
Q8]c) I drew the asymptote x= -1 , is it necessary or considered right if i drew it ?
Q10]C) I'm getting the y-axis point as (0,18) NOT (0,14) , can someone Please show me so That I know what my mistake it ? + is it necessary if I drew the vertex/minimum Point ?
Thank you in advance =]
Post by: elemis on January 02, 2011, 08:32:47 am
Post by: elemis on January 02, 2011, 08:37:31 am
980= 2A + 570
410 = 2A
A = 205
Post by: elemis on January 02, 2011, 08:38:30 am
You would lose a mark probably.
Post by: The Golden Girl =D on January 02, 2011, 08:42:01 am
Post by: The Golden Girl =D on January 02, 2011, 08:53:13 am
will check this within a couple of hours iA *gtg :-X *
Post by: elemis on January 02, 2011, 10:19:24 am
At the point where it crosses the y axis x=0
Inputting x=o into above equation gives 14
(0,14) = answer.
Post by: Weaam on January 04, 2011, 03:45:12 pm
Post by: The Golden Girl =D on January 04, 2011, 06:02:15 pm
Post by: Weaam on January 05, 2011, 08:10:48 am
I have Some on my PC , which year ?From 2005 to 2010 ? I know its alot but i REALLY need them :-\
Post by: The Golden Girl =D on January 05, 2011, 08:23:25 am
Post by: The Golden Girl =D on January 05, 2011, 08:25:30 am
2005
Post by: The Golden Girl =D on January 05, 2011, 08:27:03 am
Post by: The Golden Girl =D on January 05, 2011, 08:29:01 am
Post by: The Golden Girl =D on January 05, 2011, 08:31:32 am
Post by: Weaam on January 05, 2011, 08:48:09 am
2008THANKSS SO MUCHHH! I OWE U BIG TIMEE ;D
Post by: The Golden Girl =D on January 05, 2011, 10:01:37 am
Post by: The Golden Girl =D on January 06, 2011, 07:55:31 am
C1 book Page : 151 ,Q 10.
the curve C has Equation
y = 8x + x^2 + 9/x
the Points P and Q lie on C and have X-coordinates -3 and 1 respectively
a] find an equation of the chord PQ
b] show that the tangents to Z at the points P and Q are Parallel
the Tangent to C at P and the normal to C at Q intersect at point R(17,2)
D] find QR
E] explain why angle PQR is a right triangle and find the area of triangle PQR.
c and d - I can do them as for the rest , I haven't gotten a single thing correct ,I don't know why =[
Post by: The Golden Girl =D on January 06, 2011, 08:35:52 am
Post by: The Golden Girl =D on January 06, 2011, 10:57:02 am
Question No.8 on page 148
Attached.
Post by: Dibss on January 06, 2011, 12:05:14 pm
the curve C has Equationfor a) find coords of p by substituting x=-3 in the equation
y = 8x + x^2 + 9/x
the Points P and Q lie on C and have X-coordinates -3 and 1 respectively
a] find an equation of the chord PQ
b] show that the tangents to Z at the points P and Q are Parallel
y=-24+9-3
y=-18
P(-3,-18)
for q, put x=1 in the eqtn.
y=8+1+9
y=18
Q(1,18)
To find eq of PQ, find grad
m = -36/-4
=9
Eq PQ => (y-18)/(x-1)=9
Solve
b) Find dy/dx
8+2x-9x^-2
to find grad at P, put in x=-3
to find grad at Q put in x=1
You should get the same answer for both. Gradients are same therefore, parallel.
I'm in school right now, but I'll try and get back to you with the rest later. (:
Post by: The Golden Girl =D on January 06, 2011, 12:34:05 pm
Post by: Dibss on January 06, 2011, 02:50:11 pm
E] explain why angle PQR is a right triangle and find the area of triangle PQR.
Eq of PQ was y=9x+9 therefore grad = 9
Calculate grad of QR => you get -1
Calculate grad of PR => you get 1
Since the product of gradients of QR and PR is -1 it proves they are perpendicular to each other. Therefore PQR is a right-angled triangle.
(:
Post by: Dibss on January 06, 2011, 03:02:15 pm
Another Question.
Question No.8 on page 148
a=400
d= x or 30
a) U10= a + (n-1)d
=400+9(30)
=670
b)S10= n/2 [2a + (n-1)d]
10/2 [2 x 400 + 9 x 30]
=5350
c)Equate S10 formula to 6000
10/2 [800 +9x] = 6000
800+9x=1200
x=44.4
Obviously he can't produce 44.4 cars every year so to reach the target, the minimum he must produce is 45.
Clear? (:
Post by: The Golden Girl =D on January 06, 2011, 03:11:22 pm
Post by: Dibss on January 06, 2011, 03:17:55 pm
one se can , you tell me how to find the AREA for PQR ?
Find length of PR and then the length of QR.
^use formula square root of [(x1-x2)2+ (y1-y2)2)
then use area = 1/2 x b x h
Or you can use the other method that involves listing coordinates and the criss cross lines (I don't know how to explain it by typing :P)
clear ;D
I saw your previous post ;] Tried to quote it and then saw this. I'm glad you got it. (:
Post by: The Golden Girl =D on January 06, 2011, 03:30:24 pm
Post by: Weaam on January 09, 2011, 05:13:29 am
Post by: Malak on January 09, 2011, 04:36:50 pm
I need the mark scheme for June 2009, i cant find it anywhere ! :o
Post by: Dibss on January 09, 2011, 04:47:36 pm
+Rep. (:
Post by: Weaam on January 09, 2011, 05:04:23 pm
And Good Luck to everyone who have their C1 exam tommorow ;D
Post by: elemis on January 09, 2011, 05:24:43 pm
Post by: The Golden Girl =D on January 09, 2011, 05:28:51 pm
Post by: The Golden Girl =D on January 09, 2011, 05:43:48 pm
Q8] b)
Thanks In Advance.
Post by: elemis on January 09, 2011, 06:08:33 pm
can someone do this for me in detail. :-X
Q8] b)
Thanks In Advance.
Doing it.
Post by: elemis on January 09, 2011, 06:12:43 pm
Forming an arithmetic progression formula :
56 +(n-1)8 = 144 We are trying to determine how many numbers in between 56 and 144 inclusive are divisible by 8
n = 12 Thus, there are 12 numbers in between 50 and 150 that are divisible by 8
12/2 (2*56 + 11*8 ) = 1200
Post by: The Golden Girl =D on January 09, 2011, 06:18:15 pm
thanks.
Post by: 8T on January 09, 2011, 06:54:09 pm
Can someone enlighten me :P
(a) Describe fully a single transformation that maps the graph of y = 1/ x
onto the
graph of y = 3/ x
. (2)
(b) Sketch the graph of y = 3/ x
and write down the equations of any asymptotes. (3)
(c) Find the values of the constant c for which the straight line y = c - 3x is a
tangent to the curve y = 3/ x
.
I can't think of a way to solve a nor c :L
Post by: 8T on January 09, 2011, 07:01:56 pm
Post by: The Golden Girl =D on January 09, 2011, 07:06:21 pm
good luck ;)
Post by: Dibss on January 09, 2011, 07:07:25 pm
Anyone or is it too late :l?
Give me a few minutes. (:
Post by: 8T on January 09, 2011, 07:14:18 pm
Post by: Dibss on January 09, 2011, 07:17:56 pm
(a) Describe fully a single transformation that maps the graph of y = 1/ x
onto the graph of y = 3/ x
-
(c) Find the values of the constant c for which the straight line y = c - 3x is a
tangent to the curve y = 3/ x
I can't think of a way to solve a nor c :L
Part a) is like y=f(x) --> y=3f(x) therefore it's a vertical stretch of scale factor 3.
Part c) find dy/dx of both the equations and then equate them since gradients must be equal for it to be a tangent.
1)dy/dx=-3
2)y=3x-1
dy/dx=-3x-2
Therefore 3x-2 = 3
x=0
Nevermind, I seem to be doing the wrong thing :|
Wait x is not zero it's -1 & + 1Whoops :P
okay so you get x
put it in the second equation
you get y= + or - 3
Now put that in the first equation
c-3(1) =3
c=6
or
c-3(-1)=-3
c=-6
Hope it's the rigth answer. (:
Post by: 8T on January 09, 2011, 07:32:02 pm
Post by: 8T on January 09, 2011, 07:34:24 pm
Help ?
Post by: Dibss on January 09, 2011, 07:40:40 pm
Wait x is not zero it's -1 & + 1
Edited my post.
Quote
Q 9
Help ?
Trying now. (:
Post by: Dibss on January 09, 2011, 07:50:57 pm
f ?(x) = 6 ? 4x ? 3x2,
(a) find an expression for y in terms of x, (5)
(b) show that AB = k 7 , where k is an integer to be found.
---
a)To find y, integrate f'(x)
To find c use (0,0) since the curve passes though origin
? c = 0
y = 6x ? 2x^2 ? x^3
b) Find the roots of the eq or set y=0
x(6 ? 2x ? x2) = 0
Use quadratic formula to solve for x coord at A and B
A (?1 ? root7 , 0), B (?1 + root7 , 0)
to find AB the length, just subtract.
? AB = (?1 + root7 ) ? (?1 ? root7 ) = 2root7
(:
Post by: 8T on January 09, 2011, 08:28:53 pm
Now I'll read ur solving for 9 (:
Post by: 8T on January 09, 2011, 08:38:07 pm
Given thatThank YOU TAHNK YOU THANK YOU :D
f ?(x) = 6 ? 4x ? 3x2,
(a) find an expression for y in terms of x, (5)
(b) show that AB = k 7 , where k is an integer to be found.
---
a)To find y, integrate f'(x)
To find c use (0,0) since the curve passes though origin
? c = 0
y = 6x ? 2x^2 ? x^3
b) Find the roots of the eq or set y=0
x(6 ? 2x ? x2) = 0
Use quadratic formula to solve for x coord at A and B
A (?1 ? root7 , 0), B (?1 + root7 , 0)
to find AB the length, just subtract.
? AB = (?1 + root7 ) ? (?1 ? root7 ) = 2root7
(:
OKAY now can I please have question paPer 2010 january (: & the mark schme for JUNE 2010
Post by: 8T on January 09, 2011, 08:42:13 pm
I know how I just want to make sure I got the final answer is possible to offer help (: ?
Post by: 8T on January 09, 2011, 08:47:04 pm
Oh & is anyone sending me jan 2010 question paer
marsk scheme june 2010 ?
Post by: astarmathsandphysics on January 09, 2011, 10:43:43 pm
Post by: The Golden Girl =D on February 11, 2011, 01:46:17 pm
C2 book page 159 Qs 6] a) and 10]b)
Q6]a) find ,without using calculator -.- ,the values of :
sin theta ,cos theta ,given tan theta = 5/12 and theta is acute.
Q10]b)
in each of the following , eliminate theta to give an equation relating to X and Y;
x = sin theta , y= 2cos theta.
I'm having trouble in solving ALL parts of Q 6 and 10 :S
Post by: elemis on February 11, 2011, 01:56:42 pm
Post by: The Golden Girl =D on February 11, 2011, 01:59:07 pm
Post by: elemis on February 11, 2011, 02:00:46 pm
the following Questions ;
C2 book page 159 Qs 6] a) and 10]b)
Q6]a) find ,without using calculator -.- ,the values of :
sin theta ,cos theta ,given tan theta = 5/12 and theta is acute.
Q10]b)
in each of the following , eliminate theta to give an equation relating to X and Y;
x = sin theta , y= 2cos theta.
I'm having trouble in solving ALL parts of Q 6 and 10 :S
Question 6 A
Draw the triangle as shown with 5 and 12 representing sides of the triangle as per the tan ratio given.
Use opposite/hypotenuse to find the sine and adjacent/hypotenuse to find the cosine ratios in fractions.
As all are acute then all of the above angles are in the FIRST QUADRANT and hence ALL ARE POSITIVE.
Post by: elemis on February 11, 2011, 02:07:06 pm
Using
We first divide the second equation by two to remove the 2 from
Hence,
Next, we square equation 1 and equation 2 to get :
Adding the two gives :
Post by: The Golden Girl =D on February 11, 2011, 02:17:04 pm
Question 6 A
Draw the triangle as shown with 5 and 12 representing sides of the triangle as per the tan ratio given.
Use opposite/hypotenuse to find the sine and adjacent/hypotenuse to find the cosine ratios in fractions.
As all are acute then all of the above angles are in the FIRST QUADRANT and hence ALL ARE POSITIVE.
I got sin theta = 5/7 and cos theta =12/7 o.O answer is sin theta = 5/13 and cos theta 12/13 :S
Post by: elemis on February 11, 2011, 02:20:54 pm
I got sin theta = 5/7 and cos theta =12/7 o.O answer is sin theta = 5/13 and cos theta 12/13 :S
I forgot to attache the picture in my first post.
Check now.
Post by: The Golden Girl =D on February 11, 2011, 03:31:55 pm
Post by: **RoRo** on February 13, 2011, 03:40:24 pm
The circle C, with centre at the point A, has equation x2+y2-10x+9=0.
Find
(a) the coordinates of A, [2 marks]
(b) the radius of C [2 marks]
Can someone please explain to me how do you arrive to the following answers?
(a) x=5, y=0
(b) radius = 4
Post by: narnia on February 15, 2011, 06:25:46 am
cos3x=-1/2
iv done this but dont really know if my answers are correct...
i got 40, 80, 160, 200, 280 and 320 =\
please need the confirmation urgently =)
Post by: elemis on February 15, 2011, 06:56:32 am
Solve for 0<=x<360,
cos3x=-1/2
iv done this but dont really know if my answers are correct...
i got 40, 80, 160, 200, 280 and 320 =\
please need the confirmation urgently =)
If you re-insert your answers back into the original equation you should get -0.5.
Try it and see.
Post by: narnia on February 15, 2011, 07:10:14 am
If you re-insert your answers back into the original equation you should get -0.5.yea thts right on cal...
Try it and see.
buh then the answers given are 240 and 120 only...after multiplying x=40 and x=80 by 3...
what about the rest values i got? :S
Post by: narnia on February 15, 2011, 07:01:27 pm
Post by: astarmathsandphysics on February 16, 2011, 09:23:21 am
Post by: narnia on February 16, 2011, 06:22:32 pm
herethank you =)
Post by: narnia on February 17, 2011, 05:38:30 pm
where X=A+B and Y=A-B.
(b)HEnce or otherwise, solve for 0<=theta<360
sin40 + sin20 =0
iv done the (a) part....buh gettin confused for (b).. =\
Q2. 2sin2(theta) -2sin(theta)=cos2(theta) in the range -180<=theta<180.
have done this too and got +-160.5, +-19.5 and +-90...
buh few of them are wrong when i apply it to the equation....please can someone do the whole of this question and tell me? :S
Post by: astarmathsandphysics on February 18, 2011, 01:39:05 pm
sin (A-B)=sinA cos B-cos A sin B
Add these two equations and put X=A+B, Y=A-B
b)Put A=3 theta B=theta to get
sin 4 theta +sin 2 theta =2 sin3theta cos theta=0
sin sin 3theta =0 or cos theta =0
3theta =0,180,360,540, 720,900,1080 or theta =0,180,360
so theta =0,60,120,180,240,300,360
2. 2sin^2 x -2 sin =cos^2 x =1-sin^2 x so 3sin^2 x -2sin x -1=0 so (3sin x+1)(sinx -1)=
3sinx+1 =0 so x =-sin^-1 (1/3)=-19.5, 180--19.5=199.5, -180+19.5=-160.5,360+-19.5=340.5
or sinx=1 so x=90
Post by: The Golden Girl =D on February 19, 2011, 01:27:45 pm
theta is between 360 and o degrees in BOTH these Qs !
y=3 sin (x + 60) - 2
y = 4 cos3X - 2
I know such a Question won't come in the official exam but it will in MY coming test T_T
Thank you in advance
Post by: elemis on February 19, 2011, 01:47:41 pm
can someone Please do the following graphs in detail -cuz I don't know how to get them -__-
theta is between 360 and o degrees in BOTH these Qs !
y=3 sin (x + 60) - 2
y = 4 cos3X - 2
I know such a Question won't come in the official exam but it will in MY coming test T_T
Thank you in advance
Post by: The Golden Girl =D on February 19, 2011, 04:09:29 pm
graph
Thank you Ari but I want the Steps of How to do it bot what it'll look like :-\
Post by: elemis on February 19, 2011, 04:46:50 pm
Thank you Ari but I want the Steps of How to do it bot what it'll look like :-\
Draw the normal sine curve. Then move it 60 degrees to the left horizontally.
Then multiply the Y coordinates by 3 followed by a vertical translation of 2 units downwards.
Post by: The Golden Girl =D on February 25, 2011, 01:15:35 pm
f(x) = 2x3 + 3x2 -6x + 1
a} ...
B] i ) ....
b]ii) Solve equation , f(x) = 0 ---> How do I do that , can you show me in Steps
Thanks in advance =]
Post by: elemis on February 25, 2011, 01:20:23 pm
I NEED this URGENTLY !!
f(x) = 2x3 + 3x2 -6x + 1
a} ...
B] i ) ....
b]ii) Solve equation , f(x) = 0 ---> How do I do that , can you show me in Steps
Thanks in advance =]
Plug in values for x, start methodically from zero until you get a number. In this case I got x=1 as a solution.
Hence, (x-1) is a solution.
Perform algebraic division of the original equation with (x-1) and find the resultant quadratic equation.
Factorise the quadratic.
State the solutions.
Post by: The Golden Girl =D on February 25, 2011, 01:30:39 pm
Plug in values for x, start methodically from zero until you get a number. In this case I got x=1 as a solution.
Hence, (x-1) is a solution.
Perform algebraic division of the original equation with (x-1) and find the resultant quadratic equation.
Factorise the quadratic.
State the solutions.
Part in bold : you mean I try random numbers until I get the answer as Zero , right ?
Post by: elemis on February 25, 2011, 05:17:38 pm
Part in bold : you mean I try random numbers until I get the answer as Zero , right ?
Yes.
Post by: narnia on March 06, 2011, 10:24:14 am
i need an explanation...all the steps =S
Q2)Given that p=logq16, express in terms of p;
logq(8q)
Post by: narnia on March 08, 2011, 01:21:56 pm
Post by: elemis on March 08, 2011, 01:27:55 pm
can someone pleasee answer :(
5 minutes.
Post by: elemis on March 08, 2011, 01:30:46 pm
Q1)Find the possible values of x for which 22x+1=3(2x)-1
i need an explanation...all the steps =S
Q2)Given that p=logq16, express in terms of p;
logq(8q)
1) Your equation can be re-written as : (2x)2*2 = 3(2x)-1
Let y = 2x
Hence, 2y2 -3y +1=0
Solve this quadratic. You will find the value of 2x. Use logs to find the value of x.
Post by: elemis on March 08, 2011, 01:35:18 pm
Q1)Find the possible values of x for which 22x+1=3(2x)-1
i need an explanation...all the steps =S
Q2)Given that p=logq16, express in terms of p;
logq(8q)
2) logq(8q)
By the law of logs :
Hence,
4/3 p +1
Post by: narnia on March 08, 2011, 01:43:50 pm
2) logq(8q)where did the pwr 3/4 come from?
By the law of logs :and using the power rule :
Hence,
0.75p +1
Post by: elemis on March 08, 2011, 02:11:44 pm
where did the pwr 3/4 come from?
I made a small typo. I edited my post.
16^4/3 = 8
Post by: narnia on March 08, 2011, 02:34:16 pm
I made a small typo. I edited my post.yea got it now...
16^4/3 = 8
Thanks a lot
Post by: narnia on April 04, 2011, 07:54:01 pm
Post by: iluvme on April 04, 2011, 08:27:15 pm
Hope you got it :)
Post by: narnia on April 17, 2011, 03:11:04 pm
so yet another question..
Use your calculator to convert the following angles to degrees giving your answer to nearest 0.1 degree.
(i)0.46c
(ii)(root)3c
havnt done this in school i dont know why?!! :S but its theres in textbook :/
Post by: iluvme on April 17, 2011, 06:31:17 pm
It would help if you remember that 1 radian= 180/pie
Post by: Amii on May 11, 2011, 01:34:48 pm
Find the range of values of p for which the roots of the equation px2+px-2=0
The answer given is p<= -8. According to her the answer is supposed to be p< -8 and p >0
Could someone please help?
Thank you :)
Post by: narnia on May 17, 2011, 05:15:57 am
help plz..=/
And 7th question, part (d) of the attached worksheet...how to do??!
Post by: Shazizzzle on May 17, 2011, 09:04:01 am
Find the sum of all the integers between 1 and 1000 which are divisible by 7.
help plz..=/
And 7th question, part (d) of the attached worksheet...how to do??!
Sum of integers from 1 to 1000 divisible by 7, right?
We know that the 7 times table is 7, 14, 21 etc.
So consider the sequence
7 + 14 + 21 + ...
Now we need to know how many terms are divisible by 7 in 1000 integers, so simply divide 1000 by 7 using long division. You get 142.8, so basically only 142 terms in 1000 are divisible by 7.
Find the sum of those 142 terms!
a=7, d=7, n=142
Sn = 142/2 (2(7) + (142-1) (7))
Sn = 71,071.
Post by: 8T on May 23, 2011, 07:01:30 pm
Can anyone explain it pleaseee ? :D
Post by: Malak on May 23, 2011, 09:33:47 pm
JUNE 08 q.9 C2It would take really long to write the whole thing so I have uploaded my answers
Can anyone explain it pleaseee ? :D
If u dont understand any thing in it then i can surely help :)
for part a since x-20 was in a bracket i found the angles and then at the end subtract the 20, i dont know if there is any other way to solve it but i do it this way
same idea is for part b
In part b u times the range by 3 as the angle given is 3x, always do that when the angle is a multiple like 2x or 3x or whatever and at the end MAKE SURE to divide all your found angles by 3 to find x ('cause you found angle for 3x before)
Post by: 8T on May 24, 2011, 02:06:11 pm
It would take really long to write the whole thing so I have uploaded my answersThank you
If u dont understand any thing in it then i can surely help :)
for part a since x-20 was in a bracket i found the angles and then at the end subtract the 20, i dont know if there is any other way to solve it but i do it this way
same idea is for part b
In part b u times the range by 3 as the angle given is 3x, always do that when the angle is a multiple like 2x or 3x or whatever and at the end MAKE SURE to divide all your found angles by 3 to find x ('cause you found angle for 3x before)
Post by: Malak on May 24, 2011, 02:17:21 pm
Thank youWelcome :)
Post by: Malak on May 24, 2011, 11:40:29 pm
Thx
Post by: wakemeup on May 25, 2011, 09:57:34 am
Q10 part c
Thx
Ah yes, I appeared in this exam and I remember feeling particularly frustrated in this Q, even though it's quite easy now that I think about it. All you have to do is actually draw the circle. You already know that the line passes through the midpoint of the radius and you know that it' parallel to the tangent i.e it'd perpendicular to the radius. Using Pythagorean theorem, (take half the radius as perpendicular and the radius as the hypotenuse) you can now easily find the length of the line.
Hope that helps
Post by: mdwael on May 25, 2011, 10:19:45 am
Post by: Ghost Of Highbury on May 25, 2011, 10:26:00 am
Can someone please explain how to solve this question? I have attached a picture with the question in it. Its page 140 question 28 in our C2 textbook and I cant seem to get it right. I think it is flawed in the first place... ???
3/x = x+8/3
x^2+8x = 9
x^2+8x-9=0
x^2-x+9x-9=0
x(x-1)+9(x-1)=0
x = -9, x = 1
common ratio = when x =1 -> 3 ; when x = -9 -> -1/3
sum to infinity existsif |r| < 1
that means r = -1/3
fifth term = ar^4 = 3
a = 3/(-1/3)^4 = 243
sum to infinity = a/1-r = 182.25
Post by: mdwael on May 25, 2011, 10:36:18 am
3/x = x+8/3
x^2+8x = 9
x^2+8x-9=0
x^2-x+9x-9=0
x(x-1)+9(x-1)=0
x = -9, x = 1
common ratio = when x =1 -> 3 ; when x = -9 -> -1/3
sum to infinity existsif |r| < 1
that means r = -1/3
fifth term = ar^4 = 3
a = 3/(-1/3)^4 = 243
sum to infinity = a/1-r = 182.25
What you have said is partially corret according to the text book.. I got the same working as yours but the text book says: when X=1 r=1/3 and when X= -9 r= -3
your first term is correct.. 243..
Sum of infinity according to the txt book is 364.5 not 182.25 as you said and what I am getting aswell..
Post by: The Golden Girl =D on May 25, 2011, 11:08:19 am
Q10 part c
Thx
I know My paper is messy but I hope you get it :)
Attached.
I hope I helped :)
Post by: mdwael on May 25, 2011, 11:19:19 am
Post by: **RoRo** on May 25, 2011, 01:45:11 pm
Post by: Malak on May 25, 2011, 03:49:32 pm
Do we need to know how to prove the sine rule and cosine rule as well as the trigonometric angles (30, 45 and 60) ?I dont think so u need to know how to prove them, only how to apply them..and about the angles, if ur talking about the ratio/triangle thingi, i guess same is for that, just how to use it.
And Thx both of you, i understood :D
Post by: **RoRo** on May 25, 2011, 04:20:29 pm
I dont think so u need to know how to prove them, only how to apply them..and about the angles, if ur talking about the ratio/triangle thingi, i guess same is for that, just how to use it.
And Thx both of you, i understood :D
Alright, thanks! :)
Post by: **RoRo** on May 25, 2011, 05:13:21 pm
f(x)=x3+ax+b, where a and b are constants.
When f(x) is divided by (x?4) the remainder is 32.
When f(x) is divided by (x+2) the remainder is ?10.
(a) Find the value of a and the value of b.
(b) Show that (x?2) is a factor of f(x).
In the CD, it shows for part (a)
f(-2) = (-2)3 + (-2)a + b = 32 as well, isn't that wrong?
Post by: The Golden Girl =D on May 25, 2011, 05:16:49 pm
Post by: **RoRo** on May 25, 2011, 05:23:23 pm
I'm sorry I don't get which part you don't understand. *Plus what does the ? stand for *
That's because I just copy pasted the question, I'm sorry, the ? stands for - [minus sign]
My question is, you need to form 2 equations, right? Speaking about part a)
The first one is 4a+b=-32
The second one, according to me should be b-2a=-2, but the CD says it's b-2a=40, because they took f(-2) to be equal to 32 instead of -10, isn't that wrong?
Post by: The Golden Girl =D on May 25, 2011, 05:52:28 pm
That's because I just copy pasted the question, I'm sorry, the ? stands for - [minus sign]
My question is, you need to form 2 equations, right? Speaking about part a)
The first one is 4a+b=-32
The second one, according to me should be b-2a=-2, but the CD says it's b-2a=40, because they took f(-2) to be equal to 32 instead of -10, isn't that wrong?
Since it's mentioned in the Question that it's -10 then it surely is wrong.
Post by: **RoRo** on May 25, 2011, 06:04:34 pm
Another question, Exercise 5A, Q8, Find the term independent of x in the expansion of (x2 - 2/2x)3.
What's that? :o
Post by: iluvme on May 25, 2011, 06:29:07 pm
Alright, thanks! :)
Another question, Exercise 5A, Q8, Find the term independent of x in the expansion of (x2 - 2/2x)3.
What's that? :o
A term without x.
Post by: **RoRo** on May 25, 2011, 06:35:36 pm
A term without x.
okay, thank you! :)
Post by: iluvme on May 25, 2011, 06:37:16 pm
okay, thank you! :)
Welcome. :)
You want me to do the expansion?
Post by: **RoRo** on May 25, 2011, 07:02:04 pm
Welcome. :)
You want me to do the expansion?
I got it, you first expand the term and then you see where you can cancel the x terms and then you just get the value.
Post by: iluvme on May 25, 2011, 07:05:44 pm
I got it, you first expand the term and then you see where you can cancel the x terms and then you just get the value.
Yup :)
Post by: **RoRo** on May 25, 2011, 07:15:18 pm
Yup :)
Can you please tell me how do you do this tabular method for differentiation? Or just get me a link that explains it, because it's not explained in the textbook and when I google, I'm getting Integration by parts, which is actually in C4!
Post by: iluvme on May 25, 2011, 07:19:59 pm
Can you please tell me how do you do this tabular method for differentiation? Or just get me a link that explains it, because it's not explained in the textbook and when I google, I'm getting Integration by parts, which is actually in C4!
Tabular method for differentiation? Never heard of it TBH. :-\
Post by: **RoRo** on May 25, 2011, 07:30:21 pm
Tabular method for differentiation? Never heard of it TBH. :-\
Check the pink box on page 153 in your C2 textbook!
Post by: iluvme on May 25, 2011, 07:33:21 pm
Check the pink box on page 153 in your C2 textbook!
Don't do Edexcel.
Post by: The Golden Girl =D on May 25, 2011, 08:15:03 pm
Check the pink box on page 153 in your C2 textbook!
I do Edexcel and that's the first time I hear that o.O
@ iluvme Hint : in this case you need to use the tabular method and consider the gradient on either side of the stationary point.
Post by: Malak on May 25, 2011, 08:59:51 pm
For example tan(theta + pie/3) =1 in range 0< theta < 2pie
Tyvm
Post by: iluvme on May 25, 2011, 10:04:15 pm
How do u solve trig eq. in radians
For example tan(theta + pie/3) =1 in range 0< theta < 2pie
Tyvm
Tan(theta+pie/3)=1
Get the new range, pie/3< theta+pie/3< 2pie+pie/3
Theta +pie = Tan-1(1)
Since tan-1is positive, you get the values in the first and third quadrants.
theta+pie/3= 0.78, (Pie+0.78), (2pie+0.78)
Theta + pie/3 is not equal to 0.78 because of the new range we got, Pie/3<theta +pie/3 < 2pie+pie/3
Therefore,
Theta + pie/3 = 3.93, 7.06
Theta = 2.88, 6.01
Remember to change your calculator mode to Radians
Hope you got it :)
Post by: ~ Miss Relina ~ on June 19, 2011, 12:23:53 pm
Post by: Malak on June 19, 2011, 01:04:04 pm
guys do i have to memorise anything for C1 paper like square root 2Sorry, What do you mean? Are you asking about the formulas?
Post by: ~ Miss Relina ~ on June 19, 2011, 11:19:31 pm
Sorry, What do you mean? Are you asking about the formulas?no i mean if r not going to use a calculator so do we have to memorise for example something like squre root of 2 as we would not be able to use calculator for it and also we can not get it by mind like for example square root of 4
Post by: Malak on June 20, 2011, 06:24:27 pm
no i mean if r not going to use a calculator so do we have to memorise for example something like squre root of 2 as we would not be able to use calculator for it and also we can not get it by mind like for example square root of 4For C1 just leave it in the simplified surd form (e.g. Root 2), usually the Q. states that give answer in exact form which means you have to give it in surd form. You definitely don't have to remember the decimal answer.
If for example its Sq. root 12, you can simplify further to give 2sq. root3
Post by: ~ Miss Relina ~ on June 21, 2011, 02:00:01 pm
For C1 just leave it in the simplified surd form (e.g. Root 2), usually the Q. states that give answer in exact form which means you have to give it in surd form. You definitely don't have to remember the decimal answer.well Thanks alot
If for example its Sq. root 12, you can simplify further to give 2sq. root3
Post by: ~ Miss Relina ~ on August 08, 2011, 09:57:43 pm
Post by: Romeesa-Chan on August 08, 2011, 10:39:49 pm
excuse me guys can anyone give me the links for maths as-level ??
Past papers? :-\
https://studentforums.biz/pastpapers/
Post by: The Golden Girl =D on August 09, 2011, 12:51:06 pm
excuse me guys can anyone give me the links for maths as-level ??
What exactly do you want ? Please specify.
Post by: ~ Miss Relina ~ on August 09, 2011, 08:05:38 pm
What exactly do you want ? Please specify.oh sorry
i mean any websites for resources for math notes revision guides practices and etc....
Post by: ~ Miss Relina ~ on August 09, 2011, 08:09:25 pm
Past papers? :-\Thanks dear
https://studentforums.biz/pastpapers/
Post by: The Golden Girl =D on August 09, 2011, 09:12:20 pm
oh sorry
i mean any websites for resources for math notes revision guides practices and etc....
Well then I hope the following helps you ;)
Since I don't know which one you will take * that is S1 or M1 * then I'll give you the link directly for all the Past papers ;)
Past Papers :
http://www.freeexampapers.com/past_papers.php?l=Past_Papers%2FA+Level%2FMaths%2FEdexcel/ (http://www.freeexampapers.com/past_papers.php?l=Past_Papers%2FA+Level%2FMaths%2FEdexcel/)
and this
https://studentforums.biz/reference-material-83/edexcel-c1-and-c2-solution-bank-practice-papers/
(https://studentforums.biz/reference-material-83/edexcel-c1-and-c2-solution-bank-practice-papers/)
As well as the link Romeesa gave you ;)
Resources :
http://www.thestudentroom.co.uk/showthread.php?t=883493&page=3http://www.thestudentroom.co.uk/showthread.php?t=883493&page=3
(http://www.thestudentroom.co.uk/showthread.php?t=883493&page=3http://www.thestudentroom.co.uk/showthread.php?t=883493&page=3)
https://studentforums.biz/revison-notes/c2-edexcel-notes/
(https://studentforums.biz/revison-notes/c2-edexcel-notes/)
https://studentforums.biz/revison-notes/m1-mechanics-help-needed-d/
(https://studentforums.biz/revison-notes/m1-mechanics-help-needed-d/)
Revision Guides :
I'm quite sorry but I only have a Hard Copy of it. I can't scan it for many reasons including the fact that it's HUGE ::)
Don't forget to include me in your Prayers ;)
Post by: ~ Miss Relina ~ on August 11, 2011, 06:13:32 pm
check up exercise
3E question
4a and b
Post by: The Golden Girl =D on August 11, 2011, 06:43:40 pm
excuse me guys who have the c1 book plz
check up exercise
3E question
4a and b
If you have the CD , Open it and there is a solution Bank "Capital letter S" press on it and you'll get to know the answer ;)
But I already did that :P so Here is the Answer from the Cd's Solution Bank :
(a) a=1, b=-k, c=k+3
b2-4ac<0 for no real roots, so
k2-4(k+3)<0
k2-4k?12<0
(k-6)(k+2)<0
-2<k<6
(b) a=p, b=p, c=-2
b2-4ac<0 for no real roots, so
p2+8p<0
p(p+8)<0
-8<p<0
I hope I helped =D
Post by: ~ Miss Relina ~ on August 11, 2011, 09:22:46 pm
If you have the CD , Open it and there is a solution Bank "Capital letter S" press on it and you'll get to know the answer ;)
But I already did that :P so Here is the Answer from the Cd's Solution Bank :
(a) a=1, b=-k, c=k+3
b2-4ac<0 for no real roots, so
k2-4(k+3)<0
k2-4k?12<0
(k-6)(k+2)<0
-2<k<6
(b) a=p, b=p, c=-2
b2-4ac<0 for no real roots, so
p2+8p<0
p(p+8)<0
-8<p<0
I hope I helped =D
oh no unfortunately i donot have the Cd
but the thing which I donot undersatnd why did you use b2-4ac<0 ???
actually I used b2>4ac as the question asks for real roots that's my point and if u can plz explain thanks anyway
Post by: The Golden Girl =D on August 11, 2011, 09:28:36 pm
oh no unfortunately i donot have the Cd
but the thing which I donot undersatnd why did you use b2-4ac<0 ???
actually I used b2>4ac as the question asks for real roots that's my point and if u can plz explain thanks anyway
I suggest you borrow it from someone , I found it very beneficial =] Or Perhaps check the Solution Bank that Ang3l posted ;)
Delta is b2 - 4ac
There is NO such thing as B2 > 4ac
if they say REAL roots it means greater than or Equal/Greater than.
if they say NO root then it is less than
then you solve it the way it was written in the solution bank =]
I hope you got my point =]
Post by: ~ Miss Relina ~ on August 11, 2011, 09:50:39 pm
I suggest you borrow it from someone , I found it very beneficial =] Or Perhaps check the Solution Bank that Ang3l posted ;)i know i ;m annoying by askin too much but i am sorry
Delta is b2 - 4ac
There is NO such thing as B2 > 4ac
if they say REAL roots it means greater than or Equal/Greater than.
if they say NO root then it is less than
then you solve it the way it was written in the solution bank =]
I hope you got my point =]
why did you use delta b2 - 4ac
and for that one ''' There is NO such thing as B2 > 4ac'''
i got it from lesson no. 2.6 the note at the end of page 23
' all this is in your mezan 7sanatik isa jazak Allaho Kairan ''
Post by: The Golden Girl =D on August 12, 2011, 03:55:23 pm
i know i ;m annoying by askin too much but i am sorry
why did you use delta b2 - 4ac
and for that one ''' There is NO such thing as B2 > 4ac'''
i got it from lesson no. 2.6 the note at the end of page 23
' all this is in your mezan 7sanatik isa jazak Allaho Kairan ''
You're not annoying :)
There seems to be a misunderstanding here.
To be honest I never referred to the book if I misunderstood something since I ended up getting even more confused ,I always asked my private tutor or my teacher. So I'll take a Pic of the notes that talk about this and hopefully it'll clear your doubt =]
Post by: ~ Miss Relina ~ on August 12, 2011, 06:10:09 pm
You're not annoying :)
There seems to be a misunderstanding here.
To be honest I never referred to the book if I misunderstood something since I ended up getting even more confused ,I always asked my private tutor or my teacher. So I'll take a Pic of the notes that talk about this and hopefully it'll clear your doubt =]
oh now I am okay thanks alot dear
Post by: The Golden Girl =D on August 12, 2011, 08:08:17 pm
' all this is in your mezan 7sanatik isa jazak Allaho Kairan ''
Ameen mate =] ... Wa Jazaki :) , Are you an Arab ? ::)
oh now I am okay thanks alot dear
You're Welcome =]
Post by: ~ Miss Relina ~ on August 13, 2011, 05:44:49 pm
Ameen mate =] ... Wa Jazaki :) , Are you an Arab ? ::)yes I'm an ARabian
You're Welcome =]
but here is another problem
in 3F Q 9
the method iI used to get the answer was b2- 4ac>0
and it works anyways thhanks alot for previous notes posts they were really helpful
JAzakiAllah Kairan
Post by: ~ Miss Relina ~ on August 13, 2011, 08:19:08 pm
or should I make the two equation equal to get the answer ???
Thanks in advance
Post by: Malak on August 13, 2011, 08:52:50 pm
well in C1 sometimes they ask for finding the coordinates of points of intersection of cubic and quadratic equations so do I have to exactly the graph for each equation toget the points of intersectionNo, you dont have to sketch the graph exactly.
or should I make the two equation equal to get the answer ???
Thanks in advance
If you could actually post the Question then I can help better.
But check this website for great explanations: http://examsolutions.co.uk/A-Level-maths-tutorials/Edexcel/C1/index.php (http://examsolutions.co.uk/A-Level-maths-tutorials/Edexcel/C1/index.php)
It helped me a lot with C1.
Post by: The Golden Girl =D on August 13, 2011, 09:34:22 pm
yes I'm an ARabian
but here is another problem
in 3F Q 9
the method iI used to get the answer was b2- 4ac>0
and it works anyways thhanks alot for previous notes posts they were really helpful
JAzakiAllah Kairan
It is Arab not Arabian ;)
I don't really get what You're saying.
You told me earlier that you wanted to solve exc 3E Q 4a and 4b by b2 > 4ac
and I told you that that concept is wrong or in other words is not used to solve such Questions as the Questions in exc 3 E and 3 F .
so as I told you ,if you see the Words "Real Roots" you use greater than or Equal/ greater than
Here is the way it is solved in the Solution Bank ;)
Quote
a=k, b=8, c=5
x={ - b plus or minus (square root ?b2?4ac) }/2a
b2-4ac greater than or equal 0 for real roots. So
82-4k * 5 greater than or equal 0
64-20k greater than or equal 0
64 less than or equal 20k
64 /20 less than or equal k
k less than or equal 3.2
I used less than or equal and greater than or equal since I wasn't able to find the Symbols =,=
I hope you got it =]
You're Welcome ;) ... Wa Jazaki :D
Post by: The Golden Girl =D on August 13, 2011, 10:16:21 pm
well in C1 sometimes they ask for finding the coordinates of points of intersection of cubic and quadratic equations so do I have to exactly the graph for each equation toget the points of intersection
or should I make the two equation equal to get the answer ???
Thanks in advance
Yes you just equate the Equations and get the Intersection Points between the TWO graphs in case there are TWO graphs to draw ;)
But if it's just ONE graph then you have to solve as follow:
Firstly, you solve the equation as though that the X = 0 ,in other words you're finding the point of intersection of the graph with the y-axis.
Secondly, You solve the equation as though the Y = 0 , in other words you're finding the point of intersection of the graph with the x-axis.
I hope I cleared your doubt :)
No, you dont have to sketch the graph exactly.
If you could actually post the Question then I can help better.
But check this website for great explanations: http://examsolutions.co.uk/A-Level-maths-tutorials/Edexcel/C1/index.php (http://examsolutions.co.uk/A-Level-maths-tutorials/Edexcel/C1/index.php)
It helped me a lot with C1.
Seconded.
It helped me with C1 , C2 as well as M1 :D ... You should check that website ;)
Post by: ~ Miss Relina ~ on August 13, 2011, 11:12:08 pm
It is Arab not Arabian ;)
I don't really get what You're saying.
You told me earlier that you wanted to solve exc 3E Q 4a and 4b by b2 > 4ac
and I told you that that concept is wrong or in other words is not used to solve such Questions as the Questions in exc 3 E and 3 F .
so as I told you ,if you see the Words "Real Roots" you use greater than or Equal/ greater than
Here is the way it is solved in the Solution Bank ;)
I used less than or equal and greater than or equal since I wasn't able to find the Symbols =,=
I hope you got it =]
You're Welcome ;) ... Wa Jazaki :D
well Thanks for the ARAB cuold you inbox me why ???
and for the main mathematical issue
what I meant is that another example which is 3F Q 9
I didnot use b2 - 4ac < 0
but used b2-4ac > 0 can you clarify that ???
Post by: ~ Miss Relina ~ on August 13, 2011, 11:16:18 pm
No, you dont have to sketch the graph exactly.
If you could actually post the Question then I can help better.
But check this website for great explanations: http://examsolutions.co.uk/A-Level-maths-tutorials/Edexcel/C1/index.php (http://examsolutions.co.uk/A-Level-maths-tutorials/Edexcel/C1/index.php)
It helped me a lot with C1.
Thanks alot well if you want to get tired no problem for me :D ;D ;)
well here we go :
aon the same axes sketch the curves given by y=(x-1)3 and y=(x-1)(1+x)
b find the coordinates of the points of intersection.
Thanks in advance
Post by: The Golden Girl =D on August 13, 2011, 11:20:03 pm
well Thanks for the ARAB cuold you inbox me why ???
and for the main mathematical issue
what I meant is that another example which is 3F Q 9
I didnot use b2 - 4ac < 0
but used b2-4ac > 0 can you clarify that ???
Sure =]
In the Question it stated the words "Real root" , and when you see that then you should know that you should undoubtedly use " > or greater than or equal" .
How did I know that , look into these notes I posted prior ;)
EDIT: Regarding your doubt about Sketching Graphs ,I found some notes that could help.
Attached.
Source of the Notes : http://getrevising.co.uk/resources/c1_sketching_curves
Post by: ~ Miss Relina ~ on August 13, 2011, 11:41:56 pm
Sure =]
In the Question it stated the words "Real root" , and when you see that then you should know that you should undoubtedly use " > or greater than or equal" .
How did I know that , look into these notes I posted prior ;)
EDIT: Regarding your doubt about Sketching Graphs ,I found some notes that could help.
Attached.
Source of the Notes : http://getrevising.co.uk/resources/c1_sketching_curves
well okay the notes say the same thing I say too
in both questions which you had solved before the question said real roots which should be greater than or equal but you used less than or equal
that's my point dear ;)
Post by: The Golden Girl =D on August 13, 2011, 11:47:17 pm
well okay the notes say the same thing I say too
in both questions which you had solved before the question said real roots which should be greater than or equal but you used less than or equal
that's my point dear ;)
At first I used Greater than or Equal but then I used less than an equal.
Clarification :
Quote
a=k, b=8, c=5
x={ - b plus or minus (square root ?b2?4ac) }/2a
b2-4ac greater than or equal 0 for real roots. So
82-4k * 5 greater than or equal 0
64-20k greater than or equal 0
64 less than or equal 20k
64 /20 less than or equal k
k less than or equal 3.2
The on in RED and BOLD. Why I changed it into less than becase if you noticed we moved the 20 K to the RIGHT side of the inequality and you should know whenever you MOVE the unkown *one with X or K example : 20 K* you CHANGE the Sign ;)
I hope you got my point ;)
Post by: ~ Miss Relina ~ on August 14, 2011, 12:01:55 am
At first I used Greater than or Equal but then I used less than an equal.
Clarification :
The on in RED and BOLD. Why I changed it into less than becase if you noticed we moved the 20 K to the RIGHT side of the inequality and you should know whenever you MOVE the unkown *one with X or K example : 20 K* you CHANGE the Sign ;)
I hope you got my point ;)
oh ralyy I've never heard about that ''whenever you MOVE the unkown *one with X or K example : 20 K* you CHANGE the Sign ''
what I know is that when we just divide or multioly an inquality we change signs only
anyways thanks
but what about the first two questions in my first post Q 3 E [4 a and b] they mentioned real roots but you used less than or equal to get the answer right ???
by the way I know your method is right ofcourse because you git the right answer but i am just confused sorry of the conflict between the two examples 3E-4a and 3 F-9
Thanks for all your efforts I have made you tired I think
Post by: The Golden Girl =D on August 14, 2011, 12:22:10 am
oh ralyy I've never heard about that ''whenever you MOVE the unkown *one with X or K example : 20 K* you CHANGE the Sign ''
what I know is that when we just divide or multioly an inquality we change signs only
anyways thanks
but what about the first two questions in my first post Q 3 E [4 a and b] they mentioned real roots but you used less than or equal to get the answer right ???
by the way I know your method is right ofcourse because you git the right answer but i am just confused sorry of the conflict between the two examples 3E-4a and 3 F-9
Thanks for all your efforts I have made you tired I think
You seem to be REALLY confused. Habeebti ,I'll explain everything inshAllah ;) ....It's Alright I'm free for now so I am not tired =]
Okay I'm going to Quote what I said and Color it in RED so that you see that there is no need to get confused AT ALL ;)
(a) a=1, b=-k, c=k+3
b2-4ac<0 for no real roots, so
k2-4(k+3)<0
k2-4k?12<0
(k-6)(k+2)<0
-2<k<6
(b) a=p, b=p, c=-2
b2-4ac<0 for no real roots, so
p2+8p<0
p(p+8)<0
-8<p<0
This is NO real roots which is why it is < ;)
Quote
a=k, b=8, c=5
x={ - b plus or minus (square root ?b2?4ac) }/2a
b2-4ac greater than or equal 0 for real roots. So
82-4k * 5 greater than or equal 0
64-20k greater than or equal 0
64 less than or equal 20k
64 /20 less than or equal k
k less than or equal 3.2
This is Real roots which is why it is greater than or equal ;)
I hope you got my point =]
Don't forget to include me in Your Prayers =D
Post by: ~ Miss Relina ~ on August 14, 2011, 07:54:19 pm
well anyone knows any information about How to find the vertical and horizontal asymptotes of the reciprocal 1/x ??? :)
Post by: The Golden Girl =D on August 14, 2011, 08:24:23 pm
I didn't get your request , can you elaborate please ?
Post by: ~ Miss Relina ~ on August 15, 2011, 06:41:11 pm
You're welcome .. don't forget to include me in your prayers ;)
I didn't get your request , can you elaborate please ?
never mind I searched and got it Allhamduliallah
Post by: The Golden Girl =D on August 15, 2011, 09:22:37 pm
Post by: ~ Miss Relina ~ on September 17, 2011, 09:04:37 pm
my problem is that i donot know how to ge the r2 however Iwas able to get the centre point right I found in the back of the book answers but I was totally confused and don't know the exact method
ahhhh.....By the way you don't need to do the whole exercise just one example would be fine
Thanks in advance :D
Post by: Malak on September 18, 2011, 04:50:37 pm
Here is what you do.
When a quadratic equation is given use completing the square to find the center and radius and then find the equation of circle.
See this for the best explanation :P
http://www.examsolutions.co.uk/maths-revision/core-maths/coordinate-geometry/circle/centre-radius/tutorial-1.php (http://www.examsolutions.co.uk/maths-revision/core-maths/coordinate-geometry/circle/centre-radius/tutorial-1.php)
Post by: ~ Miss Relina ~ on September 18, 2011, 06:48:26 pm
Okay sorry for the previous reply ::)
Here is what you do.
When a quadratic equation is given use completing the square to find the center and radius and then find the equation of circle.
See this for the best explanation :P
http://www.examsolutions.co.uk/maths-revision/core-maths/coordinate-geometry/circle/centre-radius/tutorial-1.php (http://www.examsolutions.co.uk/maths-revision/core-maths/coordinate-geometry/circle/centre-radius/tutorial-1.php)
well thanks for the link I'll check it
but now you made me curious ;D about ur previous reply as I didn't see it ......I'm going to accept it as long as u are helping me thank you very much :) hope Allah help you with anything you want :)
Post by: ~ Miss Relina ~ on September 19, 2011, 11:54:30 am
x2 + y2 + 4x + 9y +3 = 0
my answer is :
(x+2)2-4 + (y + 4.5 )2 -20.25 +3 = 0 then adding the numbers will give -13.25 so the equation will be :
(x+2)2 + (y+ 4.5 )2 = 13.25
so centre point should be ( -2 ,-4.5 ) and radius= square root 13.25
right ?????
why the book did it like that :
(x+2)2 + (y + 3 )2 -10= 0
(x+2)2 + (y + 3 )2 = 10
so centre point is ( -2,-3) and radius=square root10
well.............the answer to the question shouldn't be like that unless there is a mistake and instead of 9y it should be 6y
plz guys try it and tell me as I found all the other examples like that >:( >:( >:(
Post by: Arthur Bon Zavi on September 19, 2011, 12:52:24 pm
well.............the answer to the question shouldn't be like that unless there is a mistake and instead of 9y it should be 6y
Yes, it should be 6y and not 9y.
Post by: ~ Miss Relina ~ on September 19, 2011, 01:09:19 pm
Yes, it should be 6y and not 9y.thanks a lot for your respond can you check others too :(
I've attached questions and book answers and first one done but others not can u plz check them too
sorry for disturbing :-[
Post by: Arthur Bon Zavi on September 19, 2011, 01:16:21 pm
thanks a lot for your respond can you check others too :(
I've attached questions and book answers and first one done but others not can u plz check them too
sorry for disturbing :-[
You can do the rest by applying the same equation of the circle.
Post by: ~ Miss Relina ~ on September 19, 2011, 02:15:23 pm
Post by: Arthur Bon Zavi on September 19, 2011, 03:19:58 pm
I know but can it be that all the questions are wrong ??? ??? ??? ??? ??? ??? ???
It can be that. Can you post the answers for the parts (b), (c) and (d) ?
Post by: ~ Miss Relina ~ on September 19, 2011, 06:06:33 pm
thanks a lot for your respond can you check others too :(they 're attached in my previous post https://studentforums.biz/math-146/c1-doubts-here!!!!!/480/?topicseen (https://studentforums.biz/math-146/c1-doubts-here!!!!!/480/?topicseen)
I've attached questions and book answers and first one done but others not can u plz check them too
sorry for disturbing :-[
Post by: Arthur Bon Zavi on September 19, 2011, 06:29:46 pm
they 're attached in my previous post https://studentforums.biz/math-146/c1-doubts-here!!!!!/480/?topicseen (https://studentforums.biz/math-146/c1-doubts-here!!!!!/480/?topicseen)
For me to see, they should be clearly visible right ? ::)
Post by: ~ Miss Relina ~ on September 20, 2011, 01:54:29 pm
For me to see, they should be clearly visible right ? ::)
okay if you have c2 book the question are on page 66 & answers at the back of the book
if you don't inform me so that I would find another way :D
Thanks in advance
Post by: Arthur Bon Zavi on September 20, 2011, 04:52:25 pm
okay if you have c2 book the question are on page 66 & answers at the back of the book
if you don't inform me so that I would find another way :D
Thanks in advance
I do CIE. Now straightly post the answers.
Post by: ~ Miss Relina ~ on September 20, 2011, 04:57:52 pm
I do CIE. Now straightly post the answers.well I will post it
(b) centre (-5/2 , 3/2) radius 37/2
(c) centre (-2,-15/4) raduis 309/2square root 2
(d)centre (2,-2) radius square root 13/square root 2
these are the last steps not the whole just plz do it and when your final answers don't agree like me it would be a book mistake
Post by: Arthur Bon Zavi on September 20, 2011, 05:13:45 pm
well I will post it
(b) centre (-5/2 , 3/2) radius 37/2
(c) centre (-2,-15/4) raduis 309/2square root 2
(d)centre (2,-2) radius square root 13/square root 2
these are the last steps not the whole just plz do it and when your final answers don't agree like me it would be a book mistake
Verify your answers with me. If they match, the printing in the book has mistakes.
(a) coordinates (-2.5 , 1.5) and radius =
(b) coordinates (-2 , -3.75) and radius =
(c) coordinates (2, -2) and radius =
Post by: ~ Miss Relina ~ on September 20, 2011, 06:31:03 pm
Verify your answers with me. If they match, the printing in the book has mistakes.
(a) coordinates (-2.5 , 1.5) and radius =
(b) coordinates (-2 , -3.75) and radius =
(c) coordinates (2, -2) and radius =
yaaaaaaaaaay my answers are exactly the same as you
so we are right and book is wrong ;)
Post by: Arthur Bon Zavi on September 20, 2011, 06:34:41 pm
yaaaaaaaaaay my answers are exactly the same as you
so we are right and book is wrong ;)
Now stop worrying and work on the rest.
Post by: ~ Miss Relina ~ on September 20, 2011, 06:43:32 pm
EDIT : I added one more
Post by: Arthur Bon Zavi on September 20, 2011, 06:47:35 pm
here is another question show me steps plz ;D ;D
Is it urgent ? Can you wait till tomorrow ?
Post by: ~ Miss Relina ~ on September 20, 2011, 06:49:05 pm
Is it urgent ? Can you wait till tomorrow ?no dear take your time
Post by: astarmathsandphysics on September 22, 2011, 11:50:10 am
q1 and q2 above
Post by: ~ Miss Relina ~ on September 22, 2011, 12:49:23 pm
hereoh thanx alot for your reply you revived me again ;D ;D
q1 and q2 above
I've been waiting for an answer and asked in other forums too Thanks alot for doing the steps tooooo :D :D :D :D
Post by: astarmathsandphysics on September 23, 2011, 10:56:37 am
Post by: ~ Miss Relina ~ on September 23, 2011, 01:29:45 pm
second one was tricky.ya you know that I've finshed the papers in my home for doing it only ;D
Post by: WARRIOR on September 23, 2011, 03:58:56 pm
use algebra ro find the se of values for x for which x (x-5 ) > 36 [this one is easy i got the answer right x<-4,x>9]
use ur answer from part a , find the set of values of y for which y^2 ( y^2 -5 ) > 36
how??
thanks in advanced. seems easy but not for me.
Post by: Arthur Bon Zavi on September 23, 2011, 04:24:33 pm
how?
use algebra ro find the se of values for x for which x (x-5 ) > 36 [this one is easy i got the answer right x<-4,x>9]
use ur answer from part a , find the set of values of y for which y^2 ( y^2 -5 ) > 36
how??
thanks in advanced. seems easy but not for me.
Let y4 be x2.
Now try and get the values of x and then of y by :
x2 = y4
Post by: astarmathsandphysics on September 27, 2011, 02:55:58 pm
Solve
Solution
Hence
Post by: MKh on September 28, 2011, 01:17:14 pm
This is quite a trickly log question
Solve
Solution
Henceor
Is it from the C2 specification? and which chapter?
Thanks.
Post by: astarmathsandphysics on September 28, 2011, 09:31:33 pm
Post by: WARRIOR on October 01, 2011, 01:30:29 pm
a show that A = .25 ( 10 + 3x - x^2) --- easy.
b.use the method of completing the square or otherwise , to find the maximum value of A and give the corresponding value of x.
i really dont understand HOW THE ANSWER IS 3 1/16 WHEN X = 1/2
this is exactly the same.
In triangle ABC , AB= (2-X) BC = (X+1) ANGLE B = 120
show that AC^2 = x^2 - x + 7---easy
find the value of x for which AC has a minium value. The answer is 1/2
thanks
Post by: astarmathsandphysics on October 01, 2011, 02:41:56 pm
0.25(10+3x-x^2)=0.25(10+(-1.5^2+3x-x^2)+1.5^2)=0.25(10-(x-1.5)^2-2.25)=0.25(12.25-(x-1.5)^2)
Put x=1.5 then x-1.5=0 and 0.25(12.25-(x-1.5)^2)=0.25*12.25=3 1/16
For the second question
x^2-x+7=(x-0.5)^2-0.5^2+7=(x-0.5)^2+6.75
min when (x-0.5)^2=0 so x=1/2
Post by: WARRIOR on October 05, 2011, 06:18:38 am
Good morning !
When drawing SKETCH graphs for logarithms (EXPONENTIAL EQUATIONS ) do we make a simple table of values? because they always if us a range of values of x?
PS : SORRY I JUST REALISED IT SAID DRAW AN ACCURATE GRAPH.
Post by: ~ Miss Relina ~ on October 05, 2011, 02:42:43 pm
a question to sove plz
click on the image for larger view
Post by: MKh on October 05, 2011, 03:16:23 pm
I have a few doubts in the Edexcel C1 textbook.
Edexcel C1 textbook - Chapter 5: Coordinate Geometry in the (x,y) plane - Mixed Exercise (5F) :-
Page 88, Question no. 8, part (c) --> Calculate the area of triangle ACD.
Page 89, question 12, part (b) --> find the area of triangle OAB.
Both triangles are on the (x,y) plane and the coordinates of all three vertices of each of them is known.
We have already studied the formula for finding the area of a triangle provided the length of two of its sides and the angle between them. We will also meet the same formula in the C2 module. Here, I don't think we have to use the same formula for finding the area of a triangle that we came across in IGCSE math.
I will look through the step-by-step solutions available with me but, for the moment, it would be great if someone who have solved this could help out.
Thank you.
Post by: astarmathsandphysics on October 05, 2011, 03:26:55 pm
Post by: WARRIOR on October 05, 2011, 03:44:21 pm
LOG(base 2)X + LOG (base 4 ) = 2 Answer is 2.52 how?
Post by: WARRIOR on October 05, 2011, 03:50:58 pm
I don't have the textbook. can you scan the questions
Q8 = The straight line L1 passes through the points A and B with coordinates ( 0,-2 ) and (6,7) respectively.
a. find the equation of L1 in the form of y= mx + c
The straight line L2 with equation x + y = 8 cuts the y-axis at point C. The lines L1 and L2 intersect t the point D.
b. Calculate the coordinated of point D
c. Calculate the area of ACD
Q 12
Find an equation of a straight line passing through the points with coordinates (-1,5) (4,-2) , giving your answer in the form of ax+ by +c=0
where a , b , c are integers.
The line crosses the x-axis at the point A and the y -axis at the point B , and O is the origin
b.find the area of OAB.
thse are mkH's questions.
Post by: ~ Miss Relina ~ on October 05, 2011, 04:57:49 pm
hi i need help in a log question
LOG(base 2)X + LOG (base 4 ) = 2 Answer is 2.52 how?
there's something miissing in this question ???
Post by: ~ Miss Relina ~ on October 05, 2011, 04:58:21 pm
here we go :D
a question to sove plz
click on the image for larger view
anyone answer me plz :(
Post by: The Golden Girl =D on October 05, 2011, 05:04:10 pm
here we go :D
a question to sove plz
click on the image for larger view
I've got that Question solved somewhere in my C2 copybook ... I'll upload it tomorrow when I come back from School ' I'm Busy now =,='
Remind me because I tend to forget ;)
Post by: ~ Miss Relina ~ on October 05, 2011, 05:10:55 pm
I've got that Question solved somewhere in my C2 copybook ... I'll upload it tomorrow when I come back from School ' I'm Busy now =,='
Remind me because I tend to forget ;)
I've got that simple answerin my C2 book too which I don't understand and I need steps to be done for me
Thanks in advance
Post by: The Golden Girl =D on October 05, 2011, 05:13:14 pm
I've got that simple answerin my C2 book too which I don't understand and I need steps to be done for me
Thanks in advance
Okay I'll see what I kind do tomorrow inshAllah =]
Post by: WARRIOR on October 05, 2011, 05:26:22 pm
there's something miissing in this question ???
yea sorry
LOG (base 2 )X + LOG(base 4) X = 2
i have another question
given that p = log( base q) 16 , express in terms of p
log ( base q) 8Q
this is in C2 , PAGE 48 , 3 , B
Post by: astarmathsandphysics on October 05, 2011, 10:56:06 pm
you have to change the base from 4 to 2
use log_a x = (log_b x)/(log_b a) so log_4 x = (log_2 x)/(log_2 4)=(log_2 x)/2
then log_2 x +1/2 log?2 x =2
3/2 log_2 x = 2 so log_2 x =4/3 so x =2^(4/3)
2nd question
log_q 8q = log_q 8 +log_q q = log_q 2^3 + 1 = log_q (16^(1/4))^3 + 1 = log_q 16^(3/4) +1 = 3/3 log_q 16 +1 =3/4 p +1
Post by: WARRIOR on October 06, 2011, 09:52:56 am
LOG (base 2 )X + LOG(base 4) X = 2I undestood them now thanks to you. Thanks a milllion sir !
you have to change the base from 4 to 2
use log_a x = (log_b x)/(log_b a) so log_4 x = (log_2 x)/(log_2 4)=(log_2 x)/2
then log_2 x +1/2 log?2 x =2
3/2 log_2 x = 2 so log_2 x =4/3 so x =2^(4/3)
2nd question
log_q 8q = log_q 8 +log_q q = log_q 2^3 + 1 = log_q (16^(1/4))^3 + 1 = log_q 16^(3/4) +1 = 3/3 log_q 16 +1 =3/4 p +1
Post by: astarmathsandphysics on October 06, 2011, 10:55:50 am
Post by: The Golden Girl =D on October 06, 2011, 01:31:17 pm
here we go :D
a question to sove plz
click on the image for larger view
Next time you post a C2 Question make sure you mention which Exercise and which page it's on plus which number it is ,it'll make it much easier for us to answer you 'those who do have the C2 book'.
HERE you go :D
P = 100 cm
A = A cm2
a) P = 2r + r theta => P stand for Perimeter ;)
100 = 2r + r theta
r theta = 100 - 2r
theta = (100/r) - 2
Area = 0.5 * r2 * theta
= 0.5 * r2 ( (100/r) - 2 )
= 0.5 (100 r - 2r2)
So, A = 50r - r2
b) i) r = (-b/2a) = (-50/2(-1)) = 25 So , r = 25
A is Maximum Value because a = -1 < 0 => I don't know what my teacher meant by that ' Sorry I can't remember what he said about this thing =,= '
ii) theta = (100/25) -2 = 4 - 2 = 2 rad So , theta = 2 rad
iii)
Area = 50(25) - (25)2
= 1250 - 625
So , A = 625 cm2
I'll attach An Image so you understand what my teacher did to the drawing so check Attachment =]
Post by: astarmathsandphysics on October 06, 2011, 02:04:09 pm
Post by: The Golden Girl =D on October 06, 2011, 02:05:36 pm
Post by: MKh on October 07, 2011, 02:08:44 pm
I don't have the textbook. can you scan the questions
Thanks for replying Sir. I am srry i don't have a scanner. But Kimo Jesus has typed the questions here.
Post by: ~ Miss Relina ~ on October 07, 2011, 06:24:32 pm
Next time you post a C2 Question make sure you mention which Exercise and which page it's on plus which number it is ,it'll make it much easier for us to answer you 'those who do have the C2 book'.
HERE you go :D
P = 100 cm
A = A cm2
a) P = 2r + r theta => P stand for Perimeter ;)
100 = 2r + r theta
r theta = 100 - 2r
theta = (100/r) - 2
Area = 0.5 * r2 * theta
= 0.5 * r2 ( (100/r) - 2 )
= 0.5 (100 r - 2r2)
So, A = 50r - r2
b) i) r = (-b/2a) = (-50/2(-1)) = 25 So , r = 25
A is Maximum Value because a = -1 < 0 => I don't know what my teacher meant by that ' Sorry I can't remember what he said about this thing =,= '
ii) theta = (100/25) -2 = 4 - 2 = 2 rad So , theta = 2 rad
iii)
Area = 50(25) - (25)2
= 1250 - 625
So , A = 625 cm2
I'll attach An Image so you understand what my teacher did to the drawing so check Attachment =]
well thanx alot
:D
Post by: ~ Miss Relina ~ on October 19, 2011, 06:44:41 pm
attached
if u have C2 book then its exercise 10D Q1 H page 167
Thanks in advance
Post by: astarmathsandphysics on October 19, 2011, 09:42:03 pm
4 sin x = sin x/cos x
Sin x (4 -1/cos x)=0
Sin x =0 so x =0,180,360
4-1/cos x =0 so cos x =1/4 so x =cos^-1 1/4 or 180-cos^-1 1/4
Post by: ~ Miss Relina ~ on October 22, 2011, 02:20:19 pm
4 sin x = tan x
4 sin x = sin x/cos x
Sin x (4 -1/cos x)=0
Sin x =0 so x =0,180,360
4-1/cos x =0 so cos x =1/4 so x =cos^-1 1/4 or 180-cos^-1 1/4
Thanks alot Sir for your support :)
Post by: astarmathsandphysics on October 23, 2011, 10:07:13 am
Post by: ~ Miss Relina ~ on October 25, 2011, 03:33:44 pm
for who have C2 book its on page 193 Q 6 in the revision exercise 3
what I don't understand is that the book's answer is 4 and when I did integration of the two parts of my sketch , one area was -ve and other was +ve so I added them ignoring the -ve sign so...............are these steps right or I've just missed somethnig up ??? cause my anwser was ''9.5''
Thanks in advance ;)
Post by: Arthur Bon Zavi on October 25, 2011, 03:53:35 pm
x4
--- - x3
4
[(x4/4) - x3]42
[{((256/4) - 64) - (4)3) - {(16/4) - (2)3}]42
| [(0) - (-4)] | = 4
Post by: ~ Miss Relina ~ on October 25, 2011, 10:33:07 pm
Integration gives you :
x4
--- - x3
4
[(x4/4) - x3]42
[{((256/4) - 64) - (4)3) - {(16/4) - (2)3}]42
| [(0) - (-4)] | = 4
Thanks
but have you done a sketch ??? as when I've done a sketch the I found two parts above and under the curve that's why I used two intergration formulas one with upperlimit 4 and lowerlimt 3 (where the curve intersect the x-axis) and other intergration with the upperlimit 3 and lowerlimit 2 but i got it wrong ...........so my sketch may be wrong or what ???
anyways Thanks you very much ;)
Post by: Arthur Bon Zavi on October 26, 2011, 07:08:51 am
Thanks
but have you done a sketch ??? as when I've done a sketch the I found two parts above and under the curve that's why I used two intergration formulas one with upperlimit 4 and lowerlimt 3 (where the curve intersect the x-axis) and other intergration with the upperlimit 3 and lowerlimit 2 but i got it wrong ...........so my sketch may be wrong or what ???
anyways Thanks you very much ;)
Yes, your sketch may be incorrect.
Post by: ~ Miss Relina ~ on October 26, 2011, 08:20:21 pm
Yes, your sketch may be incorrect.
but I think it may be right sorry for disturbing you but I'll write the steps if anything wrong tell me
x3-3x2=0
x2(x-3)=0
thus x = 0 or 3
so it may look like that attached
Post by: astarmathsandphysics on October 26, 2011, 09:29:14 pm
Post by: ~ Miss Relina ~ on October 26, 2011, 10:13:42 pm
Exactly right.then what's wrong with that integration question what is my mistake ???? I 'm really confused
anyways Thanks Sir
Post by: astarmathsandphysics on October 26, 2011, 11:33:17 pm
Integrate between 2 and 3. This will be negative because it is below the x axis, so make it positive.
Integrate between 3 and 4. This will be positive.
Add the two areas to give the answer.
Post by: ~ Miss Relina ~ on October 26, 2011, 11:36:19 pm
The graph is part below and part above the x axis.
Integrate between 2 and 3. This will be negative because it is below the x axis, so make it positive.
Integrate between 3 and 4. This will be positive.
Add the two areas to give the answer.
I did so but my answer turned to be wrong :( :(
and when Arthur did it with integration as a whole of upperlimit 4 and lowerlimit 2 it gave the book answer and that's what I'm confused about
Post by: astarmathsandphysics on October 27, 2011, 11:06:20 am
Post by: astarmathsandphysics on October 27, 2011, 11:13:41 pm
Post by: ~ Miss Relina ~ on October 28, 2011, 07:20:26 pm
Here it is
exactly the same as my answer ;)
but the problem is that the book's answer is 4 may be its wrong ??? ??? ::)
Quote
Integration gives you :
x4
--- - x3
4
[(x4/4) - x3]42
[{((256/4) - 64) - (4)3) - {(16/4) - (2)3}]42
| [(0) - (-4)] | = 4
that's Arthur answer
so what's your opinion Sir may be that book is wrong ::)
Post by: astarmathsandphysics on October 28, 2011, 07:26:29 pm
Post by: Arthur Bon Zavi on October 29, 2011, 09:10:29 am
exactly the same as my answer ;)
but the problem is that the book's answer is 4 may be its wrong ??? ??? ::)
so what's your opinion Sir may be that book is wrong ::)
What I think is you have always got some or the other problem in your books/reference-material and not in the doubts itself. :P
Post by: MKh on October 29, 2011, 10:56:51 am
exactly the same as my answer ;)
but the problem is that the book's answer is 4 may be its wrong ??? ??? ::)
so what's your opinion Sir may be that book is wrong ::)
Check the solutionsbank - it has got all the step-by-step solutions, and u will know where u hav slipped.
Post by: ~ Miss Relina ~ on October 29, 2011, 06:23:09 pm
What I think is you have always got some or the other problem in your books/reference-material and not in the doubts itself. :P
I totally agree with you it's the book's fault we're so innocent ;D
Post by: ~ Miss Relina ~ on October 29, 2011, 06:23:52 pm
Check the solutionsbank - it has got all the step-by-step solutions, and u will know where u hav slipped.
I downloaded it but unfortunately not working :-\ :(
Post by: The Golden Girl =D on October 29, 2011, 07:23:34 pm
You know the books' answer pages do go 'Koko' sometimes ::)
Post by: ~ Miss Relina ~ on October 29, 2011, 07:38:51 pm
^ try downloading it from a different source or simply try once again from the same source.
You know the books' answer pages do go 'Koko' sometimes ::)
well I did but I think my computer is the problem
Will I be annoying if I asked someone to make a printscreen or whatever for that answer .................
well actually I think I've lost my trust in these books chem.... did so what's the use of a book which drives you mad ....????? ;)
Post by: astarmathsandphysics on October 29, 2011, 07:52:36 pm
Post by: The Golden Girl =D on October 29, 2011, 08:38:34 pm
@ Sir Pauk : LOL that was funny =D
Post by: ~ Miss Relina ~ on October 29, 2011, 10:27:05 pm
Books are useful because you can eat them if you're hungry.
Looooool nice one ;D didn't came to my mind that point ;)
Post by: ~ Miss Relina ~ on October 29, 2011, 10:28:58 pm
@ Relina : I'll try to but I don't guarantee you that I'll come online the next few days ( seems my teachers remembered giving us some Quizzes before Eid holiday -.- )
@ Sir Pauk : LOL that was funny =D
okay no problem ...............yeah quizzes......loadz of work............no time........I know that feeling very well :D
Post by: WARRIOR on October 30, 2011, 07:32:52 pm
thanks in advanced
Post by: astarmathsandphysics on October 30, 2011, 07:35:22 pm
tan(-45)=-1
Post by: WARRIOR on October 30, 2011, 09:32:17 pm
Do you mean -45?no , the question is just draw tan ( theta - 45 *)
tan(-45)=-1
Post by: astarmathsandphysics on October 30, 2011, 09:49:48 pm
http://astarmathsandphysics.com/a_level_maths_notes/C1/a_level_maths_notes_c1_transformation_of_graphs.html
Post by: The Golden Girl =D on October 31, 2011, 12:27:46 pm
question in attached
thanks in advanced
I did face such a Question in Chapter 6 C3 ( I think it's the same one)
I asked my teacher where he got the -1 from and this is what he did :
He substituted (theta = 0 {x = 0} ) in the Equation, in other words:
y = tan (theta -45)
y = tan (0 - 45)
y = tan (-45)
y = -1
Which means it has to intersect with y-axis at that point i.e. ( -1 , 0 )
I hope you got it ;)
...................................................................
another question to bother this thread about is here
for who have C2 book its on page 193 Q 6 in the revision exercise 3
what I don't understand is that the book's answer is 4 and when I did integration of the two parts of my sketch , one area was -ve and other was +ve so I added them ignoring the -ve sign so...............are these steps right or I've just missed somethnig up ??? cause my anwser was ''9.5''
Thanks in advance ;)
Here is the Answer from the Solution Bank :
Quote
Area = integration sign thingy with base 2 4x3-3x2dx Remember integration sign thingy axndx=(axn+1) /(n+1 )
Anyhoo here it is
=[(x4/4) - x3]24 }=> The 4 is supposed to be in the TOP and 2 at the BOTTOM but it didn't turn out to be perfect so thought of Mentioning it incase any confusion was caused =]
=[((4)4 /4 )-(4)3)] -[(2)4/4) -(2)3) ]
=(64-64) - [(16 /4) -8]
=0-(4-8)
=0-(-4)
=4
Relina : I just copied and pasted it over here (didn't even read it thoroughly) ,I hope you get it =]
~If I helped you, Please don't forget to include me in your Prayers ;)
Post by: ~ Miss Relina ~ on November 01, 2011, 07:43:51 am
Post by: The Golden Girl =D on November 01, 2011, 02:43:23 pm
Welcome =]
Post by: YU-GI-OH! on November 18, 2011, 10:47:52 pm
Post by: Arthur Bon Zavi on November 19, 2011, 05:22:14 am
Can someone please help me with question 8 part (b) on the mixed exercise for chapter 6 (C1). It's on page 109. Thank you in advance.
Some people do not have access to that book, so I suggest you to type the question down.
Post by: ~ Miss Relina ~ on December 04, 2011, 03:11:25 pm
and for who has the C1 book its chapter 6 page 109 question 8 mixed exercise 6H
here
Post by: MKh on December 04, 2011, 06:48:57 pm
Can someone please help me with question 8 part (b) on the mixed exercise for chapter 6 (C1). It's on page 109. Thank you in advance.
Check the solutionbank - it has the steps.
Post by: ~ Miss Relina ~ on December 04, 2011, 09:37:54 pm
Check the solutionbank - it has the steps.
well it isn't working at allllllllll................I really want it as I've loaddzzzzzzz of doubts in C1 C2 S1 and my exams in JAN but solutionbank not working although I downloaded it million times........
do you know why this is happening>???
Post by: MKh on December 05, 2011, 12:04:33 pm
well it isn't working at allllllllll................I really want it as I've loaddzzzzzzz of doubts in C1 C2 S1 and my exams in JAN but solutionbank not working although I downloaded it million times........
do you know why this is happening>???
When u click on the solutionbank it comes out in a new internet explorer window, ryt?
But then there comes a yellow coloured pop-up on top wich says something like 'Internet explorer has restricted this webpage from running scripts or ActiveX controls that could access yur computer' , ryt?
Click (or right-click) on that yellow pop-up thing and select 'Allow blocked content'. Then there comes a box saying something like 'Are you sure you want to let ActiveX scripts run on your computer?'
Then click yes. The solutionbank then appears. Just try doing this and see if it works.
Post by: ~ Miss Relina ~ on December 05, 2011, 10:14:24 pm
When u click on the solutionbank it comes out in aexplorer windo new internew, ryt?
But then there comes a yellow coloured pop-up on top wich says something like 'Internet explorer has restricted this webpage from running scripts or ActiveX controls that could access yur computer' , ryt?
Click (or right-click) on that yellow pop-up thing and select 'Allow blocked content'. Then there comes a box saying something like 'Are you sure you want to let ActiveX scripts run on your computer?'
Then click yes. The solutionbank then appears. Just try doing this and see if it works.
do you mean after I download the soluitonbank ????
if yes ...... then it doesn't open in a internet window
do u know why ???
Post by: MKh on December 06, 2011, 11:55:38 am
do you mean after I download the soluitonbank ????
if yes ...... then it doesn't open in a internet window
do u know why ???
Umm....i dunno why this happens...but try downloading the C1 and C2 ActiveBooks and then select the SolutionBank icon in it which for me then opens up in a new internet explorer window. Try downloading the Activebooks instead of the SolutionBanks - hop u get it.
If this doesn't work, let me know what doubts u hav in C1 and C2 because i m also sitting for the two modules in jan and i will try my best to help u out. This will help my understandng and urz as well.
Post by: ~ Miss Relina ~ on December 06, 2011, 03:09:51 pm
Umm....i dunno why this happens...but try downloading the C1 and C2 ActiveBooks and then select the SolutionBank icon in it which for me then opens up in a new internet explorer window. Try downloading the Activebooks instead of the SolutionBanks - hop u get it.
If this doesn't work, let me know what doubts u hav in C1 and C2 because i m also sitting for the two modules in jan and i will try my best to help u out. This will help my understandng and urz as well.
ok plz can you givethe link for activeebooks ..........and do it differ from the actual book ???
well Thanks alot for your help
unti now the first doubt was the one above attached in my previous post ???
Post by: astarmathsandphysics on December 06, 2011, 08:06:47 pm
Post by: MKh on December 07, 2011, 10:19:09 am
ok plz can you givethe link for activeebooks ..........and do it differ from the actual book ???
well Thanks alot for your help
unti now the first doubt was the one above attached in my previous post ???
Like Sir astarmathsandphysics said, all ActiveBooks r free on freetextbooks.com, downalod them from there. And no, the Activebooks r not different from the textbooks - they r just to make revision easier.
Ohh mention not, do let me know what other doubts u hav soon. Solving now....will let u know later today. Meanwhile, here is the hint for this question from the C1 SolutionBank:
Hint:
Write out the first three or four terms of the calculation to help you work out what is happening.
Sorry, i m unable to copy/paste the actual solution to this question from the SolutionBank, but will understand the answer myself and then explain to u
Take care.
Post by: astarmathsandphysics on December 07, 2011, 01:03:15 pm
Am posting short and punch videos there now
Post by: ~ Miss Relina ~ on December 11, 2011, 10:27:56 pm
Like Sir astarmathsandphysics said, all ActiveBooks r free on freetextbooks.com, downalod them from there. And no, the Activebooks r not different from the textbooks - they r just to make revision easier.
Ohh mention not, do let me know what other doubts u hav soon. Solving now....will let u know later today. Meanwhile, here is the hint for this question from the C1 SolutionBank:
Hint:
Write out the first three or four terms of the calculation to help you work out what is happening.
Sorry, i m unable to copy/paste the actual solution to this question from the SolutionBank, but will understand the answer myself and then explain to u
Take care.
did you forget me :'(
Post by: astarmathsandphysics on December 11, 2011, 10:34:14 pm
http://freeetextbooks.com/torrents.php?cat=56
Post by: ~ Miss Relina ~ on December 13, 2011, 03:52:54 pm
Miss Relina go to freeetextbooks.com , register then click this link
http://freeetextbooks.com/torrents.php?cat=56
I'll try ;)
Post by: ~ Miss Relina ~ on December 13, 2011, 08:02:03 pm
by the way I found that solomon paper F
Post by: The Golden Girl =D on December 13, 2011, 09:07:16 pm
Post by: ~ Miss Relina ~ on December 13, 2011, 09:19:52 pm
Yes. I remeber I used to have "issues" with Cubic Equations,so yes it is included. (I don't know how to solve it though -.- )
but the syllabus didn't include any thing about cubic eqaution except in transformations and that one requires solving it .................are that solomon papers according to our syllabus exactly or not ??? as I didin't find a question like that before in the pastpapers??? i
???
Post by: The Golden Girl =D on December 13, 2011, 09:23:11 pm
I suggest you learn the technique just to be on the Safe side ;)
Post by: astarmathsandphysics on December 13, 2011, 09:23:55 pm
Post by: ~ Miss Relina ~ on December 14, 2011, 02:50:46 pm
I studied from the CGP Revision Guide last year and it was explained there in detail and in a pretty easy way but did I encounter such a Question in the OFFICIAL Jan Paper => NO. did I face that Question in OUR School mock (past paper) => Yes.
I suggest you learn the technique just to be on the Safe side ;)
ohaa....okay lol you're hilarious in expressing situations ;D ;D
Will I be annoying if I wanned to make sure about the simple method you mentioned above ???
here is the method which I found in my book '' by graphing in C1 or by factor theorem C2 '' are those fine :-X :-\
Thanks in advance ::)
Post by: ~ Miss Relina ~ on December 14, 2011, 02:51:09 pm
cubic equations is in c1, foe edexcel anywayThanks
Post by: MKh on December 15, 2011, 09:48:16 am
did you forget me :'(
Ofcourse not - just a bit busy dear.
Post by: ~ Miss Relina ~ on December 15, 2011, 10:00:59 am
Ofcourse not - just a bit busy dear.
sorry for disturbing you ......take your time :D
Post by: The Golden Girl =D on December 15, 2011, 12:20:49 pm
(Part 1 )
Post by: The Golden Girl =D on December 15, 2011, 12:29:31 pm
Post by: ~ Miss Relina ~ on December 15, 2011, 12:42:44 pm
thank you very much for taking time to do so......I know that these words don't even compare with your help at all ........
and all what I could do is to pray for you that may Allah help you out of any problem you face in your whole life ;)
Jazaki Allah Alfffffff Kair my beloved :)
Post by: The Golden Girl =D on December 15, 2011, 12:47:39 pm
Allah yiwaf2eek :D
Post by: ~ Miss Relina ~ on December 19, 2011, 09:01:41 am
if you need the answer notify me
Post by: astarmathsandphysics on December 19, 2011, 11:46:57 am
Post by: ~ Miss Relina ~ on December 19, 2011, 12:27:46 pm
use 1-cos^2 x = sin^2 x in last part. Attached
shoudn't part (b)
be like that (0 ,pi/9 ,2pi/9 , pi/3 )
as he asked for three equal intervals
Post by: astarmathsandphysics on December 19, 2011, 08:56:59 pm
Post by: ~ Miss Relina ~ on December 21, 2011, 09:45:34 am
They asked for three ordinates - an ordinate is an x value. 3 ordinates means two interval, since each interval has an order at each end. Lot of marks lost cos of this.
Now I got it thank you Sir
but I know it might be annoying to ask more about it but here is my doubt in your worked solution
I'm really sorry for disturbing :-[ :-[
Post by: astarmathsandphysics on December 21, 2011, 11:20:46 am
Post by: ~ Miss Relina ~ on December 21, 2011, 12:16:22 pm
I used the identity 1-cos^2 x + sin^2 x and rearranged it to give sin^2 x = 1-cos^2 x
why the integration of the 1 in( 1-cos^2x) not X following the rule (x^n+1)/(n+1)
Post by: astarmathsandphysics on December 21, 2011, 09:42:05 pm
Post by: ~ Miss Relina ~ on December 22, 2011, 08:27:48 pm
write 1 as x^0 then add one to the power and divide by the new power to give x^1/1=x
okay now I got it alll thaaaaaaaaaank you :D
Post by: astarmathsandphysics on December 22, 2011, 08:39:43 pm
http://www.astarmathsandphysics.com/test_centre/
Post by: MKh on December 23, 2011, 12:49:22 pm
no probs. Try the test centre on my website. Only GCSE/IGCSE tests up now but more soon
http://www.astarmathsandphysics.com/test_centre/
Thanks for making this test centre. It sounds really interesting.
But what exactly is it like? We can take tests there?
And by when will u be able to add the A/L tests?
Thanks once again.
Post by: ~ Miss Relina ~ on December 24, 2011, 09:57:59 am
no probs. Try the test centre on my website. Only GCSE/IGCSE tests up now but more soon
http://www.astarmathsandphysics.com/test_centre/
Thanks Sir
unfortunately I finished my IGCSE maths last year .........but I'll tell my friends about it ...... ;D
;)
Post by: astarmathsandphysics on December 24, 2011, 01:59:38 pm
Post by: ~ Miss Relina ~ on December 24, 2011, 04:01:13 pm
Post by: astarmathsandphysics on December 24, 2011, 07:17:16 pm
Post by: ~ Miss Relina ~ on December 24, 2011, 07:47:10 pm
yes you can relable the axis as 0 to 2pi then graph becomes cos x
so you mean that it wouldn't be the right graph ???
you mean it won't be cos 2x
Post by: astarmathsandphysics on December 24, 2011, 08:48:51 pm
If you relable the x axis so all the x's are doubled, it becomes the graph of cos x
Post by: ~ Miss Relina ~ on December 24, 2011, 08:54:41 pm
it is the graph of cos 2x
If you relable the x-axis so all the x's are doubled, it becomes the graph of cos x
ahhaaaaa so I should not change the range right ???
Post by: astarmathsandphysics on December 28, 2011, 10:38:37 pm
Post by: ~ Miss Relina ~ on December 29, 2011, 07:33:06 am
Some trig proofs and quadritic roots worked examples.
Thanks alot Sir .....I really appreciate your help ;)
Post by: ~ Miss Relina ~ on December 29, 2011, 07:37:22 am
Thanks alot Sir .....I really appreciate your help ;D ;D ::)
but wait a second......is that for me ??? ::) ::) ::)
I think you posted the answers for questions in that thread.....lol ;D ;D ;D
https://studentforums.biz/math-146/quadratics-and-trigonometry-problems!/
Post by: whale95 on December 29, 2011, 08:41:09 am
Q) The equation of the curve is y = 3cos2x. The equation of a line is x+2y=? (pi) . On the same Diagram, sketch the curve and the line for 0<x<? .
Find the first 3 terms in the expansion of (2-x)[2] in ascending power of x.
I did the graph but didnt know how to do the line! :S
Q)2cos? = 1/tan? 0<?<2?
I did upto cos?(2sin?-1)=0
so cos?=0 so 0
and 2sin?-1=0 so pi?/6 and pi5?/6
so now I have 3 answers which are in bold! how did pi?/2 and 3pi?/2 come??
Q)sinx=3/5 and angle is obtuse. Find tanx and cosx
Q) sin[2]theta/1-costheta = 1 + costheta Identities! Sorry if I copied it wrong! next time I will use attachments! to make it easier!
thanks alot i really benefited from the 1st answers! :D
Post by: ~ Miss Relina ~ on December 29, 2011, 10:32:51 pm
Post by: astarmathsandphysics on December 30, 2011, 11:56:34 am
Please modify your post
Post by: astarmathsandphysics on December 30, 2011, 12:48:10 pm
Post by: ~ Miss Relina ~ on December 30, 2011, 01:49:05 pm
that population question attachedokay Thanks
plz if you know (b) part as when I asked on TSR they gave me the answer but it didn't work
also for who have the solution bank you can help me plz by just printscreening the answer and post it
it's Q8 page 116 C2 book
Thanks in advance
Post by: whale95 on December 30, 2011, 05:06:52 pm
Post by: astarmathsandphysics on December 30, 2011, 07:31:21 pm
Post by: ~ Miss Relina ~ on December 31, 2011, 07:54:08 am
the series question answered
My life saver
Thanks a bunch ;D
Post by: astarmathsandphysics on December 31, 2011, 09:23:15 am
Post by: whale95 on January 01, 2012, 09:07:51 am
Post by: Master_Key on January 01, 2012, 11:41:07 am
Check the attachments please.. Questions!Answers for Q 4 5 6 :) ;)
Post by: Master_Key on January 01, 2012, 11:52:09 am
Check the attachments please.. Questions!
Q 1
6/x^4 - 5/x^3 + 2/x^2
Post by: astarmathsandphysics on January 01, 2012, 11:39:23 pm
See attachments.
Post by: ~ Miss Relina ~ on January 06, 2012, 07:15:24 am
Thanks in advance ;)
Post by: astarmathsandphysics on January 06, 2012, 11:47:29 pm
Post by: astarmathsandphysics on January 06, 2012, 11:51:40 pm
2nd question all correct
Post by: ~ Miss Relina ~ on January 07, 2012, 06:52:26 am
okay Thanks I corrected my mistake Thanks alot Sir
:D
any advice for exam after five days :o :o :o
Post by: astarmathsandphysics on January 07, 2012, 11:18:32 pm
Post by: MKh on January 08, 2012, 12:02:39 pm
Only that you don't stop
Don't stop......practicing past exam question papers. This is what u mean? If yes, i totally agree.
Post by: ~ Miss Relina ~ on January 08, 2012, 07:16:55 pm
Only that you don't stop
okay I'll try isA ;)
Post by: ~ Miss Relina ~ on January 08, 2012, 07:18:03 pm
hope to get an answer soon isA
Post by: astarmathsandphysics on January 08, 2012, 09:52:46 pm
Post by: The Golden Girl =D on January 10, 2012, 12:21:37 pm
heyyy i need ur help for c2 :( PLZZ 2011 june c2 the randians Q :/
Can someone Answer his doubt please :)
Thanks in Advance =D
Post by: The Golden Girl =D on January 10, 2012, 03:28:49 pm
Post by: MKh on January 11, 2012, 01:51:28 pm
Please find attached a part of my Edexcel C2 Revision Checklist from the Trigonometry chapter. I am still a bit uncomfortable about these parts of the C2 spec. so if u guyz could please help me by explaining what you know about these parts, it would be very helpful of u.
Exams r almost upon us so please help. C2 exam is on the 13th of jan. Good Luck to all and thanks a lot.
Post by: Malak on January 11, 2012, 04:47:15 pm
Assuming that you do, everything is basically the same.
For example 1,
you take the sin inverse of 3/4 and solve for (x + pi/2) as you normally would for x
when you have found the angles using quadrant rule/graph just subtract pi/2 in the end from those angles
thats it.
Simple rule: When you have something with x in bracket, leave it as it is and only in the end subtract/add or whatever.
for example 4 in the file, you have to write all the terms in the same trig. function.
some examples here: http://www.examsolutions.co.uk/maths-revision/core-maths/trigonometry/equations/squared-types/example-1.php (http://www.examsolutions.co.uk/maths-revision/core-maths/trigonometry/equations/squared-types/example-1.php)
if you still have a problem then let me know :D
---------------------------
Please don't double post again and also I am going to merge this thread with math doubts.
Post by: The Golden Girl =D on January 11, 2012, 05:08:18 pm
Post by: Malak on January 11, 2012, 05:09:20 pm
Can someone answer the doubt I posted earlier ,Please.Which one :S
Post by: The Golden Girl =D on January 11, 2012, 05:11:37 pm
heyyy i need ur help for c2 PLZZ 2011 june c2 the randians Q :/
Post by: Malak on January 11, 2012, 06:00:54 pm
Post by: The Golden Girl =D on January 11, 2012, 06:12:42 pm
Q.5 Please =]
Post by: astarmathsandphysics on January 11, 2012, 11:44:58 pm
Post by: astarmathsandphysics on January 11, 2012, 11:53:12 pm
Post by: astarmathsandphysics on January 11, 2012, 11:55:19 pm
Post by: The Golden Girl =D on January 12, 2012, 12:32:19 pm
Post by: ~ Miss Relina ~ on January 12, 2012, 03:46:55 pm
also those are from edexcel book so who have the book and can answer any question tell me plz tp post page or something
Post by: astarmathsandphysics on January 13, 2012, 01:37:32 pm
Post by: ~ Miss Relina ~ on January 14, 2012, 10:26:09 pm
Sorry about delay. Forum wasn't working for me last night
no probs I already have done my exam ...anyways I hope other memebers benefit from them isA..and all credit goes to you :D
well plz have a look at my doubts on S1 thread plz
Thanks in advance ;)
Post by: astarmathsandphysics on February 26, 2012, 05:28:21 pm
Sir astarmathand physics,
I have come across some difficulty to solver the sums under Core Mathematics 2 under Differentiation.
The question is -
1. A square-based pyramid has slanting edges of length 6cm. The volume of such a pyramid is 1/3 xsquare h, where x is the length of the side of the base and h is the height. Show that this volume may be written in the form 24h-2/3h to the power of 3 (24h-2/3hcube). Prove that the greatest volume will occur when h= 2squareroot 3cm.
Please solve this sums in step by steps.. Please include ur working to get the formula and to prove// Pls. DONT JUST GIVE THE ANSWER ONLY///
Pls include ur working along with diagram or what and what variable are u using to make me to understand much better.
Pls do it that urgenlty.
Thanx inadvance.
Post by: SZM on February 26, 2012, 05:56:05 pm
The questions are-
1. A rectangular garden is fenced on three sides, and the house forms the fourth side of the rectangle. Given that the total length of the fence is 80m show that the area A, of the garden is given by the formula A= y(80-2y), where y is the distance from the house to the end of the garden. Given that the area is a maximum for this length of fence, find the dimensions of the enclosed garden, and the area which is enclosed.
Sir please do not solve this sums in a rush. If u have to solve this sum in a rush, please make sure that ur working shuld be nealty and clear, I dont like to see the sums,where u r scribbling here and there. Pls do it nealty and clear.
Sir just because the question is asked to show the formula, please i want u to write ur working on how to show this formula. the best thing is if u draw the diagram again and label it and then write in details. Only then I wil be able to understand and also can solve others sums as well. Please even in diagram draw it neatly. If ur work is nealty, I wil get more attracted to it and every working wil grasp into my mind. SO do that Sir please.
Hope this sums u r going to be more clearful , as I am expecting from u.
need ur reply asap. I wil post another few sums more in another post.
Post by: SZM on February 26, 2012, 06:40:41 pm
1. A closed cylinder has total surface area equals to 600pi. Show that the volume , V cmcube, of this cylinder is given by the formula V= 300pi r- pi rcube, where r cm is the radius of the cylinder. Find the maximum volume of such a cylinder.
2. A sector of a circle has area 100cmsquare. Show that the perimeter of this sector is given by the formula
P=2r + 200/r , r> squareroot 100/pi. The radius of the sector is r cm.
Find the minimum value for the perimeter of such a sector.
3. A shape consists of a rectangular base with a semicircular top, as shown. Given that the perimeter of the shape is 40cm, show that its area, A cmsquare, is given by the formula A= 40r- 2rsquare- pi rsquare/2 where r cm is the radius of the semicircle. Find the maximum value for this area.
4. The shape shown is a wire frame in the form of a large rectangle split by parallel lengths of wire into 12 smaller equal-sized rectangles. Each smaller rectangle, the length is y mm and the width of that smaller rectangle is x mm. Given that the total length of wire used to complete the whole frame is 1512mm, show that the area of the whole shape is Ammsquare, where A = 1296x - 108xsquare/7. Find the maximum area which can be closed in this way.
5. The fixed point A has coordinates ( 8, -6, 5) and the variable point P has coordinate ( t, t, 2t).
a.) Show that APsquare= 6tsquare - 24t - 125.
b.) hence find the value of t for which the distance AP is least.
c.) Determine this least distance.
6. A wire is bent into the plane shape ABCDEA. Shape ABDE is a rectangle and BCD is a semicircle with diameter BD. The area of the region enclosed by the wire is R msquare, AE= x metres, AB= ED= y metres.The total length of the wire is 2 m.
a.) Find an expression for y in terms of x.
b.) Prove that R = x/8 (8 - 4x - pix)
Given that x can vary, using calculus and showing your working,
c.) find the maximum value of R. ( you do not have to prove that the value you obtain is a maximum)
7. A cylindrical biscuit tin has a close-fitting lid which overlaps the tin by 1cm. The radii of the tin and the lid are both x cm. The tin and the lid are made from a thin sheet of metal of area 80pi cmsquare and there is no wastage. The volume of the tin is V cmcube.
a.) Show that V = pi ( 40x - xsquare - xcube).
Given that x can vary:
b.) Use differentiation to find the positive value of x for which V is stationary
c.) Prove that this value of x gives a maximum value of V.
d.) Find this maximum value of V.
e.) Determine the percentage of the sheet metal used in the lid when V is a maximum.
8. The part of the curve with equation y= 5- 1/2xsquare for which y> 0. The point P (x,y) lies on the curve and 0 is the origin.
a.) Show that OPsquare = 1/4xsquare - 4xsquare + 25.
Taking f(x)= 1/4xto the power of 4 - 4xsquare + 25.
b.) Find the values of x for which dy/dx = 0.
c.) hence , or otherwise, find the minimum distance from 0 to the curve, showing that your answer is a minimum.
9. The part of the curve with equation y= 3 + 5x + xsquare + xcube. The curve touches the x-axis at C. The points A and B are stationary points on the curve.
a. ) Show that C has coordinates (3, 0)
b. ) Using calculus and showing all your working, find the coordinates of A and B.
10. An open tank for storing water, ABCDEF. The sides ABFE and CDEF are rectangles. The triangular ends ADE and BCF are isosceles, and angle at AED is equal to angle at BFC which is 90 degree. The ends ADE and BCF are vertical and EF is horizontal.
Given that AD= x metres:
a.) Show that the area of triangle ADE is 1/4xsquare msquare.
Given also that the capacity of the container is 4000 mcube and that the total area of the two triangular and two rectangular sides of the container is S msquare.
b.) Show that S = xsquare/2 + 16000squareroot 2 / x
Given that x can vary:
c.) Use calculus to find the minimum value of S.
d.) Justify that the value of S you have found is a minimum.
There are 10 sums above and sir please each and every sum, I am expecting from u to show ur working nealty and in details and very clearful as well along with neatly diagrams.. Pls Pls Pls Pls.
Sir why these sums are said to be a toughest and difficult part than others?? is there any tips that I have to learn or know ?? do u have any tricky tips which helps to overcome the difficulties of the sums with different situation?? Sir those types of questions,the most difficult part is how to form a formula that they are given. I know there is a diagram// But using the diagram I have to make a formula, but to make a formula with different variables/terms, i found difficult// I dont understand in some formula why they put minus and plus within a formula and how it becomes minus and plus?????????? Pls i need ur best solution to this matter
i need ur reply asap.
thanx in advance
Post by: astarmathsandphysics on February 26, 2012, 06:56:14 pm
Post by: astarmathsandphysics on February 26, 2012, 07:02:42 pm
Post by: SZM on February 26, 2012, 07:23:18 pm
Sorry SZM here are p1 and p2
Sir you dont have to say sorry for me. I am just telling for our benefits, otherwise it wil be a waste for me, when I dont understand ur working at all as well as for u the amount of time u spent to solve my sums. There shuld be benefit. right???
I am thanking u as u have done with what I have asked u to do. The steps and the diagram on how to get the formula. Same things please do the rest of the 11 sums as I have mentioned .. I want all 11 sums with workings n answers as well by 2day if possible. Please Please again I am telling u , I dont want u to scribble, here and there as it looks ugly. Pls make ur working neatly and clear with different colours not one colour inorder to get more attractive... Please leave one gap and then write the next gap, sothat I can read and understand til I get the point. If u write inbetween, I cant understand and read as well.
Hope I wil get nealty work from u the rest of the sums,as I am expecting it.
need ur reply asap.
Post by: astarmathsandphysics on February 26, 2012, 08:30:51 pm
Post by: SZM on February 27, 2012, 07:49:57 am
Ok szm
Sir astarmathandphysics, where are ur answers for the rest of the 11 sums i have mentioned?? when will u give ur answer???
I am luking for the answers as well as each and every working on how did u approach to an answer// where are ur answers??
Need ur reply urgentlyy.
Post by: astarmathsandphysics on February 27, 2012, 09:24:08 am
Will make videos to answer the rest today and post them on theeducationchannel.info
Post by: astarmathsandphysics on February 27, 2012, 09:35:20 am
Post by: astarmathsandphysics on February 27, 2012, 10:59:58 am
http://theeducationchannel.info/videos/14/A-Level-Maths-C1/most_recent/all_time/
Post by: SZM on February 27, 2012, 12:55:59 pm
just made two lovely c1 videos. Will take some hours to upload.
http://theeducationchannel.info/videos/14/A-Level-Maths-C1/most_recent/all_time/
Sir the link u mentioned here is not working , as it says internal server error// I have no idea why??
Sir thanx alot for spending ur time to solve my sums. May God Bless u.
Sir u want diagram for question 3 and 4??
But I dont know where can I draw diagram and post here??
Sir what about quesiton no. 8.) , 9.) and 10.)???
waitin for ur reply//
Post by: astarmathsandphysics on February 27, 2012, 01:45:42 pm
it is working now but I have not uploaded the video yet.
It is in a queue.
Post by: astarmathsandphysics on February 27, 2012, 01:47:49 pm
I will look at 3 and 4 and see if I can do without the diagram
Post by: astarmathsandphysics on February 27, 2012, 02:00:00 pm
For q4 I really need a diagram. In the question it says there is one.
Post by: SZM on February 27, 2012, 03:06:01 pm
q3 for SZM
For q4 I really need a diagram. In the question it says there is one.
Sir question no. 4), the diagram that u have drawn is little correct. I wil give u an extra information about diagram. The diagram says that there 6 rectangular boxes in a row side from left to the right and there are 2 rectangular boxes in column sides from top to bottom. The variable that u have labelled is correct. I have seen ur diagram under question no 3.) where u have scribbled.
Hope this information wil make u understand.... Pls let me know if u dont understand the information//
Thanx for solving the question no 3.) . May God Bless u.
Question no 9.), I have corrected this question again,
The part of the curve with equation y= 3 + 5x + xsquare - xcube. The curve touches the x-axis at A and crosses the x-axis at C. The points A and B are stationary points on the curve.
a. ) Show that C has coordinates (3, 0)
b. ) Using calculus and showing all your working, find the coordinates of A and B.
hope I have replied with what u have asked me for..
Waitn 4 ur reply.
Post by: astarmathsandphysics on February 27, 2012, 03:16:03 pm
Post by: YU-GI-OH! on April 27, 2012, 06:29:35 pm
Thanks in advance :)
Post by: xainer on August 25, 2012, 11:51:16 am
Each year, for 30 years, David will pay money into a savings scheme. In the first year he pays 800 pounds. His payments then increase by 60 pounds each year, so that he pays 860, 920 in the third year and so on.
a) Find how much David will pay in the 30th year.
b) Find the total amount that David will pay in over the 30 years.
c) David will retire when his savings reach 65000. Find out how much longer he will have to work.
Can u tell the answer for the last part? it's where i'm stuck. Because of so big numbers, it's difficult to solve the quadratic equation which i'm getting. tried reducing it also.
Post by: Saladin on August 25, 2012, 03:13:04 pm
(a) Common difference = 60
Therefore, value on 30th year = 60(30) + 740
(b) (n/2)(2a + (n-1)d)
= (30/2)(2(800) + (29)60)
= 50100
(c) You just use the equation in (b) and equate it to 6500, and solve for n. You get something like 35.45..so 36...
Also, if the person does not increase his amount after the 30th year, it will still take 36 years.
Read the question again, maybe that will help you out.
Post by: Saladin on August 25, 2012, 03:14:14 pm
May someone please solve question 5b of the June 2011 C2 paper for me.
Thanks in advance :)
Don't have the paper on me. Next time, please attach the paper.
Post by: xainer on August 25, 2012, 03:53:01 pm
@xanier:okay. will try. thanks for the quick reply! :)
(a) Common difference = 60
Therefore, value on 30th year = 60(30) + 740
(b) (n/2)(2a + (n-1)d)
= (30/2)(2(800) + (29)60)
= 50100
(c) You just use the equation in (b) and equate it to 6500, and solve for n. You get something like 35.45..so 36...
Also, if the person does not increase his amount after the 30th year, it will still take 36 years.
Read the question again, maybe that will help you out.
Post by: xainer on August 25, 2012, 04:44:14 pm
@xanier:can u tell hw to do that qustn without using calculator? the numbers are very big n using the quadratic formula doesn't give a whole number inside the square root. i checked using calculator...
(a) Common difference = 60
Therefore, value on 30th year = 60(30) + 740
(b) (n/2)(2a + (n-1)d)
= (30/2)(2(800) + (29)60)
= 50100
(c) You just use the equation in (b) and equate it to 6500, and solve for n. You get something like 35.45..so 36...
Also, if the person does not increase his amount after the 30th year, it will still take 36 years.
Read the question again, maybe that will help you out.
Post by: xainer on August 27, 2012, 04:04:24 pm
Post by: astarmathsandphysics on August 27, 2012, 04:24:30 pm
Post by: xainer on August 27, 2012, 05:43:22 pm
http://www.astarmathsandphysics.com/a_level_maths_notes/C1/a_level_maths_notes_c1_differentiation.htmlThank you!!
Post by: astarmathsandphysics on January 24, 2013, 11:22:16 am
1)
Prove that sin2x /(1+cos2x) = tanx
2)
Solve the eqn 2sec2X + tanX - 3 = 0 (X stands for theeta and 0<=X<=180)