Post by: sweetsh on May 27, 2009, 05:05:08 pm
Thank you
Post by: shan2391 on May 27, 2009, 05:14:53 pm
Post by: sweetsh on May 27, 2009, 05:16:28 pm
Post by: Ghost Of Highbury on May 27, 2009, 05:17:57 pm
i have it in my school
its fun
@shan
plz post a question in permutation and combination
so e can help
as very rightly said by sweetsh
Post by: sweetsh on May 27, 2009, 05:23:22 pm
Post by: Anonymous on May 27, 2009, 05:28:00 pm
Post by: Ghost Of Highbury on May 27, 2009, 06:20:16 pm
i found this question in a guide and coul not solve it
hope u help me in this
Q. A train 150m long is running with a speed of 68mph. In whate time will it pass a man who is running at 8mph in the same direction which the train is going.
Q. A and B undertake to do a peice of work for $600. A alone can do it in 6 days while B alone can do it in 8 days. With the help of C
they finish it in 3 days. Find the share of each?
i hope u help me.
thanks in advance
Post by: SGVaibhav on May 27, 2009, 08:26:44 pm
Post by: sweetsh on May 27, 2009, 08:28:37 pm
Post by: fREnZy on May 28, 2009, 04:26:56 am
Whenever the order is necessary, use permutations. Whenever you just have to pick, in no order whatsoever, use combinations.
Post by: divineobsidian on May 28, 2009, 05:46:44 am
like if there are 3 blue, 3 red, 4 greeen and you pick 5 how many perms/coms can there be?
thanks
Post by: vanibharutham on May 28, 2009, 09:52:31 am
a) The producers of a play require a total cast of 5, of which 3 are actors and 2 are actresses. He auditions 5 actors and 4 actresses for the cast. Find the total number of ways in which the cast can be obtained.
b) Find how many different odd 4-digit numbers LESS than 4000 can be made from the digits 1,2,3,4,5,6,7 if no digit may be repeated.
The FIRST thing to consider when solving Permutations and Combinations questions is recognizign what the hell you are doing...
ALWAYS REMEMBER:
PERMUTATIONS HAVE TO DO WITH ARRANGEMENTS OF THINGS
COMBINATIONS HAVE TO DO WITH HOW MANY POSSIBILITIES ARE THERE, WITHOUT WORRYING ABOUT ARRANGEMENTS
if we follow that rule...
If for example, we had three things.... A,B,and C....
the combination ABC would be the same as BAC or CAB or BAC etc.
However the permutations are different...
Back to the question...
a) The producers of a play require a total cast of 5, of which 3 are actors and 2 are actresses. He auditions 5 actors and 4 actresses for the cast. Find the total number of ways in which the cast can be obtained.
This is obviously worried about combinations, not arrangements...
This my method of solving this question:
From: 5 actors 4 actresses
Choose: 3 actors 2 actresses
=> 5C3 x 4C2 = 10 x 6 = 60 combinations
b) Find how many different odd 4-digit numbers LESS than 4000 can be made from the digits 1,2,3,4,5,6,7 if no digit may be repeated.
Obviously this is a permutation question :) its got to with arrangements....
This is my method in solving this;
First, find out how many 4-digit odd numbers you can have...
well
_ x _ x _ x _
We have 4 digits to fill from 7 numbers.... however for it to be an odd number, the last digit NEEDS to be either 1,3,5 or 7...
and that gives us four digits on the last one....
_ x _ x _ x 4
What about the rest.... well.....
Obviously you will only have 6 choices for your first one, 5 for the next, and 4 for the next...
which gives us
6 x 5 x 4 x 4 = 480 odd numbers....
To find out how numbers from those odd numbers are LESS than 4000...
THEY NEED TO START WITH A 1, 2 or a 3... . and end with a 1,3,5 or 7...
First lets see all possibilites starting with 2....
Obviously because its starting with 2, we only have 1 option for the start.
1 x _ x _ x _
once again, we have 1,3,5,7 for the last option.... that gives us ... 4
1 x _ x _ x 4
how many numbers do we have left... 5 to choose from...
therefore.... 1 x 5 x 4 x 4 = 80 odd numbers Starting with 2, and that are less than 4000....
Now for numbers starting with 1 or 3....
we have two options for the first one....
2 x _ x _ x _
and we only have 3 options for the last one.... 1,3,5,7 but because we are already using one of them as our first number, we can only choose 3....
2 x _ x _ x 3
how many numbers do we have left? well we have.... 5...
therefore...
2 x 5 x 4 x 3 = 120....
120 + 80 = 200
We have 200, four digit odd numbers less than 4000...
Question 5 from October/November 2002 Paper 2...Cambridge
Post by: vanibharutham on May 28, 2009, 11:43:29 am
im happy to help...
Post by: sweetsh on May 28, 2009, 11:44:44 am
Post by: Ghost Of Highbury on May 28, 2009, 01:55:25 pm
Q. A and B undertake to do a peice of work for $600. A alone can do it in 6 days while B alone can do it in 8 days. With the help of C
they finish it in 3 days. Find the share of each??
hope u help..
Post by: vanibharutham on May 29, 2009, 07:58:54 am
have no idea how to solve them
Post by: Padapop on May 29, 2009, 12:17:14 pm
Post by: astarmathsandphysics on May 29, 2009, 02:03:53 pm
Post by: astarmathsandphysics on May 29, 2009, 07:15:00 pm
Q. A train 150m long is running with a speed of 68mph. In whate time will it pass a man who is running at 8mph in the same direction which the train is going.
Q. A and B undertake to do a peice of work for $600. A alone can do it in 6 days while B alone can do it in 8 days. With the help of C
they finish it in 3 days. Find the share of each??
hope u help..
man a does $100 of work a day so in 3 days earns $300
man b does $75 dollars of work a day so earns $225
Mean c earns $600 -$225-$300=$75
ans is 300:225:75 or 4:3:1
Post by: vanibharutham on May 29, 2009, 08:22:24 pm
Closing speed is68-8 equals 60 so t equals d/v=2.5
how can u be sure that they start from the same position?
isnt this relative velocity?
Surely if The train starts off at lets say (10i + 20j) and has a velocity of 68m/s
Its position would be (10i + 20j) + 68*t
Isnt the whole thing dependent on position?
i dont know? ??? could you explain
Post by: astarmathsandphysics on May 29, 2009, 09:06:13 pm
Post by: lil^$tar on May 29, 2009, 10:41:47 pm
Post by: ramezamgad on May 30, 2009, 01:21:36 am
there are only 2 probabilities
both are moving in the same direction so their velocities are subtracted
they are moving towards each other so their velocities are added
waiting for ur reply lil^$tar
Post by: astarmathsandphysics on May 30, 2009, 04:36:59 am
Post by: Ghost Of Highbury on May 30, 2009, 11:09:05 am
:)
@astarmathandphysics
do u have the timetable for igcse november 2009 final..
if so can u plz upload it here.,.
or send it to eddie_adi619@hotmail.com
thanks in advance
Post by: sweetsh on May 30, 2009, 11:11:38 am
Post by: Anonymous on May 30, 2009, 11:15:45 am
Post by: Ghost Of Highbury on May 30, 2009, 11:17:12 am
i got hold of the provisional one..but no final??
Post by: lil^$tar on May 30, 2009, 05:26:12 pm
it simply means the velocity of 1 particle relative to another 1ok thanx can u giv me a question on it so dat i cud see how gud i understand it
there are only 2 probabilities
both are moving in the same direction so their velocities are subtracted
they are moving towards each other so their velocities are added
waiting for ur reply lil^$tar
Post by: vanibharutham on May 30, 2009, 07:57:41 pm
i was doing the following question:
The function f(x) = [ (ex + 1) / 4 ] for the domain x ? 0
i) Obtain an expression for f-1
ii) State the domain and range for f-1
The first part was easy...
i got the answer, and when i checked it with the mark scheme i got the right answer, which was:
ln(4x-1)
The second part of my answer didn't match the mark scheme, and i dont seem to understand the why.
The domain, according to me should be:
x > 1/4
The range should be:
f-1(x) > -3.2188
However, the mark scheme says that the answers are:
DOMAIN: x > 1/2
RANGE: f-1(x) > 0
could some one please explain my mistake?
Post by: ramezamgad on May 30, 2009, 08:21:02 pm
what is the relative velocity of two particles A and B moving
a) towards each other A with velocity 10 and B with velocity 15????
b) in the same direction A with velocity 10 and B with velocity 15????
c) in the same direction A with velocity 15 and B with velocity 15????
vanibharutham
what is the domain of the first eq.?
Post by: vanibharutham on May 30, 2009, 08:32:53 pm
sorry the symbol dint come up on there
Post by: ramezamgad on May 30, 2009, 08:37:20 pm
y=ln(4x-1)
u have to find what value of x makes y=0
ln1=0
so 4x-1=1
4x=2
x=1/2
u made 4x-1=0 which is a common mistake
Post by: vanibharutham on May 30, 2009, 08:43:18 pm
right, and then for the domain x > 1/2
you'll have to find the range, which is y > 0
THANKS SO MUCH...
i always make that mistake with domain and range when it comes to e^x and ln(x)
Post by: ramezamgad on May 30, 2009, 08:48:53 pm
and the range of f(x) is the domain of f-1
Post by: vanibharutham on May 30, 2009, 08:53:49 pm
thanks for that :P
remind me thank you over nd over agen :P
Post by: lil^$tar on May 30, 2009, 09:08:53 pm
lil^$tara)25 b)-5 c)0
what is the relative velocity of two particles A and B moving
a) towards each other A with velocity 10 and B with velocity 15????
b) in the same direction A with velocity 10 and B with velocity 15????
c) in the same direction A with velocity 15 and B with velocity 15????
vanibharutham
what is the domain of the first eq.?
Post by: ramezamgad on May 30, 2009, 09:14:16 pm
u just need to know that B will move faster than A and the distance between them will decrease until they become side to side then the distance between them will increase
hope u find that easy
Post by: lil^$tar on May 30, 2009, 09:26:12 pm
Post by: Ghost Of Highbury on May 31, 2009, 08:33:09 am
Hi,
i was doing the following question:
The function f(x) = [ (ex + 1) / 4 ] for the domain x ? 0
i) Obtain an expression for f-1
ii) State the domain and range for f-1
The first part was easy...
i got the answer, and when i checked it with the mark scheme i got the right answer, which was:
ln(4x-1)
The second part of my answer didn't match the mark scheme, and i dont seem to understand the why.
The domain, according to me should be:
x > 1/4
The range should be:
f-1(x) > -3.2188
However, the mark scheme says that the answers are:
DOMAIN: x > 1/2
RANGE: f-1(x) > 0
could some one please explain my mistake?
which paper did this question appear in..
year/month/paper..
Post by: Ghost Of Highbury on May 31, 2009, 10:15:21 am
Post by: vanibharutham on May 31, 2009, 03:17:16 pm
May/ June 2003 Paper 2 Cambridge
Post by: shan2391 on May 31, 2009, 04:45:15 pm
Ref: Q.8 OCt/NOv08 Paper2
Post by: vanibharutham on May 31, 2009, 07:46:55 pm
The question states that
f(x) = 3 + 5 sin2x
Im sure you can easily rearrange that to make it simpler:
f(x) = 5sin2x + 3
The amplitude is always going to be the number infront the trignometric function...
In this case the amplitude will 5 :)
simple,
if forexample f(x) was 25000sin2x, your amplitude would be 25000
Your amplitude involves the actual trignometric identity... in this case... 2x.
Always remember this rule:
To find the period of any trignometric graph, divide the coefficient of x by 360 (but in this case 2pi as we are dealing with radians)
Therefore our period is 180 degrees or simply pi.
Hope that helped
Post by: shan2391 on May 31, 2009, 08:55:48 pm
Shan2391Thanks a lot man
The question states that
f(x) = 3 + 5 sin2x
Im sure you can easily rearrange that to make it simpler:
f(x) = 5sin2x + 3
The amplitude is always going to be the number infront the trignometric function...
In this case the amplitude will 5 :)
simple,
if forexample f(x) was 25000sin2x, your amplitude would be 25000
Your amplitude involves the actual trignometric identity... in this case... 2x.
Always remember this rule:
To find the period of any trignometric graph, divide the coefficient of x by 360 (but in this case 2pi as we are dealing with radians)
Therefore our period is 180 degrees or simply pi.
Hope that helped
but what does +3 show in a graph
Post by: jkiam11 on June 01, 2009, 04:44:17 am
Eg. for 3 + 5 sin2x
so the maximum point would be 5+3=8
and the minimum point would be -5+3=-2
Post by: waiwai on June 01, 2009, 09:26:12 am
Can you please tell me the usual A* grade boundary for Additional Mathematics?
Thank You
Post by: Ghost Of Highbury on June 01, 2009, 09:29:16 am
For A it is generally paper 1 -56 (max 68) (+/- 4)
paper 2 -66 (max68) (+/4)
Post by: waiwai on June 01, 2009, 09:33:02 am
no idea for A* as it does not ecist in the individual component..
For A it is generally paper 1 -56 (max 68) (+/- 4)
paper 2 -66 (max68) (+/4)
Thank you
So is A maximum of 68% ?
Thanks again
Post by: Ghost Of Highbury on June 01, 2009, 09:34:09 am
not 68%
it is 85%
Post by: vanibharutham on June 01, 2009, 09:39:15 am
it will be somewhere around 85%
and they will move the A grade lower to around 80%??
do u think thats a fair claim?
Post by: Padapop on June 01, 2009, 10:04:17 am
I heard cambridege added the A* for this course
it will be somewhere around 85%
and they will move the A grade lower to around 80%??
do u think thats a fair claim?
Where did u hear that from? Can someone provide links/evidence etc?
It would be awesome if a* = 85% and not a= 85% cause i think thats a bit hard.
Post by: archangel on June 01, 2009, 10:06:59 am
the kind where they ask u to express y in terms of x... then u have to make it y=mx+c
Post by: vanibharutham on June 01, 2009, 11:21:04 am
Straight line exponential graphs ...
Right :P
here we go...
Normally the question will give you a table of values containing two variables from which a scientists carried a certain experiment. When they plotted their values they don't get a straight line.
You will either be told two things:
1) a formula by which the variables are supposed to be connected by
2) to plot something against something
So first, they will give you the table of values:
x: 2 3 4 5 6
y: 9.2 8.8 9.4 10.4 11.6
If they tell you to plot something against something, it is much easier. For the above table of value draw the graph of xy against x2
From the above information you know you need to find the values of x2 and the values of xy.
Ok,
so lets find the values of x2 first:
well simple
4, 9 , 16, 25, and 36
The values of xy
18.4, 26.4, 37.6, 52 and 69.6
Thus giving us the follow coordinates to plot:
(4,18.4) , ( 9,26.4), (16, 37.6), (25, 52) and (36, 69.6)
All you have to now do is plot those points do convert your exponential graph into a straight line graph...
Then you can easily find the gradient of the graph by using
y2-y1
x2-x1
basically the gradient will be:
26.4 - 18.4 = > 8/5 or 1.6
9 - 4
To find the c - value just read off your graph
In this case, your c value should approximately be 12.2
However let me show you the algebric way of solving to find c...
Use any co ordinate, with your y - iontercept co ordinate
Which means ( 25, 52 ) and ( 0 , c )
Find the gradient of those two points
you will end up with
c - 52
0-25
But that equals to 1.6 (beecause we have already found the value for m)
therefore, c - 52 = - 40
c = 12
Now you can simply replace your Y = mX + C formula
which is
xy = 1.6x2+12
just rearrange to get y in terms of x
(question taken from may/june 2003 paper 1 cambridge )
Post by: archangel on June 01, 2009, 12:08:43 pm
Archangel
Straight line exponential graphs ...
Right :P
here we go...
Normally the question will give you a table of values containing two variables from which a scientists carried a certain experiment. When they plotted their values they don't get a straight line.
You will either be told two things:
1) a formula by which the variables are supposed to be connected by
2) to plot something against something
So first, they will give you the table of values:
x: 2 3 4 5 6
y: 9.2 8.8 9.4 10.4 11.6
If they tell you to plot something against something, it is much easier. For the above table of value draw the graph of xy against x2
From the above information you know you need to find the values of x2 and the values of xy.
Ok,
so lets find the values of x2 first:
well simple
4, 9 , 16, 25, and 36
The values of xy
18.4, 26.4, 37.6, 52 and 69.6
Thus giving us the follow coordinates to plot:
(4,18.4) , ( 9,26.4), (16, 37.6), (25, 52) and (36, 69.6)
All you have to now do is plot those points do convert your exponential graph into a straight line graph...
Then you can easily find the gradient of the graph by using
y2-y1
x2-x1
basically the gradient will be:
26.4 - 18.4 = > 8/5 or 1.6
9 - 4
To find the c - value just read off your graph
In this case, your c value should approximately be 12.2
However let me show you the algebric way of solving to find c...
Use any co ordinate, with your y - iontercept co ordinate
Which means ( 25, 52 ) and ( 0 , c )
Find the gradient of those two points
you will end up with
c - 52
0-25
But that equals to 1.6 (beecause we have already found the value for m)
therefore, c - 52 = - 40
c = 12
Now you can simply replace your Y = mX + C formula
which is
xy = 1.6x2+12
just rearrange to get y in terms of x
(question taken from may/june 2003 paper 1 cambridge )
AHH THANK YOU SO MUCH! +rep
so basically, its just substituting and altering the eqn in the form y=mx+c
man, exams are 2 days away, and i havent opened my book yet :P
im going for intensive tution for 4 hours tmrw... get that brain working. :)
Post by: fREnZy on June 01, 2009, 02:29:24 pm
There isn't much time left, you should be doing past papers now. And if you don't know how to get things the mark scheme will give you a fair idea.
You'll waste a lot of time on the tuitions, learning a lot of things that you already know.
Here's good advice though. Visualize everything in your head as a graph or like it's actually happening in case of kinematics.
Like in relative velocity, picture it. It's the only thing that'll give you a fair idea. And also, don't get confused between your velocity and distance diagrams, they can look worlds apart.
Post by: vanibharutham on June 01, 2009, 05:56:09 pm
Post by: Anonymous on June 01, 2009, 05:57:46 pm
Post by: vanibharutham on June 01, 2009, 06:02:56 pm
I can finish the whole maths paper in an hour and fifteen minutes MAX. ...
but when it comes to checking my work, i tend to think my first work was right...
like in my mock exam i got 79 / 80 because i did this:
c + 6 = 9
c = 15
-.-
its those careless miistakes that are going to kill me
Post by: Anonymous on June 01, 2009, 06:08:03 pm
Post by: shan2391 on June 01, 2009, 06:47:40 pm
Post by: vanibharutham on June 01, 2009, 08:45:37 pm
The question is dealing with vectors and states this:
The points P, Q and R are such that QR = 4PQ. Given that the position vectors of P and Q relative to an
origin O are (6i + 7j) and (9i + 20j) respectively, find the unit vector parallel to OR.
Really simple question.
Let me break it down for you,
first they are asking you to find, the unit vector parallel to OR....
ok, so lets find the position vector of OR first,
To make the question easier,
lets just call it ( xi + yj )
The question also states that QR = 4PQ
What is QR ?
( xi + yj ) - ( 9i + 20j )
This statement can join to give single component vectors:
QR => (x - 9)i + (y - 20)j
Next, what is 4PQ?
4[ ( 9i + 20j ) - (6i + 7j) ]
=> 4[3i + 13j]
12i + 52j
ok,
now equate the equation:
(x - 9)i + (y - 20)j = 12i + 52j
and solve for the i's and j's separetely
therefore,
x - 9 = 21
x = 21
and
y - 20 = 52
y = 72
Giving us a position vector for R ( 21 i + 72 j)
Finding the unit vector is the easy bit
1/Magnitude of the vector * (21 i + 72 j )
=> (21 i + 72 j) / 75
Post by: shan2391 on June 01, 2009, 10:04:36 pm
Post by: Padapop on June 02, 2009, 10:01:44 am
Post by: sweetsh on June 02, 2009, 11:03:46 am
Post by: someone on June 02, 2009, 11:04:56 am
nC2=45
shuld we just try any combination using the calculator, or is there any other way to solve this? ::)
thx :D
Post by: vanibharutham on June 02, 2009, 12:39:58 pm
and we have to find n
Really simple again
we know that nC2 = >
n(n-1)
2
and that equals to 45
so therefore,
n(n-1) = 90
n2-n-90 = 0
(n+9)(n-10)=0
n = -9 and n = 10
we cant any have any negative numbers in combinations, therefore n = 10 only
and then u can check this using your calculator:
10C2 = 45
You must remember the formula for combinations, though i think it is in the front page...
nCr =
n!
(n-r)!r!
therefore, when we have nC2 =>
n(n-1)(n-2)......etc
(n-2)(n-3)...etc * 2!
which simplyly leaves us with
n(n-1)
2
hope that helped
Post by: Padapop on June 02, 2009, 12:47:17 pm
Eg Due west - GOING TOWARDS WEST ?
From north west eg from 135 degree angle or something?
Post by: vanibharutham on June 02, 2009, 12:59:02 pm
But you have to break it down into simple and usable steps...and make sure u can work out all the angles
If it says,
The wind is blowing from the south-east, you know it will be blowing on a bearing of 045, because south-east is exactly that.
If it blows due north, then it is perpendicular
Post by: divineobsidian on June 02, 2009, 01:00:07 pm
Post by: igexam on June 02, 2009, 01:38:36 pm
If so, I would be very glad if you could send them to divakarbala@gmail.com
Post by: vanibharutham on June 02, 2009, 01:47:35 pm
is anyone good on those relative velocity questions with the river. usually when its aircraft bearings are always used from north right? but whenever i meet a river question i never really know which angle the question means >.>
Divine obsidian,
Its not only aircraft bearings that are taken from the north. Any BEARING will always be taken from the north.
For the river questions,
if you are unsure, i suggest you find all possible angles, there seems no way i can explain through this forum...
Post by: EVIL DOCTOR on June 02, 2009, 01:52:06 pm
Post by: Padapop on June 02, 2009, 02:17:08 pm
Question 4
(1 + sec x) (cosec x - cot x) = tan x
Prove the identity. Shoot thats hard.
Post by: Padapop on June 02, 2009, 02:23:42 pm
Post by: Sweet_03 on June 02, 2009, 02:43:46 pm
I have a Question .. and I hope You Can Help Me
Paper 1 May-June 2005
Question 6 :
Given that the following functions is defined for the domain -2 ? x ? 3, Find The Range of
(i) f:x ? 2-3x
(ii) g:x ? |2-3x|
(iii) f:x ? 2-|3x|
Ok .. For the First one (i) i know it .. its -7 ? f(x) ? 8
But how does do i work it out for absolute values ?
Post by: astarmathsandphysics on June 02, 2009, 02:50:31 pm
Hello ,i)2-3*3,2-3*-2 ie range is -7 to 8
I have a Question .. and I hope You Can Help Me
Paper 1 May-June 2005
Question 6 :
Given that the following functions is defined for the domain -2 ? x ? 3, Find The Range of
(i) f:x ? 2-3x
(ii) g:x ? |2-3x|
(iii) f:x ? 2-|3x|
Ok .. For the First one (i) i know it .. its -7 ? f(x) ? 8
But how does do i work it out for absolute values ?
ii)0 to 8 since|2-3x| cannot be -ve but is zero at x=2/3
ii)-7 to 2 sonve x<=2 and is -7 at x=3
Post by: sweetsh on June 02, 2009, 02:53:10 pm
Post by: astarmathsandphysics on June 02, 2009, 02:59:05 pm
Paper 2 May june 2004
Question 4
(1 + sec x) (cosec x - cot x) = tan x
Prove the identity. Shoot thats hard.
(1 + sec x) (cosec x - cot x)=(cosx + 1)/cosx* (1 - cosx)/sinx=(1-cos^2x)/(c0sxsinx)=sin^2x/cosxsinx=sinx/cosx=tanx
Post by: vanibharutham on June 02, 2009, 03:01:26 pm
Post by: sweetsh on June 02, 2009, 03:01:53 pm
Post by: astarmathsandphysics on June 02, 2009, 03:07:01 pm
you read the domain off the x axis and the range off the y axis
Post by: moraesikae on June 02, 2009, 03:08:12 pm
Post by: vanibharutham on June 02, 2009, 03:10:26 pm
the range only go down to the minimum or up to maximum point
with straight line graphs you can subsitite values
however, with quadratics or cubes or absolute values, you need to basically find that turning point within your given domain
Post by: moraesikae on June 02, 2009, 03:13:29 pm
be careful with quadratics,
the range only go down to the minimum or up to maximum point
with straight line graphs you can subsitite values
however, with quadratics or cubes or absolute values, you need to basically find that turning point within your given domain
ohh thx.. so for quadratics,, you find the minimum point,, and then the maximum point of the curve by substituting to get the largest value of the domain??
Post by: astarmathsandphysics on June 02, 2009, 03:13:47 pm
be careful with quadratics,
the range only go down to the minimum or up to maximum point
with straight line graphs you can subsitite values
however, with quadratics or cubes or absolute values, you need to basically find that turning point within your given domain
No you complete the square egy=x^2+6x-1
=(x+3)^2-3^2-1=(x+3)^2-10 so range is y>=-10
Post by: moraesikae on June 02, 2009, 03:16:41 pm
be careful with quadratics,
the range only go down to the minimum or up to maximum point
with straight line graphs you can subsitite values
however, with quadratics or cubes or absolute values, you need to basically find that turning point within your given domain
No you complete the square egy=x^2+6x-1
=(x+3)^2-3^2-1=(x+3)^2-10 so range is y>=-10
so if the domain is given,, e.g -2<x<4??
what would the range be>?
Post by: shan2391 on June 02, 2009, 03:24:59 pm
3^2y +5(3^y-10)=0
Post by: astarmathsandphysics on June 02, 2009, 03:38:01 pm
be careful with quadratics,
the range only go down to the minimum or up to maximum point
with straight line graphs you can subsitite values
however, with quadratics or cubes or absolute values, you need to basically find that turning point within your given domain
No you complete the square egy=x^2+6x-1
=(x+3)^2-3^2-1=(x+3)^2-10 so range is y>=-10
so if the domain is given,, e.g -2<x<4??
what would the range be>?
In the example I gave the minimum of the graph was at x=-3. This is included in my domain but not the one you give -2<x<4. In this case you would sub x=-2 and x=4 to get the range. In the case -4<x<4 x=-3 is included so the range woul be from -10 to (4+3)^2-10=39
Post by: moraesikae on June 02, 2009, 03:41:44 pm
o rightttt!!! thx v much!!!!!!!be careful with quadratics,
the range only go down to the minimum or up to maximum point
with straight line graphs you can subsitite values
however, with quadratics or cubes or absolute values, you need to basically find that turning point within your given domain
No you complete the square egy=x^2+6x-1
=(x+3)^2-3^2-1=(x+3)^2-10 so range is y>=-10
so if the domain is given,, e.g -2<x<4??
what would the range be>?
In the example I gave the minimum of the graph was at x=-3. This is included in my domain but not the one you give -2<x<4. In this case you would sub x=-2 and x=4 to get the range. In the case -4<x<4 x=-3 is included so the range woul be from -10 to (4+3)^2-10=39
Post by: astarmathsandphysics on June 02, 2009, 03:42:42 pm
HOW do u get y
3^2y +5(3^y-10)=0
sub p=3^y to get p2+5(p-10)=0 so p2+5p-50=(p+10)(p-5)=0 so p=-10 or 5 so y=(logp)/log3 =(log-10)/log3 impossible since log-10 does not exist or y=(log5)/log3
Post by: shan2391 on June 02, 2009, 04:39:11 pm
By the way i was reading the syllabus n did not find velcity n relative velocity in it
Post by: sweetsh on June 02, 2009, 04:43:59 pm
?
Post by: sweetsh on June 02, 2009, 05:04:10 pm
Post by: shan2391 on June 02, 2009, 05:09:59 pm
sinx=5cosx
sinx/cosx=5
tanx=5
x=78.69
Post by: sweetsh on June 02, 2009, 05:13:39 pm
Post by: moraesikae on June 02, 2009, 05:16:03 pm
3sinx-3cosx=2sinx+2cosxthere is more to it,,,it could be either 78.89 or 256.7
sinx=5cosx
sinx/cosx=5
tanx=5
x=78.69
Post by: shan2391 on June 02, 2009, 05:17:29 pm
Post by: moraesikae on June 02, 2009, 05:18:56 pm
sorry i mean 78.69.3sinx-3cosx=2sinx+2cosxthere is more to it,,,it could be either 78.89 or 256.7
sinx=5cosx
sinx/cosx=5
tanx=5
x=78.69
Post by: moraesikae on June 02, 2009, 05:19:31 pm
GUys can any 1 help me with how to find domain and range of functions
read previous pages.. it's all there :)
Post by: Sweet_03 on June 02, 2009, 05:22:45 pm
Post by: ramezamgad on June 02, 2009, 05:46:35 pm
hey sweetshsorry i mean 78.69.3sinx-3cosx=2sinx+2cosxthere is more to it,,,it could be either 78.89 or 256.7
sinx=5cosx
sinx/cosx=5
tanx=5
x=78.69
the method is correct but after obtaining the eq tanx=5 u should see the range of values of x written in the quest. it is generally 0<x<360
so u have to think when tanx is +ve and tan is +ve in the 1st and 3rd quadrats (ASTC rule)
so after getting x of the 1st quadrat u calculate the other value in the 3rd quadrat which is 180+x
finally x={78.7,258.7}
Post by: shan2391 on June 02, 2009, 06:10:11 pm
AM really struggling wid this
Post by: ramezamgad on June 02, 2009, 06:38:31 pm
the domain is the values of x and they produce values of y called the range
eg find the range for the eq. f(x)=y=3x+1 0=<x<5
y=1 at x=0 y=16 at x=5
so 1=<f(x)<16
quadratic functions:
the previous method cannot be used because it is in a shape of a curve and it has a max or min value
eg f(x)=y=x^2 + 2 -2<x<4
if u solve it in the prev. method ----> y=6 y=18 and will consequently write that 6<f(x)<18 which is not right
u have to determine the turning point of the curve
the most straightforward method is to find dy/dx and equating it to 0 which will give u the value of x to get the max. or min point
in the example dy/dx=2x
0=2x -----> x=0
y=2 at x=0 therefore the right range is 2=<f(x)<18
hope u find that useful
Post by: vanibharutham on June 02, 2009, 07:43:16 pm
Post by: shan2391 on June 02, 2009, 07:48:11 pm
Post by: vanibharutham on June 02, 2009, 07:48:47 pm
Here's a common Permuatations and Combinations Questions:
a) The producers of a play require a total cast of 5, of which 3 are actors and 2 are actresses. He auditions 5 actors and 4 actresses for the cast. Find the total number of ways in which the cast can be obtained.
b) Find how many different odd 4-digit numbers LESS than 4000 can be made from the digits 1,2,3,4,5,6,7 if no digit may be repeated.
The FIRST thing to consider when solving Permutations and Combinations questions is recognizign what the hell you are doing...
ALWAYS REMEMBER:
PERMUTATIONS HAVE TO DO WITH ARRANGEMENTS OF THINGS
COMBINATIONS HAVE TO DO WITH HOW MANY POSSIBILITIES ARE THERE, WITHOUT WORRYING ABOUT ARRANGEMENTS
if we follow that rule...
If for example, we had three things.... A,B,and C....
the combination ABC would be the same as BAC or CAB or BAC etc.
However the permutations are different...
Back to the question...
a) The producers of a play require a total cast of 5, of which 3 are actors and 2 are actresses. He auditions 5 actors and 4 actresses for the cast. Find the total number of ways in which the cast can be obtained.
This is obviously worried about combinations, not arrangements...
This my method of solving this question:
From: 5 actors 4 actresses
Choose: 3 actors 2 actresses
=> 5C3 x 4C2 = 10 x 6 = 60 combinations
b) Find how many different odd 4-digit numbers LESS than 4000 can be made from the digits 1,2,3,4,5,6,7 if no digit may be repeated.
Obviously this is a permutation question :) its got to with arrangements....
This is my method in solving this;
First, find out how many 4-digit odd numbers you can have...
well
_ x _ x _ x _
We have 4 digits to fill from 7 numbers.... however for it to be an odd number, the last digit NEEDS to be either 1,3,5 or 7...
and that gives us four digits on the last one....
_ x _ x _ x 4
What about the rest.... well.....
Obviously you will only have 6 choices for your first one, 5 for the next, and 4 for the next...
which gives us
6 x 5 x 4 x 4 = 480 odd numbers....
To find out how numbers from those odd numbers are LESS than 4000...
THEY NEED TO START WITH A 1, 2 or a 3... . and end with a 1,3,5 or 7...
First lets see all possibilites starting with 2....
Obviously because its starting with 2, we only have 1 option for the start.
1 x _ x _ x _
once again, we have 1,3,5,7 for the last option.... that gives us ... 4
1 x _ x _ x 4
how many numbers do we have left... 5 to choose from...
therefore.... 1 x 5 x 4 x 4 = 80 odd numbers Starting with 2, and that are less than 4000....
Now for numbers starting with 1 or 3....
we have two options for the first one....
2 x _ x _ x _
and we only have 3 options for the last one.... 1,3,5,7 but because we are already using one of them as our first number, we can only choose 3....
2 x _ x _ x 3
how many numbers do we have left? well we have.... 5...
therefore...
2 x 5 x 4 x 3 = 120....
120 + 80 = 200
We have 200, four digit odd numbers less than 4000...
Question 5 from October/November 2002 Paper 2...Cambridge
Read that for permutation
Post by: vanibharutham on June 02, 2009, 07:51:52 pm
However, to solve them , you need to break them down logically...
Relative velocity literally means velocity relative to another....
If they give you an airplane and wind question, you will basically need to know that
V(plane) = V(plane in still air) + V(wind)
This statement holds true if you think it through logically...
In still air, there is no wind, thereforoe your plane will have a certain velocity
However when the wind vector has been added, you need add that to your velocity in still air to calculate your overall velocity of the plane
Post by: shan2391 on June 02, 2009, 08:25:47 pm
By the way giv a shot at this 1
Q7(iii) Nov 08 p1
Post by: IO4567 on June 02, 2009, 08:26:47 pm
thank you and good luck!
Post by: vanibharutham on June 02, 2009, 08:30:37 pm
and our school let us take it because we finished our IGCSE and GCSE maths a year earlier than normal... its much harder
Post by: shan2391 on June 02, 2009, 09:45:46 pm
Q.10 is quite tricky 4 me can u giv it a shot.
paper attached
Post by: ineedaz on June 02, 2009, 09:46:29 pm
Post by: shan2391 on June 02, 2009, 09:50:13 pm
Particle mechanics and vector questions are my worst enemies! oh and permutations...i thnk combinations are ok!! anyone know the grade thresholds, by any chance??ya approx 80% or higher will earn u an A
Post by: ineedaz on June 02, 2009, 09:54:59 pm
cool...wat about an A*?? thats wat im aiming for...Particle mechanics and vector questions are my worst enemies! oh and permutations...i thnk combinations are ok!! anyone know the grade thresholds, by any chance??ya approx 80% or higher will earn u an A
Post by: shan2391 on June 02, 2009, 09:57:40 pm
hey vanibharuthamany 1 please solve the Q.
Q.10 is quite tricky 4 me can u giv it a shot.
paper attached
Post by: ineedaz on June 02, 2009, 10:08:46 pm
hey vanibharuthamQ10
Q.10 is quite tricky 4 me can u giv it a shot.
paper attached
(i) Let the velocity be = xi+yj
so:
x^2 + y^2 = (15(2)^0.5)^2 = 450
but the angle between x and y is 45 because of the (2)^0.5
so x = y
therfore 2(x)^2 = 450
x^2 = 225
x = 15
therfore the velocity = 15i+15j
MAke sense so far??
Post by: shan2391 on June 02, 2009, 10:19:44 pm
Thanks
Post by: vanibharutham on June 02, 2009, 10:42:10 pm
ineedaz has got it :)
Post by: shan2391 on June 02, 2009, 10:50:14 pm
sorry wasnt there....ya but can u finsh off because i am totally lost
ineedaz has got it :)
Post by: vanibharutham on June 03, 2009, 05:43:23 am
The ship sails north-east with a speed of 15(2)0.5 km h–1.
(i) Find, in terms of i and j, the velocity of the ship.
Ok, we need to be able to find a vector (xi + yj) whose magnitude is 15(2)0.5 .
From the question we are given the ship sails north-east i.e. on a bearing of 45 degrees...hence giving us an isoceles triangle of 90,45,45. Because it is isoceles, we know that xi = yj
Therefore, just xi = xj
x2 + x2 = (15(2)0.5 )2
x = 15
Therefore velcity of ship is 15i + 15j
(ii) Show that the ship will be at the point with position vector (24.5i + 25.5j) km at 1030 hours.
We are given that "At 0900 hours a ship sails from the point P with position vector (2i + 3j) km ". And we have also worked out that the ship is a velocity of (15i + 15j) km/h
In order to find the positiion of ship at 10:30
we know 10:30 is 1.5 hours after 9:00
Therefore
FINAL POSITION = INITIAL POSITION + (VELOCITY * TIME)
FINAL POSITION = (2i + 3j) + (15i + 15j)1.5
=> 2i + 3j + 22.5i + 22.5j
=> 24.5i + 25.5j
(iii) Find, in terms of i, j and t, the position of the ship t hours after leaving P.
Using the same formula as in the last part
FINAL POSITION = INITIAL POSITION + (VELOCITY * TIME)
Therefore in terms of i,j, and t
=> (2i + 3j) + (15i + 15j)t
=> 2i + 15ti + 3j + 15tj
=> (2+15t)i + (3+15t)j
At the same time as the ship leaves P, a submarine leaves the point Q with position vector
(47i – 27j) km. The submarine proceeds with a speed of 25 km h–1 due north to meet the ship.
(iv) Find, in terms of i and j, the velocity of the ship relative to the submarine.
The submarine travels due north... therefore it has a velocity of a (25j)km/h
The velocity of ship relative to submarine
=> 15i + 15j - 25j
=> 15i - 10j
(v) Find the position vector of the point where the submarine meets the ship.
In order for the ship and their submarine to intersect, their final positions must equal
Therefore
FINAL POSITION = INITIAL POSITION + (VELOCITY * TIME)
INITIAL POSITION + (VELOCITY * TIME) = INITIAL POSITION + (VELOCITY * TIME)
(2+15t)i + (3+15t)j = (47i - 27j) + (25j)t
(2+15t)i + (3+15t)j = (47)i + (25t - 27)j
Take the i's sepeartely first
2 + 15t = 47
15t = 45
t = 3
According to this they should meet when t = 3 hours
We have to find it for j as well, and if j equals to 3, we know it will intercept
3+15t = 25t - 27
30 = 10t
t = 3
Yes therefore they meet when t = 3 hours
To find the position vector at t = 3 hours, simply substitute
(2+15t)i + (3+15t)j
(2 + 45)i + (3 + 45)j
=> 47i + 48j
Post by: me...myself...n...i on June 03, 2009, 06:50:44 am
Post by: vanibharutham on June 03, 2009, 07:00:02 am
The diagram shows a large rectangular television screen in which one corner is taken as the origin O
and i and j are unit vectors along two of the edges. In a game, an alien spacecraft appears at the point
A with position vector 12j cm and moves across the screen with velocity (40i + 15j) cm per second. A
player fires a missile from a point B; the missile is fired 0.5 seconds after the spacecraft appears on the
screen. The point B has position vector 46i cm and the velocity of the missile is (ki +30j) cm per second,
where k is a constant. Given that the missile hits the spacecraft,
(i) show that the spacecraft moved across the screen for 1.8 seconds before impact,
(ii) find the value of k.
i) First lets identify our position vectors for each of our two things on the screen after t seconds....
First the Alien Spacecraft:
FINAL POSITION = INITIAL POSITION + (VELOCITY * TIME)
=> (12j) + (40i + 15j)(t)
=> (40t)i + (12 + 15t)j
Then the missile,
FINAL POSITION = INITIAL POSITION + (VELOCITY * TIME)
=> (46i) + (ki +30j)(t-0.5)
=> (46 + k(t - 0.5))i + (30(t-0.5))j
t has to be t - 0.5 here because it left 0.5 seconds after the alien spacecraft, so it the alien spacecraft left at t, then the missile leaves at t - 0.5
Now make your positions equal each other
(40t)i + (12 + 15t)j = (46 + k(t - 0.5))i + (30(t-0.5))j
Take the i's and j's sepeartely
therefore,
12 + 15t = (30(t-0.5))
12 + 15t = 30t - 15
27 = 15t
t = 1.8
ii) (40t) = (46 + k(t - 0.5))
We know that t = 1.8 from last question
Therefore,
(40*1.8) = 46 + k(1.8-0.5)
72 = 46 + 1.3k
1.3k = 26
k = 20
hope that helps
Post by: me...myself...n...i on June 03, 2009, 08:57:57 am
Post by: archangel on June 03, 2009, 10:40:18 am
At 0900 hours a ship sails from the point P with position vector (2i + 3j) km relative to an origin O.
The ship sails north-east with a speed of 15(2)0.5 km h–1.
(i) Find, in terms of i and j, the velocity of the ship.
Ok, we need to be able to find a vector (xi + yj) whose magnitude is 15(2)0.5 .
From the question we are given the ship sails north-east i.e. on a bearing of 45 degrees...hence giving us an isoceles triangle of 90,45,45. Because it is isoceles, we know that xi = yj
Therefore, just xi = xj
x2 + x2 = (15(2)0.5 )2
x = 15
Therefore velcity of ship is 15i + 15j
(ii) Show that the ship will be at the point with position vector (24.5i + 25.5j) km at 1030 hours.
We are given that "At 0900 hours a ship sails from the point P with position vector (2i + 3j) km ". And we have also worked out that the ship is a velocity of (15i + 15j) km/h
In order to find the positiion of ship at 10:30
we know 10:30 is 1.5 hours after 9:00
Therefore
FINAL POSITION = INITIAL POSITION + (VELOCITY * TIME)
FINAL POSITION = (2i + 3j) + (15i + 15j)1.5
=> 2i + 3j + 22.5i + 22.5j
=> 24.5i + 25.5j
(iii) Find, in terms of i, j and t, the position of the ship t hours after leaving P.
Using the same formula as in the last part
FINAL POSITION = INITIAL POSITION + (VELOCITY * TIME)
Therefore in terms of i,j, and t
=> (2i + 3j) + (15i + 15j)t
=> 2i + 15ti + 3j + 15tj
=> (2+15t)i + (3+15t)j
At the same time as the ship leaves P, a submarine leaves the point Q with position vector
(47i – 27j) km. The submarine proceeds with a speed of 25 km h–1 due north to meet the ship.
(iv) Find, in terms of i and j, the velocity of the ship relative to the submarine.
The submarine travels due north... therefore it has a velocity of a (25j)km/h
The velocity of ship relative to submarine
=> 15i + 15j - 25j
=> 15i - 10j
(v) Find the position vector of the point where the submarine meets the ship.
In order for the ship and their submarine to intersect, their final positions must equal
Therefore
FINAL POSITION = INITIAL POSITION + (VELOCITY * TIME)
INITIAL POSITION + (VELOCITY * TIME) = INITIAL POSITION + (VELOCITY * TIME)
(2+15t)i + (3+15t)j = (47i - 27j) + (25j)t
(2+15t)i + (3+15t)j = (47)i + (25t - 27)j
Take the i's sepeartely first
2 + 15t = 47
15t = 45
t = 3
According to this they should meet when t = 3 hours
We have to find it for j as well, and if j equals to 3, we know it will intercept
3+15t = 25t - 27
30 = 10t
t = 3
Yes therefore they meet when t = 3 hours
To find the position vector at t = 3 hours, simply substitute
(2+15t)i + (3+15t)j
(2 + 45)i + (3 + 45)j
=> 47i + 48j
whoa. this looks freakishly familiar...
does it ring a bell in anyone? :P
Post by: ineedaz on June 03, 2009, 06:21:45 pm
indeed!!! :D :PAt 0900 hours a ship sails from the point P with position vector (2i + 3j) km relative to an origin O.
The ship sails north-east with a speed of 15(2)0.5 km h–1.
(i) Find, in terms of i and j, the velocity of the ship.
Ok, we need to be able to find a vector (xi + yj) whose magnitude is 15(2)0.5 .
From the question we are given the ship sails north-east i.e. on a bearing of 45 degrees...hence giving us an isoceles triangle of 90,45,45. Because it is isoceles, we know that xi = yj
Therefore, just xi = xj
x2 + x2 = (15(2)0.5 )2
x = 15
Therefore velcity of ship is 15i + 15j
(ii) Show that the ship will be at the point with position vector (24.5i + 25.5j) km at 1030 hours.
We are given that "At 0900 hours a ship sails from the point P with position vector (2i + 3j) km ". And we have also worked out that the ship is a velocity of (15i + 15j) km/h
In order to find the positiion of ship at 10:30
we know 10:30 is 1.5 hours after 9:00
Therefore
FINAL POSITION = INITIAL POSITION + (VELOCITY * TIME)
FINAL POSITION = (2i + 3j) + (15i + 15j)1.5
=> 2i + 3j + 22.5i + 22.5j
=> 24.5i + 25.5j
(iii) Find, in terms of i, j and t, the position of the ship t hours after leaving P.
Using the same formula as in the last part
FINAL POSITION = INITIAL POSITION + (VELOCITY * TIME)
Therefore in terms of i,j, and t
=> (2i + 3j) + (15i + 15j)t
=> 2i + 15ti + 3j + 15tj
=> (2+15t)i + (3+15t)j
At the same time as the ship leaves P, a submarine leaves the point Q with position vector
(47i – 27j) km. The submarine proceeds with a speed of 25 km h–1 due north to meet the ship.
(iv) Find, in terms of i and j, the velocity of the ship relative to the submarine.
The submarine travels due north... therefore it has a velocity of a (25j)km/h
The velocity of ship relative to submarine
=> 15i + 15j - 25j
=> 15i - 10j
(v) Find the position vector of the point where the submarine meets the ship.
In order for the ship and their submarine to intersect, their final positions must equal
Therefore
FINAL POSITION = INITIAL POSITION + (VELOCITY * TIME)
INITIAL POSITION + (VELOCITY * TIME) = INITIAL POSITION + (VELOCITY * TIME)
(2+15t)i + (3+15t)j = (47i - 27j) + (25j)t
(2+15t)i + (3+15t)j = (47)i + (25t - 27)j
Take the i's sepeartely first
2 + 15t = 47
15t = 45
t = 3
According to this they should meet when t = 3 hours
We have to find it for j as well, and if j equals to 3, we know it will intercept
3+15t = 25t - 27
30 = 10t
t = 3
Yes therefore they meet when t = 3 hours
To find the position vector at t = 3 hours, simply substitute
(2+15t)i + (3+15t)j
(2 + 45)i + (3 + 45)j
=> 47i + 48j
whoa. this looks freakishly familiar...
does it ring a bell in anyone? :P
Post by: vanibharutham on June 03, 2009, 07:12:16 pm
Post by: Anonymous on June 03, 2009, 07:18:02 pm
Post by: vanibharutham on June 03, 2009, 07:27:03 pm
Post by: ineedaz on June 03, 2009, 08:30:28 pm
SIMPLEST paper ive done so far... hope paper 2 stays that way...
PREDICTIONS FOR PAPER 2 anyone??
- permutations and combinations
- matrices
- sets
- functions
- more calculus (DUHH!!) perhaps approximate change, more area, probably another kinematics question(which will not be optional),
- simple vectors
- trigonometry equations
- graph sketching
- inequalities
- binomial expansions
don't forget to study what came in paper 1 cause they may bring it again!!! for instance, i think they might bring circular measure again cause the one in paper 1 was tooooo easy!!
Post by: sweetsh on June 03, 2009, 08:31:09 pm
Post by: Anonymous on June 03, 2009, 08:32:24 pm
And was the position vector for part ii) 16i + 28j?
Post by: ineedaz on June 03, 2009, 08:36:41 pm
was the relative velocity like 2i + 4j?yup thats what i got!! :)
And was the position vector for part ii) 16i + 28j?
Post by: ineedaz on June 03, 2009, 08:38:00 pm
I think it will be easier than Paper 1 right?let us hope so... but i wouldn't bet on it!! :-[
Post by: sweetsh on June 03, 2009, 08:38:32 pm
Post by: ineedaz on June 03, 2009, 08:48:17 pm
and sometimes, permutations can be a b**** if u know what i mean!! and so can matrices...pretty much everythng, depending on the way questions are set..
Post by: sweetsh on June 03, 2009, 08:49:30 pm
Post by: divineobsidian on June 04, 2009, 08:01:00 am
sorry for requesting so much but maybe others also have problems ;)
all help for last paper was truly appreciated
Post by: vanibharutham on June 04, 2009, 08:32:41 am
Inverse functions are really simple, but you have to follow a step-byy-step method.
If you are given an f(x) = 2x + 19
And you are told to find f-1(x) you can do it by this method:
STEP 1: write out your f(x) formula, replacing the f(x) with y
Therefore, y = 2x + 19
STEP 2: Swap all your x's and y's
Therefore, x = 2y + 19
STEP 3, Make y the subject
Therefore,
2y = x - 19
y = (x-19)/2
STEP 4, Replace the y, with f-1(x)
Therefore,
f-1(x) = (x-19)/2
They might give you a harder question involving quadratics...
Lets say
f(x) = 2x2+18
y = 2x2+18
x = 2y2+18
2y2 = x - 18
y2 = (x - 18) / 2
y = + or - SQUARE ROOT of (x - 18) / 2
Be careful with the inverse in the quadratics, because you have to make sure the inverse fits within the domain or range given of the function
Post by: Ghost Of Highbury on June 04, 2009, 08:39:16 am
------------------------
the power is usually low in additional math for binomial theorm as in 1<= n <= 10
some of us who find combination method difficult use the triangle law
its the pascals triangle
its like this
1 1
1 2 1
1 3 3 1
1 4 6 4 1
and so on ...keep on adding the middle numbers of the previous row..to get the next one..
for example if they say to (a+b)^4 --> the power is 4 soo chk out d pattern in the triangle
its 1 4 6 4 1
these numbers will act as coeffecients..
therefore (a+b)4 =a4 + 4a3b + 6a2b2 + 4ab3 + b4
---------
u put the coefficients one by one..and keep on decreasing the power of a and increasing b's power...
------
there is also a combination methon for that..
i'll explain it later
:)
Post by: vanibharutham on June 04, 2009, 08:52:09 am
First, lets establish what roots are :P
If we are given the equation
x2-4x+4 = 0
you might straight away know that the answer is
(x-2)2=0
x = 2
But that x = 2 is a root.... it is where the curve crosses the axis
REAL ROOTS mean that the equation has roots that real numbers
and there is only two possible ways for it to have real roots
EITHER - (USING ax2+bx+c)
b2 > 4(a)(c)
OR
b2 = 4(a)(c)
If we were given another equation
Lets say
x2 - 15x + 31 = 0
And we were asked to find what type of roots it had
Then simply:
Find b2 and find 4ac
b2 = (-15)2 = 225
4ac = 4(1)(31) = 124
Here,
b2 > 4ac
And hence we have real roots
Post by: Padapop on June 04, 2009, 01:20:18 pm
Ok question on intergrating virtually anything:
if intergrating sin/cos like we had to in 12 EITHER in paper 1, i think i got the area wrong because i forgot he +c, gosh my area was like 0 ahaha failed. I do remember my answer to the previous question was 1. something and 7. something.
And intergrating sin/cos/exponential/normal gets kinda confusing sometimes.
Post by: archangel on June 04, 2009, 02:36:38 pm
i suck at the ones which like digits and how many numbers can be formed :-\
like this one:
Each of seven cards has on it one of the digits 1,2,3,4,5,6,7; no two cards have the same digit.
Four of these cards are selected and arranged to form a 4-digit number.
i) How many different 4-digit numbers can be formed this way?
ii) How many of these 4-digit numbers begin and end with an even digit?
for my ans:
i) i wrote 4! x 7
but the answer is 840... help? :-[
ii) i wrote 2 x 4 x 3 x 3
but the answer is 240.... ahhh :-[
Post by: shan2391 on June 04, 2009, 03:13:03 pm
leave spaces as there r 4digits to b made _*_*_*_
now the first space has sven options so 7*_*_*_
as we have used 1 option or digit we r left with 6 so 7*6*_*_
as we have used 2 options or digits we r left with 5 so 7*6*5*_
as we have used 3 options or digits we r left with 4 so 7*6*5*4=840
got it
Post by: divineobsidian on June 04, 2009, 03:15:54 pm
unfortunately i get 120 on the 2nd question ,not a lot of help but meh. ???
Post by: shan2391 on June 04, 2009, 03:17:41 pm
_*_*_*_ as we have to make a 4digit no.
for the first as it has to b even we have 3 options so 3*_*_*_
for the last as it has to b even we have 2 options left as we used 1 even digit b4 so 3*_*_*2
now we r left with 5 options or digits so 3*5*_*2
now v r left with 4 digits so 3*5*4*4=120
Post by: shan2391 on June 04, 2009, 03:18:47 pm
yea what shan said or you can do 7P4 if you have that on your calculatori thnk the ans should b 120
unfortunately i get 120 on the 2nd question ,not a lot of help but meh. ???
Post by: Padapop on June 04, 2009, 03:25:55 pm
_ _ _ _
lets say first _ is 2, so thats one number
2nd and 3rd _ is 5 P 2 = 20
4th can then be 4 or 6 - 2 numbers
1 * 20 * 2 = 40
Remember though that 4 and 6 can also start first so:
40*3 = 120.
I think thats right, no way that can be 240.
Post by: shan2391 on June 04, 2009, 03:29:58 pm
Post by: archangel on June 04, 2009, 03:35:41 pm
so u mainly have to use permutations huh? okayyy
+ rep for all of u guys :D
well that was what the answer at the back says =
maybe its the wrong answer printed... lol
Post by: divineobsidian on June 04, 2009, 03:47:27 pm
Nov 08 P2 Q3
Find the set of values for m for which the line y=mx+2 does not meet the curve y=x^2-5x+18
thanks, all help appreciated
Post by: Padapop on June 04, 2009, 04:15:57 pm
can anyone still help me with this real root stuff?
Nov 08 P2 Q3
Find the set of values for m for which the line y=mx+2 does not meet the curve y=x^2-5x+18
thanks, all help appreciated
b^2 - 4ac < 0
All u need to know i think :D
Post by: archangel on June 04, 2009, 04:22:20 pm
can anyone still help me with this real root stuff?
Nov 08 P2 Q3
Find the set of values for m for which the line y=mx+2 does not meet the curve y=x^2-5x+18
thanks, all help appreciated
yess =)
so the eqn will be joined together, since y is the common subject
becomes x2-5x+18=mx+2
then bring everything over
x2-5x-mx+18-2=0
x2-x(5+m)+16=0
so here the components of the quad eqn is fulfilled =)
so ax2+bx+c=0
a=1 b=(5+m) c=16
since tangent never touches the curve,
we use b2-4(a)(c) < 0
so substitute....
(5+m)2-4(1)(16)<0
25+10m+m2-64<0
m2+10m-29<0
solve quadratically, im sure u'll get ure answer :D
Post by: shan2391 on June 04, 2009, 04:38:17 pm
ahhhh thanks you guys!as far as i know it when the Q. states select use combination otherwise permutation if arrangement is important
so u mainly have to use permutations huh? okayyy
+ rep for all of u guys :D
well that was what the answer at the back says =
maybe its the wrong answer printed... lol
Post by: shan2391 on June 04, 2009, 04:55:29 pm
Nov08 Q4ii
Post by: vanibharutham on June 04, 2009, 05:01:27 pm
Post by: shan2391 on June 04, 2009, 05:12:05 pm
paper 1 or 2?oops P2
Post by: vanibharutham on June 04, 2009, 05:12:30 pm
i) Differentiate x ln x with respect to x
ii) Hence find {ln x dx.
First bit is easy
y = x lnx
Dy/dx = (1)(ln x) + (x)(1)
x
=> ln x + 1
The second part is asking you to use your first answer to help you find your integral of {ln x dx
ok
Well
we know that
{ln x + 1 dx = x lnx
therefore {ln x dx => x lnx - {1 dx
=> x lnx - x + c
To check if you answer is correct simply differentiate it
Y = x lnx - x + c
dY/dx = (1)(ln x) + (x)(1/x) - 1
= ln x + 1 - 1
=> ln x
Hence we are correct
Post by: sweetsh on June 04, 2009, 05:15:50 pm
So guys I lost the prediction topics, what were the predictions?
Post by: vanibharutham on June 04, 2009, 05:16:57 pm
yea i got 1330 aswell... thnk the vector was 30 something i and 40 somethng j
SIMPLEST paper ive done so far... hope paper 2 stays that way...
PREDICTIONS FOR PAPER 2 anyone??
- permutations and combinations
- matrices
- sets
- functions
- more calculus (DUHH!!) perhaps approximate change, more area, probably another kinematics question(which will not be optional),
- simple vectors
- trigonometry equations
- graph sketching
- inequalities
- binomial expansions
don't forget to study what came in paper 1 cause they may bring it again!!! for instance, i think they might bring circular measure again cause the one in paper 1 was tooooo easy!!
Post by: sweetsh on June 04, 2009, 05:19:01 pm
Post by: shan2391 on June 04, 2009, 05:19:31 pm
Post by: vanibharutham on June 04, 2009, 05:32:20 pm
Post by: sweetsh on June 04, 2009, 08:52:46 pm
Post by: shan2391 on June 04, 2009, 10:42:30 pm
What about relative velocity? Let's do an example from past papersgud idea bringem on
Post by: shan2391 on June 04, 2009, 11:11:17 pm
Post by: astarmathsandphysics on June 04, 2009, 11:34:49 pm
guys help Q.2 in this paper(attached)
c=3 since (0,0) nusves to (0,3) b=1 since there is no scaling in the x direction and c=2 since tanpi/4=1 but 5-3=2 so there is a scaing factor 2 in the y direction
Post by: divineobsidian on June 05, 2009, 12:07:20 am
Quote
yess =)
so the eqn will be joined together, since y is the common subject
becomes x2-5x+18=mx+2
then bring everything over
x2-5x-mx+18-2=0
x2-x(5+m)+16=0
so here the components of the quad eqn is fulfilled =)
so ax2+bx+c=0
a=1 b=(5+m) c=16
since tangent never touches the curve,
we use b2-4(a)(c) < 0
so substitute....
(5+m)2-4(1)(16)<0
25+10m+m2-64<0
m2+10m-29<0
solve quadratically, im sure u'll get ure answer :D
k thanks arc sorry for long reply i was sleeping in asia
edit:yeah By the way vanib i did OR on the P1 but i think i kinda messed up
Post by: Sweet_03 on June 05, 2009, 04:42:54 am
If You Have a vector AB
and You need to find th unit vector
is this the right formula ?
1 / |AB| x AB
and to find out the magnitude ..
SQUARE ROOT of a2 + b2
Is this Right ?
Post by: vanibharutham on June 05, 2009, 05:19:24 am
Post by: Sweet_03 on June 05, 2009, 05:33:36 am
Thnnnx aLot
and Good Luck everyone
Post by: vanibharutham on June 05, 2009, 06:00:31 am
Post by: Ghost Of Highbury on June 05, 2009, 07:04:37 am
how do u find the value for a and b?"??guys help Q.2 in this paper(attached)
c=3 since (0,0) nusves to (0,3) b=1 since there is no scaling in the x direction and c=2 since tanpi/4=1 but 5-3=2 so there is a scaing factor 2 in the y direction
plzz explain clearly
Post by: astarmathsandphysics on June 05, 2009, 07:12:08 am
Post by: Ghost Of Highbury on September 25, 2009, 10:38:29 am
Post by: nid404 on September 25, 2009, 10:43:27 am
blowing from the north-east. Given that the wind has a constant speed of 12 m/s, find the direction
from which the wind is blowing.
Have a look aadi.....if you get the other part lemme know
Post by: Ghost Of Highbury on September 25, 2009, 10:45:18 am
Post by: nid404 on September 25, 2009, 10:49:45 am
shudnt 135 come between 7 and 12?
cos inverse of 7/12 =
Post by: Ghost Of Highbury on September 25, 2009, 11:20:17 am
Post by: nid404 on September 25, 2009, 11:24:22 am
but 135 shud come between 7 and 12 righ?
not possible....how r u saying that??
Post by: Ghost Of Highbury on September 25, 2009, 11:27:35 am
not possible....how r u saying that??
arre...the 12 is the wind and 7 is the cyclist..
the angle between them is "NORTH EAST"
north + 0.5 * north to north east = 90 +45 = 135
Post by: nid404 on September 25, 2009, 11:29:04 am
arre...the 12 is the wind and 7 is the cyclist..
the angle between them is "NORTH EAST"
north + 0.5 * north to north east = 90 +45 = 135
check the diagram dude...what r u saying?
Post by: Ghost Of Highbury on September 25, 2009, 11:30:18 am
Post by: nid404 on September 25, 2009, 11:39:56 am
arre...i am saying that the 135 degree position is incorrect...it shud be between 7 and 12....as the wind appears to come from north east for the cyclist.....so the angle is between the cyclist and the wind na?
draw and show re.....
Post by: Ghost Of Highbury on September 25, 2009, 11:43:13 am
Post by: nid404 on September 25, 2009, 11:53:46 am
ok re...
our diagrams differ....how do you know how the diagram should be??
Post by: Ghost Of Highbury on September 25, 2009, 11:57:56 am
because they r talking abt the cyclist and te wind...so the angle shud be in between 7(cyclist) and 12(wind) na?
Post by: nid404 on September 25, 2009, 11:59:57 am
i guess it shud be like this..
because they r talking abt the cyclist and te wind...so the angle shud be in between 7(cyclist) and 12(wind) na?
we gotta find the angle na re.....and ur totally confusing me.....
my vector diagram has right angled triangle unlike urs....which one to draw? vague questions
Post by: Ghost Of Highbury on September 25, 2009, 12:01:01 pm
Post by: nid404 on September 25, 2009, 12:02:42 pm
no right angle...
how can u say that??
Post by: Ghost Of Highbury on September 25, 2009, 12:04:50 pm
Post by: nid404 on September 25, 2009, 12:06:02 pm
the markscheme too doesnt have a right angle triangle
i don't understand how they r assuming it the way they want to.....CIE needs to confirm it's mark schemes
Post by: Ghost Of Highbury on September 25, 2009, 12:07:15 pm
Post by: nid404 on September 25, 2009, 12:09:33 pm
hahahha...no re..there markschemes are obviously correct...its just that we cant figure it out...astar or slvri can answer it...
maybe :-[
Post by: sameer210394 on October 10, 2009, 11:36:26 am
i'm accepting a lot of questions related to Additional maths portions 4 2009.
Post by: sameer210394 on October 11, 2009, 02:56:22 pm
ok re...
yaar, ur drawing is wrong, ill show u the right 1.
in my diagram, the 4 sections show north East(+,+) , North West(-,+) , South East(+,-) , South West(-,-).
as the question says the wind appears to be blowing from north east, means that the wind is coming from (+,+) 'B'. stress on this line again
and at the same time the boy is traveling south at 'A'.
centre = 0
now the relative velocity of '0A' and 'BA' = 'B0' therefore the the angle B0A is 1350 and angle 0AB is not..
tu hi soch, itna chota angle in ur drawin, how can it be 1350
i hope this was clear.
Post by: Ghost Of Highbury on October 11, 2009, 06:33:08 pm
yaar, ur drawing is wrong, ill show u the right 1.
in my diagram, the 4 sections show north East(+,+) , North West(-,+) , South East(+,-) , South West(-,-).
as the question says the wind appears to be blowing from north east, means that the wind is coming from (+,+) 'B'. stress on this line again
and at the same time the boy is traveling south at 'A'.
centre = 0
now the relative velocity of '0A' and 'BA' = 'B0' therefore the the angle B0A is 1350 and angle 0AB is not..
tu hi soch, itna chota angle in ur drawin, how can it be 1350
i hope this was clear.
abbe haaann...hell's bell...thansk a lot man!
Post by: .......dA dEviL..... on October 14, 2009, 03:21:23 am
Post by: Ghost Of Highbury on October 14, 2009, 06:21:39 am
guys...if any candidate giving exam in 2009 oct/nov session.....please be aware of the fact that there is 100% probability that this relative velocity question will come.....and more over its the relative velocity river question....take care...n best of luck 4 the upcoming 0 "zero" level exam hat will partly help us to shape our future studies....... ;D
n how can u say that??
Post by: .......dA dEviL..... on October 14, 2009, 03:11:00 pm
Post by: sameer210394 on October 14, 2009, 07:24:49 pm
in dat case...please check the pattern of hw questionz come....u wl find dat alwayz dey r da same...if may/june session vector velocity comes.....den in oct/nov session relative velocity comes....n vice versa.....seeing da pattern of da last few years......nw dey r giving more river velocity questionz...n hence ma deduction..... 8) 8)
relative velocity is much simple, gud thought ........dA dEvil.... wat u said is true for most of the cases.
Post by: astarmathsandphysics on October 14, 2009, 07:35:00 pm
Post by: sameer210394 on October 14, 2009, 08:51:36 pm
I will do something on relative veolcity and vectors for my website tomorrow.
yes, that would be very helpful. Till yesterday i didn't know that Loga(b) could be written as 1/(Logb(a))
i knew it yesterday when referring your website, very helpful.
Du u take tuitions for students after IB...
Post by: .......dA dEviL..... on October 14, 2009, 08:54:04 pm
Post by: astarmathsandphysics on October 14, 2009, 10:21:19 pm
http://www.astarmathsandphysics.com/igcse_maths_notes/igcse_maths_notes_relative_velocities_and_relative_positions.html
Post by: BlackBunny103 on April 16, 2010, 09:24:54 am
I'm having add maths exam for this year. Kinda worried about the velocity stuff :(
Post by: SGVaibhav on April 16, 2010, 10:19:42 am
i know all differentiation / integration/partial derivatives/directional derivatives and soo much with partial derivatives and applications in 3D.. but i dont know this ??? ??? ??? ??? ???
Post by: nid404 on April 18, 2010, 08:22:48 am
omg, i dont know all this
i know all differentiation / integration/partial derivatives/directional derivatives and soo much with partial derivatives and applications in 3D.. but i dont know this ??? ??? ??? ??? ???
haaawwww....so? :P
Post by: BlackBunny103 on May 30, 2010, 08:30:44 am
So if you are stuck with anything, post your doubts here and Add Maths students will help as soon as they can.
Good luck to all of us :D
Post by: J.Darren on May 30, 2010, 10:02:44 am
Post by: holtadit on May 30, 2010, 10:05:49 am
I am experiencing some major difficulty in finding the inverse of a fuction, domain and range, would anyone be kind enough to help out ?
I can help with inverse of function - thats within my syllabus.
Post by: J.Darren on May 30, 2010, 10:18:46 am
I can help with inverse of function - thats within my syllabus.Hmm ... what about exponential functions ?
Post by: holtadit on May 30, 2010, 10:19:23 am
Hmm ... what about exponential functions ?
Let me see an example.
Post by: J.Darren on May 30, 2010, 10:21:00 am
Let me see an example.A function f is defined by f (x) = e^(x–1), where x > 0.
(i) State the range of f.
(ii) Find an expression for f –1.
(iii) State the domain of f –1.
Post by: holtadit on May 30, 2010, 10:26:33 am
I do normal mathematics. Wish I could help you out.
Just hang in there; someone will be along to help you out ;)
Post by: cooldude on May 30, 2010, 10:55:26 am
Sorry Darren, but this is way out of my depth.
I do normal mathematics. Wish I could help you out.
Just hang in there; someone will be along to help you out ;)
l) okay substitute the value of 0 in the function-->
therefore f(0)=e^-1=0.368
therefore f(x)>0.368
we substitute the value of 0 because this is the least value in the domain, i know that 0 is not included, you can take something like 0.000000000000000001, but 0 is better. or we can use differentiation to find the minimum value of the function, d/dx (e^(x-1))=e^(x-1), therefore at 0 it is 0.368, therefore since this is the minimum value every value of y will be greater than 0.368, therefore the range is f(x)>0.368
ii) let f(x)=y,
now y=e^(x-1)
take log on both sides (to find the inverse we have to make x the subject)
ln y=x-1
x=ln y + 1
therefore f^-1(x)=ln x + 1
iii) the domain of f^-1(x)=range of f(x), therefore the domain is x>0.368
Post by: J.Darren on May 30, 2010, 11:08:09 am
l) okay substitute the value of 0 in the function-->Thanks mate, although I don't quite get the final bit RE: domain of the inverse function = range of the function ...
therefore f(0)=e^-1=0.368
therefore f(x)>0.368
we substitute the value of 0 because this is the least value in the domain, i know that 0 is not included, you can take something like 0.000000000000000001, but 0 is better. or we can use differentiation to find the minimum value of the function, d/dx (e^(x-1))=e^(x-1), therefore at 0 it is 0.368, therefore since this is the minimum value every value of y will be greater than 0.368, therefore the range is f(x)>0.368
ii) let f(x)=y,
now y=e^(x-1)
take log on both sides (to find the inverse we have to make x the subject)
ln y=x-1
x=ln y + 1
therefore f^-1(x)=ln x + 1
iii) the domain of f^-1(x)=range of f(x), therefore the domain is x>0.368
By the way what would the inverse function look like if the original function is, say for example, ln (x-1) ?
Post by: J.Darren on May 30, 2010, 11:10:34 am
https://studentforums.biz/index.php/topic,8457.0.html
Post by: Ghost Of Highbury on May 30, 2010, 11:21:52 am
Thanks mate, although I don't quite get the final bit RE: domain of the inverse function = range of the function ...
By the way what would the inverse function look like if the original function is, say for example, ln (x-1) ?
the inverse of any graph f(x) is the graph under reflection along y=x.
Post by: cooldude on May 30, 2010, 11:22:41 am
Thanks mate, although I don't quite get the final bit RE: domain of the inverse function = range of the function ...
By the way what would the inverse function look like if the original function is, say for example, ln (x-1) ?
yep, the same applies (vice-versa) --> the range of an inverse function = to the domain of the function
and the reason is given by aadi
Post by: Ghost Of Highbury on May 30, 2010, 11:24:50 am
yep, as aadi says and the same applies the range of an inverse function is the domain of the function and the reason is given by aadi
exactly. wen is ur add. math exam darren?
Post by: J.Darren on May 30, 2010, 11:40:27 am
exactly. wen is ur add. math exam darren?The week after, same as you guys I believe ...
Post by: cooldude on May 30, 2010, 11:41:56 am
The week after, same as you guys I believe ...
lol dude, aadi gave his add maths in nov 09 and i didnt give it all :D
Post by: syedz123 on May 31, 2010, 04:21:56 pm
I still have accouting,chems litt n econs n add math left so i wont b able to do alot of pp for add maths...can any1 plz recommend which yrs i shuld do..like gud hard papers...atleast 3 yrs plz apart 4rm 09 Thanks
Post by: J.Darren on May 31, 2010, 05:00:52 pm
am quite a week student in add math...can sum1 plz xplain question 12 second one on velocity for me...its may/june 09 p1...If I were you I would do Add Maths over the others, Economics paper 2 is all about recital, Chemistry Paper 3 - You either know or you don't, understanding the concept is the key.
I still have accouting,chems litt n econs n add math left so i wont b able to do alot of pp for add maths...can any1 plz recommend which yrs i shuld do..like gud hard papers...atleast 3 yrs plz apart 4rm 09 Thanks
Post by: cooldude on May 31, 2010, 05:29:29 pm
am quite a week student in add math...can sum1 plz xplain question 12 second one on velocity for me...its may/june 09 p1...
I still have accouting,chems litt n econs n add math left so i wont b able to do alot of pp for add maths...can any1 plz recommend which yrs i shuld do..like gud hard papers...atleast 3 yrs plz apart 4rm 09 Thanks
i) 0=kcos4t
cos4t=0
90=4t
t=22.5 sec
ii) dv/dt=a=-4ksin4t
iii) 12=4k as sin(4*(3/8))=-1
k=3
iv) u just have to sketch the graph of 3cos4t
v) integrate the expression for velocity-->
x=3sin4t/4+k
however since at the origin the displacement is 0, therefore k=0
now substitute t=pi/24
x=0.375 m
Post by: Ghost Of Highbury on May 31, 2010, 05:41:21 pm
am quite a week student in add math...can sum1 plz xplain question 12 second one on velocity for me...its may/june 09 p1...
I still have accouting,chems litt n econs n add math left so i wont b able to do alot of pp for add maths...can any1 plz recommend which yrs i shuld do..like gud hard papers...atleast 3 yrs plz apart 4rm 09 Thanks
i) substitute 0 for velocity.. 0 = cos 4t
4t = y
cos y = 0
y = cos-1 0 = 90
here its 90 degrees, but v gotta take it in radians, so Pi radians = 180 degree
4t = 0.5pi radians
t = pi/8
t = pi/4
ii) differentiate v = k cos 4t
k = a
cos 4t = b
d/dx(a) = 0
d/dx(b) = -4sin4t
dv/dx = a(db) + b(da) = -4ksin4t + 0 = -4k sin 4t
iii)
k = 3
iv) make a table and plot the points.
v) integration of 3cos4t with limits pi/24 and 0
Post by: syedz123 on May 31, 2010, 05:52:07 pm
Post by: J.Darren on May 31, 2010, 06:23:57 pm
Cosine equals to 0 at 90 degreees (first instantaneously at rest) / 0.5 pi radian, 90 / 4 = 22.5 degrees / 0.125 pi radian :o
here its 90 degrees, but v gotta take it in radians, so Pi radians = 180 degree
4t = 180
t = pi/4
Post by: Ghost Of Highbury on May 31, 2010, 06:29:29 pm
Cosine equals to 0 at 90 degreees (first instantaneously at rest) / 0.5 pi radian, 90 / 4 = 22.5 degrees / 0.125 pi radian :o
my bad.edited. thanks
Post by: syedz123 on May 31, 2010, 09:31:14 pm
i) substitute 0 for velocity.. 0 = cos 4t
4t = y
cos y = 0
y = cos-1 0 = 90
here its 90 degrees, but v gotta take it in radians, so Pi radians = 180 degree
4t = 0.5pi radians
t = pi/8
t = pi/4
ii) differentiate v = k cos 4t
k = a
cos 4t = b
d/dx(a) = 0
d/dx(b) = -4sin4t
dv/dx = a(db) + b(da) = -4ksin4t + 0 = -4k sin 4t
iii)
k = 3
iv) make a table and plot the points.
v) integration of 3cos4t with limits pi/24 and 0
sorri A@di bt i still dnt get da second one...it says find da position vector of P at 12:00...so do we jst leave da answer as -4k sin 4t? dont we hv to use da '12:00' in this and da answer shuld b in a vector f0rm?? c0z its findin da position vector...sorri i told u am a weak add math student :-\
Post by: Ghost Of Highbury on June 01, 2010, 06:02:49 am
Post by: syedz123 on June 01, 2010, 04:35:23 pm
Dude the question u r referring to is question 9...the one with the quote is different answer which is question 12...
omgness!! LOL wts happning to me...sorri man lol Thanks agen :D
Post by: syedz123 on June 02, 2010, 12:59:28 pm
Find the coefficient of x4 in the expansion of
(i) (1-x/4)(1+2x)^6
and integration 4rm q7 (ii) may/june 09 p1
plz explain step by step
Post by: cooldude on June 02, 2010, 01:20:58 pm
can someone plz help me wiv this
Find the coefficient of x4 in the expansion of
(i) (1-x/4)(1+2x)^6
k the first q-->
first expand (1+2x)^6--> 1+12x+60x^2+160x^3+240x^4........
now we need the coefficient of x^4--> (1-(x/4))(1+12x+60x^2+160x^3+240x^4........)
now look at the above line ive typed, look for powers of x which multiply to give x^4, we see that 1 and 240x^4 multiply to give x^4 i.e. 240x^4, we also see that the product of -x/4 and 160x^3 is -40x^4 which again has x^4, so now we have 2 terms which have x^4, -40x^4 and 240x^4, so when we add them we get 200x^4, and the coefficient of this is 200, the answer required, the way i told u of finding terms which multiply to give x^4 is much easier than expanding the whole thing, u can do that for this q, as it will give u a clear idea of what ive done, and ive only expanded (1+2x)^6 uptil x^4 as that is what we need and not the whole expansion
Post by: Ghost Of Highbury on June 02, 2010, 01:26:43 pm
(....160x^3 + 240x^4)(1-x/4)
160x^3 - (160x^4/4) + 240x^4 = ....240x^4 - 40x^4.... = 200x^4
hence, 200
Post by: syedz123 on June 02, 2010, 01:34:03 pm
Post by: cooldude on June 02, 2010, 01:41:04 pm
in the first part we found that the derivate is 3xe^3x
the simplest explanation i can give is that the integral of xe^3x will be 1/3 of the expression given to differentiate in part i is cuz the derivate we found in the second part is 3 times the integrand in part ii, thus since differentiation is the opposite of integration we divide it by 3 as the derivate we got in part i is 3 times the integrand in part ii, the other way i knw to find the integral is integration by parts but im not sure whether u add maths guys knw that as ive never done add maths myself so i have no idea what the syll is :D , neway it wont make a diff since the q says hence, so just try and understand what i said or wait until sum1 can give a better explanation
Post by: J.Darren on June 02, 2010, 03:18:32 pm
There you go.
Post by: jellybeans on June 02, 2010, 04:35:52 pm
Thanks :D
i knw you integrate the velocity to get the displacement but what values do you sub in? :o
thanks again. XD
Post by: cooldude on June 02, 2010, 04:42:29 pm
HIIII, um, i need help with Q12ii) pleaseeee :)
Thanks :D
i knw you integrate the velocity to get the displacement but what values do you sub in? :o
thanks again. XD
ii) integrate a to find v->
v=1.4t-0.3t^2+k
v=0.5, when t=0, thus k=0.5
thus v=1.4t-0.3t^2+0.5
x=0.7t^2-0.1t^3+0.5t
to find the dist u just have to integrate between the limits, 10 and 0, u dont have to worry about the arbitary constant as when we evaluate definite integrals we dont consider it, so now just substitute the value of 10 in the equation for x as substituting 0 wud give the displacement as 0,
x=70-100+5=-25m
Post by: jellybeans on June 02, 2010, 04:45:18 pm
ii) integrate a to find v->
v=1.4t-0.3t^2+k
v=0.5, when t=0, thus k=0.5
thus v=1.4t-0.3t^2+0.5
x=0.7t^2-0.1t^3+0.5t
to find the dist u just have to integrate between the limits, 10 and 0, u dont have to worry about the arbitary constant as when we evaluate definite integrals we dont consider it, so now just substitute the value of 10 in the equation for x as substituting 0 wud give the displacement as 0,
x=70-100+5=-25m
mmm thats what i did in the first place & got the same answer. but the MS had a different answer ... explanation ? ;D hehs
*whats with the 2x7.5 ? :-\
thanks
Post by: J.Darren on June 02, 2010, 04:48:50 pm
HIIII, um, i need help with Q12ii) pleaseeee :)(http://img710.imageshack.us/img710/8306/question12.png)
Thanks :D
i knw you integrate the velocity to get the displacement but what values do you sub in? :o
thanks again. XD
When you are dealing with displacement question, one must be very careful with the velocity, if between the given time frame, the particle experiences a change in the direction of travel (forwards / backwards), then you must do partial integration.
Post by: jellybeans on June 02, 2010, 04:52:23 pm
(http://img710.imageshack.us/img710/8306/question12.png)
When you are dealing with displacement question, one must be very careful with the velocity, if between the given time frame, the particle experiences a change in the direction of travel (forwards / backwards), then you must do partial integration.
ohohhh, but why is it 2x7.5? and not just 7.5+25?
Post by: cooldude on June 02, 2010, 04:53:31 pm
mmm thats what i did in the first place & got the same answer. but the MS had a different answer ... explanation ? ;D hehs
*whats with the 2x7.5 ? :-\
thanks
k sorry, its about the particle travelling 7.5m in the positive direction and then going back thus we multiply by 2 and then add the 25 m in the negative direction to find the total dist travelled, first the particle travels in the positive x direction and travels 7.5 m, and then for the next 5 seconds it travels 7.5 m back to the origin and then another 25 m in the negative x direction, thus a total distance of 7.5+7.5+25=40 m, sorry i misread the q, i thought it was askin the displacement not the distance travelled, neway j darren answered ur q
Post by: J.Darren on June 02, 2010, 05:03:29 pm
k sorry, its about the particle travelling 7.5m in the positive direction and then going back thus we multiply by 2 and then add the 25 m in the negative direction to find the total dist travelled, first the particle travels in the positive x direction and travels 7.5 m, and then for the next 5 seconds it travels 7.5 m back to the origin and then another 25 m in the negative x direction, thus a total distance of 7.5+7.5+25=40 m, sorry i misread the q, i thought it was askin the displacement not the distance travelled, neway j darren answered ur qThe question did not mention anything the particles going back after travelling 7.5m in the positive direction, the acceleration gradually slows down after passing through O until it becomes zero at five, from that point onwards the particle is travelling in the negative direction until t = 10 ...
Post by: jellybeans on June 02, 2010, 05:06:41 pm
The question did not mention anything the particles going back after travelling 7.5m in the positive direction, the acceleration gradually slows down after passing through O until it becomes zero at five, from that point onwards the particle is travelling in the negative direction until t = 10 ...
so why do you have to times 7.5 by twooooo? :'( .. i mean how* do you know you have to times it by 2...
is it cuz it travels and then stops.. so you double the distance?
Post by: J.Darren on June 02, 2010, 05:14:44 pm
so why do you have to times 7.5 by twooooo? :'( .. i mean how* do you know you have to times it by 2...I have no idea, but it seems that the marking scheme accepts both ...
is it cuz it travels and then stops.. so you double the distance?
Post by: jellybeans on June 02, 2010, 05:18:56 pm
I have no idea, but it seems that the marking scheme accepts both ...
lol, okay then. thanks! :)
& thanks to cooldude aswellll, :) :)
... another one ... Q9iii) ? :-[
Post by: cooldude on June 02, 2010, 05:23:01 pm
The question did not mention anything the particles going back after travelling 7.5m in the positive direction, the acceleration gradually slows down after passing through O until it becomes zero at five, from that point onwards the particle is travelling in the negative direction until t = 10 ...
actually the q wont mention that the displacement is 7.5m u have to find this by calculation, ill show u what i mean, at t=5 sec the displacement is +7.5m
x=0.7t^2-0.1t^3+0.5t
when t=5, x=+7.5m
then the displacement is 0 at-->
0=0.7t^2-0.1t^3+0.5t
at t=7.65 the displacement is again 0 proving that the particle did travel +7.5m in the positive direction between 0 and 5, and then between 0 and 7.65 it again travels back to the origin a further 7.5m, the dist travelled=15m
now for the last 2.35 sec, u can find the displacement, u can integrate between 10 and 7.65 in the displacement, we get -25m, proving that the total dist is 7.5+7.5+25=40 , this should clear things up, if u need more explanation ill be happy to give it
Post by: jellybeans on June 02, 2010, 05:31:14 pm
actually the q wont mention that the displacement is 7.5m u have to find this by calculation, ill show u what i mean, at t=5 sec the displacement is +7.5m
x=0.7t^2-0.1t^3+0.5t
when t=5, x=+7.5m
then the displacement is 0 at-->
0=0.7t^2-0.1t^3+0.5t
at t=7.65 the displacement is again 0 proving that the particle did travel +7.5m in the positive direction between 0 and 5, and then between 0 and 7.65 it again travels back to the origin a further 7.5m, the dist travelled=15m
now for the last 2.35 sec, u can find the displacement, u can integrate between 10 and 7.65 in the displacement, we get -25m, proving that the total dist is 7.5+7.5+25=40 , this should clear things up, if u need more explanation ill be happy to give it
so you have to find out t when displacement = 0 ? ..which will then tell you that it returns to its origin.. and so .. its 7.5 x 2 for going there and back?
:o
Post by: J.Darren on June 02, 2010, 05:33:33 pm
Y = ln y
X = x
m = ln b
c = ln a
ln y = ln A + k ln x
Y = ln y
X = ln x
m = k
c = ln a
Post by: jellybeans on June 02, 2010, 05:35:23 pm
ln y = ln a + x ln b
Y = ln y
X = x
m = ln b
c = ln a
ln y = ln A + k ln x
Y = ln y
X = ln x
m = k
c = ln a
wait sorry.. is that the 3rd bit? :O
cuz for the first two i used logs instead of ln , ... :-\
anyways; for iii) i did:
px+qy=xy
xy-qy = px
(x-q)y = px
y = px / (x-q)
lol .. and yh. thats how far i got for 9iii) Xb
Post by: cooldude on June 02, 2010, 05:40:19 pm
so you have to find out t when displacement = 0 ? ..which will then tell you that it returns to its origin.. and so .. its 7.5 x 2?
:o
yeah, if u get more than one value it basically tells u that the particle passes the origin at more than one point, i.e. it moves first in the positive direction and then in the negative direction, and then u have to find the distance travelled in the positive and the negative direction, and theres also one more way to find if the particle goes in the positive direction, if u find the maximum displacement and it comes to a value not equal to ur final displacement, i.e. in this q if the maximum displacement is not -25m then u get to knw that the particle has travelled in the positive direction also, for example in this u get the maximum displacement at t=5, i.e. x=7.5m, thus u get to knw that the particle has also travelled in the positive direction and u can take the movement in the positive direction into account
Post by: J.Darren on June 02, 2010, 05:41:40 pm
wait sorry.. is that the 3rd bit? :Ox - q = px / y
cuz for the first two i used logs instead of ln , ... :-\
px+qy=xy
xy-qy = px
(x-q)y = px
y = px / (x-q)
lol .. and yh. thats how far i got for 9iii) Xb
(x - q) / x = (px / y) * (1 / x)
(x - q) / x = p / y
1 - q / x = p / y
y = p / (1 - q / x)
would it works ?
Post by: jellybeans on June 02, 2010, 05:46:40 pm
yeah, if u get more than one value it basically tells u that the particle passes the origin at more than one point, i.e. it moves first in the positive direction and then in the negative direction, and then u have to find the distance travelled in the positive and the negative direction, and theres also one more way to find if the particle goes in the positive direction, if u find the maximum displacement and it comes to a value not equal to ur final displacement, i.e. in this q if the maximum displacement is not -25m then u get to knw that the particle has travelled in the positive direction also, for example in this u get the maximum displacement at t=5, i.e. x=7.5m, thus u get to knw that the particle has also travelled in the positive direction and u can take the movement in the positive direction into account
hahaha, okay! thanks ;D
x - q = px / y
(x - q) / x = (px / y) * (1 / x)
(x - q) / x = p / y
1 - q / x = p / y
y = p / (1 - q / x)
would it works ?
lol sry umm ..checkk teh ms.. :-\
Post by: J.Darren on June 02, 2010, 05:52:34 pm
Post by: jellybeans on June 02, 2010, 05:56:28 pm
No idea lol, but I don't presume those sort of Qs would appear in this year's paper, somewhat beyond our scope ...
LOL? are you sureeeee.. ? hahaha thanks anyways! ;) ;)
Post by: Ghost Of Highbury on June 02, 2010, 06:01:46 pm
qy = xy-px
qy = x(y-p)
x = (qy)/(y-p)
(reciprocal of both sides)
1/x = (y-p)/(qy)
1/x = y/qy - p/qy
1/x = 1/q - (p/q)(1/y)
1/x = (1/y)(-p/q) + 1/q
y = xm + c
y = 1/x
x = 1/y
m = (-p/q)
c = (1/q)
Post by: cooldude on June 02, 2010, 06:04:27 pm
hahaha, okay! thanks ;D
lol sry umm ..checkk teh ms.. :-\
the first one is impossible, the one above the table, ill show u the explanation (ill go backwards)-->
1/y=-9/px+1/p
reciprocate--> y=-px/9+p
write as single fraction--> y=(-px+9p)/9
cross multiply--> 9y=9p-px
rearrange--> 9y+px=9p
now the first one in the table-->
y=qy/x+p
now just multiply evrything by x ull get it
now the 2nd one-->
y/x=y/q-(p/q)
multiply evrything by q--> qy/x=y-p
cross multiply--> qy=xy-px
rearrange ull get it
and blah blah blah
Post by: jellybeans on June 02, 2010, 06:10:50 pm
the first one is impossible, the one above the table, ill show u the explanation (ill go backwards)-->:o :o :o ... which table did you get those values from? :o :o :o lol
1/y=-9/px+1/p
reciprocate--> y=-px/9+p
write as single fraction--> y=(-px+9p)/9
cross multiply--> 9y=9p-px
rearrange--> 9y+px=9p
now the first one in the table-->
y=qy/x+p
now just multiply evrything by x ull get it
now the 2nd one-->
y/x=y/q-(p/q)
multiply evrything by q--> qy/x=y-p
cross multiply--> qy=xy-px
rearrange ull get it
and blah blah blah
Post by: J.Darren on June 02, 2010, 06:13:18 pm
:o :o :o ... which table did you get those values from? :o :o :o lolThe marking scheme thingie ...
Post by: Ghost Of Highbury on June 02, 2010, 06:13:29 pm
Post by: jellybeans on June 02, 2010, 06:15:05 pm
the first one is impossible, the one above the table, ill show u the explanation (ill go backwards)-->
1/y=-9/px+1/p
reciprocate--> y=-px/9+p
write as single fraction--> y=(-px+9p)/9
cross multiply--> 9y=9p-px
rearrange--> 9y+px=9p
now the first one in the table-->
y=qy/x+p
now just multiply evrything by x ull get it
now the 2nd one-->
y/x=y/q-(p/q)
multiply evrything by q--> qy/x=y-p
cross multiply--> qy=xy-px
rearrange ull get it
and blah blah blah
The marking scheme thingie ...
Cooldude went backwards, check my post , ive gone forwards. the same thing....
OH. okay, THANKS GUYS! thaaaank you so much :P
Post by: jellybeans on June 03, 2010, 06:09:33 pm
i need help with 8ii) D:
thanks :) :)!
Post by: Ghost Of Highbury on June 03, 2010, 06:51:54 pm
Post by: astarmathsandphysics on June 03, 2010, 07:37:39 pm
Post by: Dibss on June 03, 2010, 08:45:25 pm
lol.. hi again :)Check the attachment. Hope it helps =]
i need help with 8ii) D:
thanks :) :)!
Post by: astarmathsandphysics on June 03, 2010, 10:03:03 pm
Post by: Dibss on June 03, 2010, 10:12:36 pm
it says C intersection d=165000nope, it says people who don't have both a comp and a dishwasher so that would be everything outside of (C intersection D) i think...
ok, let me go check the mark scheme and i'll get back to you!
Post by: Dibss on June 03, 2010, 10:17:27 pm
ok, let me go check the mark scheme and i'll get back to you!it's from M/J 05 P2 =]
Post by: astarmathsandphysics on June 04, 2010, 12:07:58 am
Post by: jellybeans on June 04, 2010, 03:15:04 am
INCOMPLETE Q...post the paper link and ms..ohh sry.
D: its incomplete?
answer to the venn diagram question
Check the attachment. Hope it helps =]
THANK YOU ;D ;D ;D !!
Post by: jellybeans on June 05, 2010, 04:47:33 am
thank you& its NOT an incomplete question :)
Post by: Ghost Of Highbury on June 05, 2010, 06:21:09 am
there are two different cases you need to consider......the first is that he selects 3 sweets all of different kinds......u knw that he can choose any 3 flavours frm the 6 available.....the number of selections is 6C3=20
the second case is that 2 of the sweets are the same and the other one is of a different flavour. when he chooses the first sweet, he has a choice between 6 different flavours. when he chooses the second sweet he still has a choice between 6 flavours since he can select another of that
flavour. but when he selects the third sweet he has a choice of 5 flavours since two sweets have already been taken form the sixth flavour and he
cannot take a third. the number of selections is 6*5=30
so the total is 30+20=50
Post by: jellybeans on June 05, 2010, 06:33:36 am
oh damn..........this is such a bloody hard q......i remember givin this to ma teacher and he couldnt solve it..........:o :o :o :o OMG LOL i asked my teacher as well and i think he explained it 4 times but i still had NO IDEA what he was talking about. :P i did get it for a second but then i forgot D:
there are two different cases you need to consider......the first is that he selects 3 sweets all of different kinds......u knw that he can choose any 3 flavours frm the 6 available.....the number of selections is 6C3=20
the second case is that 2 of the sweets are the same and the other one is of a different flavour. when he chooses the first sweet, he has a choice between 6 different flavours. when he chooses the second sweet he still has a choice between 6 flavours since he can select another of that
flavour. but when he selects the third sweet he has a choice of 5 flavours since two sweets have already been taken form the sixth flavour and he
cannot take a third. the number of selections is 6*5=30
so the total is 30+20=50
ANYWAYS;
'the second case is that 2 of the sweets are the same and the other one is of a different flavour. when he chooses the first sweet, he has a choice between 6 different flavours. when he chooses the second sweet he still has a choice between 6 flavours since he can select another of that
flavour. but when he selects the third sweet he has a choice of 5 flavours since two sweets have already been taken form the sixth flavour and he
cannot take a third. the number of selections is 6*5=30'
ummm i get the first part but for the second case, why is it not 6*6*5 if he has a chance of choosing 6 diff flavours the first time, 6 the next, & 5 on his third choice? :-[
thankyouuuus.
Post by: Ghost Of Highbury on June 05, 2010, 06:41:50 am
Post by: jellybeans on June 05, 2010, 06:42:51 am
Told ya, its a crappy q..!! i might have to look at the ms...can u post the ms link plz? ?
haha :P
Post by: Ghost Of Highbury on June 05, 2010, 06:46:17 am
s 6*5...
Post by: Dibss on June 05, 2010, 01:05:16 pm
(i) Evaluate f^2(0). [3]
(ii) Obtain an expression for f^-1. [2]
(iii) State the domain and the range of f^-1. [2]
I don't understand how to get the correct answer for part (iii) so could someone please help me?
Mark scheme attached.
Post by: cooldude on June 05, 2010, 01:26:35 pm
5 A function f is defined by for the domain x 0.
(i) Evaluate f^2(0). [3]
(ii) Obtain an expression for f^-1. [2]
(iii) State the domain and the range of f^-1. [2]
I don't understand how to get the correct answer for part (iii) so could someone please help me?
Mark scheme attached.
first of all the range of f^-1(x)=domain of f(x) and range of f(x)=domain of f^-1(x)
this will answer the part of the q asking the range of f^-1 wich is f^-1(x)=>0
now we use the part that range of f(x)=domain of f^-1(x) , when we subsitute 0 in (e^x)+1/4 we get 1/2, we substitute 0 as this is the minimum value in the domain given, so thus we get the range as f(x)=>0.5 and this is equal to the domain of f^-1(x), i.e. x=>0.5
Post by: Dibss on June 05, 2010, 01:31:45 pm
first of all the range of f^-1(x)=domain of f(x) and range of f(x)=domain of f^-1(x)
this will answer the part of the q asking the range of f^-1 wich is f^-1(x)=>0
now we use the part that range of f(x)=domain of f^-1(x) , when we subsitute 0 in (e^x)+1/4 we get 1/2, we substitute 0 as this is the minimum value in the domain given, so thus we get the range as f(x)=>0.5 and this is equal to the domain of f^-1(x), i.e. x=>0.5
Oh, i get it!
Thank you =]
+ REP
Post by: cooldude on June 05, 2010, 01:39:06 pm
Oh, i get it!
Thank you =]
+ REP
np and Thanks for the rep ;D
Post by: jellybeans on June 06, 2010, 12:24:35 pm
Post by: J.Darren on June 06, 2010, 12:43:15 pm
Post by: Dibss on June 06, 2010, 03:58:03 pm
I'm REALLY sorry I don't have the paper to attach :|
Post by: Ghost Of Highbury on June 06, 2010, 04:47:20 pm
October/Nov 2003 P2 Q8 iii) anyone?
I'm REALLY sorry I don't have the paper to attach :|
iii) Find no. of ways where there is an absence of rose bush of each color.
total ways = 210
no. of ways in which there is no yellow rose bush = 8C6 = 28
no. of ways in which there is no pink bush = 7C6 = 7
Red bushes have to be dere in all selections as there are 5 of them
so 210 - 28 - 7 = 175
Post by: Dibss on June 06, 2010, 06:11:06 pm
iii) Find no. of ways where there is an absence of rose bush of each color.That makes sense.
Thank you =]
+REP
Post by: Ghost Of Highbury on June 06, 2010, 08:38:13 pm
Post by: slvri on June 07, 2010, 08:12:24 am
Post by: BlackBunny103 on June 07, 2010, 09:19:16 am
Thanxx :D
Post by: slvri on June 07, 2010, 09:32:17 am
y2=4x-x2
substitute the first equation in the second
(kx-2)2=4x-x2
k2x2-4kx+4=4x-x2
k2x2+x2-4kx-4x+4=0
(k2+1)x2+(-4k-4)x+4=0
now the line meets the curve provided that for the above equation the discriminant or b2-4ac>=0
where a=k2+1,b=-4k-4 and c=4
b2-4ac>=0
(-4k-4)2-4(k2+1)(4)>=0
16k2+32k+16-16k2-16>=0
32k>=0
k>=0
i hope this helped, and is this the correct answer? :)
Post by: BlackBunny103 on June 07, 2010, 09:48:01 am
y=kx-2
y2=4x-x2
substitute the first equation in the second
(kx-2)2=4x-x2
k2x2-4kx+4=4x-x2
k2x2+x2-4kx-4x+4=0
(k2+1)x2+(-4k-4)x+4=0
now the line meets the curve provided that for the above equation the discriminant or b2-4ac>=0
where a=k2+1,b=-4k-4 and c=4
b2-4ac>=0
(-4k-4)2-4(k2+1)(4)>=0
16k2+32k+16-16k2-16>=0
32k>=0
k>=0
i hope this helped, and is this the correct answer? :)
Oh great, thanxx +rep. Yes this is the correct answer ;)
By the way are you taking Add Maths this year?
Post by: BlackBunny103 on June 07, 2010, 11:05:59 am
Thanxx :D
Post by: astarmathsandphysics on June 07, 2010, 12:24:07 pm
Post by: jellybeans on June 07, 2010, 02:11:31 pm
ummm i need help with Q2 please? :) i can find the first two values but... how do you find the rest? :o and how do you know how many 'values' can fit in within the range? :-\
THANKS IN ADVANCE ;D
Post by: BlackBunny103 on June 07, 2010, 02:48:47 pm
here
Thanxx a lot, Mr Astar :D
Post by: slvri on June 07, 2010, 03:25:40 pm
3sin(x/2-1)=1
sin(x/2-1)=1/3
convert the given range for x into a range for (x/2-1) which is the basic angle here
0<x<6pi
0<x/2<3pi
-1<x/2-1<3pi-1
-1<x/2-1<8.425
now denote x/2-1 by a(alpha but i cant write that here so a it is)
sin a is positive in 1st and 2nd quadrants
a=sin-1(1/3)=0.3398
in the 1st quadrant a=0.3398
in 2nd a=pi-0.3398=2.802
but we can go till 8.425 as indicated in the range
so another value of a=2pi+0.3398=6.623
but no other value for 2nd quadrant cause it goes outside range(2.802+2pi=9.085>8.425
so a=0.3398,2.802,6.623
x/2-1=0.3398,2.802,6.623
x/2=1.3398,3.802,7.623
x=2.6796,7.604,15.246
x=2.68,7.60,15.2 correct to 3sf
is this correct?
Post by: jellybeans on June 07, 2010, 03:32:32 pm
okay
3sin(x/2-1)=1
sin(x/2-1)=1/3
convert the given range for x into a range for (x/2-1) which is the basic angle here
0<x<6pi
0<x/2<3pi
-1<x/2-1<3pi-1
-1<x/2-1<8.425
now denote x/2-1 by a(alpha but i cant write that here so a it is)
sin a is positive in 1st and 2nd quadrants
a=sin-1(1/3)=0.3398
in the 1st quadrant a=0.3398
in 2nd a=pi-0.3398=2.802
but we can go till 8.425 as indicated in the range
so another value of a=2pi+0.3398=6.623
but no other value for 2nd quadrant cause it goes outside range(2.802+2pi=9.085>8.425
so a=0.3398,2.802,6.623
x/2-1=0.3398,2.802,6.623
x/2=1.3398,3.802,7.623
x=2.6796,7.604,15.246
x=2.68,7.60,15.2 correct to 3sf
is this correct?
YEP its correct ;) Thank Youuuu :D !
btways if the range is from 0<x< 6pi radians..
isnt the max supposed to be = 18.85? :( why is it ...8.425? :o :o :o :o
youre right ;D ..but why isnt it 6pi = 18.85? :-\ thanks!!!!!
Post by: slvri on June 07, 2010, 03:39:25 pm
that means for the limits given (0 and 6pi) you divide by 2 to get x/2 and then subtract 1 to get x/2-1 which is the required range
and by the way i gave add math in june 2009 and i gave a level math in june 2010
so yeah just felt like helping others out for the add math exam tomorrow :)
oh and HI ASTAR....remember me?
Post by: jellybeans on June 07, 2010, 03:46:58 pm
thats because theyve given you the range for x but the angle whos sine is being taken in the question is x/2-1, not x so whatever angle you find after taking inverse sine will be for x/2-1, not x itself. thats why you have to convert the given range for x to a range for x/2-1
that means for the limits given (0 and 6pi) you divide by 2 to get x/2 and then subtract 1 to get x/2-1 which is the required range
and by the way i gave add math in june 2009 and i gave a level math in june 2010
so yeah just felt like helping others out for the add math exam tomorrow :)
oh and HI ASTAR....remember me?
:o OH, so every range they give you, you have to sub it in the ... thing like (x/2-1) to get the MAX value? does this apply to EVERY equation? thanks again :) :)
Post by: J.Darren on June 07, 2010, 03:51:31 pm
Helloooo, :Dhttp://img69.imageshack.us/img69/3933/trigonometry.png
ummm i need help with Q2 please? :) i can find the first two values but... how do you find the rest? :o and how do you know how many 'values' can fit in within the range? :-\
THANKS IN ADVANCE ;D
This is a good way for those that are not accustomed to radian calculation in attempting these sort of questions ...
Post by: BlackBunny103 on June 07, 2010, 04:06:27 pm
Thanxx :D
Post by: J.Darren on June 07, 2010, 04:13:48 pm
Guys, can you please explain to me Q1 ON 2009 Paper 2? I don't know how to get the answers as shown in the mark schemes.i ) Subsitute x = 0, the range would be greater than e^-1.
Thanxx :D
ii ) f(x) = e^(x+1)
y = e^(x+1)
ln y = (x+1) ln e
ln y = x + 1
ln x = y + 1
ln x - 1 = y
iii ) the domain of f-1(x) = the range of f(x), i.e. x > e^-1
Post by: Dibss on June 07, 2010, 04:18:11 pm
Post by: J.Darren on June 07, 2010, 04:19:08 pm
Post by: Dibss on June 07, 2010, 04:21:20 pm
https://studentforums.biz/index.php/topic,54.msg234150.html#msg234150Thank you so much! =]
Post by: BlackBunny103 on June 07, 2010, 04:24:12 pm
i ) Subsitute x = 0, the range would be greater than e^-1.
ii ) f(x) = e^(x+1)
y = e^(x+1)
ln y = (x+1) ln e
ln y = x + 1
ln x = y + 1
ln x - 1 = y
iii ) the domain of f-1(x) = the range of f(x), i.e. x > e^-1
Thanxx Darren ;). +rep
But I don't understand why from going ln y = (x+1) ln e to ln y = x + 1 why did you cancel lne but still keep ln e?
And why is the domain of f-1(x) = the range of f(x) ?
Thanxx again :D
Post by: J.Darren on June 07, 2010, 04:27:16 pm
Thanxx Darren ;). +repI pretained ln e there just so that you can comprehend what exactly has happened :D
But I don't understand why from going ln y = (x+1) ln e to ln y = x + 1 why did you cancel lne but still keep ln e?
And why is the domain of f-1(x) = the range of f(x) ?
Thanxx again :D
Bear in mind that the inverse of every function is the reflection of the original function on the line y = x, as we know plugging any value of the range of the original function in a one-to-one equation into its inverse would get us back to where we strated from ...
Post by: BlackBunny103 on June 07, 2010, 11:39:53 pm
Thanxx ;)
My exam is in 4 hours :-\ ...
Post by: astarmathsandphysics on June 07, 2010, 11:45:22 pm
Post by: astarmathsandphysics on June 07, 2010, 11:47:34 pm
Post by: BlackBunny103 on June 07, 2010, 11:54:58 pm
letter M is selcted so in fact you are only choosing 3 from 6C3
Oh right, this makes sense. Thanxx :D
Since she must select her favourite beatles cd she is only choosing 3 from 8 so 8C3I understand the concept, but I'm thinking since she's selecting 4 from 9 and there are 4 beatles cs in total, so what if all of her selections contain 4 Beatles CD? What I did was 9C4.
Post by: astarmathsandphysics on June 08, 2010, 12:01:57 am
Post by: BlackBunny103 on June 08, 2010, 12:48:16 am
Thanxx :D
Post by: xlane on June 08, 2010, 01:26:00 am
period or range for 1 cycle is 180..use 180/b...so b=1
amplitude i.e height from axis of symmetry in either directn so 5-3=2, 3-1=2..therefore a=2
Post by: Alpha on June 08, 2010, 01:28:00 am
I need help with this question, how can you find a?
Thanxx :D
a is the amplitude.
You get it just by comparing to the y = tan x curve.
tan x passes through the origin, it's passing at 3. So there has been a translation of 3. Therefore, c =3.
tan pie/4 = 1. But on the graph it's 5. You take 5 - 3 to get the amplitude. (y = 3 is the axis of curve here)
a = 2
Post by: xlane on June 08, 2010, 01:28:42 am
period or range for 1 cycle is 180..use 180/b...so b=1
amplitude i.e height from axis of symmetry in either directn so 5-3=2, 3-1=2..therefore a=2
Post by: Alpha on June 08, 2010, 01:29:28 am
The value at angle pie/4 and at -pie/4 is equal from y =3 ..therefore equatn of axis of symmetry y=3..i.e C=3
period or range for 1 cycle is 180..use 180/b...so b=1
amplitude i.e height from axis of symmetry in either directn so 5-3=2, 3-1=2..therefore a=2
:) Just. :D
Post by: xlane on June 08, 2010, 01:34:30 am
Post by: Alpha on June 08, 2010, 01:42:39 am
Lol yeah..."just" tryna help....:D:D:D...tan curve usualy never comes..so i put in da explanation 2
Yah, but nah. I meant just after or before me? :D Both ;)
Post by: xlane on June 08, 2010, 01:58:50 am
Post by: Alpha on June 08, 2010, 02:05:07 am
I was getting bored... So, came in Maths thread.
Post by: BlackBunny103 on June 08, 2010, 02:14:52 am
Post by: Alpha on June 08, 2010, 02:15:48 am
Thanxx a lot both of you :D
Most welcome. :)
From both of us.
Post by: xlane on June 08, 2010, 02:25:52 am
Post by: Alpha on June 08, 2010, 02:27:34 am
Thanks alpha :D...ur are most welcum frm both of us :P
LOL. :P
Am honoured. :D
Post by: xlane on June 08, 2010, 02:30:11 am
Post by: jellybeans on June 08, 2010, 02:43:04 am
if you rearrange for 6i) ;
isnt it 2t2 - 9t - 5 = 0
(x-5)(2x+1) = 0
x = 5 or x = -1/2
why does the MS say (2x+1) => x = 1/2
??? help pleaaaase thankssss
! and for Q7iii) when are we supposed to use the "> or equal to " and ">" ? .... :( !!! THANKS
Post by: xlane on June 08, 2010, 02:46:48 am
Post by: Alpha on June 08, 2010, 03:05:22 am
! and for Q7iii) when are we supposed to use the "> or equal to " and ">" ? .... :( !!! THANKS
When the image for f= -6 is real, or exists. Use equal to sign then. Depends on the question also.
When you use > sign, the image will exist just above this value, but not for the value itself.
Post by: jellybeans on June 08, 2010, 04:40:21 am
another one! what are the limits for part ii) ? what do they mean by fifth second... ? :o help please!!!!
THANKS !!!!!!! D: D: D:
Post by: J.Darren on June 08, 2010, 04:47:42 am
Thanks!!!Find the stationery point first, if it lies between 0 - 5 seconds, you must do a two-part integration, taking the 0 - stationery point and stationery point - 5 as the limits in the two integrals ... Gee I am currently at school and will be having my examination in an hour :(
another one! what are the limits for part ii) ? what do they mean by fifth second... ? :o help please!!!!
THANKS !!!!!!! D: D: D:
Post by: SAKS10111 on June 08, 2010, 10:41:40 am
Post by: jellybeans on June 08, 2010, 10:50:47 am
How was it?! I found it really good, we need to start discussing after the 24 hour period is up!
hahahaha it was AWESOME right? :D :D :D :D :D!!!!!!! ;D ;D ;D
oh & goodluck to you j darren!
Post by: J.Darren on June 08, 2010, 11:24:58 am
Post by: adrian1993 on June 08, 2010, 11:44:29 am
The paper was easy. The threshold would probably be 69-71/80 for an A. It has happened before in the Oct Nov Paper 2009 which I found considerably more difficult.
Post by: J.Darren on June 08, 2010, 12:09:54 pm
Not neccesarily Darren.O/N 09 is pretty easy ... My score would probably be in the 75 + range, allowing room for careless mistakes. I have paid additional care in ensuring that I have all the steps shown so I would still get a follow through mark as I won't get penalised twice for the same thing. There are some questions that I find tricky in today's paper, I will discuss them tomorrow.
The paper was easy. The threshold would probably be 69-71/80 for an A. It has happened before in the Oct Nov Paper 2009 which I found considerably more difficult.
Post by: Ghost Of Highbury on June 08, 2010, 12:46:16 pm
Are u all taking GCE-O level additional maths??
Post by: J.Darren on June 08, 2010, 12:48:24 pm
Wat abt A*s?? no ones aiming for A* here??A* in IGCSE Additional Mathematics is an unrealistic goal. A friend of mine got 10 A* 1 A last year, the A grade that she attained was in Additional Mathematics :o
Are u all taking GCE-O level additional maths??
Post by: Ghost Of Highbury on June 08, 2010, 12:55:06 pm
Post by: BlackBunny103 on June 08, 2010, 12:57:46 pm
Wat abt A*s?? no ones aiming for A* here??
Are u all taking GCE-O level additional maths??
Hey I'm aiming for A* for IGCSE Add Maths :D I scored A* with 98% for my mock exams, so I will try my best to keep that A* for the real exams as well.
By the way, my paper 1 went quite well :D
Post by: Ghost Of Highbury on June 08, 2010, 12:59:43 pm
Yea i too got A*s in my mocks...but CIE's A* is no where close to the school's A*.
Its a huge ratio of students appearing for that exam..
Last session, i was sure i wud get 75+, resulting in the excitement to receive an A* in arguable IGCSE's most difficult subject.
I got an A :-/
Post by: BlackBunny103 on June 08, 2010, 01:03:24 pm
Oh thats great, always aim high.. :)
Yea i too got A*s in my mocks...but CIE's A* is no where close to the school's A*.
Its a huge ratio of students appearing for that exam..
Last session, i was sure i wud get 75+, resulting in the excitement to receive an A* in arguable IGCSE's most difficult subject.
I got an A :-/
Well do you know what is the normal range for A* (estimate)? Since for the mock, the grade boundaries at my school was >85% for an A and >90% for A*, which I assume to be pretty high. Don't tell me CIE'S A* for Add Maths is >95% lolz :P
Post by: Alpha on June 08, 2010, 01:10:22 pm
Well do you know what is the normal range for A* (estimate)? Since for the mock, the grade boundaries at my school was >85% for an A and >90% for A*, which I assume to be pretty high. Don't tell me CIE'S A* for Add Maths is >95% lolz :P
Sometimes, YES, when the paper is simple. Too simple for Cambridge standards. ;)
Post by: adrian1993 on June 08, 2010, 02:38:23 pm
Post by: xlane on June 08, 2010, 03:40:07 pm
Post by: Ghost Of Highbury on June 08, 2010, 03:41:58 pm
Post by: xlane on June 08, 2010, 03:57:50 pm
Post by: Ghost Of Highbury on June 08, 2010, 04:00:09 pm
excuse ari ben canan...Thanks ..bt could u kindly put bk my post afta da specific period of time...
I dont think that can be done now, that is the reason i told u to edit it
Anyways, you can post again after 24 hours.
Post by: xlane on June 08, 2010, 04:25:24 pm
Post by: J.Darren on June 08, 2010, 05:07:37 pm
u guys r lazyyyDude Mods got examinations as well >:(
Post by: xlane on June 08, 2010, 05:16:35 pm
Post by: syedz123 on June 08, 2010, 08:37:18 pm
Oh thats great, always aim high.. :)
Yea i too got A*s in my mocks...but CIE's A* is no where close to the school's A*.
Its a huge ratio of students appearing for that exam..
Last session, i was sure i wud get 75+, resulting in the excitement to receive an A* in arguable IGCSE's most difficult subject.
I got an A :-/
y r people arguin ova gettin A in add math..isnt dat like da highest grade for add maths...its onli this yr that they are adding A*....
Post by: Dibss on June 08, 2010, 08:40:02 pm
y r people arguin ova gettin A in add math..isnt dat like da highest grade for add maths...its onli this yr that they are adding A*....They're adding it this year for O-levs.
A*s always existed for IGCSEs which these guys are discussing (:
Post by: SAKS10111 on June 09, 2010, 08:09:29 am
Post by: tunnellord on June 09, 2010, 08:21:06 am
Post by: Ghost Of Highbury on June 09, 2010, 08:56:55 am
Post by: adrian1993 on June 09, 2010, 10:03:05 am
Post by: jellybeans on June 09, 2010, 10:28:29 am
for the first unit vector thingo is it okay if i put 1(21)
20(20)
and the question on OXY : XABYX was it 9p : 55q?
.. lol, :-\
Post by: syedz123 on June 09, 2010, 10:32:20 am
I did one careless mistake in the vectors question, but my method was correct. The final answer was 46i plus something j. I got the j but 46 i. I assume they will only reduce 1 mark...
u guys got vectors? variant 2? i did variant 1 and quite sum no of topics didnt come lol although i made sum silly mistakes lol
permutation n combination,vectors,relative velocity,conversion to linear form,trigonometry,radians on arc n area of sectors,area under the graph n sum more cnt rmmbr didnt come in variant 1...am guessing its comin for p2 so we can xpect a tough paper comin
Post by: jellybeans on June 09, 2010, 10:38:01 am
u guys got vectors? variant 2? i did variant 1 and quite sum no of topics didnt come lol although i made sum silly mistakes lol
permutation n combination,vectors,relative velocity,conversion to linear form,trigonometry,radians on arc n area of sectors,area under the graph n sum more cnt rmmbr didnt come in variant 1...am guessing its comin for p2 so we can xpect a tough paper comin
hahaha i cant rmb what we had ??? Lol but yep permutation & combination, curve & tangent equations & stuff, radians yep.. areas yep.. area under graph ummmmm it was either THAT or the other question 12 so i guess yepppp.. maaaaaan i've got a feeling that its gonna be hard as well
Post by: xlane on June 09, 2010, 11:27:50 am
Post by: SAKS10111 on June 09, 2010, 11:36:34 am
HI, haaha okay so ive got.. questions on P1 . :D VARIANT 2.
for the first unit vector thingo is it okay if i put 1(21)
20(20)
and the question on OXY : XABYX was it 9p : 55q?
.. lol, :-\
ok i have no idea what you just did for the vector thing lol, maybe its some notation i don't know. i just found out AB and then divided that by the magnitude of AB (which was 29), so it was like 21/29 i + something/29 j i think
and yes i got 9:55 for that too, though its not 9p:55q, p IS 9 and q IS 55
Post by: jellybeans on June 09, 2010, 11:39:06 am
ok i have no idea what you just did for the vector thing lol, maybe its some notation i don't know. i just found out AB and then divided that by the magnitude of AB (which was 29), so it was like 21/29 i + something/29 j i think
and yes i got 9:55 for that too, though its not 9p:55q, p IS 9 and q IS 55
oh. Lol i didnt put it in the i & j form.. were we supposed to? :o and... crap did they say in the form of p:q ? :o :o :o :( SO ITS WRONG? :( :'( AHHHHHH
and was the acceleration -1.92 m/s2?
Post by: SAKS10111 on June 09, 2010, 11:40:36 am
u guys got vectors? variant 2? i did variant 1 and quite sum no of topics didnt come lol although i made sum silly mistakes lol
permutation n combination,vectors,relative velocity,conversion to linear form,trigonometry,radians on arc n area of sectors,area under the graph n sum more cnt rmmbr didnt come in variant 1...am guessing its comin for p2 so we can xpect a tough paper comin
i think variant 2 got slightly tougher topics, because we got vectors, permutation and combinations, trigonometry, radians (arc length and area of sectors), area under the graph... all of those topics were on variant 2, so do you think the variant 2 paper 2 will be easier? sure hope so, because then it really isn't fair lol
Post by: SAKS10111 on June 09, 2010, 11:44:07 am
oh. Lol i didnt put it in the i & j form.. were we supposed to? :o and... crap did they say in the form of p:q ? :o :o :o :( SO ITS WRONG? :( :'( AHHHHHH
and was the acceleration -1.92 m/s2?
no no i don't think you need to put it in i and j form, dont wry, thats just my way of doing it, i find it easier lol. yeah i'm pretty sure they said in the form p:q, but you got the values basically right. meaning you did the areas and found that its 9:55 and THEN did you right your final answer as 9p:55q? because then they'll see you got it right, but just put that incorrectly in the final answer, so i don't think you should lose any marks, maybe just one.
YES i got acceleration as -1.92 too! which is basically deceleration of 1.92 m/s2. Was your time 25.8 s and your velocity.. i think 6. something i can't remember what i got exactly.
Post by: BlackBunny103 on June 09, 2010, 11:50:02 am
u guys got vectors? variant 2? i did variant 1 and quite sum no of topics didnt come lol although i made sum silly mistakes lol
permutation n combination,vectors,relative velocity,conversion to linear form,trigonometry,radians on arc n area of sectors,area under the graph n sum more cnt rmmbr didnt come in variant 1...am guessing its comin for p2 so we can xpect a tough paper comin
I did variant 2. What topics came up in your variant 1 for paper 1?
Post by: J.Darren on June 09, 2010, 11:51:49 am
HI, haaha okay so ive got.. questions on P1 . :D VARIANT 2.VECTORS !? I don't recall seeing any vectors in my paper :O
for the first unit vector thingo is it okay if i put 1(21)
20(20)
and the question on OXY : XABYX was it 9p : 55q?
.. lol, :-\
9P : 55Q - Yup.
Post by: J.Darren on June 09, 2010, 11:54:05 am
no no i don't think you need to put it in i and j form, dont wry, thats just my way of doing it, i find it easier lol. yeah i'm pretty sure they said in the form p:q, but you got the values basically right. meaning you did the areas and found that its 9:55 and THEN did you right your final answer as 9p:55q? because then they'll see you got it right, but just put that incorrectly in the final answer, so i don't think you should lose any marks, maybe just one.I got -1.92 as well ...
YES i got acceleration as -1.92 too! which is basically deceleration of 1.92 m/s2. Was your time 25.8 s and your velocity.. i think 6. something i can't remember what i got exactly.
Phew, I recalled the vector question, what did you guys obtained for OC ?
Post by: BlackBunny103 on June 09, 2010, 11:54:30 am
VECTORS !? I don't recall seeing any vectors in my paper :O
9P : 55Q - Yup.
Hey Darren, you're doing variant 2 right since you also got the same ratio as I did for the 2 sector areas.
There was a vector question, it was question 3, I remember.
Post by: xlane on June 09, 2010, 11:55:15 am
Post by: Ghost Of Highbury on June 09, 2010, 11:55:47 am
could sum1 plz help me out wid this crap sum...oct/nov 2003 paper 1..12 or...hw to prove da part (i)...urgent
ok first use similar triangles to determine this (check the attached image for the similar triangles)
ABC is similar to DEC
that means
30/12 = h/(12-r)
5/2 = h/(12-r)
2h = 60-5r
h = (60-5r)/2
V = pi*r^2*h
V = pi*r^2* (60-5r)/2 = r^2pi * (60-5r)/2 = pi(30r^2 - 5r^3/2)
Post by: J.Darren on June 09, 2010, 11:56:14 am
Hey Darren, you're doing variant 2 right since you also got the same ratio as I did for the 2 sector areas.Yup I recalled the question, I am guessing that P2 is going to be easy as all the hard topics were featured in yesterday's paper ...
There was a vector question, it was question 3, I remember.
By the way what did you got for OC ?
Post by: Deniz on June 09, 2010, 11:57:23 am
Rolling Stones. She selects 4 of the CDs from her collection. Calculate the number of ways in which
she can make her selection if
(i) her selection must contain her favourite Beatles CD, [2]
(ii) her selection must contain 2 CDs by one group and 2 CDs by another.
can some1 explain how to solve this question?
Post by: BlackBunny103 on June 09, 2010, 11:58:14 am
Yup I recalled the question, I am guessing that P2 is going to be easy as all the hard topics were featured in yesterday's paper ...
By the way what did you got for OC ?
I actually forgot what I got for OC. :P What did you get? Perhaps, it'll jog my memory a bit.
Post by: J.Darren on June 09, 2010, 11:59:17 am
A student has a collection of 9 CDs, of which 4 are by the Beatles, 3 are by Abba and 2 are by thei) 5C3 * 4C1
Rolling Stones. She selects 4 of the CDs from her collection. Calculate the number of ways in which
she can make her selection if
(i) her selection must contain her favourite Beatles CD, [2]
(ii) her selection must contain 2 CDs by one group and 2 CDs by another.
can some1 explain how to solve this question?
ii) 2 Rolling Stones + 2 Others
2C2 * 7C2
2 Beatles + 2 Others
4C2 * 5C2
2 Abba + 2 Others
3C2 * 6C2
Add them all together.
Post by: J.Darren on June 09, 2010, 11:59:47 am
I actually forgot what I got for OC. :P What did you get? Perhaps, it'll jog my memory a bit.I got 63 subtracted by something for i ...
Post by: BlackBunny103 on June 09, 2010, 12:01:36 pm
I got 63 subtracted by something for i ...
Yesh ;D, I remember getting 63j - _i what I've forgotten ;D YAY
Which part did you choose for Q12? I went for the second one, it was so easy.
Post by: J.Darren on June 09, 2010, 12:03:10 pm
Yesh ;D, I remember getting 63j - _i what I've forgotten ;D YAYMost of my mates did EITHER. I did EITHER as well ... however it seems that virtually everyone has gotten a negative answer for area under the curve :o
Which part did you choose for Q12? I went for the second one, it was so easy.
Post by: BlackBunny103 on June 09, 2010, 12:04:49 pm
Most of my mates did EITHER. I did EITHER as well ... however it seems that virtually everyone has gotten a negative answer for area under the curve :o
At my school, there are some did EITHER & some did the second part. For EITHER, how is that possible? I mean area can't be negative, that's a fact :-\
Post by: SAKS10111 on June 09, 2010, 12:06:03 pm
Most of my mates did EITHER. I did EITHER as well ... however it seems that virtually everyone has gotten a negative answer for area under the curve :o
ok i did EITHER and i did not get a negative answer. i got a weird answer it was like in pi and stuff. final answer was in decimals. does anyone remember what they got for the answer?
also, what did everyone get for the permutations/combinations questions (variant 2)
Post by: Deniz on June 09, 2010, 12:07:05 pm
i) 5C3 * 4C1
ii) 2 Rolling Stones + 2 Others
2C2 * 7C2
2 Beatles + 2 Others
4C2 * 5C2
2 Abba + 2 Others
3C2 * 6C2
Add them all together.
in the answer sheet, the answer of (i)is 56. i reckon u r wrong?
Post by: BlackBunny103 on June 09, 2010, 12:08:16 pm
also, what did everyone get for the permutations/combinations questions (variant 2)
That permutation & combination question was so easy. I remember neither the question nor my answers lolz, but found it easy. What were your answers? This should bring back my goldfish memory ;D
Post by: SAKS10111 on June 09, 2010, 12:11:51 pm
That permutation & combination question was so easy. I remember neither the question nor my answers lolz, but found it easy. What were your answers? This should bring back my goldfish memory ;D
haha i have the same problem, i'll remember when someone else says the answer. um i remember the final answer being 20160 i think. don't remember the first two ones very well
Post by: xlane on June 09, 2010, 12:12:25 pm
Post by: Ghost Of Highbury on June 09, 2010, 12:22:48 pm
sum1 explain how to solve da 2nd n 3rd part of oct/nov 2002 p1..12 either..please
ii) when theta = 30
R = r(1+1/sin 30) = 3r
Area of circle = pi*r^2
Area of sector = 1/2 R^2(pi/3) = (3pir^2)/2
fraction : pir^2/(3pir^2)/2 = 2/3
iii) tan 30 = r/d (R = 15 , 15 = 3r , r = 5)
tan 30 = 5/d
d = 5/tan 30 = 8.66
arc length = r*a = 5(2pi/3) = 10.47
perimeter = 10.47 + 2(8.66) = 27.8
Post by: xlane on June 09, 2010, 01:06:17 pm
Post by: adrian1993 on June 09, 2010, 01:31:00 pm
Yeah, the permutations question was so easy. I remember the last answer was 100 for that question.
I think relative velocity would probably come out tomorrow...
Post by: jellybeans on June 09, 2010, 02:33:40 pm
i got 10C7 = 120ways for i) and 100 ways for ii) :) :) did anyone get the same answers?
Post by: BlackBunny103 on June 09, 2010, 02:35:32 pm
OH, & for variant 2, the combination question for choosing 7 pieces from 6 classical & 4 modern pieces..
i got 10C7 = 120ways for i) and 100 ways for ii) :) :) did anyone get the same answers?
Yes yes, same same with me ;D
Post by: jellybeans on June 09, 2010, 02:37:45 pm
Yes yes, same same with me ;D
HAHA , (Y) whabbout Q12? which one didyou choose? :D i chose the bottom oneee.. and the coordinates for when the tangent meets the x axis is like.. (0.231, 0) ? :P ;D
Post by: jellybeans on June 09, 2010, 02:39:51 pm
Yup I recalled the question, I am guessing that P2 is going to be easy as all the hard topics were featured in yesterday's paper ...
By the way what did you got for OC ?
for the OC thing there was a double 'minus' sign when you added OA & AC right? so it makes it all positive.. or something hahahahaa
Post by: tunnellord on June 09, 2010, 02:46:03 pm
Post by: $KD1994$ on June 09, 2010, 02:47:19 pm
HAHA , (Y) whabbout Q12? which one didyou choose? :D i chose the bottom oneee.. and the coordinates for when the tangent meets the x axis is like.. (0.231, 0) ? :P ;D
i gt d same answer.!!.....!!
Post by: jellybeans on June 09, 2010, 02:49:53 pm
i gt d same answer.!!.....!!
HAHAHA yaay! :P :P do you think P2's gonna be okayy? hahaha , :D i hope its not too hard :P ;D
Post by: Dibss on June 09, 2010, 02:54:23 pm
Hmm, can someone who did variant 1 tell us what came out in their paper 1?Variant one was possibly the easiest AddMath P1 I have ever given... Aced it for sure ;D
Which means P2's gonna be hard :o :o
@Aangel42 - My grade uses Additional Mathematics 6th Edition by JF Talbert and HH HENG
Post by: Ghost Of Highbury on June 09, 2010, 03:12:52 pm
What textbook do you guys use to study for Additional Math?
Are there any resources online I could use to study for it?
Buy redspot publications books for gce-o level additional maths (= IGCSE)
Post by: xlane on June 09, 2010, 03:27:50 pm
Post by: Ghost Of Highbury on June 09, 2010, 03:28:09 pm
Thanks ghost o highbury..bt u r hardly a ghost..u can type n ur usin a modern technology :P
LOL, call me Aadi :P
Post by: BlackBunny103 on June 09, 2010, 03:59:10 pm
Thanxx :D
Post by: adrian1993 on June 09, 2010, 04:09:24 pm
Then use b square-4ac=>0.
I need help on May June 2003 Paper 2 Question 5 Part iii.
How do I get the domain and range of the inverse? I don't quite understand. Urgent help needed.
Post by: Ghost Of Highbury on June 09, 2010, 04:11:30 pm
Hey guys, would you please help me with this question? It's from MJ 2007 P2.
Thanxx :D
y = (x2 + 4)/2
y= 3x - k
(x2 + 4 ) /2 = 3x- k
(
k = -0.5x2 + 3x - 2
-0.5x2 + 3x - 2 -k = 0
multiply both sides with -2
x^2 - 6x + (4+2k) = 0
now to intersect at a point, use b^2 - 4ac >= 0
36x^2 - 16x^2 - 8kx^2 >= 0
simplify
k =< 2.5
Post by: Ghost Of Highbury on June 09, 2010, 04:22:21 pm
Blackbunny, I presume that you have to equate both equations?
Then use b square-4ac=>0.
I need help on May June 2003 Paper 2 Question 5 Part iii.
How do I get the domain and range of the inverse? I don't quite understand. Urgent help needed.
The domain of f(x) is the range of f-1 (x)
The range of f(x) is the domain of f-1(x)
So the range of f-1(x) --> x >= 0 (domain of f(x)
and the domain of f-1(x) ---> x >= 0.5 (range of f(x)
range of f(X) = the least value of f(x) is when x = 0
when x = 0, f(x) = 0.5
so as x>=0 , f(x) >=0.5
Post by: elemis on June 09, 2010, 04:24:25 pm
Post by: J.Darren on June 09, 2010, 04:24:39 pm
ok i did EITHER and i did not get a negative answer. i got a weird answer it was like in pi and stuff. final answer was in decimals. does anyone remember what they got for the answer?I got 1.3 something.
also, what did everyone get for the permutations/combinations questions (variant 2)
Post by: J.Darren on June 09, 2010, 04:25:53 pm
OH, & for variant 2, the combination question for choosing 7 pieces from 6 classical & 4 modern pieces..10C7 for the first one and 100 for the third one, forgot what I got for the second one but it is fewer than 100 for sure. The answer in part ii adds up with two other possibilities to give 100.
i got 10C7 = 120ways for i) and 100 ways for ii) :) :) did anyone get the same answers?
Post by: jellybeans on June 09, 2010, 04:27:55 pm
10C7 for the first one and 100 for the third one, forgot what I got for the second one but it is fewer than 100 for sure. The answer in part ii adds up with two other possibilities to give 100.
YOU HAD 3 PARTS?which variant did you do...? :o i only had i) and ii) ? & mine was variant 2... :)
Post by: jellybeans on June 09, 2010, 04:29:33 pm
This stuff looks WAAAAAAY harder than my normal math. :-\
hahahahahaha.. *NODSNODSNODS.
Post by: Ghost Of Highbury on June 09, 2010, 04:29:55 pm
This stuff looks WAAAAAAY harder than my normal math. :-\
lol, it is ;)
Post by: J.Darren on June 09, 2010, 04:32:41 pm
YOU HAD 3 PARTS?which variant did you do...? :o i only had i) and ii) ? & mine was variant 2... :)Same :O
Post by: jellybeans on June 09, 2010, 04:34:48 pm
Same :O
hahaha do you rmb the questions???? all i rmb was pick 7 outta 10. & second one was .. UHHH .. more classical than modern pieces? :P I THINK. wha was youuuurs?
Post by: J.Darren on June 09, 2010, 04:37:00 pm
hahaha do you rmb the questions???? all i rmb was pick 7 outta 10. & second one was .. UHHH .. more classical than modern pieces? :P I THINK. wha was youuuurs?Part i) No restriction
ii) Two pieces of modern are selected
iii) More classical pieces than modern pieces, has to contain one modern pieces.
Post by: adrian1993 on June 09, 2010, 04:37:40 pm
Post by: J.Darren on June 09, 2010, 04:38:44 pm
Yeah there were 3 answers. 120, 36 and 100.Ditto.
Post by: jellybeans on June 09, 2010, 04:39:11 pm
Yeah there were 3 answers. 120, 36 and 100.
OH RIGHTTTT RIGHT hahahahaha, OOOPS my bad XD HAHAHAHAHA totally forgot bout the 2nd one. LOL sry! ;D
Post by: adrian1993 on June 09, 2010, 04:41:40 pm
So I have a hunch that functions, sets, trigo/proving and finding x, b square-4ac, coordinate geometry, graphs, integration for area under curve, differentiation, surds, relative velocity and logarithms are coming out tomorrow.
Post by: jellybeans on June 09, 2010, 04:43:21 pm
By the way, based on the Paper 1, I've decided to some question spotting since CIE won't be stupid enough to ask the same questions twice.
So I have a hunch that functions, sets, trigo/proving and finding x, b square-4ac, coordinate geometry, graphs, integration for area under curve, differentiation, surds, relative velocity and logarithms are coming out tomorrow.
wasnt there an area under cruve for Q12 already? :'( aw man.. i thought we did lotsa integration & differentiation in paper1.... (y)
Post by: J.Darren on June 09, 2010, 04:45:27 pm
By the way, based on the Paper 1, I've decided to some question spotting since CIE won't be stupid enough to ask the same questions twice.There were like two graphs Qs in Paper 1 Variant 2 :O (Three if you include the one re : y^2 = m sec x + c as well)
So I have a hunch that functions, sets, trigo/proving and finding x, b square-4ac, coordinate geometry, graphs, integration for area under curve, differentiation, surds, relative velocity and logarithms are coming out tomorrow.
Post by: BlackBunny103 on June 09, 2010, 04:46:59 pm
Post by: J.Darren on June 09, 2010, 04:48:52 pm
I dearly pray to GOD, though I'm Buddhist, that relative vel wouldn't come up. I HATE IT >:(LOL I am hoping for Relative Velocity ... Aliens colliding with each other on a screen, how exciting :P
Post by: adrian1993 on June 09, 2010, 04:49:51 pm
Post by: J.Darren on June 09, 2010, 04:51:22 pm
Apparently they asked the centre to provide a graph paper for the second paper as well. Area under curve then appeared in the optional section. Maybe this time they'll put it in the compulsory parts.Oh right ... That's depressing, I hate drawing graphs D: The sketch that I did for the two cos x graphs were douggy ... But at least I have managed to obtain 4 solutions so whatever ... What did you guys got for the log Q By the way ?
Post by: adrian1993 on June 09, 2010, 04:54:19 pm
Post by: jellybeans on June 09, 2010, 04:54:57 pm
Oh right ... That's depressing, I hate drawing graphs D: The sketch that I did for the two cos x graphs were douggy ... But at least I have managed to obtain 4 solutions so whatever ... What did you guys got for the log Q By the way ?
sth like 30.8 . :P and you? for Q
Post by: J.Darren on June 09, 2010, 04:57:26 pm
sth like 30.8 . :P and you? for Qnaa Q - Question ... I got 2.2 something for log p ...
Post by: jellybeans on June 09, 2010, 04:58:39 pm
naa Q - Question ... I got 2.2 something for log p ...
didnt they ask for Q when V = blahblah?
Post by: J.Darren on June 09, 2010, 04:59:14 pm
didnt they ask for Q when V = blahblah?I thought it is when v = something, q = unknown ?
Post by: Dibss on June 09, 2010, 05:03:52 pm
Post by: J.Darren on June 09, 2010, 05:04:18 pm
In variant 2 did you guys have to DRAW any graphs?Two graphs in total - log x and cos x.
Post by: jellybeans on June 09, 2010, 05:05:36 pm
I thought it is when v = something, q = unknown ?
hahaha i found logq and then q .. :P :)
Post by: adrian1993 on June 09, 2010, 05:05:44 pm
The answer is 0.833 or 5/6
Can someone please enlighten me to the below question:-
g(x)-->1-(x^2)+6x
domain is 2<=x<=7
Find range.
Post by: Dibss on June 09, 2010, 05:08:25 pm
Two graphs in total - log x and cos x.We didn't even need graph paper for V1 so I'm assuming we'll have to draw tomorrow... Eeeek :|
Post by: J.Darren on June 09, 2010, 05:09:43 pm
Oh that question. They asked for log p when v=170.I thought the question required us to find the exact value of p instead of log p ?
The answer is 0.833 or 5/6
Can someone please enlighten me to the below question:-
g(x)-->1-(x^2)+6x
domain is 2<=x<=7
Find range.
Post by: adrian1993 on June 09, 2010, 05:10:08 pm
Darren, nope. It says log p when V=170.
Post by: BlackBunny103 on June 09, 2010, 05:12:57 pm
LOL I am hoping for Relative Velocity ... Aliens colliding with each other on a screen, how exciting :P
Lolz, that'd be awesome, it's more entertaining than some random boat/plane blowing in the wind ;D
Relative velocity is on my BLACK LIST for now. I'm doing IB Maths HL next year so hopefully, no more of that ;D
Post by: jellybeans on June 09, 2010, 05:13:39 pm
I thought the question required us to find the exact value of p instead of log p ?
uhuh! it says P :) :) :) IM SURE!!!!!! cuz all my friends did p as well.... but kinda got different values cuz our gradients were kinda diff
Post by: Dibss on June 09, 2010, 05:15:01 pm
Hey dibbs, what came out for V1?From what I've been reading it was completely different from V2. Topics I remember:
Lots of calculus!
Binomial exp
Logs
Sets/Venn
Functions
Surds
What about V2?
Post by: J.Darren on June 09, 2010, 05:16:42 pm
uhuh! it says P :) :) :) IM SURE!!!!!! cuz all my friends did p as well.... but kinda got different values cuz our gradients were kinda diffKudos for us :D I found p to be 25 sth - 10^1.4 :)
Adrian : You did Variant 1 ?
I will be doing A-Level Mathematics with Further Mathematics next year, and Mathematics is no more than a repetition of what has already been covered in the IGCSE Additional Mathematics syllabus :D
Dibbs : All the topics that you have just mentioned did not appear in the Paper 2 Variant 2 at all :O
Post by: Dibss on June 09, 2010, 05:19:27 pm
Dibbs : All the topics that you have just mentioned did not appear in the Paper 2 Variant 2 at all :OReally?! :o
What did come then?
Haha what if they're planning to switch variants and papers but just with different numbers? -sigh- If only.
Post by: J.Darren on June 09, 2010, 05:22:03 pm
Really?! :oUh straight line graph in the form Y = mX + c with y^2 = Y and sec x = X
What did come then?
Haha what if they're planning to switch variants and papers but just with different numbers? -sigh- If only.
2 x 2 matrix - finding the inverse and hence solve simutaneous equation
etc etc.
Post by: adrian1993 on June 09, 2010, 05:22:13 pm
Hey dibbs, were the functions and sets tough?
Post by: Dibss on June 09, 2010, 05:25:38 pm
Hey dibbs, were the functions and sets tough?TBH I found the whole paper really easy but the rest of my class had a lot of problems. I'd say that it depends on your ability. Since you've studied, don't stress, it's not tough (:
Post by: jellybeans on June 09, 2010, 05:26:15 pm
Post by: adrian1993 on June 09, 2010, 05:27:12 pm
Post by: J.Darren on June 09, 2010, 05:27:56 pm
hahah :o oh? LOL weirdddd, my whole calss said it was Q and it was the 10^ of something thingggy to find Q or p.. lol whichever it wasI think so too :(
Yeah domain and range - Does it concern natural lograithum and exponentials ? I hate them >:(
Post by: tunnellord on June 09, 2010, 05:30:58 pm
Post by: adrian1993 on June 09, 2010, 05:31:13 pm
g(x)-->1-(x^2)+6x
domain is 2<=x<=7
Find range.
Post by: J.Darren on June 09, 2010, 05:32:55 pm
dont worry darren, u will only need to substitute values given to find the domain. Its rather dogmatic stuff.Phew ...
Adrian : Rewrite it in the form ax^ + bx + c.
Post by: adrian1993 on June 09, 2010, 05:34:40 pm
Post by: J.Darren on June 09, 2010, 06:02:56 pm
I got it as <=-6 and >9. I don't know why the mark scheme says its <=-6 but <=101-(x^2)+6x
-x^2 + 6x + 1
x = -0.16 or 6.16
dy/dx = -2x + 6
-2x + 6 = 0
x = 3
(-3)^2 + 6 (3) + 1
y = 28
maximum point (3, 28)
<= 28
Post by: adrian1993 on June 09, 2010, 06:21:28 pm
Anyways I managed to figure it out. The maximum point was (-3, 10)
So obviously the domain would be less than 10.
Post by: J.Darren on June 09, 2010, 06:30:37 pm
I don't really get what you wrote.Uh I was trying to rewrite it in the quadratic form, as the parabola is going to be an inverted bowl shape, we can conclude that the range has to be smaller than the y coordinates of the maximum point ... Obviously I have made a computation error in working out the ... maximum point.
Anyways I managed to figure it out. The maximum point was (-3, 10)
So obviously the domain would be less than 10.
Post by: J.Darren on June 09, 2010, 06:43:32 pm
Post by: adrian1993 on June 09, 2010, 06:54:14 pm
Post by: xlane on June 09, 2010, 08:31:45 pm
Post by: theaguia on June 09, 2010, 09:44:01 pm
could sum1 help me wid oct/nov 09 paper 2..or part (ii)...hw to find da E coordinates..can u post the question, i dont have the paper
Post by: xlane on June 09, 2010, 10:14:48 pm
Post by: syedz123 on June 09, 2010, 10:51:15 pm
I did variant 2. What topics came up in your variant 1 for paper 1?
ermm...sets,binomial theorems,indices(surds),functions,diferentiation,polynomial,findin a coordinate at a point of contact of the tangent,rate of change n small of change,i dnt rmmbr even d0in any integrating question..i think da onli one wuz da second one of da last q..i chose da first one which involved difrentiation
like i sed our variant 1 wasnt as tought altough i still made silly mistakes...bt i think our p2 will b harder bt da p2 of variant 2 will b as easy as our p1 to make it fair then...i advice doin da topics i stated above dat came in p1 v1 for us
Post by: jellybeans on June 10, 2010, 01:29:26 am
to find when the curve crosses the x axis, we sub in y = 0 right?
so 6 sin(3x + pi/4) = 0
sin(3x + pi/4) = 0
(3x + pi/4) = sin-1 0
(3x + pi/4) = 0
3x = -pi/4
x = -1/12pi
but why does the MS say sin-1 0 = PI? isnt it supposed to be 0?? ??? help please! thanks in advanceeeee.
Post by: BlackBunny103 on June 10, 2010, 01:50:07 am
Thanxx :D
Post by: J.Darren on June 10, 2010, 03:28:13 am
:o heyyyyyy, i've got a doubt :(You are absolutely correct.
to find when the curve crosses the x axis, we sub in y = 0 right?
so 6 sin(3x + pi/4) = 0
sin(3x + pi/4) = 0
(3x + pi/4) = sin-1 0
(3x + pi/4) = 0
3x = -pi/4
x = -1/12pi
but why does the MS say sin-1 0 = PI? isnt it supposed to be 0?? ??? help please! thanks in advanceeeee.
Post by: J.Darren on June 10, 2010, 03:30:34 am
Guys, I need help with this question.Construct a right-angle triangle with the longest side pointing north-west (this is the compensate for the tide flow from west to east). The horizontal side with have a length of 2, whereas the vertical line with have a length of 1.5 (90/60), the longest side is the speed the boat travels when it is in still water. The angle would be the one between the vertical side and the longest side.
Thanxx :D
Post by: J.Darren on June 10, 2010, 03:33:14 am
:o heyyyyyy, i've got a doubt :(Ah, I think I have figured out what has gone wrong, in this case you must use the second solution which is 360 degrees. When x = 0, 6 sin (45), for sure when y = 0, x cannot be smaller than 45 degrees :P
to find when the curve crosses the x axis, we sub in y = 0 right?
so 6 sin(3x + pi/4) = 0
sin(3x + pi/4) = 0
(3x + pi/4) = sin-1 0
(3x + pi/4) = 0
3x = -pi/4
x = -1/12pi
but why does the MS say sin-1 0 = PI? isnt it supposed to be 0?? ??? help please! thanks in advanceeeee.
Post by: jellybeans on June 10, 2010, 08:38:10 am
ADD MATH'S OVERRRRRRRR, WOOP
Post by: adrian1993 on June 10, 2010, 11:02:29 am
Post by: jellybeans on June 10, 2010, 11:13:24 am
Indeed. I am so glad the paper was fairly easy
FAIRLY. hahahahaha, i cant wait to discuss it. DAMMIT question 9 was so. AH. LOL! <3 but overall twas goooood. XD
Post by: BlackBunny103 on June 10, 2010, 11:14:52 am
Paper 2 was indeed easier than paper 1. :D
Post by: J.Darren on June 10, 2010, 11:21:43 am
FAIRLY. hahahahaha, i cant wait to discuss it. DAMMIT question 9 was so. AH. LOL! <3 but overall twas goooood. XD*sigh* 7 marks gone ... Conceded 3 marks in P1 for reversing the position of the matrix in XB-1 = A ... Would I still get an A if such is the case ???
Post by: adrian1993 on June 10, 2010, 11:22:50 am
Post by: Dibss on June 10, 2010, 03:37:35 pm
*dum dum dummmmm* ;DDONEDONEDONEDONEDONEDONEDOOOOONEE!
ADD MATH'S OVERRRRRRRR, WOOP
FREEDOM ;D
Post by: elemis on June 10, 2010, 05:29:12 pm
SEE YA NEXT SESSION SUCKERS !!! :P
Post by: nid404 on October 10, 2010, 08:31:14 am
Post by: astarmathsandphysics on October 12, 2010, 08:47:31 am