Post by: Arthur Bon Zavi on October 31, 2010, 10:13:20 am
My small Doubt :
Solve the equation 3sin2theta+4cos theta = 4, giving all the roots, in the interval
0 <= theta <= 2
Post by: Arthur Bon Zavi on October 31, 2010, 10:17:15 am
Post by: Deadly_king on October 31, 2010, 11:16:34 am
All Pure Mathematics CIE (9709) doubts here!!!
My small Doubt :
Solve the equation 3sin2theta+4cos theta = 4, giving all the roots, in the interval
0 <= theta <= 2
Hey man, that's not very difficult. I won't do it but i'll surely give you a hint ;)
Let theta be x
Just replace sin2x by 1 - cos2x
Then let cos x = y and you'll be having a quadratic formula which you solve to obtain the required solution ;D
=> 3y2 - 4y + 1 = 0
I guess that you'll be able to complete the number yourself from now ;)
Post by: Arthur Bon Zavi on October 31, 2010, 11:22:48 am
Hey man, that's not very difficult. I won't do it but i'll surely give you a hint ;)
Let theta be x
Just replace sin2x by 1 - cos2x
Then let cos x = y and you'll be having a quadratic formula which you solve to obtain the required solution ;D
=> 3y2 - 4y + 1 = 0
I guess that you'll be able to complete the number yourself from now ;)
I know that was silly. :(
Post by: Deadly_king on October 31, 2010, 11:26:46 am
I know that was silly. :(
NO worries buddy, it happens.
You should be glad that you're making mistakes now and not in exams ;)
Post by: Alpha on October 31, 2010, 11:28:06 am
Post by: Arthur Bon Zavi on October 31, 2010, 11:31:55 am
Thank you Deadly_king and Alpa (It's okay).
+rep to both. :P
Post by: Alpha on October 31, 2010, 11:54:32 am
Most welcome.
Post by: elemis on October 31, 2010, 12:12:54 pm
I pray you understand my writing. ::)
Most welcome.
Your theta's look weird :P
Post by: Alpha on October 31, 2010, 12:14:10 pm
Your theta's look weird :P
Oh yeah? :-\
Post by: elemis on October 31, 2010, 12:31:02 pm
Oh yeah? :-\
Kidding ;)
Post by: Alpha on October 31, 2010, 05:02:03 pm
Post by: Arthur Bon Zavi on November 06, 2010, 05:11:32 am
Post by: Deadly_king on November 06, 2010, 05:15:17 am
Anybody has got P1 Hugh Neil and Douglas Quadling Book?
Yeah I got it. Why?
Post by: Arthur Bon Zavi on November 06, 2010, 05:29:09 am
Yeah I got it. Why?
Chapter 10
Exercise 10 D
Q-1 (a) (i) and (ii)
try and tell me!!
Getting the answers, but not being confirmed with which are at the back.. :-\
Post by: Deadly_king on November 06, 2010, 05:53:35 am
Chapter 10
Exercise 10 D
Q-1 (a) (i) and (ii)
try and tell me!!
Getting the answers, but not being confirmed with which are at the back.. :-\
Sorry for the late reply, I was busy in the science section.
Anyway it's obvious the answers from the book are not good. ;)
Let the length to be found be
Using pythagoras theorem :
sin theta = 8.8/
NOTE : ,/ stands for square root. ;D
Post by: Arthur Bon Zavi on November 06, 2010, 05:58:54 am
Sorry for the late reply, I was busy in the science section.
Anyway it's obvious the answers from the book are not good. ;)
Let the length to be found be.
Using pythagoras theorem :
sin theta = 8.8/
NOTE : ,/ stands for square root. ;D
Yes, there is a printing mistake. Then I got same answers.
No logic has to be applied in this question, CIE is dumb. :D
Post by: Deadly_king on November 06, 2010, 06:01:40 am
Yes, there is a printing mistake. Then I got same answers.
No logic has to be applied in this question, CIE is dumb. :D
Actually the book took
Post by: Arthur Bon Zavi on November 06, 2010, 06:09:06 am
Actually the book took
What they took is not near to x's real value also.
X is 12.34
Then how can they take it as X = 6.92 ?
Post by: Deadly_king on November 06, 2010, 06:17:36 am
What they took is not near to x's real value also.
X is 12.34
Then how can they take it as X = 6.92 ?
I don't know. :-[
But according to its answers for sin, cos and tan, it seems they took 4,/3
Nevermind. Let's just forget about it. ;)
Post by: Arthur Bon Zavi on November 06, 2010, 06:43:19 am
I don't know. :-[
But according to its answers for sin, cos and tan, it seems they took 4,/3
Nevermind. Let's just forget about it. ;)
But answers given by the book are surely wrong, right ?
Post by: Deadly_king on November 06, 2010, 09:43:38 am
But answers given by the book are surely wrong, right ?
Yeah they are wrong. ;)
Post by: Arthur Bon Zavi on November 06, 2010, 01:59:27 pm
Yeah they are wrong. ;)
Others are not wrong, they are getting on well with my answers. ;)
Post by: Arthur Bon Zavi on November 07, 2010, 03:25:04 pm
What is this now ? Again some printing mistake ?
Question - 3 (c) Page Number - 152
Post by: elemis on November 07, 2010, 03:48:36 pm
Post by: elemis on November 07, 2010, 03:51:44 pm
Here's how (next post)
Post by: elemis on November 07, 2010, 03:54:53 pm
Actually the textbook is wrong :P
Post by: Arthur Bon Zavi on November 07, 2010, 03:58:48 pm
Actually the textbook is wrong :P
Howzz That ? :P
Post by: SpongeBob on November 07, 2010, 04:36:56 pm
Post by: Deadly_king on November 08, 2010, 03:54:36 am
Establish The Following :
What is this now ? Again some printing mistake ?
Question - 3 (c) Page Number - 152
This identity CAN be solved. The book is NOT wrong.
Ari's working is good until the third step, except for the minus sign which is a plus sign. ;)
Then you should replace cos2x by 1 - sin2x which is then substituted with (1 + sin x)(1 - sin x)
The you can eliminate (1 + sin x) from both the numerator and denominator to obtain the required solution. :D
Post by: Arthur Bon Zavi on November 08, 2010, 05:21:02 am
(1 + sin x) and (1 + sin x) cancel out.
Therefore :
Proved the Identity :
Post by: Arthur Bon Zavi on November 14, 2010, 09:29:45 am
(i)
(ii)
After trying much, I didn't get !
Post by: Alpha on November 14, 2010, 10:57:37 am
= 1/ cos^2 x * (cos x/ sin x)
= 1/ cos x sin x
Check the second part too... I think there should be a double angle somewhere.
Post by: Arthur Bon Zavi on November 14, 2010, 11:14:27 am
1) (1+tan^2 x)/ tan x = sec^2 x/ tan x
= 1/ cos^2 x * (cos x/ sin x)
= 1/ cos x sin x
Check the second part too... I think there should be a double angle somewhere.
You took L.H.S, I don't think so ! :-\
Post by: Alpha on November 14, 2010, 11:17:37 am
= sec^2 x/ tan x
= 1/ cos^2 x * (cos x/ sin x)
= 1/ cos x sin x
-_-
Post by: Arthur Bon Zavi on November 14, 2010, 11:22:56 am
1/ tan x + tan x = 1+tan^2 x)/ tan x
= sec^2 x/ tan x
= 1/ cos^2 x * (cos x/ sin x)
= 1/ cos x sin x
-_-
Your head must be aching. Sorry, for making you run behind this. :-X
Thank You. :D
Post by: Alpha on November 14, 2010, 11:30:26 am
Your head must be aching. Sorry, for making you run behind this. :-X
Thank You. :D
You're welcome. :)
Yes, actually, my head IS aching, ain't cause of your Maths question though. ;)
Post by: $H00t!N& $t@r on November 16, 2010, 03:17:48 pm
Post by: Arthur Bon Zavi on November 16, 2010, 03:22:28 pm
Find the value of k for which there is no term in x^2 in the expansion of (1+kx)(2-x)^6 :)
Post whole Question, from the first ! If you can, tell us the page number of the P1 book and question number.
Post by: elemis on November 16, 2010, 03:29:41 pm
Find the value of k for which there is no term in x^2 in the expansion of (1+kx)(2-x)^6 :)
Doing it.
Post by: elemis on November 16, 2010, 03:37:44 pm
64 - 192x +240x2
Multiply this by (1+kx) giving :
64 - 192x + 240x2 +64kx -192kx2 +240kx3
Since x2 should not be present the sum of the co-efficients of the x2 terms must add up to zero.
Therefore, 240 -192k = 0
k = 1.25
Post by: $H00t!N& $t@r on November 16, 2010, 03:46:52 pm
+rep
Post by: elemis on November 16, 2010, 03:48:39 pm
thanks alot Ari ;D
+rep
Anytime :)
Post by: $H00t!N& $t@r on November 17, 2010, 10:18:33 am
can someone help me with m/j 05 q7 (i) and (iii) please explain the answer in (iii) Thanks
Post by: elemis on November 17, 2010, 10:24:10 am
how do you find the range of a fuction? ??? i read the book but i dont understand it :-\
can someone help me with m/j 05 q7 (i) and (iii) please explain the answer in (iii) Thanks
Sorry, I cant do this, havent done it yet.
Post by: $H00t!N& $t@r on November 17, 2010, 10:38:28 am
Post by: $H00t!N& $t@r on November 22, 2010, 05:20:45 pm
Anyway, I have another question!
Q) The equation of a curve is y = 3 cos 2x. The equation of a line is x + 2y = pie. On the same diagram, sketch the curve and the line for 0<= x<= pie.
The question is from O/N 2009 paper
Thanks ;D
Post by: samih.mabsout on November 22, 2010, 07:11:59 pm
Post by: samih.mabsout on November 22, 2010, 07:19:35 pm
i) Find the coordinates of the stationary point on the curve and determine its nature.
ii) Find the area of the region enclosed by the curve, the x-axis and the lines x=0 and x=1
* paper 1/oct/nov/2009 question 4*
Post by: Alpha on November 23, 2010, 07:01:03 am
how do you find the range of a fuction? ??? i read the book but i dont understand it :-\
can someone help me with m/j 05 q7 (i) and (iii) please explain the answer in (iii) Thanks
(i)
f : x -> 3 - 2 sinx
Amplitude is -2. y= - 2 sinx will have maximum and minimum points at 2 and -2. And since, f : x -> 3 - 2 sinx => curve is translated 3 units upward, minimum and maximum points will become:
2+3=5
and -2+3=1.
The range in which x is found therefore is: 1 </= f(x) </= 5 (Because of the equal to sign here: 0o </= x </= 360o.)
(iii)
You must have drawn the curve. For any curve to have an inverse, it has to be a one-one function, i.e. every horizontal line drawn across it should cut at only one point. From the graph, you'll notice that as you move from 90o onwards, any horizontal line through that curve will cut twice. Therefore, the curve is one-one in the range: 0o </= x </= 90o.
=> A= 90o.
N. B: Curve is also one-one in the ranges: 90o </= x </= 270o and 270o </= x </= 360o
Post by: Alpha on November 23, 2010, 07:16:58 am
wow am I the only one asking questions? :-[
Anyway, I have another question!
Q) The equation of a curve is y = 3 cos 2x. The equation of a line is x + 2y = pie. On the same diagram, sketch the curve and the line for 0<= x<= pie.
The question is from O/N 2009 paper
Thanks ;D
I answered this question of yours already in the Add Maths thread. ;)
y=4x^4+4x+9
i) Find the coordinates of the stationary point on the curve and determine its nature.
ii) Find the area of the region enclosed by the curve, the x-axis and the lines x=0 and x=1
* paper 1/oct/nov/2009 question 4*
(i)
It's not 4x4, it's x4.
dy/dx= 4x3+4
At Stat. pt, gradient i.e. dy/dx= 0,
4x3+4 = 0
x= -1
Subs. x= -1 in eqn. of curve:
y= 1-4+9
y=6
Therefore, coor. are (-1, 6).
d2y/dx2= 12x2
Rep. x=-1, d2y/dx2= 12, which is positive.
Therefore, it's a minimum curve.
(ii)
y= x4 + 4x + 9
Integrate y w.r.t. x from x=0 to x=1.
,/ : Integration of
,/y= x5/5 + 4x2/2 + 9x
= x5/5 + 2x2 + 9x
Area = 15/5 + 2(1)2 + 9(1) - 05/5 + 2(0)2 + 9(0)
You can use your calculator. Good luck. ;)
Post by: $H00t!N& $t@r on November 25, 2010, 05:48:43 pm
Post by: Alpha on November 26, 2010, 10:13:20 am
thanks Alpha! :)
Welcome. :)
Post by: HUSH1994 on November 29, 2010, 02:40:43 pm
Post by: elemis on November 29, 2010, 03:01:52 pm
anyone has some notes or something that can help about vectors
Rather than just asking for 'notes', please specify a section or type of question you have issues dealing with.
Post by: HUSH1994 on November 29, 2010, 03:03:36 pm
Post by: The Golden Girl =D on November 29, 2010, 03:50:37 pm
Check these out :
http://www.google.com/#sclient=psy&hl=en&q=Vectors+%2B+AS+%2B+Notes&aq=f&aqi=&aql=&oq=&gs_rfai=&pbx=1&fp=9b84a0f9b294c817
http://www.kent.ac.uk/ims/teaching/modules/ma590/MA555-MA590vecnotes.pdf
http://www.freeexampapers.com/past_papers.php?l=Past_Papers/A+Level/Maths/Resources/
I don't take CIE zai ma inta 3arif bas iA yinfa3ak :)
Post by: HUSH1994 on November 29, 2010, 04:07:48 pm
Post by: The Golden Girl =D on November 29, 2010, 04:20:31 pm
i found some useful things from the webpages u gave me yesterday :D,Thank You
Anytime :)
Post by: TJ-56 on December 06, 2010, 03:55:24 pm
the question is attached
Post by: islu_jf on December 24, 2010, 08:47:04 pm
Question number one!.. i know how to do the whole thing, but i always get confused what to do with the result,
for ex.
if i got tanx=-1 limits are 0<x<360 (both equal sign but don't know how to write it)
x=-45
so then what should i do?
-45(-+)180 or 180(-+)-45?
and for 360/?
and if u can put the sinx part as well.. ??
plz help
Post by: elemis on December 25, 2010, 04:08:52 am
Post by: islu_jf on December 25, 2010, 01:45:46 pm
same paper question number 5 part two?
Post by: Vampire-Love4ever on December 27, 2010, 01:12:09 pm
Q2- Express x^2+4x in the form a-(x+a)^2, stating values of a and b.
Q3- Express 2x^2-6x in the form a(x-b)^2+c stating the numerical values of a, b and c.
Post by: Arthur Bon Zavi on December 27, 2010, 03:00:50 pm
Q1- The function 3-2x-x^2 = a-(x+b)^2, find the value of a and b and hence find it's maximum value.
Q2- Express x^2+4x in the form a-(x+b)^2, stating values of a and b.
Q3- Express 2x^2-6x in the form a(x-b)^2+c stating the numerical values of a, b and c.
I do all these by a simple method, and if you don't understand, use the formula given in the book :
ax2 + bx + c = a ( x2 + b )
( ---x )
( a )
x2 + bx + c = (x2 + bx) + c = {(x + 1/2 b)2 - 1/4 b2} + c
(1)
-x2 - 2x - 3
-(x2 + 2x - 3)
-(x2 + 2x + 1 - 4)
-(x + 1)2 + 4 ---------> a = 4; b = -1; maximum value = 4.
(2)
x2 + 4x
(x2 + 4x)
(x2 + 4x + 4 - 4)
(x + 2)2 - 4 ----------> a= -4; b= -2
(3)
2x2 - 6x
2 (x2 - 3x)
2 (x2 - 3x + 2.25 - 2.25)
2 (x - 1.5)2 - 4.5 ------------> a= 2; b= 1.5; c= -4.5
Post by: Dibss on December 27, 2010, 07:38:52 pm
Post by: Arthur Bon Zavi on December 28, 2010, 04:01:07 am
^ +Rep (:
Anytime. ;D
Post by: Vampire-Love4ever on December 28, 2010, 05:56:20 pm
I do all these by a simple method, and if you don't understand, use the formula given in the book :
ax2 + bx + c = a ( x2 + b )
( ---x )
( a )
x2 + bx + c = (x2 + bx) + c = {(x + 1/2 b)2 - 1/4 b2} + c
(1)
-x2 - 2x - 3
-(x2 + 2x - 3)
-(x2 + 2x + 1 - 4)
-(x + 1)2 + 4 ---------> a = 4; b = -1; maximum value = 4.
(2)
x2 + 4x
(x2 + 4x)
(x2 + 4x + 4 - 4)
(x + 2)2 - 4 ----------> a= -4; b= -2
(3)
2x2 - 6x
2 (x2 - 3x)
2 (x2 - 3x + 2.25 - 2.25)
2 (x - 1.5)2 - 4.5 ------------> a= 2; b= 1.5; c= -4.5
Thanks!:) [rep]
Post by: acash09 on December 28, 2010, 07:14:39 pm
Question 4 --> P1 M/J 2009 (CIE AS-LEVEL) (9709/01/M/J/09)
I need explanation on how to exactly solve this question. Help is very much appreciated.
My friend told me its about amplitude and wavelength etc.
Thanks in advance.
Acash09
Post by: Dibss on December 28, 2010, 07:18:37 pm
Post by: acash09 on December 29, 2010, 04:38:52 pm
i need full explanation too.
thanks in advance
Post by: iluvme on December 29, 2010, 07:15:48 pm
i have another question too, its attached below:
i need full explanation too.
thanks in advance
y=12/(x2+3)
y=12(x2+3)-1
dy/dx= -12(x2+3)-2(2x)
=-24x/(x2+3)2
ii) Equation of the normal, First find the gradient at P on the curve, then using m1*m2=-1 Find the gradient of the normal.
Thus,
dy/dx at P x=1
-24(1)/(12+3)2
-24/16
-3/2
Using m1*m2=-1 to find the gradient of normal,
gradient of normal =2/3
Using the formula,
(y-y1)=m(x-x1)
y-3=2/3(x-1)
y=2x/3 + 7/3
As for the third part could you confirm if the answer is -0.018?
Post by: Tohru Kyo Sohma on December 31, 2010, 03:26:49 pm
differentiate the following with respect to x;
a.
b. 1/3x-5
c. 5/root 5x-2
Post by: iluvme on December 31, 2010, 03:53:42 pm
i have a doubt
differentiate the following with respect to x;
a. root 4x-1
b. 1/3x-5
c. 5/root 5x-2
Here you go.
Hope you understood. :)
Post by: Tohru Kyo Sohma on December 31, 2010, 04:30:51 pm
Here you go.thanks alot..........i'm new at SF....i heard that they help regarding subject doubts here........thanks alot
Hope you understood. :)
Post by: iluvme on December 31, 2010, 07:41:26 pm
thanks alot..........i'm new at SF....i heard that they help regarding subject doubts here........thanks alot
Hey Welcome to SF :)
We are here to help you ;)
Feel free to post your doubts. And I am glad you understood :)
Post by: Dibss on December 31, 2010, 07:53:45 pm
Post by: iluvme on December 31, 2010, 08:15:44 pm
+ Rep for your help, Iluvme :D
Thankyou Ma'am.
Post by: Tohru Kyo Sohma on January 01, 2011, 11:51:41 am
1. the curve C has equation y=Kx^2 +4/x, where K is a constant. the tangent to C at the point with x-coordinate 2 is parallel to the line with equation y=3x+8 . find the value of K and the equation of the tangent?
2. A curve has the equation y=6/1-2x . find an expression for dy/dx. hence find the equation of the normal to the curve at the point where x=2?
3. For the following function, find the stationary point and its nature; 4x+1/x ?
4. find the coordinates of the stationary point on the curve y=x^4 -2x^2 distinguishing between maxima and minima ?
sorry for asking so many questions but i dont seem to get the answers right.............an explanation would be helpful too!! :-[
Post by: elemis on January 01, 2011, 11:56:36 am
Post by: elemis on January 01, 2011, 12:00:02 pm
First differentiate the given equation :
dy/dx = 2kx - 4x-2
Since the tangent is parallel to the line y=3x+8 it is obvious that dy/dx at x=2 will have a value 3.
Hence, 4k - 1 = 3
k = 1
I'll let you find the equation of the tangent.... thats the easy part.
Post by: elemis on January 01, 2011, 12:06:42 pm
dy/dx = 4x3 - 4x
thus, 4x3 - 4x =0 at the stationary points.
Solving gives x as 1, -1 or 0
Differentiating a second time : 12x2 - 4
Substituting x for 1 gives 8. This is the minimum point.
Substituting x for -1 gives 8. This is also a minimum point.
Substituting x for 0 gives -4. This is maximum point.
Post by: Tohru Kyo Sohma on January 01, 2011, 12:08:43 pm
Question 1thank you ari ben!!!! 8)
First differentiate the given equation :
dy/dx = 2kx - 4x-2
Since the tangent is parallel to the line y=3x+8 it is obvious that dy/dx at x=2 will have a value 3.
Hence, 4k - 1 = 3
k = 1
I'll let you find the equation of the tangent.... thats the easy part.
Post by: elemis on January 01, 2011, 12:21:58 pm
thank you ari ben!!!! 8)
You're welcome.
Please dont call me Ari Ben, if you consider the meaning of my name you will realise it makes no sense calling me Ari Ben.
Ari - Lion in Hebrew
Ben - Son Of in Hebrew
Canaan - Israel in Hebrew
Post by: Tohru Kyo Sohma on January 01, 2011, 12:30:29 pm
You're welcome.
Please dont call me Ari Ben, if you consider the meaning of my name you will realise it makes no sense calling me Ari Ben.
Ari - Lion in Hebrew
Ben - Son Of in Hebrew
Canaan - Israel in Hebrew
oppss!!!...sorry ok i will not call you ari ben...........so............then can i call you ari ben canaan???
Post by: elemis on January 01, 2011, 12:31:38 pm
Post by: Tohru Kyo Sohma on January 01, 2011, 12:43:52 pm
Yeah, call me Ari or Ari Ben Canaan.ok.......i ll call you ari.........nice to meet you!!! ;D
Post by: Tohru Kyo Sohma on January 05, 2011, 04:00:21 pm
The curve with the equation y=k(x^3 -3x^2 -24x +c)
where k and c are non-zero constants, has two stationary points. show that the coordinates of one stationary point are (4, kc-80k ), and find, in a similar form the coordinates of the second stationary point. given that the mid-point of the line joining the two stationary points lies on the x-axis, show that c=26. given further that the tangent to the curve at the point of which x=2 passes through the point (5,-140), find the exact value of k.?
Post by: Dibss on January 05, 2011, 07:11:18 pm
i have a doubt;
The curve with the equation y=k(x^3 -3x^2 -24x +c)
where k and c are non-zero constants, has two stationary points. show that the coordinates of one stationary point are (4, kc-80k ), and find, in a similar form the coordinates of the second stationary point. given that the mid-point of the line joining the two stationary points lies on the x-axis, show that c=26. given further that the tangent to the curve at the point of which x=2 passes through the point (5,-140), find the exact value of k.?
This question is pretty long. General steps you'll need to go through are listed below. Ask if you need help/elaboration of a particular step, K.
(:
find dy/dx
set it to 0
you'll get two values for x
substitute them into the original equation to get the y coordinate of that particular stationary point
Now that you the two points, find the equation of the line joining them
Find the midpoint of this line (which lies on the x-axis therefore y=0)
Quote
given further that the tangent to the curve at the point of which x=2 passes through the point (5,-140), find the exact value of k.?Find dy/dx at x =2
That's the gradient of the tangent.
Proceed as need indicates... I haven't solved the qs so I don't know.
Post by: Tohru Kyo Sohma on January 05, 2011, 08:06:35 pm
This question is pretty long. General steps you'll need to go through are listed below. Ask if you need help/elaboration of a particular step, K.i cant get the correct answer.....................any further elaboration....!!!
(:
find dy/dx
set it to 0
you'll get two values for x
substitute them into the original equation to get the y coordinate of that particular stationary point
Now that you the two points, find the equation of the line joining them
Find the midpoint of this line (which lies on the x-axis therefore y=0)
Are you certain you've copied the qs correctly?
Post by: Dibss on January 05, 2011, 08:20:01 pm
i cant get the correct answer.....................
Like I asked, which part?
-show that the coordinates of one stationary point are (4, kc-80k )
-find, in a similar form the coordinates of the second stationary point.
-show that c=26.
-find the exact value of k.?
Or do you simply want the solution to the whole question without trying for yourself if you can get the other parts by following through after your particular doubt has been cleared? :|
Post by: Tohru Kyo Sohma on January 06, 2011, 01:26:23 pm
Like I asked, which part?ok......i got it.......thanks alot whisper! :D
-show that the coordinates of one stationary point are (4, kc-80k )
-find, in a similar form the coordinates of the second stationary point.
-show that c=26.
-find the exact value of k.?
Or do you simply want the solution to the whole question without trying for yourself if you can get the other parts by following through after your particular doubt has been cleared? :|
Post by: Tohru Kyo Sohma on January 06, 2011, 01:44:14 pm
use differentiation to show that the coordinates of the stationary point are (64, 131072). show further that the stationary point is a maximum point.
b. when a particular makes of a car is driven at a speed of v km/h, where 50<=v<=100, the distance travelled, per litre of fuel, by the car is modeled by;
96v^2 -v^3 +200000/25000 kilometers. for the speed in the range 50km/h to 100 km/h , find the maximum possible speed of kilometers per litre that the car can achieve?
plss hlp me!!!
Post by: elemis on January 06, 2011, 01:46:36 pm
Post by: elemis on January 06, 2011, 01:54:12 pm
x = 0 or 64
Inputting 64 into original equation gives a y coordinate of : 131,072
Differentiating a second time : 192 - 6x
Inputting 64 in to this equation gives a value of : -192. This is less than zero indicating it is a maximum point.
b) I cant answer this part because I'm unsure of the form of the equation you given. Can you show me the question paper this is from ?
Post by: Arthur Bon Zavi on January 06, 2011, 01:57:09 pm
Post by: Tohru Kyo Sohma on January 07, 2011, 10:10:58 am
192x -3x2 = 0its not from pastpapers.......its from a book....i dont know which book because our teacher only printed the qns and gave it to us to do.
b) I cant answer this part because I'm unsure of the form of the equation you given. Can you show me the question paper this is from ?
Post by: Tohru Kyo Sohma on January 07, 2011, 10:12:15 am
What is the answer of (b) ?the answer for b is 13.......i dont know how to get to this answer.
Post by: astarmathsandphysics on January 07, 2011, 04:00:00 pm
differntiatate wrt v to get (192v-3v^2 )/25000
To find the max put this equal to zero so (192v-3v^2 )/25000= 0 so 192v-3v^2 =0 so 3v(64-v)=0 so v=64
To prove a max differentiate again to get (192-6v )/25000 =(192-6*64 )/25000<0 hence max distance
Post by: Chingoo on January 15, 2011, 08:52:14 am
=_= I need help in every bit of it.
Post by: ashish on January 19, 2011, 04:33:49 pm
M/J 2010 - P32 - Q6
=_= I need help in every bit of it.
you are talking about cie right?
xlny=2x+1
lny=2+1/x
diff w.r.t x
1/ydy/dx=0-1/x2
dy/dx = -y/x2
tht part 1
Post by: ashish on January 20, 2011, 02:26:23 am
M/J 2010 - P32 - Q6
=_= I need help in every bit of it.
xlny-2x=1
x=1/(lny-2)
when y=1
x=-0.5
dy/dx=-4
y=mx+c
1=2+c(substituting all the values)
c=-1
y=-0.5x-1 (*2 throughout)
2y=-x-2
x+2y+2=0
Post by: melony on January 26, 2011, 06:05:25 pm
I REALLY NEED A MATHEMATICIAN! ;D
Please could someone guide me through a step by step solving of questions 3 and 6 oct/nov 2009 paper 32
I attached the paper so you can download it! :)
Thank you so much <3
Post by: melony on January 29, 2011, 04:05:02 pm
Post by: elemis on January 30, 2011, 04:33:00 pm
ok i managed to solve 3 PLEASE HELP WITH 6!!! :(
I really dont know.
Post by: melony on February 02, 2011, 02:13:49 pm
I really dont know.
thanx for havn a look though
thats very considerate of you :)
Post by: elemis on February 02, 2011, 02:30:27 pm
thanx for havn a look though
thats very considerate of you :)
The forum is pretty dead right now.
I'd suggest asking your teacher or a friend.
Post by: S.M.A.T on February 02, 2011, 10:32:15 pm
ok i managed to solve 3 PLEASE HELP WITH 6!!! :(
Here
Post by: yasser37 on February 03, 2011, 07:26:45 am
Post by: Deadly_king on February 03, 2011, 07:57:15 am
can someone please tell me what the product rule is? :-[ :-[
Okay......i'll take an example to explain it.
Let's say you're given
Product rule = d
In this case
So d
Now you replace in the formula of the product rule to get d
Hope it helps :D
Post by: yasser37 on February 03, 2011, 08:20:53 am
appreciated
Post by: Deadly_king on February 03, 2011, 09:38:38 am
thanks man
appreciated
You're welcome buddy :)
Post by: aloha32 on February 21, 2011, 01:29:41 pm
This is a question from P3 Nov 2002
Please help!
The curve y= e^x + 4e^-2x has one stationary point. Find the x coordinate of the stationary point and determine whether it is max or min
I differentiated the equation and got till e^x - 8e^-2x and equated it to zero but I'm stuck from there.
It'd be great if someone could guide me
Thanks! :)
Post by: Monopoly on March 05, 2011, 03:11:35 pm
Hello
This is a question from P3 Nov 2002
Please help!
The curve y= e^x + 4e^-2x has one stationary point. Find the x coordinate of the stationary point and determine whether it is max or min
I differentiated the equation and got till e^x - 8e^-2x and equated it to zero but I'm stuck from there.
It'd be great if someone could guide me
Thanks! :)
e^x - 8e^-2x can also be written as (e^3x -8)/e^2x
Equate this to zero and u will get e^3x = 8 and then solve for x
Post by: Monopoly on March 05, 2011, 03:14:13 pm
Post by: thecandydoll on March 06, 2011, 03:05:30 am
It says in the marskschemes,but i dont get it :D
Post by: yasser37 on March 11, 2011, 03:32:06 pm
p31
question 4
question 10
Post by: elemis on March 16, 2011, 02:55:37 pm
November 09
p31
question 4
question 10
10 minutes.
Post by: elemis on March 16, 2011, 03:05:11 pm
v=tanx
du/dx = -3e-3x
dv/dx = sec2x
Hence,
dy/dx = e-3xsec2x -3e-3xtanx
e-3x(sec2x - 3tanx)
Using 1+tan2x=sec2x
e-3x(tan2x - 3tanx + 1)
Solve the part in bold as a normal quadratic equation to give :
x = 1.206,0.365
Post by: elemis on March 16, 2011, 03:11:09 pm
dA/dr = 8
This should be simple to solve now.
Post by: HUSH1994 on March 18, 2011, 01:22:24 pm
Post by: Arthur Bon Zavi on March 18, 2011, 01:45:56 pm
O\N 2009 P11,Q2,how do u the graph :S
Did the radians part ?
Remember pie=180 degrees, 2pie = 360 and so on, and so 0.5pie = 90.
Take any values (4 or 5) for x (for example x = 1/4 pie, 1/3pie, 1/2 pie and pie) and you will get the values of y, then sketch the graph. Don't forget to see the domain. In this case it is 0 <= x <= pie, so you can not take values in negative and not exceed pie (180 degrees).
Post by: HUSH1994 on March 18, 2011, 02:49:04 pm
Post by: Vin on March 18, 2011, 03:46:44 pm
Revise the formula for differentiation.
[f(x)]n = n[f(x)]n-1 . f '(x)
Post by: HUSH1994 on March 19, 2011, 08:18:47 am
Post by: Arthur Bon Zavi on March 19, 2011, 08:37:38 am
Post by: HUSH1994 on March 19, 2011, 08:44:05 am
Post by: Arthur Bon Zavi on March 19, 2011, 08:45:24 am
I finished the whole syllabus and im solving past papers,why?
Good then.
Post by: elemis on March 19, 2011, 08:46:00 am
I finished the whole syllabus and im solving past papers,why?
I suggest you revise your syllabus.
Post by: Arthur Bon Zavi on March 19, 2011, 08:56:37 am
I suggest you revise your syllabus.
Precise*
Post by: HUSH1994 on March 19, 2011, 09:09:54 am
Post by: Arthur Bon Zavi on March 19, 2011, 09:14:41 am
i will,when i finish my school exams
Maths exams left ?
Post by: HUSH1994 on March 19, 2011, 09:19:46 am
Post by: Vin on March 19, 2011, 02:49:41 pm
thanks vin,a pain in the @$$ Q is ON 2005 Q5,is where does the r^3 come from and how do u do part 1?
Check those two triangles - red and black.
Use law of similarity and change the subject of formula to h. Then substitute "h" in the formula for volume of a cylinder.
Post by: Vin on March 19, 2011, 02:51:24 pm
thanks vin,a pain in the @$$ Q is ON 2005 Q5,is where does the r^3 come from and how do u do part 1?
I suggest you revise your syllabus.
If I were you, I'd listen to him^
Post by: Arthur Bon Zavi on March 20, 2011, 09:22:47 am
You've got to take the derivative of (x^2 + 3), therefore 2x.
Sorry, but I have to correct you :
9709/11/O/N/09
Q 7 (i)
dy/dx of 12/(x2 + 3) not only x2 + 3
= 12 (x2 + 3)-1
= -12 (x2 + 3)-2 (2x)
= -24x (x2 + 3) -2
= -24x/(x2 + 3) 2
That’s it for the first question.
Post by: Vin on March 20, 2011, 11:01:47 am
Sorry, but I have to correct you :
9709/11/O/N/09
Q 7 (i)
dy/dx of 12/(x2 + 3) not only x2 + 3
= 12 (x2 + 3)-1
= -12 (x2 + 3)-2 (2x)
= -24x (x2 + 3) -2
= -24x/(x2 + 3) 2
That’s it for the first question.
I know, but i remember him askin why "2x", therefore just a short explanation ;)
Post by: iluvme on March 23, 2011, 04:27:06 pm
Question attached
Post by: elemis on March 23, 2011, 05:03:36 pm
Post by: elemis on March 23, 2011, 05:11:14 pm
Vector AB is parallel to EF.
Hence, OA = DE and AB = EF.
OA = 6i
AB = 6j
DQ = 0.5*DF
Since DF = 6i + 6j
DQ = 3i + 3j
To find the position vector of Q we must go from the origin to D then to Q.
So its simply 6k + 3i + 3j
Post by: iluvme on March 23, 2011, 05:13:47 pm
Post by: Arthur Bon Zavi on March 27, 2011, 02:02:48 pm
(10) (ii) Please ...
Post by: iluvme on March 27, 2011, 07:34:16 pm
http://www.xtremepapers.net/CIE/International%20A%20And%20AS%20Level/9709%20-%20Mathematics/9709_w07_qp_1.pdf
Post by: elemis on March 28, 2011, 04:47:08 am
Question 10
http://www.xtremepapers.net/CIE/International%20A%20And%20AS%20Level/9709%20-%20Mathematics/9709_w07_qp_1.pdf
(ii)
Solving the above gives :
Doing the cos-1 of the above gives the angle : 61.9 degrees.
(iii)
If PR = 2i + 2j + 2k
PQ = -2i + 2j +4k
Then QR = PR - PQ
Find the modulus of PQ, PR and QR. Then add them all up; you should get the answer.
Post by: ashish on March 28, 2011, 06:19:15 am
http://www.xtremepapers.net/CIE/International%20A%20And%20AS%20Level/9709%20-%20Mathematics/9709_s09_qp_1.pdf
(10) (ii) Please ...
when x=0, y=13
the curve has turning point at (3,-5)
the line of symmetry is at x=3, since we have moved a length of 3 units from (0,0)(neglecting the y coordinate) till (3,0), to get the next part of the graph we ill have to move another 3 units, which will give (6,0)
or you could just solved for y=13 which worth more than one mark :D
Post by: iluvme on March 28, 2011, 06:57:58 pm
(ii)
Solving the above gives :
Doing the cos-1 of the above gives the angle : 61.9 degrees.
(iii)
If PR = 2i + 2j + 2k
PQ = -2i + 2j +4k
Then QR = PR - PQ
Find the modulus of PQ, PR and QR. Then add them all up; you should get the answer.
Thanks Ari.
How exactly do you find PR and PQ?
Post by: HUSH1994 on March 29, 2011, 07:03:05 am
Post by: elemis on March 29, 2011, 07:09:09 am
Thanks Ari.
How exactly do you find PR and PQ?
PR = PA + 0.5 AE + 0.5 AB Read the information given in the question CAREFULLY
PQ = PO + OD + 0.5 DG
Post by: elemis on March 29, 2011, 07:16:36 am
O\N 2010 P.13 Q4 part 2 please?
Hence, equation of the straight line is :
Post by: HUSH1994 on March 29, 2011, 07:22:13 am
Post by: elemis on March 29, 2011, 07:44:07 am
Thanks man,sorry for the trouble but i dont get Q6 part 1 in the same paper :(
Hold on.
Post by: elemis on March 29, 2011, 07:49:22 am
a + bi
Then, the modulus is :
For finding the argument of a complex no. :
Read your P3 textbook. There's an explanation in there.
Post by: HUSH1994 on March 29, 2011, 08:28:59 am
Post by: elemis on March 29, 2011, 08:29:24 am
i already got that before,thanks man,but i have a prob with Q9 part b part 2
Okay...
Post by: elemis on March 29, 2011, 08:34:13 am
You already know what the first term and the common difference is from (b)(i)
Post by: HUSH1994 on March 29, 2011, 08:54:39 am
Post by: iluvme on March 29, 2011, 06:51:09 pm
PR = PA + 0.5 AE + 0.5 AB Read the information given in the question CAREFULLY
PQ = PO + OD + 0.5 DG
Oh right, Thanks again Ari.
Feeling very very stupid.
Post by: Arthur Bon Zavi on March 29, 2011, 07:11:55 pm
Oh right, Thanks again Ari.
Feeling very very stupid.
Any more doubts ? :P
Post by: iluvme on March 29, 2011, 07:14:21 pm
Any more doubts ? :P
Not right now. :P
Gotta start working again. :)
Post by: Arissa_04 on April 04, 2011, 08:19:20 pm
Using the expansion of cos(3x-x) and cos(3x+x) prove that:
1/2(cos2x - cos4x) = sin3x sinx
Its a 3 mark question so I'm pretty sure its not that complicated but I cant seem to get the answers.
Thanks in advance :)
Post by: ashish on April 07, 2011, 05:07:32 pm
Hey guys, need help in a trig question.not sure which year it is is from.Here's how the question goes:
Using the expansion of cos(3x-x) and cos(3x+x) prove that:
1/2(cos2x - cos4x) = sin3x sinx
Its a 3 mark question so I'm pretty sure its not that complicated but I cant seem to get the answers.
Thanks in advance :)
cos(3x-x)= cos2x
=cos3x * cosx -- sin3xsinx
=cos3xcosx + sin3xsinx
cos(3x+x)=cos4x
=cos3xcosx-sin3xsinx
cos2x - cos4x = cos3xcosx + sin3xsinx -(cos3xcosx - sin3xsinx)
= 2sin3xsinx
half of it is sin3xsinx
Post by: Arissa_04 on April 07, 2011, 08:34:59 pm
That was well explained..I cant believe I spent so much time on that question unnecessarily complicating things by expanding cos3x and so on..
Thank you :)
Post by: thecandydoll on April 16, 2011, 11:36:02 am
OA=i?k, OB=3i+2j?3k and OC=4i?3j+2k. The mid-point of AB is M. The point N lies on AC between A and C and is such that AN = 2NC.
(i) Find a vector equation of the line MN. [4]
(ii) It is given that MN intersects BC at the point P. Find the position vector of P. [4]
I just dont get vectors sometimes,
any notes on this any notes on argand diagrams!
thanks :D
Post by: elemis on April 16, 2011, 02:00:33 pm
With respect to the origin O, the points A, B and C have position vectors given by
OA=i?k, OB=3i+2j?3k and OC=4i?3j+2k. The mid-point of AB is M. The point N lies on AC between A and C and is such that AN = 2NC.
(i) Find a vector equation of the line MN. [4]
(ii) It is given that MN intersects BC at the point P. Find the position vector of P. [4]
I just dont get vectors sometimes,
any notes on this any notes on argand diagrams!
thanks :D
Long time since I've done any vector work.
AB = -i+k+3i+2j-3k
= 2i -2k +2j
AM = 0.5 AB
= i-k+j
OM = OA + AM
= 2i +j -2k
AC = 3i -3j +3k
AN = (2/3)AC
= 2i -2j +2k
ON = OA + AN
= 3i -2j +k
Vector equation = position vector of N +
Part (b)
Find the vector equation of BC :
r = 3i+2j-3k +
Equate the equation from (a) with this new equation.
You will have three simultaneous equations.
Solve for lambda and mu. You should find lambda = 2 and mu =2
Re-insert back into one of the other equations and find the position vector of the cross over point.
Post by: HUSH1994 on April 16, 2011, 04:38:28 pm
Post by: Aadeez || Zafar on April 16, 2011, 07:29:16 pm
MJ 2004 Q10 part one,i was always wondering why do you have to change the sign of the least value,anyone knows that?give me the link of the question paper
Post by: elemis on April 17, 2011, 05:00:24 am
MJ 2004 Q10 part one,i was always wondering why do you have to change the sign of the least value,anyone knows that?
x2 -2x >15
Hence, x2 -2x -15 > 0
Solving this gives the two critical values :
x = -3 and x = 5
Sketching a parabola with these values (see below).
Since the inequality states that x must be GREATER than zero we are only interested in x-coordinates that meet the red part of the parabola.
Hence, x<-3 , x>5
Post by: thecandydoll on April 17, 2011, 05:43:36 am
OB = 3i + 4j. The point P lies on the line AB and OP is perpendicular to AB.
(i) Find a vector equation for the line AB. [1]
(ii) Find the position vector of P. [4]
(iii) Find the equation of the plane which contains AB and which is perpendicular to the plane OAB, giving your answer in the form ax + by + c???? = d.
i dont get the second part
Post by: HUSH1994 on April 17, 2011, 12:54:21 pm
Post by: Chingoo on April 21, 2011, 05:33:27 pm
Post by: EMO123 on April 22, 2011, 04:19:02 pm
can you tell me is 2 a prime number or not
and is 1 a prime number ??? :'(
it is very simple to say this to me so any one faster as much as you can
Post by: HUSH1994 on April 22, 2011, 04:26:21 pm
Post by: EMO123 on April 22, 2011, 04:27:26 pm
yeah 1 & 2 are prime numbersthanx friend but i am just stupid for a while
but now it's clear thanx friend
Post by: HUSH1994 on April 22, 2011, 04:30:51 pm
Post by: EMO123 on April 22, 2011, 04:46:12 pm
8709/1/o/n/01
Question No. 3
Post by: elemis on April 23, 2011, 05:28:18 am
paper 1
8709/1/o/n/01
Question No. 3
The inverse function of y=cosx is x = cos-1y
x = cos-1y has a range of 0 to pi.
Since the range of the inverse function is the domain of the normal function :
k = pi
Post by: EMO123 on April 23, 2011, 12:54:17 pm
The inverse function of y=cosx is x = cos-1ythanxx ari
x = cos-1y has a range of 0 to pi.
Since the range of the inverse function is the domain of the normal function :
k = pi
Post by: m.bhaiyat on April 25, 2011, 08:02:02 pm
thanks
Post by: astarmathsandphysics on April 25, 2011, 09:23:46 pm
see attachment
Post by: SkyPilotage on April 26, 2011, 08:44:04 pm
On question number 1, its always better to make 2 fraactions into a single fraction. Here you will have in the denominator (1+sinx)(1-sinx)=(1-sin^2x)=cos^2x
and the numerator would be sinx(1+sinx)-sinx(1-sinx)=sinx+sin^2x-sinx+sin^2x=2sin^x
There you go 2tan^x.
Question 3 ii)
you will multiplay ax by the "x" in the other brackets and multiplay 1 by the x^2 term in the other brackets. Their sum must be equal to zero!
Question 4 i)
-substitue x=0 y=3 ---->c=3
-ok now the maximum value for any sin function is 1 and minimum is -1. same thing for cosine.
so when y=9 sinbx=1 so a=6
-hmm now when u have sin(2x) the x values of sinx are divided by 2. so its as if its compressed by two. Look at the graphs of sin2x and sinx and compare. Maybe someone could have a better explanation for this last part.
Post by: SkyPilotage on April 26, 2011, 11:17:21 pm
hey guys..need help on may june 2009 questions 1,3(ii), and 4.
see attachment
Oh srry, I didnt see tht it was already explained ;)
Post by: polltery on April 27, 2011, 12:53:19 am
Post by: astarmathsandphysics on April 27, 2011, 09:06:51 am
Post by: SkyPilotage on April 27, 2011, 09:52:51 am
See attachmentIs it allowed to use A2 formulas in As ?
For part i) it can be solved by as rules since tan(pi-x)=-tan(x-pi)=-tanx=-k
Can u explain part ii) using the trigonometric circle? Srry for the trouble....
Post by: astarmathsandphysics on April 27, 2011, 10:08:31 am
Post by: your sister on April 29, 2011, 12:00:32 pm
Post by: your sister on April 29, 2011, 12:02:32 pm
j05 q7 (i)
ok forget it i found out :P
Post by: polltery on May 01, 2011, 02:59:32 am
http://url3.tk/graph.tk/
http://url3.tk/graph.tk/http://url3.tk/graph.tk/
Post by: HUSH1994 on May 01, 2011, 06:56:32 am
Post by: SkyPilotage on May 01, 2011, 02:52:21 pm
MJ 2010 P11 Q1 parts 2 & 3,how are we supposed to solve such questionsit was solved by astarmathsandphysics the previous page...
Post by: EMO123 on May 01, 2011, 03:30:10 pm
MJ 2010 P11 Q1 parts 2 & 3,how are we supposed to solve such questionsthis is from Mr. Astar
and for HUsh
but one thing i dont get is that i am not familier with cot x
so can anyone explain me the concept of cot x
Post by: SkyPilotage on May 01, 2011, 04:21:26 pm
this is from Mr. Astarcotx is an A2 formula which is 1 over tanx = cosx over sinx
and for HUsh
but one thing i dont get is that i am not familier with cot x
so can anyone explain me the concept of cot x
however I prefer using the tan property which is tanx=-tan(-x)
therefore tan(pi-x)= -tan(x-pi) = -tanx = -k *P.S : tan(x+180)=tan(x-180)=tanx since the tan graph repeats itself every 180 degrees unlike the six or cosine (360)
I hope tht clears the first part! :)
Post by: EMO123 on May 01, 2011, 04:25:34 pm
cotx is an A2 formula which is 1 over tanx = cosx over sinxthanx man and +rep for the information
however I prefer using the tan property which is tanx=-tan(-x)
therefore tan(pi-x)= -tan(x-pi) = -tanx = -k *P.S : tan(x+180)=tan(x-180)=tanx since the tan graph repeats itself every 180 degrees unlike the six or cosine (360)
I hope tht clears the first part! :)
Post by: SkyPilotage on May 01, 2011, 04:29:17 pm
thanx man and +rep for the informationNo problem, Glad to help ;)
Post by: HUSH1994 on May 01, 2011, 04:46:40 pm
Post by: EMO123 on May 01, 2011, 04:47:23 pm
Thanks emo :)any time
Post by: $H00t!N& $t@r on May 01, 2011, 10:02:39 pm
Here is the link: http://www.xtremepapers.me/CIE/International%20A%20And%20AS%20Level/9709%20-%20Mathematics/9709_w10_qp_12.pdf (http://www.xtremepapers.me/CIE/International%20A%20And%20AS%20Level/9709%20-%20Mathematics/9709_w10_qp_12.pdf)
Post by: ........ on May 02, 2011, 12:35:01 am
Post by: polltery on May 02, 2011, 04:45:20 am
Post by: elemis on May 02, 2011, 08:43:01 am
6(3x-2)-0.5
Concentrate only on (3x-2)-0.5.
Add one to the power :
(3x-2)0.5
Differentiate and adjust.
Post by: Vin on May 02, 2011, 01:19:15 pm
need help with q 2, 6(ii) and 10 O/N 2010 variant 12
Here is the link: http://www.xtremepapers.me/CIE/International%20A%20And%20AS%20Level/9709%20-%20Mathematics/9709_w10_qp_12.pdf (http://www.xtremepapers.me/CIE/International%20A%20And%20AS%20Level/9709%20-%20Mathematics/9709_w10_qp_12.pdf)
Q2
Right hand side
tan2x sin2 x
=tan2x (1 - cos2 x)
=tan2 x - tan2x cos2x
As tanx = sinx / cosx
cos x = sinx/ tanx
=tan2x - tan2x *
=tan2x - sin2x
Post by: Vin on May 02, 2011, 01:27:03 pm
the marked one is done! but i need help with the second one.. cant really understand how do i integrate it :/
There's a formula to integrate functions of the form (ax + b)n
f(x) = (ax + b)n
if f(x) = y
y = (ax + b)n
= (ax + b)n+1
---------- x
n + 1
DO NOT forget to multiply by 1/a
Post by: Vin on May 02, 2011, 01:36:39 pm
need help with q 2, 6(ii) and 10 O/N 2010 variant 12
Here is the link: http://www.xtremepapers.me/CIE/International%20A%20And%20AS%20Level/9709%20-%20Mathematics/9709_w10_qp_12.pdf (http://www.xtremepapers.me/CIE/International%20A%20And%20AS%20Level/9709%20-%20Mathematics/9709_w10_qp_12.pdf)
Q6 ii)
y = kx2 +1 & y = kx
equate the two to get k = 4 (follow up from the previous- but k2 - 4k = 0 )
equate k=4 in kx2 - kx +1 = 0 to get x = 1/2
equate x = 1/2 in one of the equations to get the y coordinate
Q 10
Expressing h in terms of x
Vol = 4 cm3 (given)
vol of object = x * 0.5x * h
4 = 0.5 * h * x^2
Change the sub. of formula to h to get h = 8/x2
Total Surface area (consider each side and add them up to make things easy)
= 2xh + 2*0.5*xh + 0.5 * x* x + 4/5 x * 5/4 x
solve, and substitute h = 8/x2 to get the desired answer.
ii)
dA/ dx = 1.5 * 2 * x + 24 * -1 * * x -1-1
= 3x - 24/x^2
equate this to zero to obtain the stationary point as 2
d2A/ dx2 = 3 - 24 * -2 * x -2-1
= 3 + 48/x^3
equate x =2 to get 9
since 9 > 0, point of minima
Post by: physichemaths on May 02, 2011, 11:08:49 pm
Post by: HUSH1994 on May 03, 2011, 05:30:14 am
Post by: elemis on May 03, 2011, 06:19:25 am
help me with question j08 question8? anyone?
Since the area of the triangle equals to tanx :
0.5*TN*PN = tanx
This can be re-written as :
0.5*x*y = tanx
The gradient of the tangent is simply :
dy/dx = y/x
Finding the reciprocal of both sides :
dx/dy = x/y
Re-arranging :
y*dx/dy = x
Substituting into original equation :
0.5*y2*dx/dy = tanx
Moving everything around :
0.5*y2*dx/tanx = dy
0.5*y2*dx*cotx = dy
0.5*y2*cotx = dy/dx
Done !
Post by: SkyPilotage on May 03, 2011, 10:09:03 am
MJ 2010 P13,Q6 part 1,i did AB.BC but in the MS it was AB.CB,how do u know which order do u use?Pretty Simple,
When they ask for angle ABC you either do BA.BC or AB.CB
You just have to either go from inside to outside or outside to inside.
More ex: -Angle DAC You either do DA.CA or AD.AC
Ofcourse you have to use the corresponding modulus of each!
-Angle PQR You can use QP.QR or PQ.RQ
Hope it clears things up :)
Post by: HUSH1994 on May 03, 2011, 10:11:30 am
Post by: Arthur Bon Zavi on May 03, 2011, 10:18:53 am
By the way, VECTORS was a cakewalk for you. ::)
Post by: HUSH1994 on May 03, 2011, 10:31:59 am
Post by: Arthur Bon Zavi on May 03, 2011, 10:42:22 am
i know,tbh all the pure is a piece of cake,but there are always some questions that need thinking,i got 69/75 in this papers and i finished it in 1hour 10 mins,my biggest problem is signs,esp in binomial series :S
That is a pitfall for getting 69. Think on the question and revise till the end of second. Make it near 1 hr 30 m, and 15 m for revision. I am sure you will get 75 except for those silly mistakes.
Post by: HUSH1994 on May 03, 2011, 11:00:06 am
Post by: bulono on May 04, 2011, 11:04:10 am
Post by: elemis on May 04, 2011, 11:16:41 am
May/june 2010 question 1...help
Which variant ?
Post by: physichemaths on May 05, 2011, 06:50:52 am
Post by: elemis on May 05, 2011, 07:22:33 am
help me with question 7 november 2010 paper 13. part iii)
Determine the range of the f(x) for the individual functions.
This will be the domains for the inverse of each individual function.
Post by: $H00t!N& $t@r on May 05, 2011, 08:27:02 am
http://www.xtremepapers.me/CIE/International%20A%20And%20AS%20Level/9709%20-%20Mathematics/9709_w09_qp_12.pdf (http://www.xtremepapers.me/CIE/International%20A%20And%20AS%20Level/9709%20-%20Mathematics/9709_w09_qp_12.pdf)
Post by: elemis on May 05, 2011, 09:39:01 am
question! o/n 2009 12 q4(ii)
http://www.xtremepapers.me/CIE/International%20A%20And%20AS%20Level/9709%20-%20Mathematics/9709_w09_qp_12.pdf (http://www.xtremepapers.me/CIE/International%20A%20And%20AS%20Level/9709%20-%20Mathematics/9709_w09_qp_12.pdf)
Draw it step by step.
Sin x : Fig.1
Sin 2x (there will be one complete sine curve in the 0-pi domain): Fig. 2
-3sin(2x) [reflect the curve in the x axis and stretch it by a factor of three] : Fig 3.
5 -3sin(2x) [translate the whole curve in the positive direction by 5 units] : Fig 4.
Post by: Arthur Bon Zavi on May 05, 2011, 10:39:16 am
Draw it step by step.
Sin x : Fig.1
Sin 2x (there will be one complete sine curve in the 0-pi domain): Fig. 2
-3sin(2x) [reflect the curve in the x axis and stretch it by a factor of three] : Fig 3.
5 -3sin(2x) [translate the whole curve in the positive direction by 5 units] : Fig 4.
Which software did you use for sketching those ?
Post by: elemis on May 05, 2011, 10:44:47 am
Which software did you use for sketching those ?
MS Word 2007. There's a math Plug-in you get. You used to get it for free (which is how I got it), but now you have to pay.
Post by: Arthur Bon Zavi on May 05, 2011, 10:51:09 am
MS Word 2007. There's a math Plug-in you get. You used to get it for free (which is how I got it), but now you have to pay.
Okay.
Post by: $H00t!N& $t@r on May 05, 2011, 10:51:56 am
Post by: physichemaths on May 05, 2011, 06:56:56 pm
Determine the range of the f(x) for the individual functions.
This will be the domains for the inverse of each individual function.
part iii?
Post by: I Jimmy I on May 06, 2011, 11:32:27 am
I know we have to use the "b2-4ac" method to arive with k2+4k-12=0
And I already got the answer as k=2 and k=-6
But in the Marking Scheme it says k>2 and k<-6
So how do I know which sign to use for which number.... I mean in the exam how should I know its k>2 not k>-6 and vice versa. Could some1 explain..?
Thank You
Post by: I Jimmy I on May 06, 2011, 11:53:21 am
So I should take the range of values which give an answer of >= 0? If this is correct plz confirm :)
TYVM
Post by: elemis on May 06, 2011, 11:57:54 am
b2-4ac > 0
Post by: I Jimmy I on May 06, 2011, 12:05:56 pm
Thank You :)
Post by: I Jimmy I on May 06, 2011, 12:18:02 pm
To find the values of a,b and c
I know c is the y-intercept which is 3.
I know b is the number of cycles which is 2.
But I dunno how we got a=6 altho the maximum value is 9 and minimum value is -3
Cheers
Post by: elemis on May 06, 2011, 12:25:06 pm
Ok can some1 help me solve question 4?
To find the values of a,b and c
I know c is the y-intercept
So how do I find "a" and "b"
Cheers
360/b = the period of the curve. Since, in this case, it repeats itself every 180 degrees, b = 2.
C represents how many units the normal sine curve has been translated upwards from its normal position. Can you figure this out for yourself ?
Post by: elemis on May 06, 2011, 12:27:52 pm
But I dunno how we got a=6 altho the maximum value is 9 and minimum value is -3
'a' represents the vertical stretch factor. that has been applied to the normal sine curve.
Translate the curve in the question back to its original position and consider the max and min points..... Can you see the stretch factor ?
Post by: I Jimmy I on May 06, 2011, 12:53:12 pm
Why thank u very much sir :D
Post by: physichemaths on May 06, 2011, 02:46:30 pm
urgent. anyone please help me?
Post by: bulono on May 06, 2011, 03:37:56 pm
http://www.xtremepapers.me/CIE/International%20A%20And%20AS%20Level/9709%20-%20Mathematics/9709_w10_ms_11.pdf
Post by: Vin on May 06, 2011, 05:00:01 pm
june 2011 paper 11 question 5 part ii? :(
urgent. anyone please help me?
June 2011. :/ Right. :/
question 4i...help....
http://www.xtremepapers.me/CIE/International%20A%20And%20AS%20Level/9709%20-%20Mathematics/9709_w10_ms_11.pdf
Question paper would help better my friend.
Anywho,
LHS
Multiply numerator and denominator by 1 + cos x
Solve it carefully.
You should reach the stage where you get :
sin x tan x + sin x tan x cos x
-----------------------------
sin^2 x
separate to give two fractions and cancel out the sine
you'll get:
tan x tan x cos x
----- + ------------
sin x sin x
1/ cos x + (equate tan x = sin x / cos x) and cancel out everything to get 1
Et Voila!
One tip. Write tan x = sin x / cos x at the top of the page ALWAYS while solving identities. Helps ;)
Good Luck for Monday.
Post by: iluvme on May 06, 2011, 07:30:08 pm
Q7 iii)
I got the answer using a very very long method, does anyone have a shorter method since its 1 mark only?
Post by: physichemaths on May 06, 2011, 07:31:56 pm
question 5 part iii?
Post by: SkyPilotage on May 06, 2011, 08:08:37 pm
http://www.xtremepapers.me/CIE/International%20A%20And%20AS%20Level/9709%20-%20Mathematics/9709_s10_qp_11.pdfAfter you found a+bcos^2x use that to substitute f(x) and add 1 and equate to zero...
question 5 part iii?
-Rearrange making cos subject.. then square root (remeber its + or -) and find values of x..... P.S: if cosx>1 or cos<-1 then its rejected!!
Post by: physichemaths on May 06, 2011, 08:27:37 pm
After you found a+bcos^2x use that to substitute f(x) and add 1 and equate to zero...
-Rearrange making cos subject.. then square root (remeber its + or -) and find values of x..... P.S: if cosx>1 or cos<-1 then its rejected!!
i mean part ii? typo :/
Post by: Vin on May 06, 2011, 09:04:14 pm
i mean part ii? typo :/
Minimum value of x in the domain is 0, try that first
Substitute in 2- 5cos^2 x you get 2 - 5 = -3
try with the max value, pi, you get the same, right? since its cos^2 x, it cant take -ve values, min = 0 and max = 1 (for cos^2 x NOT fx)
So minimum value of fx is -3
max is always 2, the constant. at x = pi/2 cos is 0, so 2.
Post by: physichemaths on May 06, 2011, 09:12:46 pm
Minimum value of x in the domain is 0, try that first
Substitute in 2- 5cos^2 x you get 3 - 5 = -3
try with the max value, pi, you get the same, right? since its cos^2 x, it cant take -ve values, min = 0 and max = 1 (for cos^2 x NOT fx)
So minimum value of fx is -3
max is always 2, the constant. at x = pi/2 cos is 0, so 2.
THANK YOU! :D
Post by: doubtigetastar on May 06, 2011, 09:13:34 pm
A quantity u varies with respect to t so that du/dt = a+bt where a & b are constants.
Given that it has a maximum value of 5.5 when t=1 and that its rate of change when t=2 is -3 , find u in terms of t.
Post by: Vin on May 06, 2011, 09:51:56 pm
CIE November 2010 Variant 2
Q7 iii)
I got the answer using a very very long method, does anyone have a shorter method since its 1 mark only?
Don't you complete the square for f(x) and substitute in hg? I think that's the smallest way. :/
THANK YOU! :D
Sorry there was a typo. (corrected- bold) please check again?
Help! :
A quantity u varies with respect to t so that du/dt = a+bt where a & b are constants.
Given that it has a maximum value of 5.5 when t=1 and that its rate of change when t=2 is -3 , find u in terms of t.
Do you have the answer? Cant be sure of mine actually :/
Post by: SkyPilotage on May 06, 2011, 10:32:15 pm
Help! :
A quantity u varies with respect to t so that du/dt = a+bt where a & b are constants.
Given that it has a maximum value of 5.5 when t=1 and that its rate of change when t=2 is -3 , find u in terms of t.
-du/dt=-3 when t=2 soo : a+2b=-3
-maximum value of u is 5.5 so du/dt=0 : a+b=0
So solve the eqns you get b=-3 and a=3
Then integrate du/dt to get u= 3t-1.5t^2+c --->subsitute (1,5.5) to get c=4 -----> so u=3t-1.5t^2+4
Hope it helps...
Post by: bulono on May 06, 2011, 10:37:06 pm
http://www.xtremepapers.me/CIE/International%20A%20And%20AS%20Level/9709%20-%20Mathematics/9709_s04_qp_1.pdf
in the answer scheme the perimeter of the semi circle is written as (pie)r...dont understand
Post by: SkyPilotage on May 06, 2011, 10:44:15 pm
M/J 2004 question 8 i....Circumference of circle is 2pi(r) since its half a circle the perimeter of semicircle is pi(r)..
http://www.xtremepapers.me/CIE/International%20A%20And%20AS%20Level/9709%20-%20Mathematics/9709_s04_qp_1.pdf
in the answer scheme the perimeter of the semi circle is written as (pie)r...dont understand
Post by: bulono on May 06, 2011, 10:48:10 pm
http://www.xtremepapers.me/CIE/International%20A%20And%20AS%20Level/9709%20-%20Mathematics/9709_s04_ms.pdf
Post by: Vin on May 06, 2011, 10:54:28 pm
2(pi) r /2
= (pi) r
Post by: bulono on May 06, 2011, 10:58:24 pm
Post by: doubtigetastar on May 07, 2011, 12:16:23 am
-du/dt=-3 when t=2 soo : a+2b=-3
-maximum value of u is 5.5 so du/dt=0 : a+b=0
So solve the eqns you get b=-3 and a=3
Then integrate du/dt to get u= 3t-1.5t^2+c --->subsitute (1,5.5) to get c=4 -----> so u=3t-1.5t^2+4
Hope it helps...
Thanks :)
Post by: iluvme on May 07, 2011, 07:13:04 am
Don't you complete the square for f(x) and substitute in hg? I think that's the smallest way. :/
This was what I did,
hg(x) = f(x)
h(x) = x2+bx+c
h(x-2)= (x-2)2 + b(x-2) +c = x2-4x+7
x2-4x+4+bx-2b+c= x2-4x+7
b=0 c=3
h(x) = x2+3
Post by: Arthur Bon Zavi on May 07, 2011, 07:48:40 am
the marked one is done! but i need help with the second one.. cant really understand how do i integrate it :/
No mistake in it. It is TOTALLY correct.
Post by: SkyPilotage on May 07, 2011, 06:37:22 pm
Post by: iluvme on May 07, 2011, 06:41:59 pm
Post by: Tohru Kyo Sohma on May 07, 2011, 09:14:32 pm
PLEASE PLEASE HELP ME!!!!!!!!!!!!!!!!!!!!!this is very important for me...!!!!
and i have a few more doubts while doing the papers....how do we find whe to put <,> signs in function question...i usually get the value right but the signs are wrong!!!any tips!!
Post by: SkyPilotage on May 07, 2011, 09:31:59 pm
i have a doubt here.............pls pls pls help me because i have my exam tomorrow......its P11 oct/nov 2009 qns 2!!i found the point on the curve but i cant find the points of the line x+2y= ? ( ? that pi, just incase its not clear)-If you want to plot the point of the line just substitute x with any value from 0 to pi. ex 0.25pi+2y=pi--->0.75pi=2y find y and u have a point....
PLEASE PLEASE HELP ME!!!!!!!!!!!!!!!!!!!!!this is very important for me...!!!!
and i have a few more doubts while doing the papers....how do we find whe to put <,> signs in function question...i usually get the value right but the signs are wrong!!!any tips!!
Do the same for another value of x and u get another point then join them to draw the line...
-Regarding the signs you put them on a number line and put the values on the number...outside the range of values is the same sign as a(coefficiient of x^2)....
for example if you have : x^2-x<0 so u put in the number line x=0 and x=1 soo outside this range is +ve since x^2 is +ve and inside the range is -ve. Since the question asks for <0 (-ve) it is 0<x<1. I hope that helps.
Post by: SkyPilotage on May 07, 2011, 09:32:49 pm
Yup, and the angle between them will be zero.thanks..
Post by: Tohru Kyo Sohma on May 07, 2011, 10:12:31 pm
-If you want to plot the point of the line just substitute x with any value from 0 to pi. ex 0.25pi+2y=pi--->0.75pi=2y find y and u have a point....i still didnt get question number 2 ......can u like solve it a take a picture and upload it....i think if i see it i can understand better!! :-[
Do the same for another value of x and u get another point then join them to draw the line...
-Regarding the signs you put them on a number line and put the values on the number...outside the range of values is the same sign as a(coefficiient of x^2)....
for example if you have : x^2-x<0 so u put in the number line x=0 and x=1 soo outside this range is +ve since x^2 is +ve and inside the range is -ve. Since the question asks for <0 (-ve) it is 0<x<1. I hope that helps.
Post by: SkyPilotage on May 07, 2011, 10:15:35 pm
i still didnt get question number 2 ......can u like solve it a take a picture and upload it....i think if i see it i can understand better!! :-['
yh i know but i dont have a scanner so we will have to wait for someone else to upload the pic ..
Post by: Tohru Kyo Sohma on May 07, 2011, 10:30:35 pm
'i think i kind of got the answer but im still not sure if the graph is right..its ok!
yh i know but i dont have a scanner so we will have to wait for someone else to upload the pic ..
Post by: iluvme on May 07, 2011, 11:10:03 pm
Post by: haris94 on May 08, 2011, 12:59:14 am
12/x2-4x
Post by: ........ on May 08, 2011, 01:53:08 am
how will u differentiate
12/x2-4x
http://www.wolframalpha.com/input/?i=Differentiate+12%2F%28x^2-4x%29
Post by: haris94 on May 08, 2011, 02:33:35 am
cool website By the way
Post by: physichemaths on May 08, 2011, 05:43:27 am
question 5 (b) how we know that we use the formula a(1-r^n)/1-r? why not we use a(r^n-1)/r-1??
Post by: elemis on May 08, 2011, 05:51:41 am
http://www.xtremepapers.me/CIE/International%20A%20And%20AS%20Level/9709%20-%20Mathematics/9709_w10_ms_12.pdf
question 5 (b) how we know that we use the formula a(1-r^n)/1-r? why not we use a(r^n-1)/r-1??
Both formulas are the same thing, you can use either.
Post by: Tohru Kyo Sohma on May 08, 2011, 07:17:21 am
Here you go.thanx alot amii
Post by: Tohru Kyo Sohma on May 08, 2011, 07:25:57 am
1. may/june 2009 p1 qns 4 part 2....the answer is 7pi/12....but i cant get it right....can some one show how?
2. oct/nov 2009 p12 qns 5 part 1 ,can someone show me how to solve the inequality?
3. oct/nov 2010 p12 qns 7 part 3....???
4. oct/nov 2010 p13 qns 8 part 2 ????
PLEASE HELP ME...!!!!!EXAM TOMORROW!! :-[
Post by: elemis on May 08, 2011, 08:48:33 am
6 sin(2x) +3 = 0
sin(2x) = -1/2
sin-1(-1/2) = -30, 330, 690, 210 or 570. I prefer working in degrees then converting to radians at the end.
Dividing all the above answers by 2.
-15, 165, 345,105, 285.
-15 is not in the given range (see question), hence, we ignore it.
The smallest angles is thus 105 degrees. Converting to radians give the correct answer.
Question 2
Part 1 :- Expanding :
sinx -sin2xcosx +cosx -sinxcos2x
Substitute cos2x for sin2x and sin2x for cos2x. You should get the answer.
Part 2 :-
cos3x = 8sin3x
Solve for x.
Question 3
I'll get back to you on this one.
Post by: Tohru Kyo Sohma on May 08, 2011, 09:37:43 am
Question 1thanx alot! ;D
6 sin(2x) +3 = 0
sin(2x) = -1/2
sin-1(-1/2) = -30, 330, 690, 210 or 570. I prefer working in degrees then converting to radians at the end.
Dividing all the above answers by 2.
-15, 165, 345,105, 285.
-15 is not in the given range (see question), hence, we ignore it.
The smallest angles is thus 105 degrees. Converting to radians give the correct answer.
Question 2
Part 1 :- Expanding :
sinx -sin2xcosx +cosx -sinxcos2x
Substitute cos2x for sin2x and sin2x for cos2x. You should get the answer.
Part 2 :-
cos3x = 8sin3x
Solve for x.
Question 3
I'll get back to you on this one.
Post by: SkyPilotage on May 08, 2011, 09:40:34 am
i just have a few more questions...pls pls help me...its;Question 3:
1. may/june 2009 p1 qns 4 part 2....the answer is 7pi/12....but i cant get it right....can some one show how?
2. oct/nov 2009 p12 qns 5 part 1 ,can someone show me how to solve the inequality?
3. oct/nov 2010 p12 qns 7 part 3....???
4. oct/nov 2010 p13 qns 8 part 2 ????
PLEASE HELP ME...!!!!!EXAM TOMORROW!! :-[
Complete the square of f(x) to get (x-2)^2+3 to H(x-2)=(x-2)^2+3----> so H(x) = x^2+3.....
Question 4:
length of PQ is the (length of radius of the arc - height of triangle ) x 2
----> (5-5cos(0.6rad))x2
@ Ari: do you know how marks are distributed on sketches of graphs? (not drawings)
Post by: elemis on May 08, 2011, 09:42:50 am
Question 3:
Complete the square of f(x) to get (x-2)^2+3 to H(x-2)=(x-2)^2+3----> so H(x) = x^2+3.....
Question 4:
length of PQ is the (length of radius of the arc - height of triangle ) x 2
----> (5-5cos(0.6rad))x2
@ Ari: do you know how marks are distributed on sketches of graphs? (not drawings)
You get marks for the general shape, intersection points with coordinate axis and drawing in any asymptotes.
Post by: SkyPilotage on May 08, 2011, 09:51:16 am
You get marks for the general shape, intersection points with coordinate axis and drawing in any asymptotes.ok so if its a curve like sinx, i can draw it free hand and just show where it intersects with the axis right?
-If they ask us to sketch a graph and its domain in x>0 , i should only draw the graph on the positive x-axis; should I show him the negative axis and leave it blank?
-when they ask us abt solving eqns do we need to show the quadratic formula or shall we just write the factors then the answer??
-which is better? to write the final answer as fractions or decimals? and if there is a square root of irration number do we leave it tht way or convert to decimals?
THankS :)
Post by: Tohru Kyo Sohma on May 08, 2011, 09:56:27 am
Question 3:i understood the 4th qns.thanks ...but im still stuck at the 3rd qns i got till H(x-2)=(x-2)^2+3 myself too. but i cant get the final answer as H(x) = x^2+3
Complete the square of f(x) to get (x-2)^2+3 to H(x-2)=(x-2)^2+3----> so H(x) = x^2+3.....
Question 4:
length of PQ is the (length of radius of the arc - height of triangle ) x 2
----> (5-5cos(0.6rad))x2
can you show me the steps?? :-[
plus do you understand may/june 2010 p11 qns 1??because im so lost in that qns :o
Post by: SkyPilotage on May 08, 2011, 10:12:21 am
i understood the 4th qns.thanks ...but im still stuck at the 3rd qns i got till H(x-2)=(x-2)^2+3 myself too. but i cant get the final answer as H(x) = x^2+3Actually you have to be able to see tht what gives u f(x) when u substitute x by x-2....
can you show me the steps?? :-[
plus do you understand may/june 2010 p11 qns 1??because im so lost in that qns :o
-June 2010...Yh it was answered before in this topic..Just go back some pages..
Post by: $H00t!N& $t@r on May 08, 2011, 01:19:19 pm
Q10 (ii) and (iii) m/j 2009
(http://www.xtremepapers.me/CIE/International%20A%20And%20AS%20Level/9709%20-%20Mathematics/9709_s09_qp_1.pdf)
Post by: SkyPilotage on May 08, 2011, 01:37:20 pm
questionII)
Q10 (ii) and (iii) m/j 2009
(http://www.xtremepapers.me/CIE/International%20A%20And%20AS%20Level/9709%20-%20Mathematics/9709_s09_qp_1.pdf)
the axis of symmetry is at x=3 so the distance from both sides of the axis must be equal so A=6
III)
to find the range in the domain given, the minimum point is at the vertex(x=3) and y=-5
the largest y is when x=0 or x=6---->y=13
----->the range is -5<y<13
Post by: $H00t!N& $t@r on May 08, 2011, 03:02:43 pm
II)
the axis of symmetry is at x=3 so the distance from both sides of the axis must be equal so A=6
III)
to find the range in the domain given, the minimum point is at the vertex(x=3) and y=-5
the largest y is when x=0 or x=6---->y=13
----->the range is -5<y<13
oh! thanks :)
for the question about range... i got a bit confused because i was told that whenever you have a(x + b)^2 + c ... c is always the range. I didn't consider the part where they said "When A has this value..."
So what if they only said find the range of f. Would it just be x>= -5 ?
Post by: ........ on May 08, 2011, 03:07:45 pm
Correct me if i am wrong.
Post by: elemis on May 08, 2011, 03:12:48 pm
If you have a given domain and a equation just substitute x values into the equation to get the range.
Correct me if i am wrong.
True.
Post by: $H00t!N& $t@r on May 08, 2011, 03:19:30 pm
I just wanted to know if the statement "whenever you have a(x + b)^2 + c ... c is always the range" is right or not...
Post by: Arthur Bon Zavi on May 08, 2011, 03:22:27 pm
yeah I know that...
I just wanted to know if the statement "whenever you have a(x + b)^2 + c ... c is always the range" is right or not...
Almost all cases. Look for the domain first by the way.
Post by: $H00t!N& $t@r on May 08, 2011, 03:28:30 pm
Almost all cases. Look for the domain first by the way.
ok thanks
Post by: $H00t!N& $t@r on May 08, 2011, 04:14:18 pm
Q6 ii)
y = kx2 +1 & y = kx
equate the two to get k = 4 (follow up from the previous- but k2 - 4k = 0 )
equate k=4 in kx2 - kx +1 = 0 to get x = 1/2
equate x = 1/2 in one of the equations to get the y coordinate
Q 10
Expressing h in terms of x
Vol = 4 cm3 (given)
vol of object = x * 0.5x * h
4 = 0.5 * h * x^2
Change the sub. of formula to h to get h = 8/x2
Total Surface area (consider each side and add them up to make things easy)
= 2xh + 2*0.5*xh + 0.5 * x* x + 4/5 x * 5/4 x
solve, and substitute h = 8/x2 to get the desired answer.
ii)
dA/ dx = 1.5 * 2 * x + 24 * -1 * * x -1-1
= 3x - 24/x^2
equate this to zero to obtain the stationary point as 2
d2A/ dx2 = 3 - 24 * -2 * x -2-1
= 3 + 48/x^3
equate x =2 to get 9
since 9 > 0, point of minima
Thanks a lot Vin ;D I thought no one replied.
Post by: Vin on May 08, 2011, 04:41:44 pm
Okay, i have a stupid doubt which i cant waste my time on thinking on much.
http://www.xtremepapers.me/CIE/index.php?dir=International%20A%20And%20AS%20Level/9709%20-%20Mathematics/&file=9709_s09_qp_1.pdf
Q4 part i
-.-
Post by: SkyPilotage on May 08, 2011, 04:50:23 pm
No problem ;DAhaha it happens....Actually its not stupid :D
Okay, i have a stupid doubt which i cant waste my time on thinking on much.
http://www.xtremepapers.me/CIE/index.php?dir=International%20A%20And%20AS%20Level/9709%20-%20Mathematics/&file=9709_s09_qp_1.pdf
Q4 part i
-.-
-For c:
-----Substitute x=0 to get c=3
-For a:
Remember max value of sinx is 1 and min is -1.. so substitute sin(bx)=1 --->9=a(1)+3---->a=6
-For b:
The graph oh sinx has a whole wave from 0 to 2pi but here there are 2 curves in the domain 0 to pi...Therefore B=2...
Post by: Vin on May 08, 2011, 05:14:22 pm
Ahaha it happens....Actually its not stupid :D
-For c:
-----Substitute x=0 to get c=3
-For a:
Remember max value of sinx is 1 and min is -1.. so substitute sin(bx)=1 --->9=a(1)+3---->a=6
-For b:
The graph oh sinx has a whole wave from 0 to 2pi but here there are 2 curves in the domain 0 to pi...Therefore B=2...
You're the man. Respect. Thank you so much.
I'm off. Good luck people. Kill it.
Post by: EMO123 on May 08, 2011, 05:54:41 pm
http://www.ies.co.jp/math/java/trig/graphCosX/graphCosX.html
Post by: EMO123 on May 08, 2011, 05:56:34 pm
http://www.ies.co.jp/math/java/samples/graphSinX.html
Post by: EMO123 on May 08, 2011, 06:03:27 pm
http://www.intmath.com/trigonometric-graphs/1-graphs-sine-cosine-amplitude.php
Post by: EMO123 on May 08, 2011, 06:50:41 pm
http://www.intmath.com/
Post by: Saavya on May 08, 2011, 06:51:01 pm
hope the vectors and domain/range questions come easy !
Post by: bulono on May 09, 2011, 12:50:55 pm
Post by: Saavya on May 09, 2011, 02:05:18 pm
Post by: Arthur Bon Zavi on May 09, 2011, 03:18:09 pm
omg YES ! :S
No. Not at all. As easy as you like.
Post by: EMO123 on May 09, 2011, 03:26:51 pm
not easy not even difficult
Post by: Ghost Of Highbury on May 09, 2011, 03:29:20 pm
Post by: EMO123 on May 09, 2011, 03:33:50 pm
and also control your piss :D
Post by: ShiningStar on May 10, 2011, 09:10:09 pm
Integrate cot4x
Thanks In Advance :)
Post by: elemis on May 11, 2011, 04:24:47 am
Can any one help me in this question :-\
Integrate cot4x
Thanks In Advance :)
Can you show me the original question ?
Post by: ShiningStar on May 11, 2011, 03:08:00 pm
Find the Intergration of cot4x (Limits from pi/4 till pi/3)
Post by: thecandydoll on May 13, 2011, 07:30:45 am
Find the coordinates of the point, other than the origin, where the curve has a tangent which is
parallel to the x-axis. [5]
Post by: elemis on May 13, 2011, 08:22:46 am
The equation of a curve is x3 + 2y3 = 3xy.
Find the coordinates of the point, other than the origin, where the curve has a tangent which is
parallel to the x-axis. [5]
Use implicit differentiation and equate to zero. Solve.
Post by: thecandydoll on May 14, 2011, 07:04:31 am
The equation of a curve is xy(x + y) = 2a3, where a is a non-zero constant. Show that there is only one point on the curve at which the tangent is parallel to the x-axis, and find the coordinates of this point.
I dont get how to do this/
Post by: thecandydoll on May 14, 2011, 07:08:44 am
Q8 I don't get this either.
Someshow me the workingg?
Post by: elemis on May 14, 2011, 07:28:03 am
The equation of a curve is xy(x + y) = 2a3, where a is a non-zero constant. Show that there is only one point on the curve at which the tangent is parallel to the x-axis, and find the coordinates of this point.
I dont get how to do this/
You are told ' a' is a non-zero constant. So, if you want, you can think of 2a3 as some number. When you differentiate a number like 10, for example, you simply get zero.
Hence :
x2y + xy2 = 2a3
Use the product rule for the left hand side terms and simply differentiate 2a3 to get zero because, after all, its nothing but a number.
Post by: thecandydoll on May 14, 2011, 10:45:09 am
Post by: ihatephysics on May 15, 2011, 01:41:00 am
i have attached the paper
Post by: Nera_egypt on May 16, 2011, 02:44:41 pm
heyy! i have a vectors question! it would be awesome if someone could solve it for me!
i have attached the paper
The answer is in the attachment. Hope it helps :)
Post by: ........ on May 16, 2011, 03:26:21 pm
Looking at this year's P1 and M1 almost 70 percent of the questions came from 2001 to 2003 and some from 90s.
Post by: elemis on May 16, 2011, 03:55:42 pm
Hey guys what is the hardest past paper in P3 you have encountered ?
Looking at this year's P1 and M1 almost 70 percent of the questions came from 2001 to 2003 and some from 90s.
Integration.... not because its difficult but because its annoying.
Post by: Nera_egypt on May 16, 2011, 04:46:34 pm
Post by: melony on May 16, 2011, 09:22:20 pm
i cant find any pattern, what papers do you think i should practice?
this is my worst paper By the way :-[
Post by: leebux101 on May 17, 2011, 02:51:53 pm
Post by: elemis on May 17, 2011, 03:15:35 pm
dv/dx = eu
Integration by parts :
u2 -
Integrate 2ueu using integration by parts.
You should get the correct answer.
Post by: leebux101 on May 17, 2011, 03:35:29 pm
Post by: leebux101 on May 17, 2011, 06:12:51 pm
please asap! =D thanks a gziliion in advance! =D
Post by: ShiningStar on May 17, 2011, 06:41:04 pm
State the range of values of x for which the expansion (1-4x)-3 is valid. ( Q3 Paper 9202/1/O/N/01)
What I did is:
-1<1-4x<1
-2<-4x<0
0<x<1/2
Please correct me if there is anything wrong...
What I don't understand here is that how is 1/4 accepted, as if x=1/4 then there is a 0 in the denominator which is going to give (Math Error)?? ??? :-\
Thanks!
Post by: ShiningStar on May 17, 2011, 07:08:41 pm
may june 2003 Q5 , what are the complex numbers and how to do part i)The complex numbers are: u=2i, w=(cos(2pi/3) + sin(2pi/3)i)
please asap! =D thanks a gziliion in advance! =D
5.i) w= -1/2 + (root3/2)i, what I did is that I solved the brackets where cos(2pi/3) = -1/2 and sin(2pi/3)i = (root3/2) i
uw= 2i(-1/2 + (root3/2)i) = -root3-i
u/w = 2i/(-1/2 + (root3/2)i) then you multiply it by its conjugate (-1/2 - (root3/2)i) to get it in form x+iy
Hope it helps :)
Post by: leebux101 on May 17, 2011, 07:20:11 pm
i am sorry i cant help u in ur question.. =(
Post by: leebux101 on May 17, 2011, 07:32:31 pm
how do u get the angle at the vertices! please. =)
Post by: ShiningStar on May 17, 2011, 07:47:48 pm
also for same questio part iiiYou don't have to find the angle at the vertices, you can find the lengths of UB, UA and AB and see if they are equal then it is an equaliteral triangle..!
how do u get the angle at the vertices! please. =)
Post by: leebux101 on May 17, 2011, 08:04:42 pm
secondly, can u do q10 part iii from the same paper! please!! :'(
Post by: leebux101 on May 17, 2011, 08:15:59 pm
Post by: ShiningStar on May 17, 2011, 08:27:04 pm
in the marking scheme it says that u have to get the angle at atleast two vertices as 60Use the Identity in the 10.i) to replace cosec2x, so it will become:
secondly, can u do q10 part iii from the same paper! please!! :'(
Integration cotx - cot2x dx (with the same limits)
=(ln(sinx) - 1/2 ln(sin2x)) then substitute with the limits
Post by: leebux101 on May 17, 2011, 08:35:08 pm
+repp!! =D =D
Post by: THEIGBOY on May 17, 2011, 08:45:35 pm
question 5 paper 33 june 2010 part ii
and
question 3 paper 33 octover/nov 2010 part ii
and how to do question 10 on may june 2010 paper 33
Thank you very much
Post by: TJ-56 on May 17, 2011, 09:11:18 pm
Thanx in advance.
Post by: leebux101 on May 17, 2011, 09:31:55 pm
arii!! where u at man!
we need ya here! =P
Post by: SkyPilotage on May 17, 2011, 11:06:47 pm
can someone do ON03 q 7 iii)Its A circle with radius=2 units and center u.
arii!! where u at man!
we need ya here! =P
Post by: SkyPilotage on May 17, 2011, 11:11:56 pm
Does anyone know why the root (-1\3) in question 10 (ii) in paper May June 2008 is rejected? The mark scheme just states that “Having rejected the root (-1\3) for a valid reason" lol, well what is this reason?That is because if you substitute t=-2 and t=-1/3 in the position vector of P ; the dot product of AP.AB should be +ve since the angle is acute.. If its -ve the angle is obtuse...Therefore if you use t=-2 you will get it +ve, and t=-1/3 will be -ve meaning obstuse angle which is incorrect..
Thanx in advance.
Post by: melony on May 18, 2011, 12:35:37 am
ON10 paper 31 question 7 (ii) i do not know how to get ?=-1/6 :(
June 2004 question 6 - i know it starts of like 1/3 ln (y^3+1)=x+c but i have no idea where the 1/3 comes from (should have listened in class :'()
Post by: ........ on May 18, 2011, 01:47:21 am
awh no :-\! seems like everyone's offline, i just had two questions:
ON10 paper 31 question 7 (ii) i do not know how to get ?=-1/6 :(
June 2004 question 6 - i know it starts of like 1/3 ln (y^3+1)=x+c but i have no idea where the 1/3 comes from (should have listened in class :'()
Um i just did Oct/Nov 2010 so yea heres your solution to 2010 query
Post by: ........ on May 18, 2011, 02:00:03 am
can someone explain how to do questions like -->
question 5 paper 33 june 2010 part ii
and
question 3 paper 33 octover/nov 2010 part ii
and how to do question 10 on may june 2010 paper 33
Thank you very much
Modulas 5 is the radius and (3,4) is the center of the circle and you have to shade all the points on and inside the circle center.October/November 2010 part ii
Post by: leebux101 on May 18, 2011, 02:03:40 am
awh no :-\! seems like everyone's offline, i just had two questions:for june 2004,
ON10 paper 31 question 7 (ii) i do not know how to get ?=-1/6 :(
June 2004 question 6 - i know it starts of like 1/3 ln (y^3+1)=x+c but i have no idea where the 1/3 comes from (should have listened in class :'()
u get y2ln(y3 +1)
so when differentiating the inner content, u get y3ln(y3 +1)/3y2
y2 gets cancelled and u get u 1/3 ln(y3 +1)
please reconfirm this cud i am not sure if i am right..
can anyone reconfirm this.. ? pleaseeee!!! =(
Post by: leebux101 on May 18, 2011, 02:04:39 am
Its A circle with radius=2 units and center u.i wrote part iii not part ii, do u know part iii then?
Post by: ........ on May 18, 2011, 02:26:38 am
can someone explain how to do questions like -->question 5 paper 33 june 2010 part ii
question 5 paper 33 june 2010 part ii
and
question 3 paper 33 octover/nov 2010 part ii
and how to do question 10 on may june 2010 paper 33
Thank you very much
Post by: ........ on May 18, 2011, 02:35:13 am
awh no :-\! seems like everyone's offline, i just had two questions:
ON10 paper 31 question 7 (ii) i do not know how to get ?=-1/6 :(
June 2004 question 6 - i know it starts of like 1/3 ln (y^3+1)=x+c but i have no idea where the 1/3 comes from (should have listened in class :'()
as for the second part
listen when there is a equation like y^2/y^3+1 the only way to integrate this is when you differentiate the denominator the exact value should come in the numerator. In this case the differentiation of y^3+1 is 3y^2 but in the numerator it is y^2 so you take the 3 out by multiplying it by 1/3. so now it comes 1/3(y^2/y^3+1)and then you get 1/3ln(y^3+1)
Post by: melony on May 18, 2011, 08:08:57 am
i wish you well for the exam :)
Post by: TJ-56 on May 18, 2011, 08:21:33 am
That is because if you substitute t=-2 and t=-1/3 in the position vector of P ; the dot product of AP.AB should be +ve since the angle is acute.. If its -ve the angle is obtuse...Therefore if you use t=-2 you will get it +ve, and t=-1/3 will be -ve meaning obstuse angle which is incorrect..
Oh ok thanx.. that cleared it up,
+rep
Post by: aloha32 on May 18, 2011, 08:42:46 am
I really need help with complex numbers and argand diagrams!
Can u plz help me with these qs/
Nov 06 q9 part 4
Nov 09 q7 part4
JUNE 10 V1 Q7 Part 3
THANK U
Post by: SkyPilotage on May 18, 2011, 09:47:38 am
Hi guysSadly , i dont have a good scanner to show you the argand diagrams :( So hopefully someone will upload them for you...
I really need help with complex numbers and argand diagrams!
Can u plz help me with these qs/
Nov 06 q9 part 4
Nov 09 q7 part4
JUNE 10 V1 Q7 Part 3
THANK U
Who has the Alevel Cie Pure mathematics 2 and 3 by Hugh Neil and Douglas Quading???
Page 300 number 6 I) and page 303 number 9I)
Post by: Arthur Bon Zavi on May 18, 2011, 12:12:08 pm
Sadly , i dont have a good scanner to show you the argand diagrams :( So hopefully someone will upload them for you...
Who has the Alevel Cie Pure mathematics 2 and 3 by Hugh Neil and Douglas Quading???
Page 300 number 6 I) and page 303 number 9I)
Me. :P
Post by: ........ on May 18, 2011, 01:47:36 pm
Hi guys
I really need help with complex numbers and argand diagrams!
Can u plz help me with these qs/
Nov 06 q9 part 4
Nov 09 q7 part4
JUNE 10 V1 Q7 Part 3
THANK U
for November 06
Find the modulas of Ou which is SQR(2) and the modulas of the point (1,0) on the x axis which is 1.
therefore the least value of |z|=SQR(2)-1=0.414
Post by: physichemaths on May 18, 2011, 01:53:30 pm
how to do question 7 part ii?
Post by: physichemaths on May 18, 2011, 02:17:30 pm
I am not sure but i think this is the solution to part ii
are u referring at me? question 7 is about vector :/
Post by: ........ on May 18, 2011, 02:20:27 pm
are u referring at me? question 7 is about vector :/
sorry i was answering to aloha.
Post by: SkyPilotage on May 18, 2011, 02:23:54 pm
http://www.xtremepapers.me/CIE/International%20A%20And%20AS%20Level/9709%20-%20Mathematics/9709_w10_ms_31.pdfGeneral Point of OP is (1+t)i+(2+t)j+(2-t)k
how to do question 7 part ii?
Dot product of OP.Direction vector of AB = 0 since they are perpendicular.--> you will get t=-1/3
Substitute in GM of OP and there you have it...
Post by: ........ on May 18, 2011, 02:41:09 pm
Should have taken radius as one unit.
yeah so check the x and y values for the lowest point
y/x will give you tan of the argument
Post by: leebux101 on May 18, 2011, 02:48:25 pm
as for the second partso i am kinda right?
listen when there is a equation like y^2/y^3+1 the only way to integrate this is when you differentiate the denominator the exact value should come in the numerator. In this case the differentiation of y^3+1 is 3y^2 but in the numerator it is y^2 so you take the 3 out by multiplying it by 1/3. so now it comes 1/3(y^2/y^3+1)and then you get 1/3ln(y^3+1)
Post by: ........ on May 18, 2011, 02:50:02 pm
so i am kinda right?
Yes sir/Madam.
Post by: leebux101 on May 18, 2011, 02:54:11 pm
Yes sir/Madam.phew ! thank God, hey can u do ON 2003 Q 7 iii ) pleaseeee!!! =D
Post by: ........ on May 18, 2011, 04:03:45 pm
Post by: ........ on May 18, 2011, 04:27:55 pm
Post by: thecandydoll on May 18, 2011, 04:46:56 pm
Post by: SkyPilotage on May 18, 2011, 04:57:16 pm
so how do you get the greatest arg value and least too?The only was is by your graph!!! then You may use pythagrous theorem to find the value for greatest or least value of Z.
Post by: leebux101 on May 18, 2011, 05:02:30 pm
Please correct me if i am wrong.i owe u man!. =D
its madam By the way, ;)
Post by: leebux101 on May 18, 2011, 05:06:58 pm
Please correct me if i am wrong.By the way, how do u know its half?
Post by: ShiningStar on May 18, 2011, 05:35:28 pm
By the way, how do u know its half?as the two triangles are congruents.
Post by: leebux101 on May 18, 2011, 05:39:05 pm
Post by: ........ on May 18, 2011, 06:03:56 pm
Sorry I did a mistake.
Should have taken radius as one unit.
yeah so check the x and y values for the lowest point
y/x will give you tan of the argument
This was the second part.
Post by: leebux101 on May 18, 2011, 06:12:15 pm
Post by: leebux101 on May 18, 2011, 06:13:00 pm
ur a life saver!
Post by: ........ on May 18, 2011, 06:13:35 pm
Post by: SkyPilotage on May 18, 2011, 06:14:24 pm
also can u please do ON2010 Q1 from P32! please! please please!Just square both sides..
ur a life saver!
Post by: SkyPilotage on May 18, 2011, 06:17:10 pm
This was the second part.Actually you solved it correctly the first time...
The least value for modulus of z is where the line from the origin intersets the circle closest to the origin which is somewhere at the centre of the circle.. Check the diagram you drew again.. And isnt the modulus of u is root 2 not 2root2?
Post by: ........ on May 18, 2011, 06:18:31 pm
Just square both sides..and to find the range
just put the values -7 and 1 on a number line
take out any value less than 1 like 0 and put it into (x-1)(x+7)
take out another value between -7 and 1 and put it into ^
and finally a value less than -7 and put it into (x-1)(x+7)
and see if they are less than 0
Post by: leebux101 on May 18, 2011, 06:20:54 pm
and the answer is the same in the marking scheme
secondly >> i did square the sides! =( but i dont get it,
can u give a step by step description please! please! =(
Post by: leebux101 on May 18, 2011, 06:21:56 pm
and to find the rangehey i didnt know we could do that... interesting..
just put the values -7 and 1 on a number line
take out any value less than 1 like 0 and put it into (x-1)(x+7)
take out another value between -7 and 1 and put it into ^
and finally a value less than -7 and put it into (x-1)(x+7)
and see if they are less than 0
Post by: ........ on May 18, 2011, 06:22:45 pm
Actually you solved it correctly the first time...
The least value for modulus of z is where the line from the origin intersets the circle closest to the origin which is somewhere at the centre of the circle.. Check the diagram you drew again.. And isnt the modulus of u is root 2 not 2root2?
dude they're not asking for least value of modulus of z
they're asking for modulus of z when argument of z is least
Post by: SkyPilotage on May 18, 2011, 06:23:49 pm
which question are you on? ahaha
dude they're not asking for least value of modulus of z
they're asking for modulus of z when argument of z is least
Post by: ........ on May 18, 2011, 06:26:11 pm
Hi guys
I really need help with complex numbers and argand diagrams!
Can u plz help me with these qs/
JUNE 10 V1 Q7 Part 3
THANK U
This one
I am getting confused bro..
There were 3 questions that were being asked
Post by: SkyPilotage on May 18, 2011, 06:26:53 pm
and to find the rangeWould be easier for you if you say that outside the range, the sign is the same sign as a(coefficient of x^2) so if its <0 then its -7<x<1 ...
just put the values -7 and 1 on a number line
take out any value less than 1 like 0 and put it into (x-1)(x+7)
take out another value between -7 and 1 and put it into ^
and finally a value less than -7 and put it into (x-1)(x+7)
and see if they are less than 0
But your way is perfectly correct too...
Post by: SkyPilotage on May 18, 2011, 06:27:50 pm
This oneOH oH my bad!! I thought you were on Nov 2006 9 iii) it has also u and z-u=1 srry :D
I am getting confused bro..
There were 3 questions that were being asked
Post by: leebux101 on May 18, 2011, 06:28:45 pm
guys please i need this one ON 2010 P32 Q1
Post by: leebux101 on May 18, 2011, 06:29:38 pm
Post by: ........ on May 18, 2011, 06:31:16 pm
confusion between 3 papers! lol haha! =D
Yeah man.. I got a headache
But yea wait i have the solution somewhere
Post by: leebux101 on May 18, 2011, 06:33:55 pm
so are u gonna do On2010 P32 Q 1 for me please. =(
Post by: SkyPilotage on May 18, 2011, 06:35:35 pm
yup, maths gives u headaches, especially complex numbers..Just square both sides?? Did it work? :-\
so are u gonna do On2010 P32 Q 1 for me please. =(
Post by: ........ on May 18, 2011, 06:36:43 pm
@Skypilotage. Yo. Maybe. Try. I am doing the past papers through this method :/
Post by: SkyPilotage on May 18, 2011, 06:37:32 pm
Forgive me if i made any mistake.It does :D was being little sarcastic :P
@Skypilotage. Yo. Maybe. Try. I am doing the past papers through this method :/
However, there is a greater chance for error by squaring both sides....
Post by: leebux101 on May 18, 2011, 06:40:51 pm
i wrote for 4*9 as 54 ...
aaarrrrghhhh!!!!!!!!!!!!! >:(
*sigh* anyway thanks a billion guys! =D =D
Post by: SkyPilotage on May 18, 2011, 06:42:17 pm
i should hit myself right nowMost of it goes to Dumb Economist :D
i wrote for 4*9 as 54 ...
aaarrrrghhhh!!!!!!!!!!!!! >:(
*sigh* anyway thanks a billion guys! =D =D
Would be alot easier if i have a scanner :P
Post by: leebux101 on May 18, 2011, 06:43:21 pm
Post by: SkyPilotage on May 18, 2011, 06:44:16 pm
no honestly thankyou to both of you! =Dglad to help!!
Kill the exam tmrw :)
Post by: leebux101 on May 18, 2011, 06:45:27 pm
are u ready?
Post by: ........ on May 18, 2011, 06:45:58 pm
And Sky.Dude i learned alot from you
Thanks
And ace the exams guys :D
Post by: SkyPilotage on May 18, 2011, 06:49:55 pm
I am a average student so . I will do everything properly so i wont make mistakes in my exam :/ .You pretty much new it all before I said anything :P
And Sky.Dude i learned alot from you
Thanks
And ace the exams guys :D
IsA alll the best tmrw for all of us!!
Post by: TJ-56 on May 18, 2011, 06:53:23 pm
Thx in advance
Post by: leebux101 on May 18, 2011, 06:54:20 pm
I am a average student so . I will do everything properly so i wont make mistakes in my exam :/ .average? i reallyy doubt that. Mashallah ur smart! =D
And Sky.Dude i learned alot from you
Thanks
And ace the exams guys :D
good luck for tomorow guys, i hope P3 isnt as hard as P1... (i heard P1 was reallyy hard) =D
Post by: SkyPilotage on May 18, 2011, 06:55:25 pm
Can someone solve Oct Nov 2010 P31 Q7 (iii) ?Economist willl upload em for you :D
Thx in advance
Post by: TJ-56 on May 18, 2011, 07:05:03 pm
I did it in a few minutes, so excuse of sloppiness :)
Tell me if anythings wrong too, even though I made sure of them.
Post by: Arthur Bon Zavi on May 18, 2011, 07:29:51 pm
I am a average student so . I will do everything properly so i wont make mistakes in my exam :/ .
And Sky.Dude i learned alot from you
Thanks
And ace the exams guys :D
Don't go on his words. This man can take the crap out of mathematics. God of mathematics.
Post by: yasser37 on May 18, 2011, 08:05:39 pm
for example the one in november 09, question 7 part iv
and other example of how to draw them if possible
please help
thanks
Post by: yasser37 on May 18, 2011, 08:25:14 pm
http://www.xtremepapers.me/CIE/International%20A%20And%20AS%20Level/9709%20-%20Mathematics/9709_w03_qp_3.pdf
(iv) Using your diagram, calculate the least value of || for points on this locus. [2]
I know that it's done in a previous post but I didn't understand it
and november 06
question 9 part 4
please someone explain
thanks
Post by: SkyPilotage on May 18, 2011, 08:45:38 pm
alsoThE least value for modulus of z means find the shortest distance of a line for the origin from the circle....Then you use ur gragh to calculate it...
http://www.xtremepapers.me/CIE/International%20A%20And%20AS%20Level/9709%20-%20Mathematics/9709_w03_qp_3.pdf
(iv) Using your diagram, calculate the least value of || for points on this locus. [2]
I know that it's done in a previous post but I didn't understand it
and november 06
question 9 part 4
please someone explain
thanks
Post by: THEIGBOY on May 18, 2011, 09:21:11 pm
thanks
Post by: Reekx on May 18, 2011, 10:45:32 pm
Post by: yasser37 on May 19, 2011, 06:29:48 am
can anyone help me in Argand diagrams?
for example the one in november 09, question 7 part iv
and other example of how to draw them if possible
please help
thanks
and november 06
question 9 part 4
please someone explain
thanks
can someone help please
Post by: cs on June 14, 2011, 07:52:20 am
Q5iii
and
8iii
Thank you :)
Post by: Shoshou..Mony on June 14, 2011, 08:41:36 am
5)iii) -5 < 4sinx - 3cox < 5
1/4sinx-3cosx+6 has greatest value = 1/(-5+6)
i.e equal 1
8)b)iii) (OA/OB) = OC
Post by: cs on June 15, 2011, 06:33:27 am
@cs :
5)iii) -5 < 4sinx - 3cox < 5
1/4sinx-3cosx+6 has greatest value = 1/(-5+6)
i.e equal 1
8)b)iii) (OA/OB) = OC
Thank you so much for your help :)
+REP
Post by: Shoshou..Mony on June 15, 2011, 09:56:57 am
Thank you so much for your help :)
+REP
If you have any more doubts then post before I forget my P3 syllabus. :P
Thanks for the +rep. =]
Post by: Arissa_04 on July 25, 2011, 02:53:46 pm
Could someone please explain how to do June 2010 Paper 3 variant 1-Question 4 ii)
and also Paper 3 June 2009 Question 10(the whole question).
Thanks in advance.:)
Post by: dlehddud on August 17, 2011, 09:07:29 am
As we know that sin 3x sin x= 0.5 (cos 2x- cos 4x) from the previous part of the question, we can substitute this into our given integral:
integral sign (0.5 cos 2x- 0.5 cos 4x)
if we integrate this, we get
(sin2x)/4 - (sin4x)/8
substituting limits, we get
((sin (2pi/3))/4 - (sin(4pi/3))/8) - ((sin (2pi/6))/4 - (sin(4pi/6))/8)
=> (sqrt 3)/8 - (-(sqrt3)/16) - (sqrt 3)/8 - (-(sqrt 3)/16)
=> 2(sqrt 3)/16
=>(sqrt 3)/8
question 10 2009
i) To find x at M, we must find the maximum of the curve, which is simply the derivative of the curve equated to zero.
y=x^2*(sqrt(1-x^2))
we find the derivative by using the product rule,
dy/dx= 2x*(sqrt(1-x^2))+0.5*(-2x)/(sqrt(1-x^2))*x^2
= 2x(sqrt(1-x^2)) - x^3/(sqrt(1-x^2))
dy/dx=0
2x(1-x^2)/(sqrt(1-x^2)) - x^3/(sqrt(1-x^2))=0 ......... i've equated the denominators if u were unsure
(2x- 3x^3)/(sqrt(1-x^2))=0
Now we only require the numerator to be zero. Therefore,
2x-3x^3=0
x(2-3x^2)=0
Hence, x=0 and 2-3x^2=0
=> 3x^2=2
=> x^2=2/3
=> x= sqrt(2/3)
therefore, the x coordinate of M is sqrt(2/3). It cannot be 0, as M is obviously not centered at x=0
ii)As we are integrating by substituting, we must also change the variables around a bit:
x=sin(theta)
dx/d(theta)=cos(theta)
dx=cos(theta) d(theta)
To do this, we merely substitute in sin (theta) and the new variables to replace dx into the equation given:
Area=integral sign sin^2(theta)*sqrt(1-sin^2(theta))*cos(theta) d(theta)
Area=integral sign sin^2(theta)*sqrt(cos^2(theta))*cos(theta) d(theta)
Area=integral sign sin^2(theta)*cos(theta)*cos(theta) d(theta)
Area=integral sign 0.25*(sin2x)^2 d(theta)
the limits are given by substituting into the equation x=sin(theta)
we are finding theta, so we use theta=sin^-1 x
iii)Do not think that because they give you an equation, that you can use it straight away. We cannot integrate most trigonometric integrals if they are squared, such as this. So we convert it to a form we CAN integrate, namely, a cos4x form:
0.25*(sin^2(2theta))
cos(4theta)= 1-2*(sin^2(2theta))
thus,
2*(sin^2(2theta))=1-cos(4theta)
and so
0.25*(sin^2(2theta))=(1-cos(4theta))/8
Hence we are integrating this, NOT 0.25*(sin^2(2theta))
(1-cos(4theta))/8= 1/8 - (cos(4theta))/8
So, we integrate this:
Area= integral sign 1/8 - cos(4theta))/8 d(theta)
Area= 1/8*(theta) - (sin(4theta))/32 with limits pi/2 and 0
Area= 1/8*(pi/2) - (sin(2pi))/32 - 1/8*(0) + (sin(0))/32
Area=pi/16
Hope this wasn't too confusing =P
It is however up to the standard of working you require at CIE A-level so it shouldn't be tooo hard :)
Ask any questions if you don't get certain parts of it!
Post by: TBT on August 28, 2011, 05:31:09 pm
Post by: Tohru Kyo Sohma on August 28, 2011, 08:04:15 pm
y^2=12x intersects 3y=4x+6 at two points.
bring 3 to the other side and the formula becomes y=4/3x+2
now put this formula in the other formula [4/3x+2]^2=12x
simplify and get x and y values
x1=3 and y1=6
x2=3/4 and y2=3
distance between 2 points is=root of [((3/4)-3)^2+(3-6)^2]
and the answer is 3.75
hope u get it!!
ask me again if u dont understand! ;D
Post by: perish007 on September 09, 2011, 09:37:28 am
Post by: S.M.A.T on September 09, 2011, 08:03:00 pm
please do Q10, iii may/june 2011 32
dy/dx=(2xe^-x)-((x^2)(e^-x)) from part ii
let the tangent be y=mx
At point of intersection,
mx=(x^2)(e^-x)
after solving
m=0 or m=x(e^-x)
m=(dy/dx)
therefore,
x(e^-x)=(2xe^-x)-((x^2)(e^-x))
after solving,x=1
I think there is a short method which I don't now :-\
Post by: curiousguy on September 11, 2011, 11:13:16 am
2. Show that the points (7,12), (-3,-12), and (14,-5) lie on a circle with centre (2,0).
Please any one help me it s really urgent
Post by: Arthur Bon Zavi on September 11, 2011, 12:08:47 pm
Quote
1. Show that the triangle formed by the points (-3,-2), (2,-7) and (-2,5) is isosceles.
Using the distance formula, sq. rt [(x2-x1)2 + (y2-y1)2] and confirm that all the sides are equal in length.
Quote
2. Show that the points (7,12), (-3,-12), and (14,-5) lie on a circle with centre (2,0).
O = (2,0); A = (7,12); B = (-3,-12); C = (14,-5), then:
OA=OB=OC
Use the distance formula again to verify.
Post by: curiousguy on September 19, 2011, 02:59:05 pm
Thank u in advance
Post by: Arthur Bon Zavi on September 19, 2011, 03:26:25 pm
Guys its like in IGCSE i wasnt having this additional maths, pure maths and all when i was in the middle east. But now when i came in my country and started a-levels, people say they have all the basics in p1. So my request to u guys is my questions may be easy in ur sight, but if u guys actually help me out to gain proper concept in p1 i wud really appreciate it.
Thank u in advance
I am doing the first one, with it you can work on the rest.
2x2 + 8x - 10
2(x2 + 4x - 5)
2(x2 + 4x + 4 - 9)
2(x + 2)2 - 18 ----> (ax + b)2 + c
Post by: curiousguy on September 19, 2011, 04:44:28 pm
Post by: Arthur Bon Zavi on September 19, 2011, 04:55:13 pm
Day after 2morrow is my class test on p1 1st chapter coordinates,points and lines. Please post as many typical questions along with answers that i may get in common.wud appreciate any help(__)
Did you understand the previous sum I solved ?
Post by: curiousguy on September 20, 2011, 01:32:42 am
Post by: Arthur Bon Zavi on September 20, 2011, 12:37:57 pm
Yes its easy u know. But try poting some typical questions and answers from the 1st chapter i need to prepare my test 2omorrow
Do you use this book ? (http://www.amazon.com/Pure-Mathematics-International-Cambridge-Examinations/dp/0521530113/ref=sr_1_1?ie=UTF8&qid=1316518729&sr=8-1)
Try solving the miscellaneous exercise. That's all.
Post by: curiousguy on September 20, 2011, 02:43:46 pm
Post by: Arthur Bon Zavi on September 20, 2011, 04:51:24 pm
Yes i have it! If u have done them please scan and post it here. If that is also not possible kindly do 4,5,6,7 sums as soon as possible.
To be honest, I never did any of the exercises from the book seriously. I just read all the chapters and examples. In the end, I solved all the past papers with variants.
Miscellaneous exercise from the first chapter or the fourth ?
Post by: astarmathsandphysics on September 22, 2011, 11:52:52 am
Post by: cs on September 23, 2011, 01:54:13 am
Post by: astarmathsandphysics on September 23, 2011, 11:06:32 am
Post by: cs on September 23, 2011, 03:11:31 pm
the modulus question above - answer attacjed
thanks! we can only use graph method for this question?
Post by: cs on September 23, 2011, 03:14:30 pm
dy/dx=(2xe^-x)-((x^2)(e^-x)) from part ii
let the tangent be y=mx
At point of intersection,
mx=(x^2)(e^-x)
after solving
m=0 or m=x(e^-x)
m=(dy/dx)
therefore,
x(e^-x)=(2xe^-x)-((x^2)(e^-x))
after solving,x=1
I think there is a short method which I don't now :-\
having doubt for this question too, thanks!
having doubt for this question too, thanks!
Post by: astarmathsandphysics on September 23, 2011, 08:32:45 pm
Post by: cs on September 25, 2011, 02:02:58 pm
Post by: astarmathsandphysics on September 26, 2011, 09:25:54 am
Post by: cs on September 27, 2011, 10:45:24 am
By squaring both sides you sometimes get superfluous solutions with don't satisfy the problem. See attachment.
oh thank you, so its best to draw a graph :)
Post by: astarmathsandphysics on September 27, 2011, 10:52:26 am
Post by: curiousguy on October 03, 2011, 12:18:26 pm
Post by: curiousguy on October 03, 2011, 12:19:12 pm
Post by: astarmathsandphysics on October 03, 2011, 02:38:10 pm
Post by: ashwinkandel on October 10, 2011, 04:52:51 pm
(i) Sketch the curve y = 2 sin x for 0 <= x <= 2pi. [1]
(ii) By adding a suitable straight line to your sketch, determine the number of real
roots of the equation
2pi sin x = pi- x.
State the equation of the straight line. [3]
Post by: Arthur Bon Zavi on October 10, 2011, 04:55:31 pm
Q.
(i) Sketch the curve y = 2 sin x for 0 <= x <= 2pi. [1]
(ii) By adding a suitable straight line to your sketch, determine the number of real
roots of the equation
2pi sin x = pi- x.
State the equation of the straight line. [3]
What are the question marks ? Simply give year and the session.
Post by: Sea Lover on October 10, 2011, 09:36:23 pm
The function f is defined by
f(x) = x^2 ? 4x + 7 for x > 2.
The function g is defined by
g(x) = x ? 2 for x > 2.
The function h is such that f = hg and the domain of h is x > 0.
(iii) Obtain an expression for h(x).
In the markscheme the answer is x^2 +3, can someon explain it to me and how to get it?
Post by: astarmathsandphysics on October 10, 2011, 09:51:27 pm
what does the question mark mean
Post by: astarmathsandphysics on October 10, 2011, 09:53:28 pm
Post by: astarmathsandphysics on October 10, 2011, 09:55:39 pm
f=x^2 -4x+7=(x+2)^2 +3
take h=g^2 +3 and g=x-2
Post by: Sea Lover on October 10, 2011, 10:00:26 pm
The function f is defined by
f(x) = x^2 - 4x + 7 for x > 2.
The function g is defined by
g(x) = x - 2 for x > 2.
The function h is such that f = hg and the domain of h is x > 0.
(iii) Obtain an expression for h(x).
Post by: astarmathsandphysics on October 10, 2011, 10:03:36 pm
Post by: Sea Lover on October 10, 2011, 10:05:21 pm
Post by: Sea Lover on October 10, 2011, 10:06:30 pm
Post by: Sea Lover on October 10, 2011, 10:08:59 pm
Post by: astarmathsandphysics on October 10, 2011, 10:09:31 pm
There are any number of possible answers.
Post by: ashwinkandel on October 11, 2011, 05:09:30 am
Post by: ashwinkandel on October 11, 2011, 08:23:58 am
Post by: astarmathsandphysics on October 11, 2011, 09:10:02 am
should that be tan^-1 x
Post by: Arthur Bon Zavi on October 11, 2011, 09:26:52 am
First keep the value of x as 0 in f(x), you will get 3. Then keep a higher value, for example f(0.5 pie), you will get 1. So as you move from 0 to pie, the value of f(x) is decreasing so the range is :
(ii)
f(2/3 pie)
= 3 - 2tan(0.5 X 2/3 pie)
= 3 - 2tan(1/3 pie)
= 3 - 2 (
=
(iii)
You simply need to draw the graph. Make sure you plot the graph only for the range given in the question.
(iv)
f(x) = 3 - 2 tan(
y = 3 - 2tan(
2tan(
tan(
x = 2 X tan-1
y = 2 X tan-1
Post by: astarmathsandphysics on October 11, 2011, 09:35:14 am
Post by: Arthur Bon Zavi on October 11, 2011, 09:41:15 am
thanks
No problem, sire.
Post by: ashwinkandel on October 11, 2011, 10:16:28 am
Post by: Arthur Bon Zavi on October 11, 2011, 10:36:22 am
domain 0<=x<2pi
It's just pi not 2pi.
Post by: astarmathsandphysics on October 11, 2011, 10:37:01 am
Multiple the graph of tanx by a factor 2 in the x direction and 3 in the y direction to get 3 tan(x/2)
Post by: ashwinkandel on October 11, 2011, 11:08:20 am
Post by: astarmathsandphysics on October 11, 2011, 11:35:59 am
Post by: Arthur Bon Zavi on October 11, 2011, 02:24:08 pm
@Arthur Bon Zavi If you can , can you plz show me show the graph looks like?? This is the first time i have come through a question in CIE Maths exam where graph of tanx has been asked to plot... I know how to plot for y=tanx ... but this y=3-tan(x/2) for domain 0<=x<2pi seems complicated..
Have a look at the attachments. The red horizontal line on the positive axis is x = pi.
Post by: ashwinkandel on October 11, 2011, 04:10:06 pm
Post by: Alpha on October 14, 2011, 01:08:05 pm
http://www.xtremepapers.com/CIE/International%20A%20And%20AS%20Level/9709%20-%20Mathematics/9709_s11_qp_31.pdf
No 3, on vectors.
Points A and B have coordinates (-1, 2, 5) and (2, -2, 11) respectively. The plane p passes through
B and is perpendicular to AB.
(i) Find an equation of p, giving your answer in the form ax + by + cz = d. [3]
(ii) Find the acute angle between p and the y-axis. [4]
How would you proceed with the second part?
I got the answer, but I checked the markscheme, and don't understand the procedures there.
Post by: astarmathsandphysics on October 14, 2011, 02:54:06 pm
Post by: Alpha on October 14, 2011, 03:01:27 pm
The question is the same as finding the angle between (a,b,c) and (0,1,0)
What will you take as (a,b,c)?
I took two points on the y-axis.
I cannot understand why Cambridge has taken vectors OB and OA. Shed some light please? :-\
Post by: Deadly_king on October 16, 2011, 06:50:42 am
What will you take as (a,b,c)?
I took two points on the y-axis.
I cannot understand why Cambridge has taken vectors OB and OA. Shed some light please? :-\
I don't think I'll be of much help since I've forgotten most of my Maths by now :-\
Anyway I'll try as much as I remember though am not so sure.....
To find the angle between a line and a plane we first have to find the angle between the normal of the plane and the direction vector of the line.
We need to find the direction vector of the normal of the plane by using OA and OB -----> OB - OA
Then you use the formula to find angle ;)
Post by: astarmathsandphysics on October 16, 2011, 10:35:37 am
a=-3
b=4
c=4
Post by: Alpha on October 16, 2011, 12:24:22 pm
I don't think I'll be of much help since I've forgotten most of my Maths by now :-\
Anyway I'll try as much as I remember though am not so sure.....
To find the angle between a line and a plane we first have to find the angle between the normal of the plane and the direction vector of the line.
We need to find the direction vector of the normal of the plane by using OA and OB -----> OB - OA
Then you use the formula to find angle ;)
Normal and direction vectors, sin formula, yes, I did.:)
But in the ER, OA is multiplied by OB. Gives you the answer, but I don't understand why. :-\
No worries. Thank you though. :)
Tabke abc as (-1, 2, 5) - (2, -2, 11)=(-1-2,2--2,,5-1)=(-3,4,4)
a=-3
b=4
c=4
(-1-2,2--2,,5-11)
c= -6. ;)
I get what you mean. I did that too.
http://www.xtremepapers.com/CIE/International%20A%20And%20AS%20Level/9709%20-%20Mathematics/9709_s11_er.pdf
The examiners' report, on pg 22. It says by multiplying OA and OB. This is what I don't get. :-\
Even YA's best contributors do not. :(
Appreciate your help. Thank you Sir. :)
Post by: astarmathsandphysics on October 16, 2011, 09:34:08 pm
Post by: Alpha on October 17, 2011, 06:08:58 pm
Do thety mean the cross product?
No, scalar product is the dot product. They use |OA|.|OB| cos Angle= OA.OB
I checked this method. It does give you the answer. But I still don't find any logic. :-\
Post by: astarmathsandphysics on October 17, 2011, 10:10:16 pm
Post by: Alpha on October 18, 2011, 06:36:33 am
I recognize this question from a p4 paper I did today. Attached.
I did that too, it's the first part. :(
I asked it on YA, you might wish to check it out:
http://answers.yahoo.com/question/index;_ylt=Ag99ppCNFePLsrjxoLIt3mTsy6IX;_ylv=3?qid=20111015052949AAvMuAv
There too, my question remain unanswered though. Perhaps the ER's could be clearer. :-\
Post by: astarmathsandphysics on October 18, 2011, 08:05:02 am
Post by: Alpha on October 18, 2011, 08:07:52 am
Give me time to get up
Sure. :)
(http://www.ilovecoffeebook.com/images/clip_image085.jpg)
;D
Post by: astarmathsandphysics on October 18, 2011, 09:30:51 am
Post by: Alpha on October 18, 2011, 09:37:55 am
HERE
30.8 is the answer. You are using the same method as I did. Not the one used by Cambridge. It's okay. Most important is that I get the answer. :)
Thank you for your time +rep. :)
Post by: cs on October 26, 2011, 08:11:38 am
Paper 3 :)
Post by: cs on October 26, 2011, 12:03:48 pm
q7 part ii
which shaded area is correct, thanks
Post by: astarmathsandphysics on October 26, 2011, 09:40:33 pm
Post by: astarmathsandphysics on October 26, 2011, 09:42:27 pm
Post by: cs on October 27, 2011, 02:34:43 am
Post by: preity on October 27, 2011, 06:07:24 am
Post by: astarmathsandphysics on October 27, 2011, 11:04:49 am
Post by: curiousguy on October 30, 2011, 03:21:02 pm
Post by: astarmathsandphysics on October 30, 2011, 04:51:54 pm
http://www.astarmathsandphysics.com/a_level_maths_notes/a_level_maths_notes_c1_menu.html
Post by: Most UniQue™ on October 30, 2011, 08:28:50 pm
1.A funnel has a circular top of diameter 20cm and a height of 30cm. When the depth of liquid in the funnel is 12 cm, the liquid is dripping from the funnel at a rate of 0.2 cm^3/s. At what rate is the depth of the liquid in the funnel decreasing at this instant?
2. The sum of the first twenty terms of an arithmetic progression is 50, and the sum of the next twenty terms is -50. Find the sum of the first eight terms.
3. Determine the coefficient of p^4q^7 in the expansion of (2p-q)(p+q)^10
Thanks!:D
Post by: astarmathsandphysics on October 30, 2011, 09:50:36 pm
Post by: astarmathsandphysics on October 30, 2011, 10:04:38 pm
Post by: curiousguy on October 31, 2011, 09:04:20 am
Post by: Most UniQue™ on October 31, 2011, 11:25:56 am
here
Thanks a lot sir!
Post by: urni.nath on October 31, 2011, 01:41:08 pm
At a speed of S km per hour a car will travel y kilometers on each litre of petrol,where
y= 5 + 1/5 S - 1/800 S^2
Calculate the speed at which the car should be driven for maximum economy.
Any help is appreciated!
Thanks
Post by: astarmathsandphysics on October 31, 2011, 02:18:21 pm
1/5-s/400=0 so s=80
Post by: urni.nath on November 01, 2011, 12:02:20 pm
Post by: astarmathsandphysics on November 01, 2011, 12:33:43 pm
Post by: urni.nath on November 01, 2011, 01:15:19 pm
1) a rectangular sheet of metal measures 50cm by 40cm. Equal sides of side x cm are cut from each corner and discarded. The sheet is then folded up to make a tray of depth x cm. What is the domain of possible values of x ? Find the value of x which maximizes the capacity of the tray.
2)An open rectangular box is to be made with a square base, and its capacity is to be 4000cm^3 . Find the length of the side of the base when the amount of material used to make the box is as small as possible. (ignore "flaps")
Thank you so much for your help!
Post by: astarmathsandphysics on November 01, 2011, 01:26:54 pm
Post by: astarmathsandphysics on November 01, 2011, 03:31:07 pm
Post by: urni.nath on November 01, 2011, 03:41:59 pm
Post by: astarmathsandphysics on November 01, 2011, 03:53:11 pm
Post by: Most UniQue™ on November 14, 2011, 04:41:16 pm
Post by: Malak on November 14, 2011, 04:44:05 pm
Why has this thread been combined with P2 and P3?I have combined them because there were too many sticky threads and I thought it is confusing and besides these aren't threads which are always crowded so its better to have one thread for a subject rather then having 3 threads for each unit of a subject.
Post by: Most UniQue™ on November 14, 2011, 05:19:01 pm
I have combined them because there were too many sticky threads and I thought it is confusing and besides these aren't threads which are always crowded so its better to have one thread for a subject rather then having 3 threads for each unit of a subject.
It makes it confusing whether the question belongs to p1 p2 or p3. Ex. for those who are studying p1 and want to answer the doubts for their own revision also.
Post by: Malak on November 14, 2011, 05:35:31 pm
It makes it confusing whether the question belongs to p1 p2 or p3. Ex. for those who are studying p1 and want to answer the doubts for their own revision also.Everyone can specify which unit it is and people who have done unit 2 or 3, obv., would know that its from unit 1 ::)
Post by: silvercameron on November 24, 2011, 09:56:57 am
Write this in the form (x+p)2+q where p and q are constants to be found in terms of k.
Post by: astarmathsandphysics on November 24, 2011, 10:12:37 am
Post by: silvercameron on November 24, 2011, 02:39:12 pm
Post by: astarmathsandphysics on November 24, 2011, 02:47:04 pm
Post by: urni.nath on November 29, 2011, 03:52:59 pm
Prove : 10sin^2(x)-5cos^2(x)+2=4sin(x)
Post by: donhassan on November 29, 2011, 07:32:51 pm
could you pls solve this :10sin^2(x)-5cos^2(x)+2=4sin(x)
10sin^2(x)-5cos^2(x)+2=4sin(x)
10sin^2(x)-5(1-sin^2(x))+2=4sin(x)
10sin^2(x)-5+5sin^2(x)+2=4sin(x)
15sin^2(x)-4sin(x)-3=0
solve this equation as quadratic equation
u will get
sin(x)=0.6 and sin(x)=-1/3 => 19.47
(x)=36.9 degree =180+19.47 = 199.5
(x)=143.1 degree =360-19.47 =340.5
hope u understood this piece of cake
Post by: urni.nath on November 30, 2011, 04:36:20 am
Post by: Most UniQue™ on December 08, 2011, 07:44:57 am
constant a.
2. Check the attachment
Please help. Got exam on Saturday. Thanks
Post by: astarmathsandphysics on December 08, 2011, 10:54:43 am
Post by: Alpha on December 08, 2011, 12:02:48 pm
Post by: astarmathsandphysics on December 08, 2011, 12:15:26 pm
I am making videos there
Post by: Most UniQue™ on December 08, 2011, 06:59:20 pm
Another problem:
Post by: astarmathsandphysics on December 08, 2011, 07:14:29 pm
Post by: Most UniQue™ on December 10, 2011, 07:59:18 pm
Done step by tiny step in attachment. Am in a slow mood.
Thank you!
Post by: silvercameron on December 27, 2011, 05:05:45 pm
A rectangular box is to be made with a square base, and its capacity is to be 4000 cm^3. Find the length of the side of the base when the amount of material used to make the box is as small as possible.
Post by: astarmathsandphysics on December 28, 2011, 09:44:32 am
Post by: silvercameron on December 28, 2011, 05:14:21 pm
I got the correct answer.
Post by: astarmathsandphysics on December 28, 2011, 10:14:19 pm
Post by: White Eagle on January 25, 2012, 01:41:20 pm
I AM stuck with
Find vector equestion of cartesian equestion
x+3y=7
Post by: astarmathsandphysics on February 02, 2012, 09:33:07 am
http://astarmathsandphysics.com/a_level_maths_notes/C4/a_level_maths_notes_c4_changing_between_cartesian_and_vector_forms_of_equations_of_lines.html
Post by: donhassan on March 11, 2012, 09:56:31 pm
The planes p1 and p2 have equations X+Y-Z=0 AND 2X-4Y+Z+12=0 respectively.
1.) find the numbers p,q,r such that vector (p,q,r) is parallel to both p1 and p2
2.)hence find the equation of the plane through the point (3,8,2) which is perpendicular to both p1 and p2
answer:1.)p=1 q=1 r=2
2.)X+y+2z=15
guys pls explain me in lucid way
thanks
donhassan
Post by: astarmathsandphysics on March 12, 2012, 07:36:24 pm
find n_1 x n_2 =(-1,-1,-6)
2.-3x-3y-6z=d
Put x=3,y=8, z=2 to find d=-45 so -3x-3y-6z=-45
and dividing by -3 gives x+y+2z=15
Post by: donhassan on March 13, 2012, 07:09:33 pm
1. The normals are n_1=(1,1,-1) and n_2=(2,-4,1)
find n_1 x n_2 =(-1,-1,-6)
2.-3x-3y-6z=d
Put x=3,y=8, z=2 to find d=-45 so -3x-3y-6z=-45
and dividing by -3 gives x+y+2z=15
sir i didn't understood first part :(
Post by: astarmathsandphysics on March 13, 2012, 09:31:15 pm
A plane with equation 3x+4y+6z=2423 has normal (3,4,6)
Post by: HUSH1994 on April 06, 2012, 08:48:44 am
Post by: sabbath_92 on April 07, 2012, 06:19:34 am
P3 MJ 2008 Q3 part 1 and Q6 please :-[
a=rsin x
and the next is just implicit differentiation, equate dy/dx = 0
Post by: LekhB on April 11, 2012, 07:35:28 pm
(http://i.imgur.com/CG0xc.jpg?1)
(http://i.imgur.com/6X98f.jpg?1)
Post by: haris94 on April 11, 2012, 08:27:18 pm
Stuck with those two:
(http://i.imgur.com/CG0xc.jpg?1)
(http://i.imgur.com/6X98f.jpg?1)
Looks like your in luck. I just finished doing the first question and then saw your post....so here you go :)
Hope you're able to read my handwriting :P
First question continues on the second attachment
Post by: LekhB on August 31, 2012, 06:27:31 pm
(http://i.imgur.com/xx9SH.jpg) (http://imgur.com/xx9SH)
(http://i.imgur.com/Z3VP3.jpg) (http://imgur.com/Z3VP3)
(http://i.imgur.com/Gd0qk.jpg) (http://imgur.com/Gd0qk)
Post by: astarmathsandphysics on September 01, 2012, 07:32:26 am
Post by: astarmathsandphysics on September 01, 2012, 08:16:11 am
Post by: astarmathsandphysics on September 01, 2012, 08:21:29 am
Post by: astarmathsandphysics on September 01, 2012, 08:31:36 am
Post by: sameer210394 on October 03, 2012, 07:46:22 am
Post by: astarmathsandphysics on October 03, 2012, 09:59:02 am
m1=tan theta 1 and m2=tan theta 2
Two lines are perprendicular is theta 2 = theta 1 +90
tan theta 2 = tan (theta 1 +90) =- 1/(tan theta 1)
so m1 m2 =tan theta 1 tan theta 2 =-1
Post by: Mrsmart1017 on October 05, 2012, 05:51:15 am
Post by: sameer210394 on October 05, 2012, 06:47:54 am
Post by: sameer210394 on October 05, 2012, 06:55:17 am
(cosx + sinx) / (cosx - sinx) = root3
divide numerator & denominator by cosx
(1 + tanx) / (1 - tanx) = root3
tanx = (root3 -1) / (root3 +1)
x = 15, 195
Post by: astarmathsandphysics on October 05, 2012, 10:23:04 am
Use the inequality sinx < x < tanx for a suitable value of x to show that Pi lies between 3 and 2(root3)Take x=pi/6
sin (pi/6)<pi/6<tan (pi/6)
1/2<pi/6<sqrt(3)
Multiply by 6
3<pi<2 sqrt(3)
Post by: Mrsmart1017 on October 05, 2012, 10:34:42 am
Post by: acash09 on November 16, 2012, 07:21:42 pm
Post by: astarmathsandphysics on November 16, 2012, 11:38:58 pm
Post by: whale95 on November 17, 2012, 08:33:20 pm
Post by: astarmathsandphysics on November 18, 2012, 12:01:16 am
Post by: astarmathsandphysics on November 18, 2012, 10:22:53 am
Post by: whale95 on November 18, 2012, 03:10:08 pm
Yes please and thank you so much! :)
Post by: astarmathsandphysics on November 18, 2012, 06:55:03 pm
Post by: silvercameron on December 07, 2012, 06:50:13 am
show that the integration of (ex - e-x) dx with limits x=0 and x=1 is equal to (e-1)2/e
Post by: astarmathsandphysics on December 07, 2012, 09:15:03 pm
Post by: silvercameron on December 09, 2012, 12:34:22 pm
here silvercameronthankyou sir! Ur awesome. :D
Post by: astarmathsandphysics on December 10, 2012, 11:58:51 am
Post by: silvercameron on January 19, 2013, 04:01:53 am
how to integrate : 1/(u+u^2) du
Post by: mathsphychem on May 16, 2013, 06:50:37 am
question 2?
Post by: astarmathsandphysics on November 06, 2013, 10:22:38 am
a P3 question
how to integrate : 1/(u+u^2) du
Partial fractions