Mathematics: Post your doubts here!

[RoOkaYya G]

The latter please.It gets too confusing to work out equations here,especially due the powers and such

open this link
http://www.twiddla.com/1787175

 

[RoOkaYya G]

Right here Thanks

-___________- theres a lot to write

*sigh*
(i)
use product rule n differenciate y=xcos0.5x
dy/dx=(cos*0.5x)-(0.5*x*sin0.5x)
d2y/dx2= -0.5*sin0.5x - [ dy/dx of the part 0.5*x*sin0.5x using product rule again)

d2y/dx2 = (-0.5sin*0.5x) - (0.5sin*0.5x) - (0.25xcos0.5x)
= -sin0.5x - 0.25 xcos0.5x


replace the d2y/dx2 in the equation given...it will be = 0

(ii)
integrate the f(x) using integration by parts (LATE)
use limits on x axis which r x=0 and x=pi
replace n solve

 

[Emma23]

Hi, can someone help me with question 6..
http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_w11_qp_33.pdf

 

[Biocalife]

JazakAllah Ruqayya, tell me more about the LATE stuff ?

 

[abdul moeed rana]

need some hardcore p1 question.for practise.am appearing in oct/nov this session on 8th oct.so plzzzz help me!

 

[Thought blocker]

Hi, can someone help me with question 6..
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w11_qp_33.pdf

Whole 6?

 

[RoOkaYya G]

JazakAllah Ruqayya, tell me more about the LATE stuff ?

LATE means Logarithmic,Arithmetic,Trigonometric,Exponential
the u for the int. by parts will be arithmetic in ths case cos theres x and cos. so according to LATE. A comes before T. so the u will be x
and the dv/dx = cos 0.5x
then u carry out with the integrations

u=x dv/dx= cos*0.5x
du/dx= 1 v= -2sin*0.5x

use uv- (int) v* du/dx and replace

 

[Evil Philanthropist]

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_w13_qp_13.pdf

Hey! Can someone help me with question 8(iii)? I want to know the whole solving process.
And even question 9. How do we get + and - k?

 

[Thought blocker]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w13_qp_13.pdf

Hey! Can someone help me with question 8(iii)? I want to know the whole solving process.
And even question 9. How do we get + and - k?

I dont have much time
8iii)
use this expnsion (x + y)5 = x^5 +5x^4 y + 10x^3 y^2 +10x^2 y^3 +5xy^4 + y^5
PT : http://www.mathsisfun.com/pascals-triangle.html
9)
t has a stationary point where dy/dx = 0 so first find dy/dx
Then second derivative for nature.
And that + - k thing is something like this :¬
-k^2 (x + 2 )^(-2) + 1 = 0
-k^2 = -1 / (x + 2)^(-2)
-k^2 = -1 /[ 1 / (x + 2)^2 ]
-k^2 = - (x + 2)^2
- and - gets cancle, hence say positive,
k = Root(x + 2)^2
So square cancels out and coz we took square, we dont know if k is + or -
Hence
+/-k = x + 2

 

[RoOkaYya G]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w13_qp_13.pdf

Hey! Can someone help me with question 8(iii)? I want to know the whole solving process.
And even question 9. How do we get + and - k?

replace the bracket (x+3x^2) by another alphabet uhm lets say y

[1+]^5
using binomial expansion
=y^5 + 5y^4 + 10y^3 + 10y^2 + 5y + 1

now replace the y with the initial bracket i.e (x+3x^2)

use the answers in part (i) and part (ii) to replce the power 5 and power 4
use simple cubic n quadratic expansion for y^3 and y^2

hope u understood

for Qu.9

differenciate y
dy/dx :-

y= k^2 (x+2) ^-1 +x
dy/dx= -k^2 (x+2) ^-2 + 1
at stationary pnt, dy/dx = 0

cross mutiply
k^2 = (x+2)^2
(x+2)^2 = k^2
(x+2)= +or - k (since square root applied)
x=k-2 or x= -k-2

differenciate dy/dx again to obtain d^2y/dx^2

if d^2y/dx^2 >0 then minimum if <0 then maximum

 

[RoOkaYya G]

I dont have much time
8iii)
use this expnsion (x + y)5 = x^5 +5x^4 y + 10x^3 y^2 +10x^2 y^3 +5xy^4 + y^5
PT : http://www.mathsisfun.com/pascals-triangle.html
9)
t has a stationary point where dy/dx = 0 so first find dy/dx
Then second derivative for nature.
And that + - k thing is something like this :¬
-k^2 (x + 2 )^(-2) + 1 = 0
-k^2 = -1 / (x + 2)^(-2)
-k^2 = -1 /[ 1 / (x + 2)^2 ]
-k^2 = - (x + 2)^2
- and - gets cancle, hence say positive,
k = Root(x + 2)^2
So square cancels out and coz we took square, we dont know if k is + or -
Hence
+/-k = x + 2

for the Qu 8 u shld substitute the whole bracket it'll be easier

 

[Thought blocker]

replace the bracket (x+3x^2) by another alphabet uhm lets say y

[1+]^5
using binomial expansion
=y^5 + 5y^4 + 10y^3 + 10y^2 + 5y + 1

now replace the y with the initial bracket i.e (x+3x^2)

use the answers in part (i) and part (ii) to replce the power 5 and power 4
use simple cubic n quadratic expansion for y^3 and y^2

hope u understood

for Qu.9

differenciate y
dy/dx :-

y= k^2 (x+2) ^-1 +x
dy/dx= -k^2 (x+2) ^-2 + 1
at stationary pnt, dy/dx = 0

cross mutiply
k^2 = (x+2)^2
(x+2)^2 = k^2
(x+2)= +or - k (since square root applied)
x=k-2 or x= -k-2

differenciate dy/dx again to obtain d^2y/dx^2

if d^2y/dx^2 >0 then minimum if <0 then maximum

Yes, like that.

 

[RoOkaYya G]

Yes, like that.

yh
if u do (x+y)^5 itll be hard...itll be wiser to replace the whole bracket so tht itll be easier n u bee able to use answer from (i) and (ii)
its 4 marks u need not hack ur brain for plan action thts y theres Qu (i) (ii) of quest 8

 

[Evil Philanthropist]

Thanks alott! :')
I got a silly question, but, when we are applying square root to both sides in question 9, why is only K + and - ? Why not + and - (x+2) too?

 

[RoOkaYya G]

Thanks alott! :')
I got a silly question, but, when we are applying square root to both sides in question 9, why is only K + and - ? Why not + and - (x+2) too?

it cancels
n anyway only on one side u can put it. either on (x+2) or k side.
its like the constant c added to one side only (if theres 2 side) during integration
got it?

 

[Evil Philanthropist]

OHH-I just realized. OH MY GOD. I got it. Thanks alot!

 

[RoOkaYya G]

OHH-I just realized. OH MY GOD. I got it. Thanks alot!

welcome
any other doubt?

 

[Evil Philanthropist]

Actually, yes, but they are from the recent 2014 May June Paper.
http://justpastpapers.com/wp-content/uploads/2014/05/IMG_3474.jpg
(The papers aren't exactly out yet, so this is a best version I could find^)

Question 3 and 4 (i) part.
For Q3, I didn't get the reflex part at all. I just know how to do it with acute, obtuse and right angle. If you can help me with the concept, I will so grateful!

 

[RoOkaYya G]

Actually, yes, but they are from the recent 2014 May June Paper.
http://justpastpapers.com/wp-content/uploads/2014/05/IMG_3474.jpg
(The papers aren't exactly out yet, so this is a best version I could find^)

Question 3 and 4 (i) part.
For Q3, I didn't get the reflex part at all. I just know how to do it with acute, obtuse and right angle. If you can help me with the concept, I will so grateful!

ive already done it all
Wait ill work it out again

gceguide.com/search/label/AS %26 A Level %3A Mathematics %289709%29


Qu 3
(a) IM USING X INSTEAD OF THETA. EASIER TO TYPE xD
using identities
sin^2 x + cos^2 x = 1 ------- equ. 1
cos x=k
=> cos^2 x= k^2 --------equ. 2

replace 2 in 1
sin^2 x + k^2 =1
make sin subject of formula
sinx= square root (1-k^2)

for reflex angle; sinx is negative here (2nd quadrant)
therefore sinx = - square root (1-k^2)


(b) (i)
tanx= sinx/cosx
=[ -square root (1-k^2) ] / k

(ii) i dunno how to explain -___- ask someone else abt this bit :S
its coz of the quadrant its negative in the 2nd one (all sin tan cos)


Qu 4 (i)
area of triangle AOB = areas of AXB (semicircle)
0.5*r^2* sin theta = [0.5 * r^2 * theta ] - [0.5*r^2*sin theta]

cancel 0.5* r^2 through out i.e divide by 0.5* r^2 through out

sin theta= theta - sin theta
2sin theta=theta
==> p=2

 

[Evil Philanthropist]

ive already done it all
Wait ill work it out again

gceguide.com/search/label/AS %26 A Level %3A Mathematics %289709%29


Qu 3
(a) IM USING X INSTEAD OF THETA. EASIER TO TYPE xD
using identities
sin^2 x + cos^2 x = 1 ------- equ. 1
cos x=k
=> cos^2 x= k^2 --------equ. 2

replace 2 in 1
sin^2 x + k^2 =1
make sin subject of formula
sinx= square root (1-k^2)

for reflex angle; sinx is negative here (2nd quadrant)
therefore sinx = - square root (1-k^2)


(b) (i)
tanx= sinx/cosx
=[ -square root (1-k^2) ] / k

(ii) i dunno how to explain -___- ask someone else abt this bit :S
its coz of the quadrant its negative in the 2nd one (all sin tan cos)


Qu 4 (i)
area of triangle AOB = areas of AXB (semicircle)
0.5*r^2* sin theta = [0.5 * r^2 * theta ] - [0.5*r^2*sin theta]

cancel 0.5* r^2 through out i.e divide by 0.5* r^2 through out

sin theta= theta - sin theta
2sin theta=theta
==> p=2

I got Q3 and most of Q4, thanks alot! <3
In question 4 tho, how did you get the area of AOB? Where did r^2 come from?