Hi, can someone help me with question 6..
http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_w11_qp_33.pdf
http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_w11_qp_33.pdf
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Mathematics: Post your doubts here!
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need some hardcore p1 question.for practise.am appearing in oct/nov this session on 8th oct.so plzzzz help me!
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Whole 6?
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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
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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?
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?
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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
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replace the bracket (x+3x^2) by another alphabet uhm lets say y
[1+
]^5using 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
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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
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Yes, like that.replace the bracket (x+3x^2) by another alphabet uhm lets say y
[1+
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
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yhYes, like that.
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
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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?
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?
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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?
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OHH-I just realized. OH MY GOD. I got it. Thanks alot!
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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!
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!
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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
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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?
Emergent, please help!
I have no doubt in solving this problem as I have nearly finished the further. Yet I need to make sure I score all the method mark given. for ii), please take a look at the mark scheme as below.
Can anybody tell how to use the correct trig formulae to obtain a 3-term quadratic in tan sita, sin 2sita,cos 2sita, or tan2sita?
Very very time-limited, please reply quickly, thanks!!!!
I have no doubt in solving this problem as I have nearly finished the further. Yet I need to make sure I score all the method mark given. for ii), please take a look at the mark scheme as below.
Can anybody tell how to use the correct trig formulae to obtain a 3-term quadratic in tan sita, sin 2sita,cos 2sita, or tan2sita?
Very very time-limited, please reply quickly, thanks!!!!
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using formula 0.5 * ab * sin cI 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?
use the radius r as the side of the triangle. it is isoceles.