Mathematics: Post your doubts here!

Mustafa Malik

Mustafa Malik

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Hello everyone
I am new to this part of Xtremepapers cuz i hv just finished my O levels n in these vacations i wanna study for A levels .Although i created a thread but there was no reply. Actually i need the name of the best books for Math A levels that most renown skools follow.
 
Rutzaba

Rutzaba

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Fatima18 said:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w11_qp_12.pdf
Hi guys....For number 4 ii) I em getting 8 for K..However the marking says its 8.5...how to solve?thanx
Click to expand...

remember that wen evr it says that u need to calculate the tangent the discriminant wud be equals to 0
b^2 -4ac =0
now we make y the subject of formula and substitute it into eq of curve
2y+ x =k
2y= k-x
y= (k-x)/2 substitute the value into curve
y ^2+2x =13
((k-x)/2 )^ 2 +2x= 13
((k^2 -2kx +x^2)/4)) +2x =13
(k^2 -2kx +x^2) +8x =52
you get
x^2 + (8-2k) x + (k^2-52)
a=1 b= 8-2k c= k^2 -52
b^2 -4ac
(8-2k)^2 - 4(1)(k^2 -52) =0
64- 32k +4k^2 -4k^2 +208 =0
-32k + 272=0
32k= 272
k= 8.5
 
F

Fatima18

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Rutzaba said:
remember that wen evr it says that u need to calculate the tangent the discriminant wud be equals to 0
b^2 -4ac =0
now we make y the subject of formula and substitute it into eq of curve
2y+ x =k
2y= k-x
y= (k-x)/2 substitute the value into curve
y ^2+2x =13
((k-x)/2 )^ 2 +2x= 13
((k^2 -2kx +x^2)/4)) +2x =13
(k^2 -2kx +x^2) +8x =52
you get
x^2 + (8-2k) x + (k^2-52)
a=1 b= 8-2k c= k^2 -52
b^2 -4ac
(8-2k)^2 - 4(1)(k^2 -52) =0
64- 32k +4k^2 -4k^2 +208 =0
-32k + 272=0
32k= 272
k= 8.5
Click to expand...

Thank you so much ! :D
 
waleedsmz

waleedsmz

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Hemdon said:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s11_qp_12.pdf
Click to expand...
Hemdon said:
Question No 9.1st part.....i cant solve it.....and the mark scheme says : one limit is -1 and other is 3...
Can some one explain please...
Click to expand...

The cosine function can take up any value from -1 to 1. But since it is squared ( k = 2 ) , then it can take any value from 0 to 1. Therefore, to find the first limit, you substitute with 0 , it gives you 3. The other limit, you substitute with 1, it gives you 3-4 =-1.
 
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