IGCSE/GCSE/O & A Level/IB/University Student Forum

Qualification => Subject Doubts => GCE AS & A2 Level => Math => Topic started by: superduper2009 on May 11, 2010, 09:38:23 pm

Title: URGENT!!!!!! math paper 1 cie 2008 may june question 8
Post by: superduper2009 on May 11, 2010, 09:38:23 pm
can sum1 please do and explain question 8 in may/june 2008 paper 1
i'll b really gr8full ...
thanks alot
superduper2009
Title: Re: URGENT!!!!!! math paper 1 cie 2008 may june question 8
Post by: astarmathsandphysics on May 12, 2010, 12:23:18 am
1st thing when i get up
Title: Re: URGENT!!!!!! math paper 1 cie 2008 may june question 8
Post by: cashem'up on May 12, 2010, 05:30:16 am
Quote from: superduper2009 on May 11, 2010, 09:38:23 pm
can sum1 please do and explain question 8 in may/june 2008 paper 1
i'll b really gr8full ...
thanks alot
superduper2009
step 1  find fg(x)
2. f(9/2-x)
3  4(9/2-x) - 2k
 so u found the expansion now simplify it
 36/2-x -2k
ok now equate it to x as Q says

36/2-x -2k = x
2k= 36/2-x -x
2k = 36 - (2x -x2) / 2-x
4k -2kx = 36 - 2x + x2
x2 - 2x +2kx + 36-4k=0
x2 +(-2+2k)x + (36- 4k )

now we have a quadratic equation........ now as we know when b2 - 4ac< 0 then no roots
b2 - 4ac=0 then u get two equal roots
b2 - 4ac>0 and two distinct roots

so  here
(2k-k)2 - (4x1x(36-4k) = 0
u will then get a quadratic solve it and u will get the values

k=5 and -7

Thanks  ;D ;D ;D ;D