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 -x
2) / 2-x
4k -2kx = 36 - 2x + x
2x
2 - 2x +2kx + 36-4k=0
x
2 +(-2+2k)x + (36- 4k )
now we have a quadratic equation........ now as we know when b
2 - 4ac< 0 then no roots
b
2 - 4ac=0 then u get two equal roots
b
2 - 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