Question

The first two terms in the expansion of (1+ax/2)^10 + (1+bx)^10 in ascending powers of x., are 2 and 90x^2
Given that a is less than b find the values of the constants a and b . (11 marks)

Thanks :)

Answer 1

(1+ax/2)^10 + (1+bx)^10
= (1 + 10(ax/2) + (10)(9)/2 (ax/2)^2 + ... ) + (1 + 10bx + (10)(9)/2 (bx)^2 + ...)

= 2 + (5ax + 10bx) + (45/4) a^2 x^2 + 45b x^2 + ..

= 2 + (5a+10b)x + (45/2 a^2 + 45b^2)x^2 + ...

first term matches your given, but
the 2nd only contains first degree x , not x^2

something is awry here!

Answer 2

taking a guess, we have

45/2 a^2 + 45b^2 = 90
a^2 + 2b^2 = 4

Can't think of any nonzero integers that will do the job.

Answer 3

Unless the x term had a coeffecient of zero, then of course you wouldn't count it,

so 5a + 10b = 0 ---> a = -2b
and
45/2 a^2 + 45b^2 = 90
(45/2)(4b^2 + 45b^2 = 90
90b^2 + 45b^2 = 90
135b^2 = 90
b^2 = 90/135 = 2/3

b = ±V2/√3

if b = +√2/√2 then a = -2√2/√3
if b = -√2/√3 , then a = 2√2/√3