Let `a , b , c` be nonzero real numbers such that `int_0^1(1+cos^8x)(a x^2+b x+c)dx` `=int_0^2(1+cos^8x)(a x^2+b x+c)dx=0` Then show that the equation `a x^2+b x+c=0` will have one root between 0 and 1 and other root between 1 and 2.

Options

A. no root in (0, 2)
B. at least one root in (1, 2)
C. two roots in (0, 2)
D. two imaginary roots

Answer

B

Explanation

Consider the function `phi (x) = underset(0)overset(x)int(1+cos^(8)x)(ax^(2)+bx+c)dx`
Clearly, `phi (x)` being an integral function, is continuous and differentiable on [1, 2] and (1, 2) respectively such that `phi(1) = phi(2)` (given). Therefore, by Rolle's Theorem there exists at least one point k(1, 2) such that `phi'(k) = 0`
We have, `phi'(x) = (1+cos^(8)x)(ax^(2) + bx + c)`
`therefore" "phi'(k) = 0`
`rArr" "(1+cos^(8)k)(ak^(2) + bk +c) = 0`.
`rArr" "ak^(2) + bk + c = 0" "[1 + cos^(8) k ne 0]`
`rArr" "k` is a root of `ax^(2) + bx + c = 0`
`Also, `k in (1, 2)`