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. one root between 0 and 1 and other root between 1 and 2
B. both roots between 0 and 1
C. both the roots between 1 and 2
D. none of these

Answer

A

Explanation

Condider the function `phi(x)` given by
`phi(x)=underset(0)overset(x)int(1+cos^(8)t)(at^(2)+bt+c)dt`
`rArr phi'(x)=(1+cos^(8)x)(ax^(2)+bx+c) " "....(i)`
We observe that
`phi (0)=0`
`phi(1)=underset(0)overset(1)int(1+cos^(8)t)(at^(2)+bt+c)dt=0` [ Givev]
` and phi(2)=underset(0)overset(2)int(1+cos^(8)t)(at^(2)+bt+c)dt=0` [ Given]
Therefore, 0,1 and 2, are the root sat `phi(x)` .
By Rolle's theorem `phi'(x)=0` will hav at least one real root betwen 0 and 1 and at leat one real between 1 and 2 .