The correct option is B At least one root in (0,2)
Given ∫10(1+cos8x)(ax2+bx+c)dx=∫20(1+cos8x)(ax2+bx+c)dx⇒∫20(1+cos8x)(ax2+bx+c)dx=∫10(1+cos8x)(ax2+bx+c)dx+∫21(1+cos8x)(ax2+bx+c)dx⇒∫21(1+cos8x)(ax2+bx+c)dx=0
Now we know that if ∫βαf(x)dx=0 then it means that f(x)is +ve on some part of (α,β) and -ve on other part of (α,β).
But here 1+cos8x is always +ve.
∴ax2+bx+c is positive on some part of [1,2] and negative on other part of [1,2]
∴ax2+bx=0 has at least one root in (1,2)
⇒ax2+bx+c=0 has at least one root in (0,2)