    Question

# Let a,b,c be non – zero real numbers such that ∫10(1+cos8x)(ax2+bx+c)dx=∫20(1+cos8x)(ax2+bx+c)dx. Then the quadratic equation ax2+bx+c=0 has

A
No root in (0, 2)
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B
At least one root in (0,2)
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C
A double root in (0, 2)
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D
Two imaginary roots
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Solution

## 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)  Suggest Corrections  0      Similar questions
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