Question

Let a, b, c be non-zero real numbers such that
${\int }_0^1(1+co{s}^{8}x)(a{x}^2+bx+c)dx={\int }_0^2(1+co{s}^{8}x)(a{x}^2+bx+c)dx$
Then the quadratic equation $a{x}^2+bx+c=0$ has [IIT 1981; CEE 1993]

• A. No root in (0, 2)
• B. At least one root in (0, 2)
• C. A double root in (0, 2)
• D. None of these

Answer 1

The correct option is B

At least one root in (0, 2)

We have ${\int }_0^{2}f\left(x\right)dx={\int }_0^{1}f\left(x\right)dx+{\int }_1^{2}f\left(x\right)dx$
where $f\left(x\right)=(a{x}^2+bx+c)(1+co{s}^{8}x)$
If $f\left(x\right)>0(<0)\text{/}\vee xϵ(1,2)then{\int }_1^{2}f\left(x\right)dx>0(<0)$
Thus $f\left(x\right)=(1+co{s}^{8}x)(a{x}^2+bx+c)$ must be
positive for some value of x in [1,2] and must be
negative for some value of x in [1,2]. As $(1+co{s}^{8}x)\ge 1$
it follows that if $g\left(x\right)=a{x}^2+bx+c$, then there exist
some $\alpha ,\beta ϵ(1,2)$ such that $g\left(\alpha \right)>0$ and $g\left(\beta \right)<0$.
Since g is continuous on R, therefore there exist
some c between $\alpha$ and $\beta$ such that g(c)=0.
Thus $a{x}^2+bx+c=0$ has at least one root in (1,2) and hence in (0,2).