Question

If ${\int }_0^1(1+\sin {}^{4}x)(a{x}^2+bx+c)dx={\int }_0^2(1+\sin {}^{4}x)(a{x}^2+bx+c)$dx then the quadratic equation $a{x}^2+bc+c=0$ has

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

Answer 1

Answer: (A)

at least one root in (1,2)

Consider ${\int }_0^1(1+\sin {}^{4}x)(a{x}^2+bx+c)dx$
obviously $f\left(x\right)$ is continuous and differentiable in the inteval $[1,2]$
Also $f\left(1\right)=f\left(2\right)$
Therefore by Rolle's theorem there exists at least one point $k\in (1,2)$
such that $f^{\prime }\left(k\right)=0$
Now $f^{\prime }\left(x\right)={\int }_0^1(1+\sin {}^{4}x)(a{x}^2+bx+c)dx\Rightarrow (a{x}^2+bx+c)=0$
As $1+\sin {}^{4}x\ne 0$
Therefore k is root of $(a{x}^2+bx+c)=0$
where $k\in (1,2)$
Hence, atleast one root $\in (1,2)$