Question

The maximum value of the function $f\left(x\right)={\int }_0^{1}t\sin (x+\pi t)dt,x\in R$ is -

• A. $\frac{1}{\pi }\sqrt{{\pi }^2+4}$
• B. $\frac{1}{{\pi }^2}\sqrt{{\pi }^2+4}$
• C. $\sqrt{{\pi }^2+4}$
• D. $\frac{1}{2{\pi }^2}\sqrt{{\pi }^2+4}$

Answer 1

The correct option is B $\frac{1}{{\pi }^2}\sqrt{{\pi }^2+4}$

$f\left(x\right)={[-\frac{t}{\pi }\cos (x+t\pi \left)\right]}_0^1+\frac{1}{\pi }{\int }_0^1\cos (x+t\pi )dt$

$=\frac{1}{\pi }\cos x-\frac{2}{{\pi }^2}\sin x$

$f(x{)}_{max.}=\sqrt{\frac{1}{{\pi }^2}+\frac{4}{{\pi }^4}}=\frac{\sqrt{{\pi }^2+4}}{{\pi }^2}$