Question

The maximum value of the function
$f\left(x\right)=\underset{0}{\overset{1}{\int }}t\sin (x+\pi t)dt,x\in R$ 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 $\frac{1}{{\pi }^2}\sqrt{{\pi }^2+4}$

$f\left(x\right)={[-\frac{t}{\pi }\cos (x+\pi t\left)\right]}_0^1+\frac{1}{\pi }\underset{0}{\overset{1}{\int }}\cos (x+\pi t)dt$
$=\frac{1}{\pi }\cos x-\frac{2}{{\pi }^2}\sin x$

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