Question

Let $f\left(x\right)$ is a derivable function satisfying $f\left(x\right)=\underset{0}{\overset{x}{\int }}{e}^t\sin (x-t)dt$ and $g\left(x\right)=f^{\prime \prime }\left(x\right)-f\left(x\right),$ then the number of possible integers in the range of $g\left(x\right)$ is

Answer 1

$f\left(x\right)=\underset{0}{\overset{x}{\int }}{e}^t\sin (x-t)dt$
Applying integration by parts succesively : taking $u=\sin (x-t),v=e^t$
$\Rightarrow f\left(x\right)={\left[\sin \right(x-t\left)e^t\right]}_0^x+\underset{0}{\overset{x}{\int }}{e}^t\cos (x-t)dt=-\sin x+[{\left[\cos \right(x-t\left)e^t\right]}_0^x-\underset{0}{\overset{x}{\int }}{e}^t\sin (x-t)dt]=-\sin x+e^x-\cos x-f\left(x\right)\Rightarrow f\left(x\right)=\frac{1}{2}(-\cos x-\sin x+e^x)f^{\prime }\left(x\right)=\frac{1}{2}[\sin x-\cos x+e^x]f^{\prime \prime }\left(x\right)=\frac{1}{2}[\cos x+\sin x+e^x]g\left(x\right)=f^{\prime \prime }\left(x\right)-f\left(x\right)=\frac{1}{2}[\cos x+\sin x+e^x+\cos x+\sin x-e^x]=\frac{1}{2}\left[2\right(\cos x+\sin x\left)\right]g\left(x\right)=\cos x+\sin x$
$\Rightarrow$ Range of $g\left(x\right)$ is $[-\sqrt{2},\sqrt{2}]$
Thus, $3$ integral points in range of $g\left(x\right)$