Question

$f$ is a non-zero function such that $f\left(x\right)=\underset{0}{\overset{x}{\int }}f\left(t\right)\sin \left(k\right(x-t\left)\right)dt$ and
$f^{\prime \prime }\left(x\right)=0$. Then the value(s) of $k$ is/are

• A. $0$
• B. $1$
• C. $2$
• D. $3$

Answer 1

The correct option is B $1$

$f\left(x\right)=\underset{0}{\overset{x}{\int }}f\left(t\right)\sin \left(k\right(x-t\left)\right)dt\Rightarrow f\left(x\right)=\underset{0}{\overset{x}{\int }}f\left(t\right)[\sin kx\cos kt-\sin kt\cos kx]dt$
$\Rightarrow f\left(x\right)=\sin kx\underset{0}{\overset{x}{\int }}f\left(t\right)\cos ktdt-\cos kx{\int }_0^{x}f\left(t\right)\sin ktdt$
$\therefore f^{\prime }\left(x\right)=k\cos kx\underset{0}{\overset{x}{\int }}f\left(t\right)\cos ktdt+\sin kx\left[f\right(x\left)\cos kx\right]+k\sin kx\underset{0}{\overset{x}{\int }}f\left(t\right)\sin ktdt-\cos kx\left[f\right(x\left)\sin kx\right]$
$\Rightarrow f^{\prime }\left(x\right)=k\cos kx\underset{0}{\overset{x}{\int }}f\left(t\right)\cos ktdt+k\sin kx\underset{0}{\overset{x}{\int }}f\left(t\right)\sin ktdt$

Again differentiating wrt $x$, we get
$f^{\prime \prime }\left(x\right)=-k^2\sin kx\underset{0}{\overset{x}{\int }}f\left(t\right)\cos ktdt+k\cos kx\left[f\right(x\left)\cos kx\right]+k^2\cos kx\underset{0}{\overset{x}{\int }}f\left(t\right)\sin ktdt+k\sin kx\left[f\right(x\left)\sin kx\right]$
$\Rightarrow f^{\prime \prime }\left(x\right)=-k^{2}f\left(x\right)+kf\left(x\right)\Rightarrow k=0,1$
But $f$ is a non-zero function.
$\therefore k=1$