Question

$f\left(x\right)$ satisfies the relation $f\left(x\right)-\lambda \underset{0}{\overset{\pi /2}{\int }}\sin x\cos tf\left(t\right)dt=\sin x.$

If $\underset{0}{\overset{\pi /2}{\int }}f\left(x\right)dx=3$, then the value of $\lambda$ is

• A. $1$
• B. $\frac{3}{2}$
• C. $\frac{4}{3}$
• D. none of these

Answer 1

The correct option is C $\frac{4}{3}$

$f\left(x\right)-\lambda \underset{0}{\overset{\pi /2}{\int }}\sin x\cos tf\left(t\right)dt=\sin x\Rightarrow f\left(x\right)-\lambda \sin x\underset{0}{\overset{\pi /2}{\int }}\cos tf\left(t\right)dt=\sin x$
$\Rightarrow f\left(x\right)-A\sin x=\sin x$
where, $A=\lambda \underset{0}{\overset{\pi /2}{\int }}\cos tf\left(t\right)dt$

$\Rightarrow f\left(x\right)=(A+1)\sin x$
$\Rightarrow f\left(t\right)=(A+1)\sin t$

$\therefore A=\lambda \underset{0}{\overset{\pi /2}{\int }}(A+1)\cos t\sin tdt\Rightarrow A=\frac{\lambda (A+1)}{2}\underset{0}{\overset{\pi /2}{\int }}\sin 2tdt\Rightarrow A=\frac{\lambda (A+1)}{2}{[-\frac{\cos 2t}{2}]}_0^{\pi /2}\Rightarrow A=\frac{\lambda (A+1)}{2}\Rightarrow A=\frac{\lambda }{2-\lambda }$
So,
$f\left(x\right)=(\frac{\lambda }{2-\lambda }+1)\sin x\Rightarrow f\left(x\right)=\left(\frac{2}{2-\lambda }\right)\sin x$

Now,
$\underset{0}{\overset{\pi /2}{\int }}f\left(x\right)dx=3\Rightarrow \underset{0}{\overset{\pi /2}{\int }}\left(\frac{2}{2-\lambda }\right)\sin xdx=3\Rightarrow -\left(\frac{2}{2-\lambda }\right){\left[\cos x\right]}_0^{\pi /2}=3\Rightarrow \frac{2}{2-\lambda }=3\Rightarrow \lambda =\frac{4}{3}$