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 $f\left(x\right)=2$ has at least one real root, then

• A. $\lambda \in [1,4]$
• B. $\lambda \in [-1,2]$
• C. $\lambda \in [0,1]$
• D. $\lambda \in [1,3]$

Answer 1

The correct option is D $\lambda \in [1,3]$

$f\left(x\right)-\lambda \underset{0}{\overset{\pi /2}{\int }}\sin x\cos tf\left(t\right)dt=\sin x\Rightarrow f\left(x\right)-\lambda \sin x\underset{0}{\overset{\pi /2}{\int }}\cos tf\left(t\right)dt=\sin x$
$\Rightarrow f\left(x\right)-A\sin x=\sin x$
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,
$f\left(x\right)=2\Rightarrow \left(\frac{2}{2-\lambda }\right)\sin x=2\Rightarrow \sin x=2-\lambda \because -1\le \sin x\le 1\Rightarrow -1\le 2-\lambda \le 1\Rightarrow \lambda \in [1,3]$