Question

lf $y=\sin \left(\sin x\right)$ and $\frac{d^{2}y}{d{x}^2}+\frac{dy}{dx}\tan x+f\left(x\right)=0$, then $f\left(x\right)$ equals to

• A. ${\sin }^2$ xsin $\left(\cos x\right)$
• B. ${\sin }^{2}x\cos \left(\cos x\right)$
• C. ${\cos }^{2}x\left(\sin \right(\cos x\left)\right)$
• D. ${\cos }^2$ $x\sin \left(\sin x\right)$

Answer 1

The correct option is D ${\cos }^2$ $x\sin \left(\sin x\right)$

Given, $y=\sin \left(\sin x\right)$

$\frac{dy}{dx}=\cos \left(\sin x\right)\cos \left(x\right)$

$\frac{d^{2}y}{d{x}^2}=-\sin \left(\sin x\right){\cos }^{2}x-\cos \left(\sin x\right)\sin x$

Also given, $\frac{d^{2}y}{d{x}^2}+\frac{dy}{dx}\tan x+f\left(x\right)=0$

Therefore, $-\sin \left(\sin x\right){\cos }^{2}x-\cos \left(\sin x\right)\sin x+\cos \left(\sin x\right)\cos x\times \frac{\sin x}{\cos x}+f\left(x\right)=0$

$\Rightarrow -\sin \left(\sin x\right){\cos }^{3}x-\cos \left(\sin x\right)\sin x.\cos x+\cos \left(\sin x\right)\sin x.\cos x+f\left(x\right)\cos x=0$

$\Rightarrow -\sin \left(\sin x\right){\cos }^{2}x-\cos \left(\sin x\right)\sin x+\cos \left(\sin x\right)\sin x+f\left(x\right)=0$

$\Rightarrow f\left(x\right)=\sin \left(\sin x\right){\cos }^{2}x$