Question

If $f\left(x\right)=x+\sin x,$ then the differential coefficient of $f\left(x\right)$ at $x=0,$ is

• A. $0$
• B. $2$
• C. $1$
• D. $-1$

Answer 1

The correct option is B $2$

$R.H.D.=$
$f^{\prime }\left(0^{+}\right)=\underset{h\to 0}{lim}\frac{f(0+h)-f\left(0\right)}{h}$
$=\underset{h\to 0}{lim}\frac{(0+h)+\sin (0+h)-0}{h}$
$=\underset{h\to 0}{lim}\frac{h+\sin h}{h}=\underset{h\to 0}{lim}(1+\frac{\sin h}{h})$
$\Rightarrow f^{\prime }\left(0^{+}\right)=2$
$\text{OR}$
$L.H.D.=$
$f^{\prime }\left(0^{-}\right)=\underset{h\to 0}{lim}\frac{f(0-h)-f\left(0\right)}{-h}$
$=\underset{h\to 0}{lim}\frac{(0-h)+\sin (0-h)-0}{-h}$
$=\underset{h\to 0}{lim}\frac{-h-\sin h}{-h}=\underset{h\to 0}{lim}(1+\frac{\sin h}{h})$
$\Rightarrow f^{\prime }\left(0^{-}\right)=2$

$\Rightarrow L.H.D.=R.H.D.=f^{\prime }\left(0\right)$
$\therefore$ $f^{\prime }\left(0\right)=2$