Question

If $f\left(x\right)=\{\begin{array}{cc}1,& x<0\\ 1+\sin x,& 0\le x<\frac{\pi }{2}\end{array}$, then at $x=0$ the derivative $f^{\prime }\left(x\right)$ is

• A. $1$
• B. $0$
• C. Infinite
• D. Not defined

Answer 1

Answer: (D)

Not defined

$f\left(0\right)=1+\sin 0=1$
$RHD=f^{\prime }\left(0^{+}\right)=\underset{h\to 0}{lim}\frac{1+\sin (0+h)-1}{h}$
$=\underset{h\to 0}{lim}\left(\frac{\sin h}{h}\right)=1$
$LHD=f^{\prime }\left(0^{-}\right)=\underset{h\to 0}{lim}\left(\frac{1-1}{-h}\right)=0$
Hence, $f^{\prime }\left(0\right)$ does not exist.