Question

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

• A. $1$
• B. $0$
• C. infinite
• D. not defined

Answer 1

The correct option is A $1$

First check the differentiability at $x=0$
$\underset{x\to 0^{-}}{lim}{f}^{\prime }\left(x\right)=0$
$\underset{x\to 0^{+}}{lim}{f}^{\prime }\left(x\right)=\underset{x\to 0^{+}}{lim}\cos x=1$
Therefore, $\underset{x\to 0^{-}}{lim}{f}^{\prime }\left(x\right)\ne \underset{x\to 0^{+}}{lim}{f}^{\prime }\left(x\right)$
Hence, function is not differentiable at $x=0$
So, derivative of $f\left(x\right)$ do not defined at $x=0$