Question

At $x=1$, the function $f\left(x\right)=\{\begin{array}{cc}x-1,& x\le 1\\ x^3-1,& x>1\end{array}$ is

• A. continuous and differentiable
• B. continuous but not differentiable
• C. discontinuous and differentiable
• D. discontinuous and not differentiable

Answer 1

The correct option is B continuous but not differentiable

Given : $f\left(x\right)=\{\begin{array}{cc}x-1,& x\le 1\\ x^3-1,& x>1\end{array}$
For the given function,
$f\left(1^{+}\right)=f\left(1^{-}\right)=f\left(1\right)=0$
Hence it is continuous at $x=1$

For the given function, derivative is obtained as:
$f^{\prime }\left(x\right)=\{\begin{array}{cc}1,& x<1\\ 3x^2,& x>1\end{array}$
Clearly, at $x=1$
$L.H.D.\ne R.H.D.$
$\Rightarrow f\left(x\right)$ is not differentiable at $x=1$.
$\therefore f\left(x\right)$ is continuous but not differentiable at $x=1$