Question

The function $f(x) = \left\{ \begin{matrix} mx^{2},x \leq 1 \\ 2x,x > 1 \end{matrix} \right.\$ is

• A. Continuous everywhere but not differentiable at $x = 1$
• B. Continuous and differentiable everywhere
• C. Not continuous at $f(x) = \left\{ \begin{matrix} (\cos x)^{1/x},x \neq 0 \\ k,x = 0 \end{matrix} \right.\$
• D. None of these

Answer 1

Answer: (A)

Continuous everywhere but not differentiable at $x = 1$

Answer 2

We have,

Clearly, $f ( x )$ is continuous and differentiable for all non-zero x.

Now, $\lim _ { x \rightarrow 0 ^ { - } } f ( x ) = \lim _ { x \rightarrow 0 } e ^ { x }$= 1 and $\lim _ { x \rightarrow 0 ^ { + } } f ( x ) = \lim _ { x \rightarrow 0 } e ^ { - x } = 1$

Also, $f ( 0 ) = e ^ { 0 } = 1$

So, $f ( x )$ is continuous for all x.

(LHD at $x = 0$) = $\left( \frac { d } { d x } \left( e ^ { x } \right) \right) _ { x = 0 } = \left[ e ^ { x } \right] _ { x = 0 } = e ^ { 0 } = 1$

(RHD at $x = 0$) = $\left( \frac { d } { d x } \left( e ^ { - x } \right) \right) _ { x = 0 } = \left[ - e ^ { - x } \right] _ { x = 0 } = - 1$

So, $f ( x )$ is not differentiable at $x = 0$.

Hence, is everywhere continuous but not

differentiable at This fact is also evident from the graph of the function.