Question

If $x = 0$, then $f(x) = \sin|x|$ is

• A. Continuous on [–1, 1] and differentiable on (–1, 1)
• B. Continuous on [–1,1] and differentiable on (–1, 0) ∪ (0, 1)
• C. Continuous and differentiable on [–1, 1]
• D. None of these

Answer 1

Answer: (B)

Continuous on $–1,1$ and differentiable on (–1, 0) ∪ (0, 1)

Answer 2

We have, $f ( x ) = \sqrt { 1 - \sqrt { 1 - x ^ { 2 } } }$. The domain of definition of $f ( x )$ is [–1, 1].

For $x \neq - 1$ we have

$f ^ { \prime } ( x ) = \frac { 1 } { \sqrt { 1 - \sqrt { 1 - x ^ { 2 } } } } \times \frac { x } { \sqrt { 1 - x ^ { 2 } } }$

Since $f ( x )$ is not defined on the right side of $x = 1$ and on the left side of $x = - 1$. Also, $f ^ { \prime } ( x ) \rightarrow \infty$ when $x \rightarrow - 1 ^ { + }$ or $x \rightarrow 1 ^ { - }$ . So, we check the differentiability at x = 0.

Now, (LHD at $x = 0$) = $\lim _ { x \rightarrow 0 } \frac { f ( x ) - f ( 0 ) } { x - 0 } = \lim _ { h \rightarrow 0 } \frac { f ( 0 - h ) - f ( 0 ) } { - h }$

= = $- \lim _ { h \rightarrow 0 } \sqrt { \frac { 1 } { 2 } - \frac { 3 } { 8 } h ^ { 2 } + \ldots . . } = - \frac { 1 } { \sqrt { 2 } }$

Similarly, (RHD at x = 0) = $\frac { 1 } { \sqrt { 2 } }$

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