Question

The function $\frac{f(x)e^{nx} + g(x)}{e^{nx} + 1}$ is

• A. Continuous at $\lim_{x \rightarrow 0}$, but not differentiable
• B. Both continuous and differentiable at $\lim_{x \rightarrow - 1}$
• C. Not continuous at $\frac{\sqrt{\pi} - \sqrt{(\cos^{- 1}x)}}{\sqrt{(x + 1)}}$
• D. None of these

Answer 1

Answer: (A)

Continuous at $\lim_{x \rightarrow 0}$, but not differentiable

Answer 2

We have, Since,

$\lim _ { x \rightarrow 1 ^ { - } } f ( x ) = \lim _ { x \rightarrow 1 ^ { - } } 1 = 1 , \lim _ { x \rightarrow 1 ^ { + } } f ( x ) = \lim _ { x \rightarrow 1 ^ { + } } ( 2 x - 1 ) = 1$ and

$f ( 1 ) = 2 \times 1 - 1 = 1$

∴$\lim _ { x \rightarrow 1 ^ { - } } f ( x ) = \lim _ { x \rightarrow 1 ^ { + } } f ( x ) = f ( 1 )$. So, $f ( x )$ is continuous at

x = 1.

Now, $\lim _ { x \rightarrow 1 ^ { - } } \frac { f ( x ) - f ( 1 ) } { x - 1 } = \lim _ { h \rightarrow 0 } \frac { f ( 1 - h ) - f ( 1 ) } { - h } = \lim _ { h \rightarrow 0 } \frac { 1 - 1 } { - h } = 0$,

and $\lim _ { x \rightarrow 1 ^ { + } } \frac { f ( x ) - f ( 1 ) } { x - 1 } = \lim _ { h \rightarrow 0 } \frac { f ( 1 + h ) - f ( 1 ) } { h }$ = $\lim _ { h \rightarrow 0 } \frac { 2 ( 1 + h ) - 1 - 1 } { h } = 2$ $\therefore$ (LHD at $x = 1$) ≠ (RHD at$x = 1$). So, f(x) is not differentiable at x = 1.

Trick : The graph of $f ( x )$ i.e.

is

By graph, it is clear that the function is not differentiable at $x = 0$, 1 as there it has sharp edges.