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Question: If \(- \frac{1}{4}\), then \(\frac{1}{8}\) is...

If −14- \frac{1}{4}, then 18\frac{1}{8} is

A

Continuous as well as differentiable for all x`

B

Continuous for all x but not differentiable at −18- \frac{1}{8}

C

Neither differentiable nor continuous at f(x)={xifxis rational1−xifxis irrational, f(x) = \left\{ \begin{matrix} x & \text{if}x\text{is rational} \\ 1 - x & \text{if}x\text{is irrational} \end{matrix}, \right.\

D

Discontinuous every where

Answer

Continuous for all x but not differentiable at −18- \frac{1}{8}

Explanation

Solution

f(0)=0f ( 0 ) = 0 and f(x)=xe−(1∣x∣+1x)f ( x ) = x e ^ { - \left( \frac { 1 } { | x | } + \frac { 1 } { x } \right) }

R.H.L. = lim⁡h→0(0+h)e−2/h=lim⁡h→0he2/h=0\lim _ { h \rightarrow 0 } ( 0 + h ) e ^ { - 2 / h } = \lim _ { h \rightarrow 0 } \frac { h } { e ^ { 2 / h } } = 0

L.H.L. = lim⁡h→0(0−h)e−(1h−1h)=0\lim _ { h \rightarrow 0 } ( 0 - h ) e ^ { - \left( \frac { 1 } { h } - \frac { 1 } { h } \right) } = 0

∴ f(x)f ( x ) is continuous.

Rf′(x)R f ^ { \prime } ( x ) at (x=0)=lim⁡h→0f(0+h)−f(0)h( x = 0 ) = \lim _ { h \rightarrow 0 } \frac { f ( 0 + h ) - f ( 0 ) } { h }= lim⁡h→0he−2/hh=e−∞=0\lim _ { h \rightarrow 0 } \frac { h e ^ { - 2 / h } } { h } = e ^ { - \infty } = 0

Lf′(x)L f ^ { \prime } ( x ) at (x=0)=lim⁡h→0f(0−h)−f(0)−h( x = 0 ) = \lim _ { h \rightarrow 0 } \frac { f ( 0 - h ) - f ( 0 ) } { - h }= lim⁡h→0−he−(1h−1h)−h\lim _ { h \rightarrow 0 } \frac { - h e ^ { - \left( \frac { 1 } { h } - \frac { 1 } { h } \right) } } { - h } = +1

⇒ Lf′(x)≠Rf′(x)L f ^ { \prime } ( x ) \neq R f ^ { \prime } ( x )

f(x)f ( x ) is not differentiable at x=0x = 0