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Question: The set of points where the function \(\frac{\log(1 + ax) - \log ⥄ (1 - bx)}{x}\)is differentiable...

The set of points where the function log⁡(1+ax)−log⁡⥄(1−bx)x\frac{\log(1 + ax) - \log ⥄ (1 - bx)}{x}is differentiable

A

1x−[x]\frac{1}{x - \lbrack x\rbrack}

B

f(x)={2x+1whenx<1kwhenx=15x−2whenx>1 f(x) = \left\{ \begin{matrix} 2x + 1\text{when}x < 1 \\ k\text{when}x = 1 \\ 5x - 2\text{when}x > 1 \end{matrix} \right.\

C

f(x)={sin⁡(1x),x≠0k,x=0 f(x) = \left\{ \begin{aligned} & \begin{matrix} \sin\left( \frac{1}{x} \right), & x \neq 0 \end{matrix} \\ & \begin{matrix} k, & x = 0 \end{matrix} \end{aligned} \right.\

D

None

Answer

f(x)={2x+1whenx<1kwhenx=15x−2whenx>1 f(x) = \left\{ \begin{matrix} 2x + 1\text{when}x < 1 \\ k\text{when}x = 1 \\ 5x - 2\text{when}x > 1 \end{matrix} \right.\

Explanation

Solution

Clearly, f(x)f ( x ) is differentiable for all non-zero values of x, For x≠0x \neq 0, we have f′(x)=xe−x21−e−x2f ^ { \prime } ( x ) = \frac { x e ^ { - x ^ { 2 } } } { \sqrt { 1 - e ^ { - x ^ { 2 } } } }

Now, (L.H.D. at x = 0) = lim⁡x→0−f(x)−f(0)x−0=lim⁡h→0f(0−h)−f(0)−h\lim _ { x \rightarrow 0 ^ { - } } \frac { f ( x ) - f ( 0 ) } { x - 0 } = \lim _ { h \rightarrow 0 } \frac { f ( 0 - h ) - f ( 0 ) } { - h }

= lim⁡h→01−e−h2−h=lim⁡h→0−1−e−h2h\lim _ { h \rightarrow 0 } \frac { \sqrt { 1 - e ^ { - h ^ { 2 } } } } { - h } = \lim _ { h \rightarrow 0 } - \frac { \sqrt { 1 - e ^ { - h ^ { 2 } } } } { h } = −lim⁡h→0eh2−1h2×1eh2=−1- \lim _ { h \rightarrow 0 } \sqrt { \frac { e ^ { h ^ { 2 } } - 1 } { h ^ { 2 } } } \times \frac { 1 } { \sqrt { e ^ { h ^ { 2 } } } } = - 1

and, (RHD at x = 0) = lim⁡x→0+f(x)−f(0)x−0=lim⁡h→01−e−h2−0h\lim _ { x \rightarrow 0 ^ { + } } \frac { f ( x ) - f ( 0 ) } { x - 0 } = \lim _ { h \rightarrow 0 } \frac { \sqrt { 1 - e ^ { - h ^ { 2 } } } - 0 } { h }

= lim⁡h→0eh2−1h2×1eh2=1\lim _ { h \rightarrow 0 } \sqrt { \frac { e ^ { h ^ { 2 } } - 1 } { h ^ { 2 } } } \times \frac { 1 } { \sqrt { e ^ { h ^ { 2 } } } } = 1 .

So, f(x)f ( x ) is not differentiable at x=0x = 0 , Hence, the points of differentiability of f(x)f ( x ) are (−∞,0)∪(0,∞)( - \infty , 0 ) \cup ( 0 , \infty ).