The set of points where f( x ) = x \over 1 + | x | is differentiable is

( - \infty ,0 ) \cup ( 0,\infty )

( - \infty ,1 ) \cup ( - 1,\infty )

Limits Continuity and Differentiability PYQ — JEE Main 2006 · Shift 0 · Q55 with Solution

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Correct answer: C

We are given $f(x)=\frac{x}{1+|x|}.$

To find where $f$ is differentiable, we examine the absolute value by splitting into cases.

For $x\ge 0$ , we have $|x|=x$ . Hence $f(x)=\frac{x}{1+x}, \qquad x\ge 0.$ This is a rational function with denominator $1+x>0$ for $x\ge 0$ , so it is differentiable for all $x>0$ . Its derivative is $f'(x)=\frac{(1+x)-x}{(1+x)^2}=\frac{1}{(1+x)^2}, \, x>0.$
For $x<0$ , we have $|x|=-x$ . Hence $f(x)=\frac{x}{1-x}, \qquad x<0.$ Again this is a rational function, and $1-x\neq 0$ for all $x<0$ , so it is differentiable for all $x<0$ . Its derivative is $f'(x)=\frac{(1-x)-x(-1)}{(1-x)^2}=\frac{1}{(1-x)^2}, \, x<0.$
The only point to check is $x=0$ .

First, check continuity: $f(0)=\frac{0}{1+|0|}=0.$ Also, $\lim_{x\to 0^-} \frac{x}{1-x}=0, \qquad \lim_{x\to 0^+} \frac{x}{1+x}=0.$ So $f$ is continuous at $x=0$ .

Now check differentiability at $x=0$ using one-sided derivatives.

Left-hand derivative:

=\lim_{h\to 0^-}\frac{\frac{h}{1-h}-0}{h} =\lim_{h\to 0^-}\frac{1}{1-h}=1.$$ Right-hand derivative: $$f'_+(0)=\lim_{h\to 0^+}\frac{f(h)-f(0)}{h} =\lim_{h\to 0^+}\frac{\frac{h}{1+h}-0}{h} =\lim_{h\to 0^+}\frac{1}{1+h}=1.$$ Since $$f'_-(0)=f'_+(0)=1,$$ $f$ is differentiable at $x=0$. 6. Therefore, $f$ is differentiable for all real $x$. So the set of points where $f$ is differentiable is $$(-\infty,\infty).$$ Hence the correct option is **C**.

More from Limits Continuity and Differentiability

View written solutionFree

Correct answer: C

We are given $f(x)=\frac{x}{1+|x|}.$

To find where $f$ is differentiable, we examine the absolute value by splitting into cases.

For $x\ge 0$ , we have $|x|=x$ . Hence $f(x)=\frac{x}{1+x}, \qquad x\ge 0.$ This is a rational function with denominator $1+x>0$ for $x\ge 0$ , so it is differentiable for all $x>0$ . Its derivative is $f'(x)=\frac{(1+x)-x}{(1+x)^2}=\frac{1}{(1+x)^2}, \, x>0.$
For $x<0$ , we have $|x|=-x$ . Hence $f(x)=\frac{x}{1-x}, \qquad x<0.$ Again this is a rational function, and $1-x\neq 0$ for all $x<0$ , so it is differentiable for all $x<0$ . Its derivative is $f'(x)=\frac{(1-x)-x(-1)}{(1-x)^2}=\frac{1}{(1-x)^2}, \, x<0.$
The only point to check is $x=0$ .

First, check continuity: $f(0)=\frac{0}{1+|0|}=0.$ Also, $\lim_{x\to 0^-} \frac{x}{1-x}=0, \qquad \lim_{x\to 0^+} \frac{x}{1+x}=0.$ So $f$ is continuous at $x=0$ .

Now check differentiability at $x=0$ using one-sided derivatives.

Left-hand derivative:

=\lim_{h\to 0^-}\frac{\frac{h}{1-h}-0}{h} =\lim_{h\to 0^-}\frac{1}{1-h}=1.$$ Right-hand derivative: $$f'_+(0)=\lim_{h\to 0^+}\frac{f(h)-f(0)}{h} =\lim_{h\to 0^+}\frac{\frac{h}{1+h}-0}{h} =\lim_{h\to 0^+}\frac{1}{1+h}=1.$$ Since $$f'_-(0)=f'_+(0)=1,$$ $f$ is differentiable at $x=0$. 6. Therefore, $f$ is differentiable for all real $x$. So the set of points where $f$ is differentiable is $$(-\infty,\infty).$$ Hence the correct option is **C**.

Limits Continuity and Differentiability question

Limits Continuity and Differentiability question

More from Limits Continuity and Differentiability

Limits Continuity and Differentiability question

Limits Continuity and Differentiability question

More from Limits Continuity and Differentiability