MHT CET · Maths · Differentiation
The set of all points, where the derivative of the functions \(\mathrm{f}(x)=\frac{x}{1+|x|}\) exists, is
- A \((-\infty, \infty)\)
- B \([0, \infty)\)
- C \((-\infty, 0) \cup(0, \infty)\)
- D \((0, \infty)\)
Answer & Solution
Correct Answer
(A) \((-\infty, \infty)\)
Step-by-step Solution
Detailed explanation
\(\mathrm{f}(x)\) can be written as
\(\begin{aligned}
& \mathrm{f}(x)=\left\{\begin{array}{l}
\frac{x}{1-x}, x \leq 0 \\
\frac{x}{1+x}, x>0
\end{array}\right. \\
& \mathrm{f}^{\prime}(x)= \begin{cases}\frac{(1-x)+x}{(1+x)^2}, & x \leq 0 \\
\frac{(1+x)-x}{(1+x)^2}, & x>0\end{cases} \\
& \mathrm{f}^{\prime}(x)=\frac{1}{(1+x)^2} \forall x \in(-\infty, \infty)
\end{aligned}\)
\(\therefore \quad\) Derivative of \(\mathrm{f}(x)\) exists \(\forall x \in(-\infty, \infty)\)
\(\begin{aligned}
& \mathrm{f}(x)=\left\{\begin{array}{l}
\frac{x}{1-x}, x \leq 0 \\
\frac{x}{1+x}, x>0
\end{array}\right. \\
& \mathrm{f}^{\prime}(x)= \begin{cases}\frac{(1-x)+x}{(1+x)^2}, & x \leq 0 \\
\frac{(1+x)-x}{(1+x)^2}, & x>0\end{cases} \\
& \mathrm{f}^{\prime}(x)=\frac{1}{(1+x)^2} \forall x \in(-\infty, \infty)
\end{aligned}\)
\(\therefore \quad\) Derivative of \(\mathrm{f}(x)\) exists \(\forall x \in(-\infty, \infty)\)
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