GATE CSE 2024 | Set 1 | Question: 1

Question

Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a function such that $f(x)=\max \left\{x, x^3\right\}, x \in \mathbb{R}$, where $\mathbb{R}$ is the set of all real numbers. The set of all points where $f(x)$ is NOT differentiable is

• A. $\{-1,1,2\}$
• B. $\{-2,-1,1\}$
• C. $\{0,1\}$
• D. $\{-1,0,1\}$ 

Comments

$f(x) =\left\{\begin{matrix} \large x, & x\leq1\\\large x^3, & -1 < x < 0 \\ \large{x},&0 \leq x \leq 1\\ \large{x^3},&x>1\end{matrix}\right.$

clearly $f(x)$ is continuous at every real x, when it comes to differentiability,

$ $

$f'(x) =\left\{\begin{matrix} \large 1, & x <-1\\\large 3x^2, & -1 < x < 0 \\ \large{1},&0 <  x < 1\\ \large{3x^2},&x>1\end{matrix}\right.$

$ $

At $x = -1$:

$f'(-1+) = 3(-1)^2 = 3$

$f'(-1^-) = 1$, since LHD $\neq$ RHD $f(x)$ is not differentiable at x = -1

Similaryly at $x=0 , 1$, LHD $\neq$ RHD.

So the points where $f(x)$ is not differentiable is $\set{-1 , 0 , 1}$.

So Option D is correct.

Answer 1

Answer : Option D
If we try to plot the function f(x) = max(x,x³), it will look something like the image below as we can clearly see the funtion is continuous everywhere but it is not differentiable at {-1,0,1}

Comments

But how can someone plot the graph in exam, even it's better if you mention why it's not differentiable at $\set{-1,0,1}$ (Sharp turn/corner).

Answer 2

D