Let m and n be the number of points at which the function f(x)= \ x, x^3, x^5, ., x^ 21 \ , x ∈ ℝ , is not differentiable and not continuous, respectively. Then m+n is equal to ________ .

Correct answer: 0 f(x)=\ array cc x, & x -1 x^ 21 , & -1 ≤ x 0 x, & 0 ≤ x 1 x^ 21 , & x ≥ 1 array . f(x) is continuous everywhere. aligned & n=0 & f^ (x)=\ array cc 1, & x -1 21 x^ 20 , & -1 ≤ x 0 1, & 0 x 1 21 x^ 20 , & x ≥ 1 array . aligned f(x) is non-differentiable at x=-1,0,1 aligned & m=3 & ~m+n=3 aligned

JEE Main 20254 AprilMorning Shift

Exam: JEE Main
Year: 2025
Date: 4 April
Shift: Morning Shift
Subject: Mathematics
Chapter: Continuity and Differentiability

JEE Main 2025 · 4 April · Morning Shift · Mathematics · Continuity and Differentiability

Let m and n be the number of points at which the function f(x)= \...

Let

m

and

n

be the number of points at which the function

f(\mathrm{x})=\max \left\{\mathrm{x}, \mathrm{x}^3, \mathrm{x}^5, \ldots ., \mathrm{x}^{21}\right\}, \mathrm{x} \in \mathbb{R}

, is not differentiable and not continuous, respectively. Then

\mathrm{m}+\mathrm{n}

is equal to ________ .

Correct answer

Step-by-step solution

f(x)=\left\{

\begin{array}{cc}<br/>x, & x \lt -1 \\ x^{21}, & -1 \leq x \lt 0 \\ x, & 0 \leq x \lt 1 \\ x^{21}, & x \geq 1<br/>\end{array}

\right.

f(x)

is continuous everywhere.

\begin{aligned}<br/>& \therefore \mathrm{n}=0 \\ & \mathrm{f}^{\prime}(\mathrm{x})=\left\{\begin{array}{cc}<br/>1, & \mathrm{x} \lt -1 \\ 21 \mathrm{x}^{20}, & -1 \leq \mathrm{x} \lt 0 \\ 1, & 0 \lt \mathrm{x} \lt 1 \\ 21 \mathrm{x}^{20}, & \mathrm{x} \geq 1<br/>\end{array}\right.<br/>\end{aligned}

\therefore \mathrm{f}(\mathrm{x})

is non-differentiable at

\mathrm{x}=-1,0,1

\begin{aligned}<br/>& \therefore \mathrm{m}=3 \\ & \mathrm{~m}+\mathrm{n}=3<br/>\end{aligned}

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