Let f(x) = \ array ll x ^ 2 ((1)/(x)); x ≠ 0 0; x = 0 array ), then at x = 0

Correct answer: B. f is continuous but f ^ ' is not continuous Given: f(x) = \ array ll x ^ 2 ((1)/(x)); x ≠ 0 0; x = 0 array ) _ x → 0 f(x) = _ x → 0 [ x ^ 2 ((1)/(x))] ⇒ _ x → 0 f(x) = _ x → 0 x ^ 2 × [numberbetween - 1to1] ⇒ _ x → 0 f(x) = 0 = f(0) Hence, f(x) is continuous at x = 0 . Now, RHD = f ^ ' ( 0 ^ + ) = _ h → 0 [(f(0 + h) - f(0))/(h)] ⇒ RHD = f ^ ' ( 0 ^ + ) = _ h → 0 [ h ^ 2 ((1)/(h)) - 0 h ] ⇒ RHD = f ^ ' ( 0 ^ + ) = _ h → 0 [hsin((1)/(h))] = 0 And, LHD = f ^ ' ( 0 ^ - ) = _ h → 0 [(f(0 - h) - f(0))/(- h)] ⇒ LHD = f ^ ' ( 0 ^ - ) = _ h → 0 [ - h ^ 2 ((1)/(h)) - 0 - h ] = 0 Since, LHD = RHD = finite So, f(x) is differentiable at x = 0 . Now, f ^ ' (x) = \ array ll 2xsin((1)/(x)) - ((1)/(x)); x ≠ 0 0; x = 0 array ) Now, _ x → 0 f ^ ' (x) = _ x → 0 [2xsin((1)/(x)) - ((1)/(x))] ⇒ _ x → 0 f ^ ' (x) = _ x → 0 [2x\ (1)/(x) - (1)/(3 !) 1 x ^ 3 + (1)/(5 !) 1 x ^ 5 - . . . .\ - \ 1 - (1)/(2 !) 1 x ^ 2 + (1)/(4 !) 1 x ^ 4 - . . . .\ ] ⇒ _ x → 0 f ^ ' (x) = _ x → 0 [\ 1 - (2)/(3 !) 1 x ^ 2 + (2)/(5 !) 1 x ^ 4 - . . . .\ - \ - (1)/(2 !) 1 x ^ 2 + (1)/(4 !) 1 x ^ 4 - . . . .\ ] This limit does not exists finitely, hence f ^ ' (x) is discontinuous at x = 0 .

JEE Main 202324 JanuaryMorning Shift

Exam: JEE Main
Year: 2023
Date: 24 January
Shift: Morning Shift
Subject: Mathematics
Chapter: Continuity and Differentiability

JEE Main 2023 · 24 January · Morning Shift · Mathematics · Continuity and Differentiability

Let f(x) = \ array ll x ^ 2 ((1)/(x)); x ≠ 0 0; x = 0 array ), then...

Let

f\left(x\right) = \left\{

\begin{array}{ll}{x}^{2}\sin \left(\dfrac{1}{x}\right); x \ne 0 \\ 0; x = 0\end{array}

\right), then at

x = 0

Options

A $f$ is continuous but not differentiable
B $f$ is continuous but ${f}^{'}$ is not continuous
Cboth $f$ and ${f}^{'}$ are continuous
D ${f}^{'}$ is continuous but not differentiable

Correct answer

B. $f$ is continuous but ${f}^{'}$ is not continuous

Step-by-step solution

Given:

$f\left(x\right) = \left\{$ $\begin{array}{ll}{x}^{2}\sin \left(\dfrac{1}{x}\right); x \ne 0 \\ 0; x = 0\end{array}$ $\right)$ \lim_{x \to 0}f\left(x\right) = \lim_{x \to 0}\left[{x}^{2}\sin \left(\dfrac{1}{x}\right)\right] $$ ⇒ \lim_{x \to 0}f\left(x\right) = \lim_{x \to 0}{x}^{2} \times \left[numberbetween - 1to1\right] $$ ⇒ \lim_{x \to 0}f\left(x\right) = 0 = f\left(0\right) $Hence,$ f\left(x\right) $is continuous at$ x = 0 $.Now,$ RHD = {f}^{'}\left({0}^{+}\right) = \lim_{h \to 0}\left[\dfrac{f\left(0 + h\right) - f\left(0\right)}{h}\right] $$ ⇒ RHD = {f}^{'}\left({0}^{+}\right) = \lim_{h \to 0}\left[\dfrac{{h}^{2}\sin \left(\dfrac{1}{h}\right) - 0}{h}\right] $$ ⇒ RHD = {f}^{'}\left({0}^{+}\right) = \lim_{h \to 0}\left[hsin\left(\dfrac{1}{h}\right)\right] = 0 $And,$ LHD = {f}^{'}\left({0}^{-}\right) = \lim_{h \to 0}\left[\dfrac{f\left(0 - h\right) - f\left(0\right)}{- h}\right] $$ ⇒ LHD = {f}^{'}\left({0}^{-}\right) = \lim_{h \to 0}\left[\dfrac{- {h}^{2}\sin \left(\dfrac{1}{h}\right) - 0}{- h}\right] = 0 $Since,$ LHD = RHD = finite $So,$ f\left(x\right) $is differentiable at$ x = 0 $.Now,$ {f}^{'}\left(x\right) = \left\{ $\begin{array}{ll}2xsin\left(\dfrac{1}{x}\right) - \cos \left(\dfrac{1}{x}\right); x \ne 0 \\ 0; x = 0\end{array}$ \right)

Now,

$\lim_{x \to 0}{f}^{'}\left(x\right) = \lim_{x \to 0}\left[2xsin\left(\dfrac{1}{x}\right) - \cos \left(\dfrac{1}{x}\right)\right]$

$⇒ \lim_{x \to 0}{f}^{'}\left(x\right) = \lim_{x \to 0}\left[2x\left\{\dfrac{1}{x} - \dfrac{1}{3 !}\dfrac{1}{{x}^{3}} + \dfrac{1}{5 !}\dfrac{1}{{x}^{5}} - . . . .\right\} - \left\{1 - \dfrac{1}{2 !}\dfrac{1}{{x}^{2}} + \dfrac{1}{4 !}\dfrac{1}{{x}^{4}} - . . . .\right\}\right]$

$⇒ \lim_{x \to 0}{f}^{'}\left(x\right) = \lim_{x \to 0}\left[\left\{1 - \dfrac{2}{3 !}\dfrac{1}{{x}^{2}} + \dfrac{2}{5 !}\dfrac{1}{{x}^{4}} - . . . .\right\} -$ \left\{- \dfrac{1}{2 !}\dfrac{1}{{x}^{2}} + \dfrac{1}{4 !}\dfrac{1}{{x}^{4}} - . . . .\right\}\right] $This limit does not exists finitely, hence$ {f}^{'} $\left(x\right)$ is discontinuous at $x = 0$ .

Practice Continuity and Differentiability on Quantrex Academy →

More from Continuity and Differentiability

Let ℝ denote the set of all real numbers. Let f : ℝ arrow ℝ be an arbitrary function and… JEE Advanced 2026 Consider the function f : (-(π)/(2), (π)/(2)) arrow (-∞, ∞) defined by f(x) = (|x| + |x… JEE Advanced 2026 For a real number α , let [α] denote the greatest integer less than or equal to α . For a… JEE Advanced 2026 For the function f(x) = e^ |x| - |x| , x ∈ ℝ , consider the following statements… JEE Main 2026 Let f(x) = cases (1)/(3), & x ≤ π/2 (b(1- x))/((π-2x)^2), & x > π/2 cases . If f is… JEE Main 2026 Let f(x) = cases x^3 + 8 ; & x Then the number of points, where the function g ∘ f is… JEE Main 2026 Let f(x)= cases e^ x-1 , & x<0 x^2-5x+6, & x ≥ 0 cases and g(x)=f(|x|)+|f(x)| . If the… JEE Main 2026 The number of points, at which the function f(x) = \ 6x, 2 + 3x^2\ + |x - 1|| |x^2… JEE Main 2026 The number of points in the interval [2, 4] , at which the function f(x) = [x^2 - x… JEE Main 2026 Let f(x)= _ θ arrow 0 ( π x-x^ ((2)/(θ)) (x-1) 1+x^ ((2)/(θ)) (x-1) ), x ∈ R . Consider… JEE Main 2026 Let [t] denote the greatest integer less than or equal to t . If the function f(x)=\… JEE Main 2026 If the function f(x)= e^ x (e^ x-x -1)+ _ e ( x+ x)-x x-x is continuous at x=0 , then the… JEE Main 2026 Let α, β ∈ ℝ be such that the function f(x)= cases 2 α(x^ 2 -2)+2 β x &, x be… JEE Main 2026 If f(x)=\ array cc a|x|+x^ 2 -2( |x|)( |x|) x &, x ≠ 0 b &, x=0 array . is continuous at… JEE Main 2026 Let f(x)= cases a x^ 2 +2 a x+3 4 x^ 2 +4 x-3 &, x ≠-(3)/(2), (1)/(2) ~b &, x=-(3)/(2)… JEE Main 2026 Let [·] denote the greatest integer function, and let f(x)= \ 2 x, x^ 2 \ . Let S = \ x… JEE Main 2026 Full Continuity and Differentiability list Continuity and Differentiability JEE Maths PYQs JEE Mathematics PYQs All JEE Main PYQs