Let $\mathrm{f}(\mathrm{x})=\left\{\begin{array}{cc}\mathrm{x}^{2} \sin \left(\frac{1}{\mathrm{x}}\right) & , x \neq 0 \\ 0 & , x=0\end{array}\right.$; Then at $\mathrm{x}=0$

(B) $f$ is continuous but $f^{\prime}$ is not continuous

(A) f is continuous but not differentiable

(C) f and f' both are continuous

(D) $\mathrm{f}^{\prime}$ is continuous but not differentiable

JEE Main · 2023 · 24 Jan 2023 · Shift 2 · Continuity and Differentiability

Let

f (x) = {x^{2} sin (\frac{1}{x}) 0, x \neq = 0, x = 0

; Then at

x = 0

Answer: (B)

Since

∣ x^{2} sin (1/ x) ∣ \leq x^{2} \to 0 = f (0)

, the function is continuous at

0

, and

f^{'} (0) = lim_{h \to 0} h sin (1/ h) = 0

exists. For

x \neq = 0

f^{'} (x) = 2 x sin (1/ x) - cos (1/ x)

, and as

x \to 0

the term

cos (1/ x)

oscillates without settling, so

lim_{x \to 0} f^{'} (x)

does not exist. Thus

f

is continuous but

f^{'}

is not continuous.

JEE Main · 2023 · 24 Jan 2023 · Shift 2 · Single correct option