Let $\mathrm{F}: \mathbb{R} \rightarrow \mathbb{R}$ be a function. We say that f has PROPERTY 1 if $\lim _{\mathrm{h} \rightarrow 0} \frac{\mathrm{f}(\mathrm{h})-\mathrm{f}(0)}{\sqrt{|\mathrm{h}|}}$ exists and is finite, and PROPERTY 2 if $\lim _{\mathrm{h} \rightarrow 0} \frac{\mathrm{f}(\mathrm{h})-\mathrm{f}(0)}{\mathrm{h}^{2}}$ exists and is finite. Then which of the following options is/are correct ?

(B) $\mathrm{f}(\mathrm{x})=\mathrm{x}^{2 / 3}$ has PROPERTY 1; (D) $\mathrm{f}(\mathrm{x})=|\mathrm{x}|$ has PROPERTY 1

(A) $\mathrm{f}(\mathrm{x})=\mathrm{x}|\mathrm{x}|$ has PROPERTY 2

(C) $f(x)=\sin x$ has PROPERTY 2

JEE Advanced · 2019 · 27 May 2019 · Paper 2 · Continuity and Differentiability

Let

F : R \to R

be a function. We say that f has
PROPERTY 1 if

lim_{h \to 0} \frac{f ( h ) - f ( 0 )}{∣ h ∣}

exists and is finite, and
PROPERTY 2 if

lim_{h \to 0} \frac{f ( h ) - f ( 0 )}{h ^{2}}

exists and is finite.
Then which of the following options is/are correct ?

Answer: (B), (D)

PROPERTY 1: For

f (x) = x^{2/3}

lim_{h \to 0} \frac{h ^{2/3}}{∣ h ∣} = lim_{h \to 0} \frac{∣ h ∣ ^{2/3}}{∣ h ∣ ^{1/2}} = lim_{h \to 0} ∣ h ∣^{1/6} = 0

(finite). For

f (x) = ∣ x ∣

lim_{h \to 0} \frac{∣ h ∣}{∣ h ∣} = lim_{h \to 0} ∣ h ∣ = 0

(finite). PROPERTY 2: For

f (x) = x ∣ x ∣

lim_{h \to 0} \frac{h ∣ h ∣}{h ^{2}}

gives RHL=1, LHL=-1 (not equal, so limit DNE). For

f (x) = sin x

lim_{h \to 0} \frac{s i n h}{h ^{2}}

DNE. Therefore, options (B) and (D) are correct.

JEE Advanced · 2019 · 27 May 2019 · Paper 2 · One or more correct options