JEE Advanced 2019 · Paper 2 · Mathematics Q4

MathematicsMultiple correct+4 / -1Question 40 of 54

Let $f:\mathbb R\to\mathbb R$ be a function. We say that $f$ has

PROPERTY 1 if $\displaystyle\lim_{h\to0}\frac{f(h) - f(0)}{\sqrt{|h|}}$ exists and is finite, and

PROPERTY 2 if $\displaystyle\lim_{h\to0}\frac{f(h) - f(0)}{h^2}$ exists and is finite.

Then which of the following options is/are correct?

f(x) = |x|

has PROPERTY 1

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

has PROPERTY 1

f(x) = x|x|

has PROPERTY 2

f(x) = \sin x

has PROPERTY 2

Answer and worked solution

Official answer: A, B

(A) $\dfrac{|h|}{\sqrt{|h|}}=\sqrt{|h|}\to0$ . True.
(B) $\dfrac{|h|^{2/3}}{|h|^{1/2}}=|h|^{1/6}\to0$ . True.
(C) $\dfrac{h|h|}{h^2}=\operatorname{sgn}h$ , which has no limit at 0. False.
(D) $\dfrac{\sin h}{h^2}\sim\dfrac1h$ , which is unbounded. False.

(A), (B)

Topics:Limits Differentiability

JEE Advanced 2019 · Paper 2, Mathematics question 4