QJEEAdv19Ma81bd: JEE Advanced 2019 Mathematics - Class 12 JEEAdv19Ma81bd Solution: Let f : R be a function. We say that f has PROPERTY 1 if exists …

Question

Verified Solution • 2025–26 Session moderate

Let f : R be a function. We say that f has

PROPERTY 1 if

\mathop {\lim }\limits_{h \to 0} {{f(h) - f(0)} \over {\sqrt {|h|} }}

exists and is finite, and

PROPERTY 2 if

\mathop {\lim }\limits_{h \to 0} {{f(h) - f(0)} \over {{h^2}}}

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

Accepted Answer

It is given, that f : R

\to

R and

Property 1 :

\mathop {\lim }\limits_{h \to 0} {{f(h) - f(0)} \over {\sqrt {|h|} }}

exists and finite, and

Property 2 :

\mathop {\lim }\limits_{h \to 0} {{f(h) - f(0)} \over {{h^2}}}

exists and finite.

Option A,

P2 :

\mathop {\lim }\limits_{h \to 0} {{\sin h - \sin 0} \over {{h^2}}} = \mathop {\lim }\limits_{h \to 0} {1 \over h}\left( {{{\sin \,h} \over h}} \right)

= doesn't exist.

Option B,

P1 :

\mathop {\lim }\limits_{h \to 0} {{{h^{2/3}} - 0} \over {\sqrt {|h|} }} = \mathop {\lim }\limits_{h \to 0} {h^{2/3 - 1/2}} = \mathop {\lim }\limits_{h \to 0} {h^{1/6}} = 0

exists and finite.

Option C,

P1 :

\mathop {\lim }\limits_{h \to 0} {{|h| - 0} \over {\sqrt {|h|} }} = \mathop {\lim }\limits_{h \to 0} \sqrt {|h|} = 0

, exists and finite.

Option D,

P2 :

lim_{h \to 0} \frac{h | h | - 0}{h^{2}} = lim_{h \to 0} \frac{| h |}{h} = {\begin{matrix} 1 & i f h \to 0^{+} \\ - 1 & i f h \to 0^{-} \end{matrix}

So,

\mathop {\lim }\limits_{h \to 0} {{f(h) - f(0)} \over {{h^2}}}

does not exist.

Hence, options (b) and (c) are correct.

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