JEE Advance - Mathematics (2019 - Paper 2 Offline - No. 6)
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?
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?
f(x) = sin x has PROPERTY 2
f(x) = x2/3 has PROPERTY 1
f(x) = |x| has PROPERTY 1
f(x) = x|x| has PROPERTY 2
Explanation
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 : $$\mathop {\lim }\limits_{h \to 0} {{h|h| - 0} \over {{h^2}}} = \mathop {\lim }\limits_{h \to 0} {{|h|} \over h} = \left\{ {\matrix{ 1 & {if\,h \to {0^ + }} \cr { - 1} & {if\,h \to {0^ - }} \cr } } \right.$$
So, $$\mathop {\lim }\limits_{h \to 0} {{f(h) - f(0)} \over {{h^2}}}$$ does not exist.
Hence, options (b) and (c) are correct.
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 : $$\mathop {\lim }\limits_{h \to 0} {{h|h| - 0} \over {{h^2}}} = \mathop {\lim }\limits_{h \to 0} {{|h|} \over h} = \left\{ {\matrix{ 1 & {if\,h \to {0^ + }} \cr { - 1} & {if\,h \to {0^ - }} \cr } } \right.$$
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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