JEE Advance - Mathematics (2015 - Paper 1 Offline - No. 19)
Let $$g:R \to R$$ be a differentiable function with $$g(0) = 0$$, $$g'(0) = 0$$ and $$g'(1) \ne 0$$. Let
$$f(x) = \left\{ {\matrix{ {{x \over {|x|}}g(x),} & {x \ne 0} \cr {0,} & {x = 0} \cr } } \right.$$
and $$h(x) = {e^{|x|}}$$ for all $$x \in R$$. Let $$(f\, \circ \,h)(x)$$ denote $$f(h(x))$$ and $$(h\, \circ \,f)(x)$$ denote $$f(f(x))$$. Then which of the following is (are) true?
Explanation
Rewrite f as
$$f(x) = \left\{ {\matrix{ {g(x),} & {x > 0} \cr 0 & {x = 0} \cr { - g(x),} & {x < 0} \cr } } \right.$$
We have, $$f'(x) = \left\{ {\matrix{ {g'(x),} & {x > 0} \cr { - g'(x),} & {x < 0} \cr } } \right.$$
At x = 0
$$f'(0) = \mathop {\lim }\limits_{h \to 0} {{f(0 + h) - f(0)} \over h}$$
$$ = \mathop {\lim }\limits_{h \to 0} {{g(h) - 0} \over h}$$
$$ = \mathop {\lim }\limits_{h \to 0} {{g(h) - g(0)} \over h} = g'(0)$$
$$f'(x) = \left\{ {\matrix{ {g'(x),} & {x \ge 0} \cr { - g'(x),} & {x < 0} \cr } } \right.$$
f is differentiable at x = 0.
$$\therefore$$ Option (a) is correct.
(b) $$h(x) = {e^{|x|}} = \left\{ {\matrix{ {{e^x},} & {x \ge 0} \cr {{e^{ - x}},} & {x < 0} \cr } } \right.$$
$$ \Rightarrow h'(x) = \left\{ {\matrix{ {{e^x},} & {x \ge 0} \cr { - {e^{ - x}},} & {x < 0} \cr } } \right.$$
$$ \Rightarrow h'({0^ + }) = 1$$
and $$h'({0^ - }) = - 1$$
So, h(x) is not differentiable at x = 0.
$$\therefore$$ Option (b) is not correct.
(c) $$(foh)(x) = f\{ h(x)\} $$, as $$h(x) > 0$$
$$ = \left\{ {\matrix{ {g({e^x}),} & {x \ge 0} \cr {g({e^{ - x}}),} & {x < 0} \cr } } \right.$$
$$ \Rightarrow (foh)'(x) = \left\{ {\matrix{ {{e^x}g'({e^x}),} & {x \ge 0} \cr { - {e^x}g'({e^{ - x}}),} & {x < 0} \cr } } \right.$$
$$ \Rightarrow (foh)'({0^ + }) = g'(1),(foh)'({0^ - })$$
$$ = - g'(1)$$
So, $$(hof)(x)$$ is not differentiable at
$$x = 0$$.
$$\therefore$$ Option (c) is not correct.
(d) $$(hof)(x) = {e^{\left| {f(x)} \right|}} = \left\{ {\matrix{ {{e^{\left| {g(x)} \right|}},} & {x \ne 0} \cr {{e^0} = 1,} & {x = 0} \cr } } \right.$$
Now, $$(hof)'(0) = \mathop {\lim }\limits_{h \to 0} {{{e^{\left| {g(x)} \right|}} - 1} \over x}$$
$$ = \mathop {\lim }\limits_{h \to 0} {{{e^{\left| {g(x)} \right|}} - 1} \over {|g(x)|}}\,.\,{{|g(x)|} \over x}$$
$$ = \mathop {\lim }\limits_{h \to 0} {{{e^{\left| {g(x)} \right|}} - 1} \over {|g(x)|}}\,.\,\mathop {\lim }\limits_{h \to 0} {{\left| {g(x)| - 0} \right|} \over {|x|}}\,.\mathop {\lim }\limits_{h \to 0} {{|x|} \over x}$$
$$ = 1\,.\,g'(0)\,.\,\mathop {\lim }\limits_{h \to 0} {{|x|} \over x}$$
= 0, as g'(0) = 0
$$\therefore$$ Option (d) is correct.
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