JEE Advanced 2015 · Paper 1 · Mathematics Q56

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MathematicsMultiple correct+4 / -2Question 56 of 60

Let $g:\mathbb R\to\mathbb R$ be a differentiable function with $g(0)=0$ , $g'(0)=0$ and $g'(1)\ne0$ . Let
$f(x)=\begin{cases}\dfrac{x}{|x|}g(x),&x\ne0\\0,&x=0\end{cases}$
and $h(x)=e^{|x|}$ for all $x\in\mathbb R$ . Let $(f\circ h)(x)$ denote $f(h(x))$ and $(h\circ f)(x)$ denote $h(f(x))$ . Then which of the following is (are) true?

f

is differentiable at

x=0

h

is differentiable at

x=0

f\circ h

is differentiable at

x=0

h\circ f

is differentiable at

x=0

Answer and worked solution

Official answer: A, D

(A) $\dfrac{f(h)-f(0)}h=\dfrac{\operatorname{sgn}(h)\,g(h)}h\to0$ , since $g(h)/h\to g'(0)=0$ . Differentiable. True.
(B) $e^{|x|}$ has a corner at 0. False.
(C) $f(h(x))=g(e^{|x|})$ (as $e^{|x|}>0$ ). Its one-sided derivatives at 0 are $\pm g'(1)\ne0$ . Not differentiable. False.
(D) $h(f(x))=e^{|g(x)|}$ , and $|g(x)|=o(x)$ near 0. The derivative is 0. True.

(A), (D)

Topics:Differentiability Composition and inverse functions

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JEE Advanced 2015 · Paper 1, Mathematics question 56