JEE Advanced 2015 Paper 1Question 1
Let $g:\mathbb{R}\to\mathbb{R}$ be a differentiable function with $g(0)=0$, $g'(0)=0$ and $g'(1)\ne0$. Let $$f(x)=\begin{cases}\frac{x}{|x|}g(x), & x\ne0\\ 0, & x=0\end{cases}$$ 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?
Differentiation and its Application
Solution
Concept. Check each function at $0$ with the definition of the derivative, using $g(x)=o(x)$ (since $g(0)=g'(0)=0$). Formulas used. (1) $g(0)=g'(0)=0\Rightarrow\dfrac{g(x)}{x}\to0$ as $x\to0$Step 1 - (A). $\dfrac{f(x)-f(0)}{x}=\dfrac{g(x)}{|x|}$. By (1), its absolute value tends to $0$. So $f'(0)=0$. True. Step 2 - (B). $h=e^{|x|}$ has one-sided derivatives $1$ and $-1$ at $0$. False. Step 3 - (C). $h(x)=e^{|x|}>0$, so $f(h(x))=g\left(e^{|x|}\right)$. Its one-sided derivatives at $0$ are $g'(1)$ and $-g'(1)$, which differ because $g'(1)\ne0$. False. Step 4 - (D). $h(f(x))=e^{|g(x)|}$, and $\dfrac{e^{|g(x)|}-1}{x}\approx\dfrac{|g(x)|}{x}\to0$ by (1). So it is differentiable at $0$. True. Answer: (A), (D)