JEE Advanced 2014 · Paper 1 · Mathematics Q54

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MathematicsInteger+3Question 54 of 60

Let $f:\mathbb R\to\mathbb R$ and $g:\mathbb R\to\mathbb R$ be respectively given by $f(x)=|x|+1$ and $g(x)=x^2+1$ . Define $h:\mathbb R\to\mathbb R$ by
$h(x)=\begin{cases}\max\{f(x),g(x)\}&\text{if }x\le0,\\\min\{f(x),g(x)\}&\text{if }x>0.\end{cases}$
The number of points at which $h(x)$ is not differentiable is

Your answer

Answer and worked solution

Official answer: 3

Compare $f$ and $g$ .

For $x\le0$ : $g-f=x^2+x=x(x+1)$ , so $g\ge f$ on $(-\infty,-1]$ and $f>g$ on $(-1,0)$ . The maximum is $g$ on $(-\infty,-1]$ and $f=1-x$ on $[-1,0]$ .
For $x>0$ : $g-f=x^2-x$ , so $g<f$ on $(0,1)$ and $g>f$ beyond 1. The minimum is $g$ on $(0,1]$ and $f=1+x$ on $[1,\infty)$ .

Corners. $h$ is continuous (all pieces agree at the joins). One-sided derivatives:

$x=-1$ : $g'=-2$ , $f'=-1$ . Corner.
$x=0$ : left $-1$ (from $1-x$ ), right $0$ (from $x^2+1$ ). Corner.
$x=1$ : $g'=2$ , $f'=1$ . Corner.

Answer: 3

Topics:Differentiability Graphs of functions

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JEE Advanced 2014 · Paper 1, Mathematics question 54