JEE Advanced 2020 · Paper 2 · Mathematics Q5

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MathematicsInteger+3 / -1Question 41 of 54

Let the functions $f:(-1,1)\to\mathbb R$ and $g:(-1,1)\to(-1,1)$ be defined by

$f(x) = |2x-1| + |2x+1|\quad\text{and}\quad g(x) = x - [x],$

where $[x]$ denotes the greatest integer less than or equal to $x$ . Let $f\circ g:(-1,1)\to\mathbb R$ be the composite function defined by $(f\circ g)(x) = f(g(x))$ . Suppose $c$ is the number of points in the interval $(-1,1)$ at which $f\circ g$ is NOT continuous, and suppose $d$ is the number of points in the interval $(-1,1)$ at which $f\circ g$ is NOT differentiable. Then the value of $c + d$ is ______

Your answer

Answer and worked solution

Official answer: 4

The pieces.

$g(x)=x+1$ on $(-1,0)$ and $g(x)=x$ on $[0,1)$ ; its values lie in $[0,1)$ .
On $[0,1)$ , $f(t)=2$ for $t<\frac12$ and $f(t)=4t$ for $t\ge\frac12$ .

The composite.
$f\circ g(x)=\begin{cases}2,&-1<x<-\frac12\\4(x+1),&-\frac12\le x<0\\2,&0\le x<\frac12\\4x,&\frac12\le x<1\end{cases}$

Checking the joins.

At $x=-\frac12$ : both sides give 2, so it is continuous, but the slopes 0 and 4 differ: not differentiable.
At $x=0$ : the left limit is 4 and the value is 2: discontinuous, hence not differentiable.
At $x=\frac12$ : continuous, with slopes 0 and 4: not differentiable.

So $c=1$ and $d=3$ , and $c+d=4$ .

Answer: 4

Topics:Differentiability Continuity Composition and inverse functions

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JEE Advanced 2020 · Paper 2, Mathematics question 5