JEE Main 2022 · 28 Jun · Shift 2 · Mathematics Q6

← JEE Main 2022 · 28 Jun · Shift 2

MathematicsSingle correct+4 / -1Question 6 of 90

Let $f,g:\mathbf R\to\mathbf R$ be functions defined by
$f(x)=\begin{cases}[x], & x<0\\ |1-x|, & x\ge0\end{cases}\quad\text{and}\quad g(x)=\begin{cases}e^x-x, & x<0\\ (x-1)^2-1, & x\ge0\end{cases}$
where $[x]$ denote the greatest integer less than or equal to $x$ . Then, the function $f\circ g$ is discontinuous at exactly :

one point

two points

three points

four points

Answer and worked solution

Official answer: B

$x<0$ . $g=e^x-x>1$ , so $f(g)=g-1$ , continuous, tending to 0 at $0^-$ .

$0\le x<2$ . $g=(x-1)^2-1\in[-1,0)$ for $0<x<2$ , so $f(g)=[g]=-1$ ; but $f(g(0))=f(0)=1$ .

$x\ge2$ . $g\ge0$ , $f(g)=|1-g|$ , equal to 1 at $x=2$ while the left limit is $-1$ .

Breaks at $x=0$ and $x=2$ only.

(B)

Topics:Continuity Composition and inverse functions

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JEE Main 2022 · 28 Jun · Shift 2, Mathematics question 6