JEE Main 2022 · 26 Jun · Shift 2 · Mathematics Q1

MathematicsSingle correct+4 / -1Question 1 of 90

Let $f:\mathbb R\to\mathbb R$ be defined as $f(x)=x-1$ and $g:\mathbb R-\{1,-1\}\to\mathbb R$ be defined as $g(x)=\dfrac{x^2}{x^2-1}$ .

Then the function $f\circ g$ is :

one-one but not onto

onto but not one-one

both one-one and onto

neither one-one nor onto

Answer and worked solution

Official answer: D

Compose. $f(g(x))=\dfrac{x^2}{x^2-1}-1=\dfrac1{x^2-1}$ .

One-one? It is even ( $x$ and $-x$ give the same value), so no.

Onto? $x^2-1\in[-1,0)\cup(0,\infty)$ , so the values are $(-\infty,-1]\cup(0,\infty)$ ; $(-1,0]$ is missed. Not onto.

(D)

Topics:Types of functions Composition and inverse functions

JEE Main 2022 · 26 Jun · Shift 2, Mathematics question 1