Let f, g : R rightarrow R be defined as : f( x )= x -1 and g ( x )= \ begin matrix e x , & x ge 0 \\ x +1, & x le 0 \\end matrix . . Then the function f( g ( x )) is

Mathematics · Functions

JEE Main 2024 — 5 April, Shift 2 — Question 11

Let $f, \mathrm{~g}: \mathrm{R} \rightarrow \mathrm{R}$ be defined as : $f(\mathrm{x})=|\mathrm{x}-1|$ and $\text{g}\left( \text{x} \right)=\left\{ \begin{matrix}{{\text{e}}^{\text{x}}}, & \text{x}\ge 0 \\\text{x}+1, & \text{x}\le 0 \\\end{matrix} \right.$ . Then the function $f(\mathrm{~g}(\mathrm{x}))$ is

Option A:
neither one-one nor onto.
Correct
Option B:
one-one but not onto.
Option C:
both one-one and onto
Option D:
onto but not one-one.

Answer: A

Step-by-step solution

$f\left( g\left( x \right) \right)=\left| g\left( x \right)-1 \right|$ $\text{fog(x)}=\left[ \begin{matrix}\left| {{e}^{x}}-1 \right| & x\ge 0 \\\left| x+1-1 \right| & x\le 0 \\\end{matrix} \right.$

fog(x)= $\left[ \begin{matrix}{{e}^{x}}-1 & x\ge 0 \\-x & x\le 0 \\\end{matrix} \right.$

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Exam: JEE Main 2024
Paper: 5 April, Shift 2
Subject: Mathematics
Chapter: Functions
Topic: One-One, many-one, onto, into, bijective functions

The number of solutions of sin ^2 x+ (2+2 x-x^2 ) sin x-3(x-1)^2=0 , where -pi leq x leq pi , is

Answer: A

Step-by-step solution

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