2 marks, −0.667MathematicsSets, relations and functionsOne-to-one, onto and invertible functionsModerateabout 1 min

IIT JAM 2022 Question 13: Sets, relations and functions

Let $f:[0,\infty)\to\mathbb{R}$ be defined by $f(x) = \dfrac{x+1}{x+2}$ for all $x$ in the domain. Then $f$ is

Tap an option to check your answer.

Answer and worked solution

Answer: (B)

On $[0,\infty)$ , $f'(x) = \dfrac{(x+2)-(x+1)}{(x+2)^2} = \dfrac{1}{(x+2)^2} > 0$ , so $f$ is strictly increasing and therefore one-one.

As $x$ runs over $[0,\infty)$ , $f$ increases from $f(0)=\tfrac12$ towards the limit $1$ , giving range $\left[\tfrac12, 1\right)$ . This is not all of $\mathbb{R}$ , so $f$ is not onto. The answer is (B).

The trap: (A) one-one and onto.

The codomain is all of $\mathbb{R}$ , not the range $\left[\tfrac12,1\right)$ ; forgetting that the range is a bounded interval leads to wrongly calling $f$ onto.

Plain-text version

Let f:[0,∞)→ℝ be defined by f(x) = (x+1)/(x+2) for all x in the domain. Then f is (A) one-one and onto. (B) one-one but not onto. (C) onto but not one-one. (D) neither one-one nor onto.

Also written as: Let f:[0, inf)->R be defined by f(x) = (x+1)/(x+2) for all x in the domain. Then f is

Sets, relations and functions has come up 65 times in 6 exams since 2004.

What this question tests

This question is on One-to-one, onto and invertible functions, part of Sets, relations and functions in Mathematics.

Injective, surjective and bijective functions and when an inverse exists.

TestStop has 11 past-paper questions on this subtopic.

Take this paper as a timed mock More on One-to-one, onto and invertible functions

Question from the IIT JAM 2022 paper. Solutions are TestStop's own. Not affiliated with or endorsed by any examination authority.