The modulus function f : R R, given by f(x) =, x, is : Option 1: One-one Option 2: Onto Option 3: One-one and onto both Option 4: Neither one-one nor onto

Answer: Neither one-one nor onto. Solution: Not one-one: f(-1) = 1 = f(1) but -1 ≠ 1. Not onto: range is [0, ∞), but codomain is R (negative values not attained). So f is neither one-one nor onto.

Algebra

Relations & Functions

Easy

The modulus function $f : R \to R$ , given by $f(x) = |x|$ , is :

One-one

Onto

One-one and onto both

Neither one-one nor onto

✅ Correct Option: 4

Not one-one: $f(-1) = 1 = f(1)$ but $-1 \neq 1$ .

Not onto: range is $[0, \infty)$ , but codomain is $\mathbb{R}$ (negative values not attained).

So $f$ is neither one-one nor onto.

2022: 7 Aug Shift 2

Let L be the set of all lines in a plane and R be the relation in L defined as $R = \{(l_1, l_2) : l_1 \text{ is perpendicular to } l_2, \text{ where } l_1, l_2 \in L\}$ . Choose the correct answer :

2026: 18 May Shift 1

If $f(x) = \sin^{-1}\left(\dfrac{2^{x+1}}{1+4^x}\right)$ , then

Match List-I with List-II

List-I	List-II
Value of $x$	$\dfrac{1}{2}f(x)$
(A) $x = 0$	(I) $\tan^{-1}\sqrt{2}$
(B) $x = \dfrac{1}{2}$	(II) $\tan^{-1}4$
(C) $x = 1$	(III) $\dfrac{\pi}{4}$
(D) $x = 2$	(IV) $\tan^{-1}2$

Choose the correct answer from the options given below:

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If $f\left(x+\dfrac{1}{x}\right) = x^2 + \dfrac{1}{x^2}$ then $f(x) =$

2022: 25 May Shift 1

If $f: R \to R$ is defined by $f(x) = \sin x + x$ , then $f(f(x))$ is:

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Relation R on the set $A = \{1, 2, ..., 15\}$ defined as $R = \{(x, y): y - 4x = 0\}$ is

The modulus function f:R→Rf : R \to Rf:R→R, given by f(x)=∣x∣f(x) = |x|f(x)=∣x∣, is :

Solution

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CUET Mathematics Past Year Questions (Topic-Wise)

The modulus function $f : R \to R$ , given by $f(x) = |x|$ , is :