Let A=R-\2\ and B=R-\1\ and a function f:A B defined by f(x)=(x-1)/(x-2)\ x A, then the function f(x) is : Option 1: Injective but not surjective Option 2: Surjective but not injective Option 3: Both injective and surjective Option 4: Neither injective nor surjective

Answer: Both injective and surjective. Solution: If f(x_1)=f(x_2), cross-multiplying reduces to x_1=x_2, so f is injective. Solving y=(x-1)/(x-2) gives x=(2y-1)/(y-1), which lies in A for every y≠1, so f is onto B. Hence it is bijective.

Algebra

Relations & Functions

Medium

core

Let $A=R-\{2\}$ and $B=R-\{1\}$ and a function $f:A\to B$ defined by $f(x)=\dfrac{x-1}{x-2}\ \forall x\in A$ , then the function $f(x)$ is :

Injective but not surjective

Surjective but not injective

Both injective and surjective

Neither injective nor surjective

✅ Correct Option: 3

If $f(x_1)=f(x_2)$ , cross-multiplying reduces to $x_1=x_2$ , so $f$ is injective. Solving $y=\dfrac{x-1}{x-2}$ gives $x=\dfrac{2y-1}{y-1}$ , which lies in $A$ for every $y\neq1$ , so $f$ is onto $B$ . Hence it is bijective.

2025: 2 June Shift 1

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

2026: 19 May Shift 2

Let X be a non-empty set and let P be the collection of all subsets of X. Let R be a relation in P, defined by $R = \{(A, B): A \text{ is subset of } B\}$ . Then, R is

A. Reflexive relation

B. Symmetric relation

C. Not a transitive relation

D. Transitive relation

Choose the correct answer from the options given below:

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:

2022: 17 Aug Shift 2

Let $f : [2, \infty) \to \mathbf{R}$ be a function defined by $f(x) = x^2 - 4x + 5$ . The range of $f$ is :

2025: 30 May Shift 1

Let $f: \mathbb{R} \to \mathbb{R}$ be defined as $f(x) = 100x + 1$ , where $\mathbb{R}$ is a set of real numbers, then

2022: 30 Aug Shift 1

Which of the following is true on the basis of above diagram?

Let A=R−{2}A=R-\{2\}A=R−{2} and B=R−{1}B=R-\{1\}B=R−{1} and a function f:A→Bf:A\to Bf:A→B defined by f(x)=x−1x−2 ∀x∈Af(x)=\dfrac{x-1}{x-2}\ \forall x\in Af(x)=x−2x−1​ ∀x∈A, then the function f(x)f(x)f(x) is :

Solution

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CUET Mathematics Past Year Questions (Topic-Wise)

Let $A=R-\{2\}$ and $B=R-\{1\}$ and a function $f:A\to B$ defined by $f(x)=\dfrac{x-1}{x-2}\ \forall x\in A$ , then the function $f(x)$ is :