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Q1:

2026: 31 May Shift 2

Relations & Functions

Easy

Core

A relation RR on set of real numbers (R)(\mathbb{R}) is defined as R={(a,b):a≤b,a,b∈R}R = \{(a,b): a \le b, a,b \in \mathbb{R}\}. The relation is :

Answer options
Option 3
Correct Answer
Explanation for 2026: 31 May Shift 2 MAT question 1

Q2:

2026: 31 May Shift 2

Relations & Functions

Medium

Core

The function f:R→[0,1]f: R \to [0,1] defined by f(x)=x2x2+1f(x) = \dfrac{x^2}{x^2+1} is: (Where RR is set of real numbers)

A. One-one

B. one-one and onto

C. onto

D. Not one-one

Choose the correct answer from the options given below:

Answer options
Option 4
Correct Answer
Explanation for 2026: 31 May Shift 2 MAT question 2

Q3:

2026: 30 May Shift 2

Relations & Functions

Easy

Core

The greatest integer function f:R→Rf: R\to R defined by f(x)=[x]f(x) = [x], then which of the following are true?

(A) f(x)f(x) is one - one and onto.

(B) Domain of f(x)=Rf(x) = R (the set of real numbers).

(C) f(x)f(x) is neither one-one nor onto.

(D) Range of f(x)=Zf(x) = Z (The set of integers).

Choose the correct answer from the options given below:

Answer options
Option 2
Correct Answer
Explanation for 2026: 30 May Shift 2 MAT question 3

Q5:

2026: 30 May Shift 1

Relations & Functions

Medium

Core

The relation R in the set of real numbers, defined as R = {(a, b): 1 + ab > 0}, is:

Answer options
Option 1
Correct Answer
Explanation for 2026: 30 May Shift 1 MAT question 5

Q6:

2026: 30 May Shift 1

Relations & Functions

Easy

Core

The signum function f:R→Rf:R\to R, defined by

f(x)={−1,x<00,x=01,x>0f(x)=\begin{cases}-1, & x<0\\0, & x=0\\1, & x>0\end{cases} is: (Where R is set of real numbers)

Answer options
Option 4
Correct Answer
Explanation for 2026: 30 May Shift 1 MAT question 6

Q7:

2026: 29 May Shift 2

Relations & Functions

Medium

Common

For the function f(x)=∣x−1∣+∣x−2∣+∣x−3∣f(x) = |x-1| + |x-2| + |x-3|

Match the LIST-I with LIST-II

LIST-I
Function f(x)=f(x) =
LIST-II
For Interval
A.−3x+6-3x + 6I.x≥3x \geq 3
B.−x+4-x + 4II.x<1x < 1
C.xxIII.1≤x<21 \leq x < 2
D.3x−63x - 6IV.2≤x<32 \leq x < 3

Choose the correct answer from the options given below:

Answer options
Option 1
Correct Answer
Explanation for 2026: 29 May Shift 2 MAT question 7

Q8:

2026: 29 May Shift 2

Relations & Functions

Medium

Core

Let S be the set of all real numbers and let R be a relation on S, defined by aRb⇔∣a−b∣≤1aRb \Leftrightarrow |a-b| \leq 1. Then, R is

Answer options
Option 1
Correct Answer
Explanation for 2026: 29 May Shift 2 MAT question 8

Q9:

2026: 29 May Shift 2

Relations & Functions

Medium

Core

Match the LIST-I with LIST-II (Consider N and R as set of natural numbers and set of real numbers respectively)

LIST-I
Function
LIST-II
A.f:N→N,f(x)=x2f: N \to N, f(x) = x^2 isI.Neither one-one nor onto
B.f:R→R,f(x)=3−2xf: R \to R, f(x) = 3 - 2x isII.one-one but not onto
C.f:R→R,f(x)=x2f: R \to R, f(x) = x^2 isIII.onto but not one-one
D.f:{−1,1}→{1},f(x)=∣x∣f: \{-1,1\} \to \{1\}, f(x) = \vert x \vert isIV.one-one and onto

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Answer options
Option 3
Correct Answer
Explanation for 2026: 29 May Shift 2 MAT question 9

Q10:

2026: 29 May Shift 1

Relations & Functions

Medium

Core

Let R1,R2,R3,R4R_1, R_2, R_3, R_4 be the relations defined on the set A={1,2,3}A=\{1,2,3\}. then

Match List-I with List-II

List-I (Relation)List-II (Type)
(A) R1={(1,1),(2,2),(3,3)}R_1=\{(1,1),(2,2),(3,3)\}(I) Only transitive
(B) R2={(1,2),(2,3),(1,3)}R_2=\{(1,2),(2,3),(1,3)\}(II) Equivalence relation
(C) R3={(1,2),(2,1),(1,1),(2,2)}R_3=\{(1,2),(2,1),(1,1),(2,2)\}(III) Reflexive and symmetric but not transitive
(D) R4={(1,1),(2,2),(3,3),(2,3),(3,2),(1,2),(2,1)}R_4=\{(1,1),(2,2),(3,3),(2,3),(3,2),(1,2),(2,1)\}(IV) Symmetric and transitive but not reflexive

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Q11:

2026: 29 May Shift 1

Relations & Functions

Medium

Core

Consider a function f:[−1,1]→Yf:[-1,1] \to Y defined as f(x)=xx+2f(x)=\dfrac{x}{x+2} and Y=range(f(x))Y=range(f(x)), then which of following statements are correct?

(A) Function f(x)f(x) is one-one

(B) Function f(x)f(x) is not one-one

(C) Function f(x)f(x) is onto

(D) Function f(x)f(x) is invertible

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Q12:

2026: 26 May Shift 2

Relations & Functions

Medium

Core

Let A={1,2,3}A = \{ 1,2,3 \}, then Match List-I with List-II

LIST-ILIST-II
A. R1={(1,1),(2,2),(3,3),(1,2)}R_1 = \{(1,1), (2,2), (3,3), (1,2)\}I. Symmetric but neither reflexive nor transitive
B. R2={(1,1),(3,3),(1,3),(3,1)}R_2 = \{(1,1), (3,3), (1,3), (3,1)\}II. Reflexive and transitive but not symmetric
C. R3={(1,1),(2,2),(3,3),(1,2),(2,1),(2,3),(3,2)}R_3 = \{(1,1), (2,2), (3,3), (1,2), (2,1), (2,3), (3,2)\}III. Symmetric and transitive but not reflexive
D. R4={(2,2),(3,3),(1,2),(2,1)}R_4 = \{(2,2), (3,3), (1,2), (2,1)\}IV. Reflexive and symmetric but not transitive

Choose the correct answer from the options given below:

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Q13:

2026: 26 May Shift 2

Relations & Functions

Easy

Core

If f(x+1x)=x2+1x2f\left(x+\dfrac{1}{x}\right) = x^2 + \dfrac{1}{x^2} then f(x)=f(x) =

Answer options

Q14:

2026: 25 May Shift 1

Relations & Functions

Easy

Core

Let f:N→Nf: \mathbb{N} \to \mathbb{N} defined by f(x)=3x2+2x+7f(x) = 3x^2 + 2x + 7, where N\mathbb{N} is set of natural numbers. Then f(x)f(x) is:

Answer options

Q15:

2026: 25 May Shift 1

Relations & Functions

Medium

Core

For the relations R1,R2,R3,R4R_1, R_2, R_3, R_4 defined on the set A={1,2,3}A = \{1,2,3\},

Match the LIST-I with LIST-II

LIST-ILIST-II
A. R1={(1,2),(2,1),(3,3)}R_1 = \{(1,2),(2,1),(3,3)\}I. Transitive only
B. R2={(1,1),(2,2),(3,3)}R_2 = \{(1,1),(2,2),(3,3)\}II. Neither reflexive nor symmetric nor transitive
C. R3={(1,2),(2,3),(1,3)}R_3 = \{(1,2),(2,3),(1,3)\}III. Symmetric but neither reflexive nor transitive
D. R4={(3,3),(3,1),(1,2)}R_4 = \{(3,3),(3,1),(1,2)\}IV. Equivalence

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Q16:

2026: 22 May Shift 2

Relations & Functions

Medium

Core

Consider a function f:[0,π2]→Rf:\left[0,\frac{\pi}{2}\right] \to R, given by f(x)=sin⁡xf(x)=\sin x and g:[0,π2]→Rg:\left[0,\frac{\pi}{2}\right] \to R, given by g(x)=cos⁡xg(x)=\cos x, Where R is a set of real numbers, Which of the following statements are correct?

(A) f and g are one-one

(B) f and g are not one-one

(C) f+g is not one-one

(D) f+g is one-one

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Q17:

2026: 22 May Shift 2

Relations & Functions

Easy

Core

The relation R on the set A = {1, 2, 3} defined by R = {(1,1), (2,2), (3,3), (1,3), (2,3), (1,2)} is

Answer options

Q18:

2026: 22 May Shift 1

Relations & Functions

Medium

Core

Let X be a non empty set and S be the collection of all subsets of X. A relation R in S is defined by R={(A,B):A⊆B}R=\{(A,B): A\subseteq B\}, where X⊆YX\subseteq Y means XX is proper subset of YY. Then R is:

Answer options

Q19:

2026: 22 May Shift 1

Relations & Functions

Easy

Core

Let a function f:R→Rf:\mathbb{R}\to\mathbb{R} defined as f(x)=x−[x]f(x)=x-[x], (where R\mathbb{R} is set of real numbers & [.][.] denotes greatest integer function) .Then the function:

Answer options

Q20:

2026: 21 May Shift 2

Relations & Functions

Medium

Core

If a function f:R→Tf: R \to T, defined by f(x)=3sin⁡x+cos⁡xf(x) = \sqrt{3}\sin x + \cos x is an onto function, then the interval of TT is

Answer options

Q21:

2026: 21 May Shift 2

Relations & Functions

Easy

Core

The relation R on the set N of natural numbers defined as R={(x,y):y=x+3 and x≤5}R = \{(x,y): y = x+3 \text{ and } x \le 5\} is

Answer options

Q23:

2026: 21 May Shift 1

Relations & Functions

Medium

Core

If A=R−{2}A = R - \{2\} and B=R−{1}B = R - \{1\} and function f:A→Bf: A \rightarrow B defined by f(x)=x−1x−2f(x) = \frac{x-1}{x-2}, then ff is

Answer options

Q24:

2026: 19 May Shift 2

Relations & Functions

Medium

Core

Match the LIST-I with LIST-II

[Where R and Z denotes set of real numbers and set of integers respectively]

LIST-I
Function
LIST-II
Nature
A.f:R→R,f(x)=sin⁡2x+cos⁡2xf: R \to R, f(x) = \sin^2 x + \cos^2 xI.Bijective
B.f:Z→Z,f(x)=x3f: Z \to Z, f(x) = x^3II.Surjective but not injective
C.f:R→R,f(x)=x3−xf: R \to R, f(x) = x^3 - xIII.Injective but not surjective
D.f:R→R,f(x)=3−4xf: R \to R, f(x) = 3 - 4xIV.Neither injective nor surjective

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Answer options

Q25:

2026: 19 May Shift 2

Relations & Functions

Easy

Core

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 is subset of B}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

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Q26:

2026: 19 May Shift 1

Relations & Functions

Medium

Core

For the functions f(x)=x2f(x) = x^2 and g(x)=x3g(x) = x^3, which of the following statements are TRUE?

[where N is the set of natural numbers and Z is the set of integers]

(A) f:N→Nf: N \to N is one-one but not onto.

(B) f:Z→Zf: Z \to Z is neither one-one nor onto.

(C) g:N→Ng: N \to N is one-one but not onto.

(D) g:Z→Zg: Z \to Z is one-one but not onto.

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Answer options

Q27:

2026: 19 May Shift 1

Relations & Functions

Easy

Core

Let A={0,1,2,3}A = \{0, 1, 2, 3\} and R be a relation on set A defined as R={(0,0),(0,1),(0,3),(1,0),(1,1),(2,2),(3,0),(3,3)}R = \{(0,0), (0,1), (0,3), (1,0), (1,1), (2,2), (3,0), (3,3)\}, then the relation R is

Answer options

Q28:

2026: 18 May Shift 2

Relations & Functions

Hard

Core

For the relation R defined below, where R\mathbb{R} and N denotes set of real numbers and set of natural numbers respectively, Match the LIST-I with LIST-II

LIST-I RelationLIST-II Properties of relation hold by R
A. R={(a,b):a,b∈R;a2+b2=4}R=\{(a,b): a,b\in \mathbb{R}; a^2+b^2=4\}I. Neither reflexive, nor symmetric, nor transitive
B. R={(a,b):a,b∈R;1+ab>0}R=\{(a,b): a,b\in \mathbb{R}; 1+ab>0\}II. Equivalence relation
C. R={(a,b):a,b∈N;a+2b=8}R=\{(a,b): a,b\in N; a+2b=8\}III. Symmetric but neither reflexive nor transitive
D. On N×N, relation R defined by (a,b)R(c,d)⇔a+d=b+c(a,b)R(c,d) \Leftrightarrow a+d=b+cIV. Reflexive and symmetric, but not transitive

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Q29:

2026: 18 May Shift 1

Relations & Functions

Medium

Core

If f(x)=sin⁡−1(2x+11+4x)f(x) = \sin^{-1}\left(\dfrac{2^{x+1}}{1+4^x}\right), then

Match List-I with List-II

List-IList-II
Value of xx12f(x)\dfrac{1}{2}f(x)
(A) x=0x = 0(I) tan⁡−12\tan^{-1}\sqrt{2}
(B) x=12x = \dfrac{1}{2}(II) tan⁡−14\tan^{-1}4
(C) x=1x = 1(III) π4\dfrac{\pi}{4}
(D) x=2x = 2(IV) tan⁡−12\tan^{-1}2

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Q30:

2026: 18 May Shift 1

Relations & Functions

Medium

Core

Let a relation RR be defined on a set A={1,2,3,4,5}A = \{1,2,3,4,5\} as R={(a,b):∣a−b∣ is even or zero}R = \{(a,b): |a-b| \text{ is even or zero}\}. Then which of the following statements is not TRUE?

Answer options