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Home/JEE MAINS/mathematics/relations and functions
Curated PYQ Collection

Top 50 Most Repeated RELATIONS AND FUNCTIONS PYQs | JEE MAINS

A curated collection of the most important questions from RELATIONS AND FUNCTIONS, fully solved with step-by-step concepts to prepare for JEE MAINS.

Question #1

Practice Question

A.(a, a) ∈ R for every a ∈ A.
B.(a, b) ∈ R implies (b, a) ∈ R.
C.(a, b) ∈ R and (b, c) ∈ R imply (a, c) ∈ R.
D.No element of A is related to any other element.

Concept Applied

Reflexivity requires that each element of the set be related to itself. The second option defines symmetry, the third defines transitivity, and the fo...

Read Full Step-by-Step Solution →

Question #2

Practice Question

A.f⁻¹(x)= (x+5)/3 and its domain is \(\mathbb{R}\).
B.f⁻¹(x)= (x-5)/3 and its domain is \(\mathbb{R}\).
C.f⁻¹(x)= (x+5)/3 and its domain is \(\mathbb{R}\setminus\{5\}\).
D.f⁻¹(x)= (x-5)/3 and its domain is \(\mathbb{R}\setminus\{5\}\).

Concept Applied

Set \(y=3x-5\). Solving for \(x\) gives \(x=(y+5)/3\). Since the original linear function is bijective on \(\mathbb{R}\), the inverse is defined for a...

Read Full Step-by-Step Solution →

Question #3

Practice Question

A.$f^{-1}(x)=\dfrac{x+5}{3}$ and $f$ is invertible.
B.$f^{-1}(x)=\dfrac{x-5}{3}$ and $f$ is not invertible.
C.$f^{-1}(x)=\dfrac{x+5}{3}$ and $f$ is not invertible.
D.$f^{-1}(x)=\dfrac{x-5}{3}$ and $f$ is invertible.

Concept Applied

A linear function with non‑zero slope is one‑to‑one, hence invertible. Solving $y=3x-5$ for $x$ gives $x=(y+5)/3$, so $f^{-1}(x)=(x+5)/3$....

Read Full Step-by-Step Solution →

Question #4

Practice Question

Concept Applied

Bijective functions from set to itself are permutations. For 3 elements, 3! = 6....

Read Full Step-by-Step Solution →

Question #5

Practice Question

A.Symmetric about the y‑axis
B.Symmetric about the x‑axis
C.Symmetric about the origin
D.No symmetry

Concept Applied

For an odd function, f(−x) = −f(x). Substituting −x gives (−x)³ − 4(−x) = −x³ + 4x = −(x³ − 4x) = −f(x), confirming odd symmetry. Hence the graph is s...

Read Full Step-by-Step Solution →

Question #6

Practice Question

Concept Applied

$g(x) > 1$ and $x \ne 2$ ⇒ $x < 1.5$ or $x > 3$; only $x>3$ valid....

Read Full Step-by-Step Solution →

Question #7

Practice Question

A.Reflexive
B.Symmetric
C.Transitive
D.Antisymmetric

Concept Applied

R is an equivalence relation: reflexive (a-a=0), symmetric (if 3|a-b then 3|b-a), transitive (if 3|a-b and 3|b-c then 3|a-c)....

Read Full Step-by-Step Solution →

Question #8

Practice Question

Concept Applied

$f(g(3))=16$, $g(f(3))=10$, difference is 6....

Read Full Step-by-Step Solution →

Question #9

Practice Question

A.R = {(a,b) | a - b is even}
B.R = {(a,b) | a divides b}
C.R = {(a,b) | a ≤ b}
D.R = {(a,b) | a·b > 0}

Concept Applied

An equivalence relation must be reflexive, symmetric, and transitive. The relation defined by an even difference satisfies all three properties, where...

Read Full Step-by-Step Solution →

Question #10

Practice Question

A.$\displaystyle \frac{4}{3}$ only
B.$\displaystyle \frac{5}{3}$ only
C.$\displaystyle \frac{4}{3}\text{ and }\frac{5}{3}$
D.No real solution

Concept Applied

Let $x=n+f$ with $n=\lfloor x\rfloor$ and $0\le f<1$. Then $\lfloor 2x\rfloor =2n+\lfloor 2f\rfloor$. Solving

...

Read Full Step-by-Step Solution →

Question #11

Practice Question

A.R = \{(1,1),(2,2),(3,3),(4,4),(1,2),(2,1)\}
B.R = \{(1,1),(2,2),(3,3),(4,4)\}
C.R = \{(1,2),(2,3),(3,1)\}
D.R = \{(1,1),(2,2),(3,3),(4,4),(1,3),(3,1)\}

Concept Applied

A reflexive relation must contain (a,a) for every element a. Symmetry requires that if (a,b) is present, (b,a) must also be present. The given set inc...

Read Full Step-by-Step Solution →

Question #12

Practice Question

Concept Applied

Composite function...

Read Full Step-by-Step Solution →

Question #13

Practice Question

Concept Applied

First compute \(g(2)=2^{2}-1=3\). Then evaluate \(f(3)=2\times3+3=9\). Hence \((f\circ g)(2)=9\)....

Read Full Step-by-Step Solution →

Question #14

Practice Question

A.$(-2, 2)$
B.$(-\infty, -2) \cup (2, \infty)$
C.$[-2, 2]$
D.$R$

Concept Applied

$f$ is one-one if $D \le 0$.

6-4y^2 \le 0 \implies |y| \ge 2$. Not one-one for $|y|<2$....

Read Full Step-by-Step Solution →

Question #15

Practice Question

A.(a, a) ∈ R for every a ∈ A.
B.(a, b) ∈ R implies (b, a) ∈ R.
C.(a, b) ∈ R and (b, c) ∈ R imply (a, c) ∈ R.
D.No element of A is related to any other element.

Concept Applied

Reflexivity requires that each element of the set be related to itself. The second option defines symmetry, the third defines transitivity, and the fo...

Read Full Step-by-Step Solution →

Question #16

Practice Question

A.f⁻¹(x)= (x+5)/3 and its domain is \(\mathbb{R}\).
B.f⁻¹(x)= (x-5)/3 and its domain is \(\mathbb{R}\).
C.f⁻¹(x)= (x+5)/3 and its domain is \(\mathbb{R}\setminus\{5\}\).
D.f⁻¹(x)= (x-5)/3 and its domain is \(\mathbb{R}\setminus\{5\}\).

Concept Applied

Set \(y=3x-5\). Solving for \(x\) gives \(x=(y+5)/3\). Since the original linear function is bijective on \(\mathbb{R}\), the inverse is defined for a...

Read Full Step-by-Step Solution →

Question #17

Practice Question

A.$f^{-1}(x)=\dfrac{x+5}{3}$ and $f$ is invertible.
B.$f^{-1}(x)=\dfrac{x-5}{3}$ and $f$ is not invertible.
C.$f^{-1}(x)=\dfrac{x+5}{3}$ and $f$ is not invertible.
D.$f^{-1}(x)=\dfrac{x-5}{3}$ and $f$ is invertible.

Concept Applied

A linear function with non‑zero slope is one‑to‑one, hence invertible. Solving $y=3x-5$ for $x$ gives $x=(y+5)/3$, so $f^{-1}(x)=(x+5)/3$....

Read Full Step-by-Step Solution →

Question #18

Practice Question

Concept Applied

Bijective functions from set to itself are permutations. For 3 elements, 3! = 6....

Read Full Step-by-Step Solution →

Question #19

Practice Question

A.Symmetric about the y‑axis
B.Symmetric about the x‑axis
C.Symmetric about the origin
D.No symmetry

Concept Applied

For an odd function, f(−x) = −f(x). Substituting −x gives (−x)³ − 4(−x) = −x³ + 4x = −(x³ − 4x) = −f(x), confirming odd symmetry. Hence the graph is s...

Read Full Step-by-Step Solution →

Question #20

Practice Question

Concept Applied

$g(x) > 1$ and $x \ne 2$ ⇒ $x < 1.5$ or $x > 3$; only $x>3$ valid....

Read Full Step-by-Step Solution →

Question #21

Practice Question

A.Reflexive
B.Symmetric
C.Transitive
D.Antisymmetric

Concept Applied

R is an equivalence relation: reflexive (a-a=0), symmetric (if 3|a-b then 3|b-a), transitive (if 3|a-b and 3|b-c then 3|a-c)....

Read Full Step-by-Step Solution →

Question #22

Practice Question

Concept Applied

$f(g(3))=16$, $g(f(3))=10$, difference is 6....

Read Full Step-by-Step Solution →

Question #23

Practice Question

A.R = {(a,b) | a - b is even}
B.R = {(a,b) | a divides b}
C.R = {(a,b) | a ≤ b}
D.R = {(a,b) | a·b > 0}

Concept Applied

An equivalence relation must be reflexive, symmetric, and transitive. The relation defined by an even difference satisfies all three properties, where...

Read Full Step-by-Step Solution →

Question #24

Practice Question

A.$\displaystyle \frac{4}{3}$ only
B.$\displaystyle \frac{5}{3}$ only
C.$\displaystyle \frac{4}{3}\text{ and }\frac{5}{3}$
D.No real solution

Concept Applied

Let $x=n+f$ with $n=\lfloor x\rfloor$ and $0\le f<1$. Then $\lfloor 2x\rfloor =2n+\lfloor 2f\rfloor$. Solving

...

Read Full Step-by-Step Solution →

Question #25

Practice Question

A.R = \{(1,1),(2,2),(3,3),(4,4),(1,2),(2,1)\}
B.R = \{(1,1),(2,2),(3,3),(4,4)\}
C.R = \{(1,2),(2,3),(3,1)\}
D.R = \{(1,1),(2,2),(3,3),(4,4),(1,3),(3,1)\}

Concept Applied

A reflexive relation must contain (a,a) for every element a. Symmetry requires that if (a,b) is present, (b,a) must also be present. The given set inc...

Read Full Step-by-Step Solution →

Question #26

Practice Question

Concept Applied

Composite function...

Read Full Step-by-Step Solution →

Question #27

Practice Question

Concept Applied

First compute \(g(2)=2^{2}-1=3\). Then evaluate \(f(3)=2\times3+3=9\). Hence \((f\circ g)(2)=9\)....

Read Full Step-by-Step Solution →

Question #28

Practice Question

A.$(-2, 2)$
B.$(-\infty, -2) \cup (2, \infty)$
C.$[-2, 2]$
D.$R$

Concept Applied

$f$ is one-one if $D \le 0$.

6-4y^2 \le 0 \implies |y| \ge 2$. Not one-one for $|y|<2$....

Read Full Step-by-Step Solution →

Question #29

Practice Question

A.(a, a) ∈ R for every a ∈ A.
B.(a, b) ∈ R implies (b, a) ∈ R.
C.(a, b) ∈ R and (b, c) ∈ R imply (a, c) ∈ R.
D.No element of A is related to any other element.

Concept Applied

Reflexivity requires that each element of the set be related to itself. The second option defines symmetry, the third defines transitivity, and the fo...

Read Full Step-by-Step Solution →

Question #30

Practice Question

A.f⁻¹(x)= (x+5)/3 and its domain is \(\mathbb{R}\).
B.f⁻¹(x)= (x-5)/3 and its domain is \(\mathbb{R}\).
C.f⁻¹(x)= (x+5)/3 and its domain is \(\mathbb{R}\setminus\{5\}\).
D.f⁻¹(x)= (x-5)/3 and its domain is \(\mathbb{R}\setminus\{5\}\).

Concept Applied

Set \(y=3x-5\). Solving for \(x\) gives \(x=(y+5)/3\). Since the original linear function is bijective on \(\mathbb{R}\), the inverse is defined for a...

Read Full Step-by-Step Solution →

Question #31

Practice Question

A.$f^{-1}(x)=\dfrac{x+5}{3}$ and $f$ is invertible.
B.$f^{-1}(x)=\dfrac{x-5}{3}$ and $f$ is not invertible.
C.$f^{-1}(x)=\dfrac{x+5}{3}$ and $f$ is not invertible.
D.$f^{-1}(x)=\dfrac{x-5}{3}$ and $f$ is invertible.

Concept Applied

A linear function with non‑zero slope is one‑to‑one, hence invertible. Solving $y=3x-5$ for $x$ gives $x=(y+5)/3$, so $f^{-1}(x)=(x+5)/3$....

Read Full Step-by-Step Solution →

Question #32

Practice Question

Concept Applied

Bijective functions from set to itself are permutations. For 3 elements, 3! = 6....

Read Full Step-by-Step Solution →

Question #33

Practice Question

A.Symmetric about the y‑axis
B.Symmetric about the x‑axis
C.Symmetric about the origin
D.No symmetry

Concept Applied

For an odd function, f(−x) = −f(x). Substituting −x gives (−x)³ − 4(−x) = −x³ + 4x = −(x³ − 4x) = −f(x), confirming odd symmetry. Hence the graph is s...

Read Full Step-by-Step Solution →

Question #34

Practice Question

Concept Applied

$g(x) > 1$ and $x \ne 2$ ⇒ $x < 1.5$ or $x > 3$; only $x>3$ valid....

Read Full Step-by-Step Solution →

Question #35

Practice Question

A.Reflexive
B.Symmetric
C.Transitive
D.Antisymmetric

Concept Applied

R is an equivalence relation: reflexive (a-a=0), symmetric (if 3|a-b then 3|b-a), transitive (if 3|a-b and 3|b-c then 3|a-c)....

Read Full Step-by-Step Solution →

Question #36

Practice Question

Concept Applied

$f(g(3))=16$, $g(f(3))=10$, difference is 6....

Read Full Step-by-Step Solution →

Question #37

Practice Question

A.R = {(a,b) | a - b is even}
B.R = {(a,b) | a divides b}
C.R = {(a,b) | a ≤ b}
D.R = {(a,b) | a·b > 0}

Concept Applied

An equivalence relation must be reflexive, symmetric, and transitive. The relation defined by an even difference satisfies all three properties, where...

Read Full Step-by-Step Solution →

Question #38

Practice Question

A.$\displaystyle \frac{4}{3}$ only
B.$\displaystyle \frac{5}{3}$ only
C.$\displaystyle \frac{4}{3}\text{ and }\frac{5}{3}$
D.No real solution

Concept Applied

Let $x=n+f$ with $n=\lfloor x\rfloor$ and $0\le f<1$. Then $\lfloor 2x\rfloor =2n+\lfloor 2f\rfloor$. Solving

...

Read Full Step-by-Step Solution →

Question #39

Practice Question

A.R = \{(1,1),(2,2),(3,3),(4,4),(1,2),(2,1)\}
B.R = \{(1,1),(2,2),(3,3),(4,4)\}
C.R = \{(1,2),(2,3),(3,1)\}
D.R = \{(1,1),(2,2),(3,3),(4,4),(1,3),(3,1)\}

Concept Applied

A reflexive relation must contain (a,a) for every element a. Symmetry requires that if (a,b) is present, (b,a) must also be present. The given set inc...

Read Full Step-by-Step Solution →

Question #40

Practice Question

Concept Applied

Composite function...

Read Full Step-by-Step Solution →

Question #41

Practice Question

Concept Applied

First compute \(g(2)=2^{2}-1=3\). Then evaluate \(f(3)=2\times3+3=9\). Hence \((f\circ g)(2)=9\)....

Read Full Step-by-Step Solution →

Question #42

Practice Question

A.$(-2, 2)$
B.$(-\infty, -2) \cup (2, \infty)$
C.$[-2, 2]$
D.$R$

Concept Applied

$f$ is one-one if $D \le 0$.

6-4y^2 \le 0 \implies |y| \ge 2$. Not one-one for $|y|<2$....

Read Full Step-by-Step Solution →

Question #43

Practice Question

A.(a, a) ∈ R for every a ∈ A.
B.(a, b) ∈ R implies (b, a) ∈ R.
C.(a, b) ∈ R and (b, c) ∈ R imply (a, c) ∈ R.
D.No element of A is related to any other element.

Concept Applied

Reflexivity requires that each element of the set be related to itself. The second option defines symmetry, the third defines transitivity, and the fo...

Read Full Step-by-Step Solution →

Question #44

Practice Question

A.f⁻¹(x)= (x+5)/3 and its domain is \(\mathbb{R}\).
B.f⁻¹(x)= (x-5)/3 and its domain is \(\mathbb{R}\).
C.f⁻¹(x)= (x+5)/3 and its domain is \(\mathbb{R}\setminus\{5\}\).
D.f⁻¹(x)= (x-5)/3 and its domain is \(\mathbb{R}\setminus\{5\}\).

Concept Applied

Set \(y=3x-5\). Solving for \(x\) gives \(x=(y+5)/3\). Since the original linear function is bijective on \(\mathbb{R}\), the inverse is defined for a...

Read Full Step-by-Step Solution →

Question #45

Practice Question

A.$f^{-1}(x)=\dfrac{x+5}{3}$ and $f$ is invertible.
B.$f^{-1}(x)=\dfrac{x-5}{3}$ and $f$ is not invertible.
C.$f^{-1}(x)=\dfrac{x+5}{3}$ and $f$ is not invertible.
D.$f^{-1}(x)=\dfrac{x-5}{3}$ and $f$ is invertible.

Concept Applied

A linear function with non‑zero slope is one‑to‑one, hence invertible. Solving $y=3x-5$ for $x$ gives $x=(y+5)/3$, so $f^{-1}(x)=(x+5)/3$....

Read Full Step-by-Step Solution →

Question #46

Practice Question

Concept Applied

Bijective functions from set to itself are permutations. For 3 elements, 3! = 6....

Read Full Step-by-Step Solution →

Question #47

Practice Question

A.Symmetric about the y‑axis
B.Symmetric about the x‑axis
C.Symmetric about the origin
D.No symmetry

Concept Applied

For an odd function, f(−x) = −f(x). Substituting −x gives (−x)³ − 4(−x) = −x³ + 4x = −(x³ − 4x) = −f(x), confirming odd symmetry. Hence the graph is s...

Read Full Step-by-Step Solution →

Question #48

Practice Question

Concept Applied

$g(x) > 1$ and $x \ne 2$ ⇒ $x < 1.5$ or $x > 3$; only $x>3$ valid....

Read Full Step-by-Step Solution →

Question #49

Practice Question

A.Reflexive
B.Symmetric
C.Transitive
D.Antisymmetric

Concept Applied

R is an equivalence relation: reflexive (a-a=0), symmetric (if 3|a-b then 3|b-a), transitive (if 3|a-b and 3|b-c then 3|a-c)....

Read Full Step-by-Step Solution →

Question #50

Practice Question

Concept Applied

$f(g(3))=16$, $g(f(3))=10$, difference is 6....

Read Full Step-by-Step Solution →
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