Relations and Functions Advanced Problems for JEE Main and WBJEE

Prime Maths

25 Multiple Choice Questions (MCQs) on Relations and Functions for students of class XI and XII preparing for board examinations or JEE Mains, IIT Advanced WBJEE or any other competitive entrance examination.

👨‍🏫 Author: Singh

📞 WA: +91-9038126497

Prime Maths

Advanced Algebra - Relations and Functions

Test your understanding of core concepts.

0

Done

25

Left

0%

Score

📝 Instructions

• This quiz contains 25 multiple choice questions.
• Select only one correct answer per question.
• Use the navigator to jump between questions.
• Submit when you are finished to see results.

Question Navigator

Progress 0%

Question 1

The number of reflexive relations on a set \(A\) of \(n\) elements is equal to

A\( 2^{n^2}\) B\( n^2\) C\( 2^{n(n-1)} \) D\( n^2-n\)

For the relation to be reflexive, \((a_i,a_i) \in A \times A \quad \forall a_i \in A\) So the from the remaining \(n^2 - n\) elements, any number of elements can be choosen to form a subset of \(A \times A\)

Question 2

Let \(X\) be a non-void set. If \(\rho_1\) and \(\rho_2\) be the transitive relations on \(X\), then [ \( \circ\) denotes composition ]

A \(\rho_1 \circ \rho_2\) is transitive relation B \(\rho_1 \circ \rho_2\) is not transitive relation C \(\rho_1 \circ \rho_2\) is equivalence relation D \(\rho_1 \circ \rho_2\) is not any relation on \(X\)

Let's define a set \(X = \{1, 2, 3, 4, 5\}\). Now, let's create two relations on \(X\), \(\rho_1\) and \(\rho_2\), both of which are transitive:Let \(\rho_1 = \{(1, 2), (3, 4)\}\) Let \(\rho_2 = \{(2, 3), (4, 5)\}\) (Note: These are trivially transitive). Now, let's find the composition \(\rho_1 \circ \rho_2\). By definition of composition, \((x, z) \in \rho_1 \circ \rho_2\) if there is some intermediate element \(y\) such that \((x, y) \in \rho_1\) and \((y, z) \in \rho_2\). So, the resulting composition is:\(\rho_1 \circ \rho_2 = \{(1, 3), (3, 5)\}\) which is clearly not transitive.

Question 3

If one root of \(x^2 + px - q^2 = 0\), \(p\) and \(q\) are real, be less than \(2\) and other be greater than \(2\), then

A \( 4+2p+q^2 > 0 \) B \( 4+2p+q^2 \\< 0 \) C \( 4+2p-q^2 > 0 \) D \( 4+2p-q^2 \\< 0 \)

Let \(y=x^2+px-q^2\) as \(x \rightarrow \pm \infty, y \rightarrow \infty\) thus the resulting parabola is open in the positive direction of \(y\)-axis. As \(2\), lies between the roots, \(y\) is less than \(0\) at \(x=2\).

Question 4

If \(R\) and \(Q\) are equivalence relations on set \(A\), then which of the following is not an equivalence relation

A\(R^{-1} \cap Q^{-1}\) B\(Q^{-1}\) C\(R \cup Q\) D\(R \cap Q\)

Let \( A=\{1,2,3\}, P =\{(1,1),(2,2),(3,3),(2,3),(3,2)\}\) and \(Q =\{(1,1),(2,2),(3,3),(1,2),(2,1)\}\). It is easy to see that both \(P \text{ and } Q\) are equivalence relation on \(A\) but \(P \cup Q\) is not an equivalence relation.

Question 5

Let \( \rho \) be a relation defined on set of natural numbers \( \mathbb{N} \), as \( \rho = \{(x, y) \in \mathbb{N} \times \mathbb{N} : 2x + y = 41\} \). Then domain A and range B are

A\( A \subset \{x \in \mathbb{N} : 1 \le x \le 20\} \) and \( B \subset \{y \in \mathbb{N} : 1 \le y \le 39\} \) B\( A = \{x \in \mathbb{N} : 1 \le x \le 15\} \) and \( B = \{y \in \mathbb{N} : 2 \le y \le 30\} \) C\( A \equiv \mathbb{N}, B \equiv \mathbb{Q} \) D\( A \equiv \mathbb{Q}, B \equiv \mathbb{Q} \)

See that \(y= 41 - 2x\) and for \(x \in \{1,2,...,20 \}\), \( y \in \{1,3,5,...,39\}\)

Question 6

In \(\mathbb{R}\), a relation \(\rho\) is defined as follows: \(\forall a,b \in \mathbb{R}\), \(a\rho b \quad \text{holds iff} \quad a^2 -4ab+3b^2=0 \)

A\(\rho\) is equivalence relation B\(\rho\) is only symmetric C\(\rho\) is only reflexive D\(\rho\) is only transitive

See that \(a^2-4ab+3b^2 =0 \implies a=b\) or \(a=3b\)

Question 7

Let \( f : \mathbb{R} \to \mathbb{R} \) be a function defined by \( f(x) = \frac{e^{|x|} - e^{-x}}{e^x + e^{-x}} \), then

A \( f \) is both one-one and onto B \( f \) is one-one but not onto C \( f \) is onto but not one-one D \( f \) is neither one-one nor onto

\Injectivity (One-One): For \( x < 0 \), \( |x| = -x \). $$ f(x) = \frac{e^{-x} - e^{-x}}{e^x + e^{-x}} = 0 $$ Since \( f(-1) = f(-2) = 0 \), multiple inputs yield the same output. Therefore, \( f \) is not one-one. 2. Surjectivity (Onto): For \( x \ge 0 \), \( |x| = x \). $$ f(x) = \frac{e^x - e^{-x}}{e^x + e^{-x}} $$ The range for \( x \ge 0 \) is \( [0, 1) \) ( use limits), the total range of \( f \) is \( [0, 1) \). Since the range \( [0, 1) \neq \) codomain \( \mathbb{R} \), \( f \) is not onto.

Question 8

Let \( f : \mathbb{R} \to \mathbb{R} \) be a function defined by \( f(x) = 2026x^3+2025x^2+2024x+2023 \), then

A \( f \) is both one-one and onto B \( f \) is one-one but not onto C \( f \) is onto but not one-one D \( f \) is neither one-one nor onto

It is easy to see that \(f\) is injective. For surjectivity, think about the nature of roots of a cubic equation.

Question 9

Let \(u + v + w = 3\), \(u, v, w \in \mathbb{R}\) and \(f(x) = ux^2 + vx + w\) be such that \(f(x + y) = f(x) + f(y) + xy\), \(\forall x, y \in \mathbb{R}\). Then \(v\) is equal to

A\(\frac{5}{2}\) B\(\frac{1}{2}\) C\(\frac{1}{\sqrt{2}}\) D\(3\)

Start by substituting \(f(x) = ux^2 + vx + w\) directly into the given equation \(f(x + y) = f(x) + f(y) + xy\). Expand the left side, and then compare the coefficients of the \(xy\) terms and the constant terms on both sides. This will reveal the values of \(u\) and \(w\) first, which you can then use with the condition \(u + v + w = 3\) to find \(v\).

Question 10

Let \(X\) and \(Y\) be two non-empty sets. Let \(f:X \rightarrow Y\) be a function. For \(A \subset X\) and \(B \subset Y\) , define \(f(A)=\{f(x):x \in A\}\) and \(f^{-1}(B)=\{x \in X : f(x) \in B \}\), then

A\(f(f^{-1}(B))=B\) B\(f(f^{-1}(B)) \subset B\) C\(f^{-1}(f(A))=A\) D\(f^{-1}(f(A)) \subset A\)

Collection of all pre-images will be mapped into \(B\). Not every element in \(Y\) has a pre-image. Additional exercise, what will happen if the function is surfective?

Question 11

Range of the function \(f(x)= \frac{x^2+x+2}{x^2+x+1}\ \: x \in \mathbb{R}\)

A\((1,\infty)\) B\((1, 1.5]\) C\((1,\frac{7}{3}]\) D\([1,\frac{7}{3}]\)

Rewrite the function as \(f(x)=1+\frac{1}{x^2+x+1}\)

Question 12

Let \(X=\{v,i,n,o,d\}\) and \(Y=\{p,m\}\). The number of onto ( surjective) functions from \(X\) to \(Y\) is

A\(32\) B\(28 \) C\(30\) D\(36\)

To find the number of onto functions, first calculate the total number of unrestricted mappings from \(\{v, i, n, o, d\}\) to \(\{p, m\}\), which is \(2^5\). Then, simply subtract the cases where the mapping is not onto—specifically, the exact number of scenarios where all \(5\) elements map exclusively to just a single output in the codomain.

Question 13

Find the natural number \(a\) for which \(\sum_{k=1}^{n} f(a+k) = 16(2^n-1)\) where the function \(f\) satisfies the relation \(f(x+y)=f(x)f(y)\) for all natural numbers \(x,y\) and further , \(f(1)=2\).

A\(11\) B\(5\) C\(7\) D\(3\)

Hint not yet added.

Question 14

The domain of the definition of the function \(f(x)=\frac{1}{4-x^2} + log_{10} (x^3-x)\) is

A \( (-1,0) \cup (1,2) \cup (3,\infty) \) B \( (-2,-1) \cup (-1,0) \cup (2,\infty) \) C \( (-1,0) \cup (1,2) \cup (2,\infty) \) D \( (1,2) \cup (2,\infty)\)

Critical points of \(x^3-x\) are \(-1,0\) and \(1\)

Question 15

Let \(f(x)= a^x \quad (a > 0) \) be written as \(f(x)=f_1(x)+f_2(x)\), where \(f_1(x)\) is an even function and \(f_2(x)\) is an odd function. Then \(f_1(x+y)+f_1(x-y)\) equals

A\(2f_1(x+y)f_2(x-y)\) B\(2f_1(x+y)f_1(x-y)\) C\(2f_1(x)f_2(x)\) D\(2f_1(x)f_1(x)\)

see that \(f(x)=a^x=f_1(x)+f_2(x)\) and \(\f(-x)=a^{-x}=f_1(x)-f_2(x))

Question 16

Let \(g(x) = 1 + x - [x]\) and \(f(x) = \begin{cases} -1, & x < 0 \\ 0, & x = 0 \\ 1, & x > 0 \end{cases}\), then for all \(x\), \(f[g(x)]\) is equal to

A\(x\) B\(1\) C\(f(x)\) D\(g(x)\)

Note that \(x-[x] = \{x\}\) ( fractional part of \(x\) )

Question 17

Let \(\mathbb{N}\) be the set of natural numbers and two functions \(f\) and \(g\) defined as \(f,g:\mathbb{N} \rightarrow \mathbb{N}\) such that \(f(x) = \begin{cases} \frac{n+1}{2}, & \text{ if n is odd} \\ \frac{n}{2}, & \text{ if n is even} \end{cases}\) and \(g(n)=n-(-1)^n\). Then \(f\circ g\) is

Aone-one but not onto Bonto but not one-one Cboth one-one and onto Dneither one-one nor onto

Start by evaluating the composite function for the first couple of natural numbers. What exactly are the outputs for \(n = 1\) and \(n = 2\) when you calculate \((f \circ g)(n)\)? If you choose an even natural number for your initial input, say \(n = 2k\), what does the composite function \((f \circ g)(2k)\) ultimately simplify to?

Question 18

If \(f\) is an even function defined on the interval \((-5, 5)\), the four real values of \(x\) satisfying the equation: \(f(x) = f\left(\frac{x+1}{x+2}\right)\) are

A\(x = \frac{-1 \pm \sqrt{7}}{2} \quad \text{and} \quad x = \frac{-3 \pm \sqrt{5}}{2}\) B\(x = \frac{-1 \pm \sqrt{5}}{2} \quad \text{and} \quad x = \frac{-1 \pm \sqrt{5}}{2}\) C\(x = \frac{-2 \pm \sqrt{5}}{2} \quad \text{and} \quad x = \frac{-3 \pm \sqrt{5}}{2}\) D\(x = \frac{-1 \pm \sqrt{5}}{2} \quad \text{and} \quad x = \frac{-3 \pm \sqrt{5}}{2}\)

Since the function \(f\) is an even function, the condition \(f(a) = f(b)\) implies that either \(a = b\) or \(a = -b\).

Question 19

Let \(f:X \rightarrow X\) be such that \(f(f(x))=x\), for all \(x \in X\) and \( X \subset R \), then

A\(f\) is one-to-one B\(f\) is onto C\(f\) is one-to-one not onto D\(f\) is onto but not one-to-one

Hint not yet added.

Question 20

Let \( f : R \rightarrow R \) be such that \( f \) is injective and \( f(x)f(y) = f(x+y) \) for \( \forall x, y \in R \). If \( f(x) \), \( f(y) \), \( f(z) \) are in G.P., then \( x, y, z \) are in

AAP always and \(f(0)=1\) BGP always and \(f(0)=0\) CAP depending on the value of \( x, y, z \) DGP depending on the value of \( x, y, z \)

Hint not yet added.

Question 21

The domain of definition of \( f(x) = \sqrt{\frac{1 - |x|}{2 - |x|}} \) is

A\( (-\infty, -1) \cup (2, \infty) \) B\( [-1, 1] \cup (2, \infty) \cup (-\infty, -2) \) C\( (-\infty, 1) \cup (2, \infty) \) D\( [-1, 1] \cup (2, \infty) \)

Hint not yet added.

Question 22

Let \( f(x) = ax^2 + bx + c \), \( g(x) = px^2 + qx + r \) such that \( f(1) = g(1) \), \( f(2) = g(2) \) and \( f(3) - g(3) = 2 \). Then \( f(4) - g(4) \) is

A4 B5 C6 D7

Hint not yet added.

Question 23

Let \( R \) be the set of real numbers and the functions \( f : R \rightarrow R \) and \( g : R \rightarrow R \) be defined by \( f(x) = x^2 + 2x - 3 \) and \( g(x) = x + 1 \). Then, the value of \( x \) for which \( f(g(x)) = g(f(x)) \) is

A\( -1 \) B\( 0 \) C\( 1 \) D\( 2 \)

Hint not yet added.

Question 24

The minimum value of the function \(f(x)=2|x-1|+|x-2|\) is

A\( 0 \) B\( 1 \) C\( 2 \) D\( 3 \)

Sketch the graph.

Question 25

Let \(\mathrm{A}=\{-3,-2,-1,0,1,2,3\}\) and \(R\) be a relation on \(A\) defined by \(x R y\) if and only if \(2 x-y \in\{0,1\}\). Let \(l\) be the number of elements in \(R\). Let \(m\) and \(n\) be the minimum number of elements required to be added to \(R\) to make it a reflexive and symmetric relation, respectively. Then \(l+\mathrm{m}+ \mathrm{n}\) is equal to :-

A\(18\) B\(15\) C\(16\) D\(17\)

Hint not yet added.

Prime Maths - Excellence in Mathematical Education

© 2026 Prime Maths | Created by Vinod Singh

Prove that cos pi/11 + cos 3pi/11 + cos 5pi/11 + cos 7pi/11 + cos 9pi/11...

A elegant solution for a difficult trigonometric problem using complex numbers. Many problem of trigonometry and even algebra can be beautifully solved using complex numbers. The use of different formulas from trigonometry can be avoided. #CBSE #ISC #IIT #ISI #wbjee #wbchse #jeeadvanced Prove that cos pi/11 + cos 3pi/11 + cos 5pi/11 + cos 7pi/11 + cos 9pi/11 =1/2

Integrate max { x+|x|,x-[x] } from -n to n JEE Main and Advance and Indian Statistical Institute B.Math & B.Stat

Integrate max { x+|x|,x-[x] } from -n to n Difficult Integration for JEE Main and Advance and Indian Statistical Institute Involving Limits Integration for Indian Statistical Institute B.Math & B.Stat : Integration, JEE Main and Advance, WBJEE #CBSE #ISC #wbchse #HS #jeeadvanced

WBJEE MATHEMATICS PAPER 2020 SOLVED

 The West Bengal Joint Entrance Examinations Board

The West Bengal Joint Entrance Examinations Board (WBJEEB) was established in 1962 by Government of West Bengal in exercise of the powers conferred under article 162 of the Constitution of India in pursuant to No. 828-Edn(T), dated 02.03.1962.

Subsequently in 2014, the Government of West Bengal enacted the West Bengal Act XIV of 2014 to form The West Bengal Joint Entrance Examinations Board and empowered it to conduct Common Entrance Examinations for selection of candidates for admission to undergraduate and postgraduate Professional, Vocational and General Degree Courses in the State of West Bengal and to conduct on-line counselling process or otherwise adopting a single-window approach.

WBJEEB has been instrumental in the admission process based on online application and allotment through e-Counselling since 2012. It advocates fairness and transparency, ensures no-error, and adopts state-of-the-art technology.

WBJEE 2022 Mathematics Syllabus

S.No.	Topics
1	Algebra Arithmetic Progression G.P., H.P Sets, Relations and Mappings Logarithms Complex Numbers Permutation and combination Polynomial equation Principle of mathematical induction Matrices Binomial theorem (positive integral index) Statistics and Probability
2	Trigonometry
3	Coordinate geometry of two dimensions
4	Coordinate geometry of three dimensions
5	Differential calculus Calculus Integral calculus Application of Calculus Differential Equations Vectors