CBSE Class 12 Science Mathematics · Ch 1: Relations and Functions

Topics in this test

Difficulty: 3 easy · 9 medium · 18 hard

Sample questions with answers

The first 5 questions of this free test.

1. Q1. Let $A = \{1, 2, 3\}$. The number of equivalence relations on $A$ that contain the pair $(1, 2)$ is:

   • A. $3$
   • B. $2$
   • C. $5$
   • D. $1$
   Solution

   An equivalence relation containing $(1,2)$ must also contain $(2,1)$, so either $\{1,2\}$ stay in one class with $3$ separate, or all three $\{1,2,3\}$ form one class — giving exactly $2$ such relations.

2. Q2. On the set $\mathbb{Z}$ of integers, define $R=\{(a,b):a-b\text{ is divisible by }2\}$. The number of distinct equivalence classes determined by $R$ is

   • A. $2$
   • B. $1$
   • C. infinitely many
   • D. $3$
   Solution

   Two integers are related exactly when they have the same parity, so the classes are the even integers and the odd integers — just $2$ classes.

3. Q3. Match each function in Column I with its nature in Column II. Column I: (P) $f:\mathbb{R}\to\mathbb{R},\,f(x)=x^2$; (Q) $f:\mathbb{R}\to\mathbb{R},\,f(x)=x^3$; (R) $f:\mathbb{N}\to\mathbb{N},\,f(x)=x+1$. Column II: (1) one-one and onto; (2) one-one but not onto; (3) neither one-one nor onto. The correct matching is

   • A. P-3, Q-2, R-1
   • B. P-1, Q-3, R-2
   • C. P-3, Q-1, R-2
   • D. P-2, Q-1, R-3
   Solution

   $x^2$ gives $f(-1)=f(1)$ and never negative values (neither), $x^3$ is a bijection on $\mathbb{R}$ (one-one and onto), and $x+1$ on $\mathbb{N}$ is one-one but $1$ has no pre-image (not onto).

4. Q4. Consider the following statements. Statement I: Every identity relation on a set is an equivalence relation. Statement II: Every reflexive relation on a set is an equivalence relation. Which is correct?

   • A. Statement I is false and Statement II is true
   • B. Both statements are false
   • C. Both statements are true
   • D. Statement I is true and Statement II is false
   Solution

   The identity relation is reflexive, symmetric and transitive, so it is an equivalence relation; but a reflexive relation need not be symmetric or transitive, so Statement II is false.

5. Q5. Assertion (A): The relation $R=\{(a,b):a\le b\}$ on the set of real numbers is reflexive and transitive but not symmetric. Reason (R): $a\le a$ holds for every real $a$, and $a\le b,\,b\le c\Rightarrow a\le c$, but $a\le b$ does not imply $b\le a$.

   • A. A is false but R is true
   • B. Both A and R are true but R is not the correct explanation of A
   • C. A is true but R is false
   • D. Both A and R are true and R is the correct explanation of A
   Solution

   Both statements are true, and the reason correctly explains why $R$ is reflexive and transitive but fails symmetry.

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