Class 12 Relations and Functions Mock Test (CBSE)

Topics in Relations and Functions

• Types of relations
• Reflexive, symmetric and transitive relations
• Equivalence relations and classes
• One-one functions
• Onto functions
• Bijective functions

Relations and Functions practice questions with answers

1. Q1Multiple choice·1 mark
   Let $R$ be the relation in the set $A = \{1, 2, 3\}$ given by $R = \{(1, 1), (2, 2), (3, 3), (1, 2), (2, 3), (1, 3)\}$. Choose the correct statement.

   1. (a)Reflexive and transitive, but not symmetric
   2. (b)Reflexive and symmetric, but not transitive
   3. (c)Symmetric and transitive, but not reflexive
   4. (d)An equivalence relation
   Show answer and step marks

   1. Correct option: (a) Reflexive and transitive, but not symmetric+1

   Answer: (a)
2. Q2Multiple choice·1 mark
   Let $R$ be a relation defined on the set of real numbers $\mathbb{R}$ as $R = \{(a, b) : a \le b\}$. Then the relation $R$ is:

   1. (a)An equivalence relation
   2. (b)Reflexive and symmetric, but not transitive
   3. (c)Reflexive and transitive, but not symmetric
   4. (d)Symmetric and transitive, but not reflexive
   Show answer and step marks

   1. Correct option: (c) Reflexive and transitive, but not symmetric+1

   Answer: (c)
3. Q3Long answer·5 marks
   Let $f: \mathbb{R} \to \mathbb{R}$ be defined by $f(x) = x^2 - 6x + 13$. (a) Write $f(x)$ in the form $(x - h)^2 + k$ and state the values of $h$ and $k$. [1] (b) Find the set $B$ such that $f: \mathbb{R} \to B$ is onto. [2] (c) Show that $3$ is not an element of the set $B$ found in part (b). [1] (d) Is $f: \mathbb{R} \to \mathbb{R}$ onto? Give a reason. [1]
   Show answer and step marks

   1. $f(x) = (x^2 - 6x + 9) + 4 = (x - 3)^2 + 4$+0.5
   2. Hence $h = 3$ and $k = 4$.+0.5
   3. Since $(x - 3)^2 \ge 0$ for all real $x$, $f(x) \ge 4$, so the range of $f$ lies in $[4, \infty)$.+1
   4. Take $B = [4, \infty)$: every $y \ge 4$ has the real pre-image $x = 3 \pm \sqrt{y - 4}$, so $f: \mathbb{R} \to B$ is onto.+1
   5. Put $f(x) = 3$ to get $(x - 3)^2 = -1$.+0.5
   6. A square cannot be negative, so there is no real $x$ with $f(x) = 3$; hence $3 \notin B$.+0.5
   7. Since the range $[4, \infty)$ is not equal to the codomain $\mathbb{R}$ (for example $3 \in \mathbb{R}$ has no pre-image), $f: \mathbb{R} \to \mathbb{R}$ is not onto.+1

   Answer: (a) $h = 3$, $k = 4$, so $f(x) = (x - 3)^2 + 4$. (b) $B = [4, \infty)$. (c) $(x-3)^2 = -1$ has no real solution, so $3 \notin B$. (d) No, $f: \mathbb{R} \to \mathbb{R}$ is not onto, because the range is $[4, \infty)$, not $\mathbb{R}$.

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