This page provides comprehensive Class 11 Maths Exemplar Chapter 2 Exercise 2.1 Solutions. Detailed step-by-step solutions for Class 11 Maths NCERT Exemplar Chapter 2 Relations and Functions Exercise 2.1. Free PDF download and interactive practice.
Exercise 2.1
NCERT Exemplar Class 11 - Relations and Functions (MCQs)
Q1
Let $n(A) = m$, and $n(B) = n$. Then the total number of non-empty relations that can be defined from $A$ to $B$ is
The total number of relations from set A to set B is the number of subsets of A × B.
Since n(A × B) = mn, the total number of relations is $2^{mn}$.
The number of non-empty relations is $2^{mn} - 1$ (excluding the empty set).
$2^{mn} - 1$
Q2
If $f(x) = x^3 - \frac{1}{x^3}$, then $f(x) + f(\frac{1}{x})$ is equal to
Given $f(x) = x^3 - \frac{1}{x^3}$.
Then $f(\frac{1}{x}) = (\frac{1}{x})^3 - \frac{1}{(1/x)^3} = \frac{1}{x^3} - x^3$.
Taking the square root of both sides, $\sqrt{1 + x^2y^2} \le \sqrt{1 + x^2 + y^2 + x^2y^2}$.
Thus, $f(xy) \le f(x)f(y)$.
$f(xy) \le f(x)f(y)$
Q6
Let $f(x) = ax + b$, where $a$ and $b$ are integers. If $f(-1) = -5$ and $f(3) = 3$, then $a$ and $b$ are equal to
$f(-1) = a(-1) + b = -a + b = -5$ ... (i)
$f(3) = a(3) + b = 3a + b = 3$ ... (ii)
Subtracting equation (i) from (ii): $(3a+b) - (-a+b) = 3 - (-5) \Rightarrow 4a = 8 \Rightarrow a = 2$.
Substituting $a=2$ in (i): $-2 + b = -5 \Rightarrow b = -3$.
$a = 2, b = -3$
Q7
The domain of the function $f$ defined by $f(x) = \sqrt{4 - x} + \frac{1}{\sqrt{x^2 - 1}}$ is equal to
For $\sqrt{4-x}$ to be defined, we need $4-x \ge 0 \Rightarrow x \le 4$. This gives the interval $(-\infty, 4]$.
For $\frac{1}{\sqrt{x^2-1}}$ to be defined, we need $x^2 - 1 > 0 \Rightarrow (x-1)(x+1) > 0$.
This inequality holds when $x < -1$ or $x > 1$. This gives the interval $(-\infty, -1) \cup (1, \infty)$.
The domain of the function is the intersection of both conditions: $((-\infty, 4]) \cap ((-\infty, -1) \cup (1, \infty))$.
Combining these, we get $x \in (-\infty, -1) \cup (1, 4]$.
$(-\infty, -1) \cup (1, 4]$
Q8
The domain and range of the real function $f$ defined by $f(x) = \frac{4 - x}{x - 4}$ is given by
Domain: The function is defined for all real numbers except where the denominator is zero. $x - 4 \ne 0 \Rightarrow x \ne 4$. So, Domain = $R - \{4\}$.
Range: For any $x \ne 4$, we can simplify the function: $f(x) = \frac{-(x-4)}{x-4} = -1$.
Since the function's value is always -1, the Range is the set $\{-1\}$.
$Domain = $R - \{4\}$, Range = $\{-1\}$$
Q9
The domain and range of the real function $f$ defined by $f(x) = \sqrt{x - 1}$ is given by
Domain: For the square root to be defined, the expression inside must be non-negative. $x - 1 \ge 0 \Rightarrow x \ge 1$. So, Domain = $[1, \infty)$.
Range: The output of a square root function is always non-negative. Since $\sqrt{x-1} \ge 0$, the Range = $[0, \infty)$.
$Domain = $[1, \infty)$, Range = $[0, \infty)$$
Q10
The domain of the function $f$ given by $f(x) = \frac{x^2 + 2x + 1}{x^2 - 8x + 12}$ is
The function is undefined when the denominator is zero. $x^2 - 8x + 12 = 0$.
Factoring the quadratic: $(x-6)(x-2) = 0$.
This gives $x = 6$ or $x = 2$.
Therefore, the domain is all real numbers except 2 and 6. Domain = $R - \{2, 6\}$.
$R - \{2, 6\}$
Q11
The domain and range of the function $f$ given by $f(x) = 2 - |x - 5|$ is
Domain: The function is defined for all real numbers $x$. So, Domain = $R$.
Range: We know that $|x-5| \ge 0$ for all $x$.
Multiplying by -1 reverses the inequality: $-|x-5| \le 0$.
Adding 2 to both sides: $2 - |x-5| \le 2$.
So, $f(x) \le 2$. The range is $(-\infty, 2]$.
$Domain = $R$, Range = $(-\infty, 2]$$
Q12
The domain for which the functions $f(x) = 2x^2 - 1$ and $g(x) = 1 - 3x$ are equal is
Set $f(x) = g(x) \Rightarrow 2x^2 - 1 = 1 - 3x$.
Rearrange into a quadratic equation: $2x^2 + 3x - 2 = 0$.
Factor the quadratic: $2x^2 + 4x - x - 2 = 0 \Rightarrow 2x(x+2) - 1(x+2) = 0$.
This gives $(2x-1)(x+2) = 0$.
The solutions are $x = 1/2$ and $x = -2$. The domain is the set $\{1/2, -2\}$.
$\{1/2, -2\}$
Q13
If $f(x) = |x - 2|$, then range of $f$ is
The absolute value function $|x-2|$ is always non-negative for any real number $x$.
Therefore, the range of $f(x)$ is all non-negative real numbers, which is $[0, \infty)$.
$[0, \infty)$
Q14
Let $f(x) = \frac{1}{\sqrt{x + |x|}}$. The domain of $f$ is
For the function to be defined, the expression inside the square root in the denominator must be strictly positive: $x + |x| > 0$.
Case 1: If $x > 0$, then $|x| = x$. The expression becomes $x + x = 2x$. Since $x > 0$, $2x > 0$. This is valid.
Case 2: If $x = 0$, then $|x| = 0$. The expression is $0 + 0 = 0$. This is not allowed as it makes the denominator zero.
Case 3: If $x < 0$, then $|x| = -x$. The expression is $x + (-x) = 0$. This is not allowed.
So, the domain is all positive real numbers, $(0, \infty)$.
$(0, \infty)$
Q15
The range of the function $f(x) = \frac{|x-4|}{x-4}$ is
The function is not defined at $x=4$.
Case 1: If $x > 4$, then $x-4 > 0$, so $|x-4| = x-4$. The function becomes $f(x) = \frac{x-4}{x-4} = 1$.
Case 2: If $x < 4$, then $x-4 < 0$, so $|x-4| = -(x-4)$. The function becomes $f(x) = \frac{-(x-4)}{x-4} = -1$.
The function only takes two possible values, 1 and -1. The range is the set $\{1, -1\}$.
$\{1, -1\}$
Q16
The domain of the function $f(x) = \log_{10}(1-x)$ is
The logarithm function is defined only for positive arguments.
Therefore, we must have $1 - x > 0 \Rightarrow 1 > x \Rightarrow x < 1$.
The domain is $(-\infty, 1)$.
$(-\infty, 1)$
Q17
The range of the function $f(x) = 1 + \sin x$ is
The range of the sine function is $-1 \le \sin x \le 1$.
Adding 1 to all parts of the inequality: $-1 + 1 \le 1 + \sin x \le 1 + 1$.
This simplifies to $0 \le 1 + \sin x \le 2$.
Therefore, the range of $f(x)$ is $[0, 2]$.
$[0, 2]$
Q18
Let $A = \{1, 2, 3\}$ and $B = \{a, b\}$. The total number of functions from $A$ to $B$ is
The number of functions from a set A with m elements to a set B with n elements is given by $n^m$.
Here, $n(A) = m = 3$ and $n(B) = n = 2$.
The total number of functions is $2^3 = 8$.
$2^3$
Q19
Let $R$ be a relation on $N$ defined by $x + 2y = 8$. The domain of $R$ is
The relation is $x + 2y = 8$, where $x, y$ are natural numbers (N). We can write $x = 8 - 2y$.
Since $x \in N$, we must have $x > 0$, so $8 - 2y > 0 \Rightarrow 8 > 2y \Rightarrow 4 > y$.
Since $y \in N$, the possible values for y are 1, 2, 3.
If $y=1$, $x = 8 - 2(1) = 6$. The pair is (6, 1).
If $y=2$, $x = 8 - 2(2) = 4$. The pair is (4, 2).
If $y=3$, $x = 8 - 2(3) = 2$. The pair is (2, 3).
The domain of R is the set of all first elements (x-values): $\{2, 4, 6\}$.
$\{2, 4, 6\}$
Q20
Let $f$ and $g$ be two real functions given by $f = \{(0, 1), (2, 0), (3, -4), (4, 2), (5, 1)\}$ and $g = \{(1, 0), (2, 2), (3, -1), (4, 4), (5, 3)\}$. Then the domain of $f \cdot g$ is given by _________.
Domain of $f$, $D_f = \{0, 2, 3, 4, 5\}$.
Domain of $g$, $D_g = \{1, 2, 3, 4, 5\}$.
The domain of $f \cdot g$ is the intersection of their domains: $D_f \cap D_g$.
$D_f \cap D_g = \{2, 3, 4, 5\}$.
$\{2, 3, 4, 5\}$
Q21
Let $f = \{(2, 4), (5, 6), (8, -1), (10, -3)\}$ and $g = \{(2, 5), (7, 1), (8, 4), (10, 13), (11, 5)\}$ be two real functions. Match the following: (a) $f - g$ (b) $f + g$ (c) $f \cdot g$ (d) $f/g$