Get detailed, step-by-step solutions for NCERT Class 11 Maths Chapter 2 Exercise 2.3 . Learn the critical rules to identify a function from a relation and practice finding the Domain and Range of various real-valued functions, including quadratic, square root, and absolute value functions. Master function notation and practical applications like the Celsius-Fahrenheit conversion function. Essential for advanced study of algebra and calculus.
This exercise focuses on identifying functions from relations and determining the domain and range of real-valued functions.
A function is a relation where every element in the domain has exactly one image (output) in the range.
1. Identifying Functions
A relation is a function if no two ordered pairs have the same first element (domain element) but different second elements (range elements).
(i) $R = \{(2, 1), (5, 1), (8, 1), (11, 1), (14, 1), (17, 1)\}$
- Is it a Function? Yes. Every first element (2, 5, 8, 11, 14, 17) is unique. It’s a many-one function.
- Domain: The set of first elements.$$\mathbf{\text{Domain} = \{2, 5, 8, 11, 14, 17\}}$$
- Range: The set of second elements.$$\mathbf{\text{Range} = \{1\}}$$
(ii) $R = \{(2, 1), (4, 2), (6, 3), (8, 4), (10, 5), (12, 6), (14, 7)\}$
- Is it a Function? Yes. Every first element (2, 4, 6, 8, 10, 12, 14) is unique.
- Domain:$$\mathbf{\text{Domain} = \{2, 4, 6, 8, 10, 12, 14\}}$$
- Range:$$\mathbf{\text{Range} = \{1, 2, 3, 4, 5, 6, 7\}}$$
(iii) $R = \{(1, 3), (1, 5), (2, 5)\}$
- Is it a Function? No. The first element 1 is related to two different second elements (3 and 5). This violates the definition of a function.
2. Finding Domain and Range of Real Functions
(i) $f(x) = -|x|$
- Domain: Since $x$ can be any real number, the domain is the set of all real numbers, $\mathbf{R}$.
- Range: The absolute value $|x|$ is always $\ge 0$. Therefore, $-|x|$ is always $\le 0$.$$\mathbf{\text{Range} = \{y \in \mathbf{R} : y \le 0\} \text{ or } (-\infty, 0]}$$
(ii) $f(x) = \sqrt{9 – x^2}$
- Domain: For $f(x)$ to be a real number, the term inside the square root must be non-negative:$$9 – x^2 \ge 0$$$$x^2 \le 9$$$$-3 \le x \le 3$$$$\mathbf{\text{Domain} = [-3, 3]}$$
- Range: We evaluate the value of $f(x)$ for $x$ in the domain $[-3, 3]$.
- Minimum value: Occurs when $x^2$ is maximum (i.e., $x=\pm 3$). $f(\pm 3) = \sqrt{9 – 9} = 0$.
- Maximum value: Occurs when $x^2$ is minimum (i.e., $x=0$). $f(0) = \sqrt{9 – 0} = 3$.$$\mathbf{\text{Range} = [0, 3]}$$
3. Evaluating Function $f(x) = 2x – 5$
(i) $f(0)$
$$f(0) = 2(0) – 5 = 0 – 5 = \mathbf{-5}$$
(ii) $f(7)$
$$f(7) = 2(7) – 5 = 14 – 5 = \mathbf{9}$$
(iii) $f(-3)$
$$f(-3) = 2(-3) – 5 = -6 – 5 = \mathbf{-11}$$
4. Temperature Conversion Function $t(C) = \frac{9C}{5} + 32$
(i) $t(0)$ (Celsius to Fahrenheit at freezing point)
$$t(0) = \frac{9(0)}{5} + 32 = 0 + 32 = \mathbf{32}$$
(ii) $t(28)$
$$t(28) = \frac{9(28)}{5} + 32 = \frac{252}{5} + 32 = 50.4 + 32 = \mathbf{82.4}$$
(iii) $t(-10)$
$$t(-10) = \frac{9(-10)}{5} + 32 = 9(-2) + 32 = -18 + 32 = \mathbf{14}$$
(iv) Find $C$ when $t(C) = 212$
$$212 = \frac{9C}{5} + 32$$
$$212 – 32 = \frac{9C}{5}$$
$$180 = \frac{9C}{5}$$
$$9C = 180 \times 5 = 900$$
$$C = \frac{900}{9} = \mathbf{100}$$
5. Finding the Range of Functions
(i) $f(x) = 2 – 3x, \quad x \in \mathbf{R}, x > 0$
We analyze the effect of $x > 0$ on the expression $2 – 3x$.
- Since $x > 0$, multiplying by $-3$ reverses the inequality:$$-3x < 0$$
- Adding 2 to both sides:$$2 – 3x < 2 + 0$$$$f(x) < 2$$$$\mathbf{\text{Range} = \{y \in \mathbf{R} : y < 2\} \text{ or } (-\infty, 2)}$$
(ii) $f(x) = x^2 + 2, \quad x$ is a real number
We analyze the possible values of $x^2 + 2$.
- Since $x$ is a real number, the square $x^2$ is always non-negative:$$x^2 \ge 0$$
- Adding 2 to both sides:$$x^2 + 2 \ge 2$$$$f(x) \ge 2$$$$\mathbf{\text{Range} = \{y \in \mathbf{R} : y \ge 2\} \text{ or } [2, \infty)}$$
(iii) $f(x) = x, \quad x$ is a real number
This is the Identity Function. For every input $x$, the output $f(x)$ is exactly $x$.
$$\mathbf{\text{Range} = \mathbf{R} \text{ or } (-\infty, \infty)}$$