Functions
NCERT Class 11 Maths • Exercise 2.3
1. Identifying Functions
[Image of Function Mapping Diagram Examples]Q1: Which relations are functions? Determine Domain & Range.
(i) $R = \{(2,1), (5,1), (8,1), (11,1), (14,1), (17,1)\}$
Reason: Every element in the domain has a unique image (it’s okay if multiple inputs go to the same output “1”).
Function
Function
Domain: $\{2, 5, 8, 11, 14, 17\}$
Range: $\{1\}$
(ii) $R = \{(2,1), (4,2), (6,3), (8,4), (10,5), (12,6), (14,7)\}$
Reason: Every element in the domain has a unique, distinct image.
Function
Function
Domain: $\{2, 4, 6, 8, 10, 12, 14\}$
Range: $\{1, 2, 3, 4, 5, 6, 7\}$
(iii) $R = \{(1,3), (1,5), (2,5)\}$
Reason: The element 1 corresponds to two different images (3 and 5).
Not a Function
Not a Function
2. Domain & Range of Real Functions
Q2(i): Find Domain and Range of $f(x) = -|x|$.
[Image of graph of f(x) = -|x|]
Analysis: $|x|$ is defined for all real numbers.
Domain $R$ (Set of Real Numbers)
For Range: We know $|x| \ge 0$ for all $x \in R$.
Multiplying by -1 reverses the inequality:
$-|x| \le 0$.
Range $(-\infty, 0]$
For Range: We know $|x| \ge 0$ for all $x \in R$.
Multiplying by -1 reverses the inequality:
$-|x| \le 0$.
Range $(-\infty, 0]$
Q2(ii): Find Domain and Range of $f(x) = \sqrt{9 – x^2}$.
Domain (Inputs): The term under the square root must be non-negative.
$9 – x^2 \ge 0$
$x^2 \le 9 \Rightarrow -3 \le x \le 3$.
Domain: $[-3, 3]$
$x^2 \le 9 \Rightarrow -3 \le x \le 3$.
Domain: $[-3, 3]$
Range (Outputs): Let $y = \sqrt{9-x^2}$. Since it’s a square root, $y \ge 0$.
Squaring both sides: $y^2 = 9 – x^2 \Rightarrow x^2 + y^2 = 9$.
This is a circle with radius 3.
Since $y \ge 0$ and the max value of $x^2$ is 9 (making $y=0$), and min value of $x^2$ is 0 (making $y=3$).
Range: $[0, 3]$
This is a circle with radius 3.
Since $y \ge 0$ and the max value of $x^2$ is 9 (making $y=0$), and min value of $x^2$ is 0 (making $y=3$).
Range: $[0, 3]$
3. Function Evaluation
Q3: $f(x) = 2x – 5$. Find values.
- (i) $f(0)$: $2(0) – 5 = -5$
- (ii) $f(7)$: $2(7) – 5 = 14 – 5 = 9$
- (iii) $f(-3)$: $2(-3) – 5 = -6 – 5 = -11$
4. Temperature Mapping
Q4: $t(C) = \frac{9C}{5} + 32$. Find values.
- (i) $t(0)$: $\frac{9(0)}{5} + 32 = 32^\circ F$
- (ii) $t(28)$: $\frac{9(28)}{5} + 32 = \frac{252}{5} + 32 = 50.4 + 32 = 82.4^\circ F$
- (iii) $t(-10)$: $\frac{9(-10)}{5} + 32 = -18 + 32 = 14^\circ F$
- (iv) Value of C when $t(C) = 212$:
$\frac{9C}{5} + 32 = 212$
$\frac{9C}{5} = 180$
$9C = 900 \Rightarrow C = 100^\circ C$ (Boiling point of water)
5. Finding Range of Functions
Q5: Find the range of the following.
(i) $f(x) = 2 – 3x, \quad x \in R, x > 0$
Start with domain condition: $x > 0$
Multiply by -3: $-3x < 0$ (inequality flips)
Add 2: $2 – 3x < 2$
So, $f(x) < 2$.
Range: $(-\infty, 2)$
Multiply by -3: $-3x < 0$ (inequality flips)
Add 2: $2 – 3x < 2$
So, $f(x) < 2$.
Range: $(-\infty, 2)$
(ii) $f(x) = x^2 + 2, \quad x \in R$
We know $x^2 \ge 0$ for any real number.
Add 2: $x^2 + 2 \ge 2$.
So, $f(x) \ge 2$.
Range: $[2, \infty)$
Add 2: $x^2 + 2 \ge 2$.
So, $f(x) \ge 2$.
Range: $[2, \infty)$
(iii) $f(x) = x, \quad x \in R$
This is the Identity Function.
Output is exactly the input.
Since domain is $R$, the output covers all real numbers.
Range: $R$ (Set of Real Numbers)
Output is exactly the input.
Since domain is $R$, the output covers all real numbers.
Range: $R$ (Set of Real Numbers)