Anand Classes brings you a free downloadable PDF of the NCERT Solutions for Class 11 Maths Chapter 2 – Relations and Functions, Exercise 2.3, covering topics such as identifying functions, finding domains and ranges, algebra of real functions and different types of functions like identity, constant, modulus, signum and greatest integer functions. The material is carefully aligned with the latest CBSE syllabus and is ideal for both board exam preparation and strengthening conceptual clarity. Click the print button to download study material and notes.
NCERT Question 1 : Which of the following relations are functions? Give reasons. If it is a function, determine its domain and range.
(i) $\{(2, 1), (5, 1), (8, 1), (11, 1), (14, 1), (17, 1)\}$
(ii) $\{(2, 1), (4, 2), (6, 3), (8, 4), (10, 5), (12, 6), (14, 7)\}$
(iii) $\{(1, 3), (1, 5), (2, 5)\}$
Solution :
(i) $\{(2, 1), (5, 1), (8, 1), (11, 1), (14, 1), (17, 1)\}$
Each element of the first set (domain) has a unique image.
So, this relation is a function.
- Domain $= \{2, 5, 8, 11, 14, 17\}$
- Range $= \{1\}$
(ii) $\{(2, 1), (4, 2), (6, 3), (8, 4), (10, 5), (12, 6), (14, 7)\}$
Each element in the domain has a unique image.
So, this relation is a function.
- Domain $= \{2, 4, 6, 8, 10, 12, 14\}$
- Range $= \{1, 2, 3, 4, 5, 6, 7\}$
(iii) $\{(1, 3), (1, 5), (2, 5)\}$
Here, the element $1$ corresponds to two different images $3$ and $5$.
So, this relation is not a function.
Continue learning relations and functions with high-quality notes from Anand Classes — perfect for CBSE & JEE preparation!
NCERT Question 2 : Find the domain and range of the following real function:
(i) $f(x) = -|x|$
(ii) $f(x) = \sqrt{9 – x^2}$
Solution :
We know,
$$|x| =
\begin{cases}
x, & x \ge 0 \ \\
-x, & x < 0
\end{cases}$$
So,
$$f(x) = -|x| =
\begin{cases}
-x, & x \ge 0 \ \\
x, & x < 0
\end{cases}$$
As $f(x)$ is defined for $x ∈ R$, the domain of $f$ is $R$.
It is also seen that the range of $f(x) = –|x|$ is all real numbers except positive real numbers.
Since $f(x)$ is defined for all real $x$:
- Domain $= \mathbb{R}$
Also, $-|x|$ is always less than or equal to 0:
- Range $= (-\infty, 0]$
(ii) $f(x) = \sqrt{9 – x^2}$
For the square root to be defined:
$$9 – x^2 \ge 0$$
$$x^2 \le 9$$
$$|x| \le 3$$
$$-3 \le x \le 3$$
Thus:
- Domain $= [-3, 3]$
Since square root values are always non-negative. For any value of $x$ in the range $[–3, 3]$, the value of $f(x)$ will lie between 0 and 3.
$$0 \le \sqrt{9 – x^2} \le 3$$
- Range $= [0, 3]$
Master relations, functions, and domain-range concepts with top-quality math notes from Anand Classes — helpful for CBSE & competitive exams like JEE!
NCERT Question 3 : A function $f$ is defined by $f(x) = 2x – 5$. Write down the values of:
(i) $f(0)$, (ii) $f(7)$, (iii) $f(-3)$
Solution:
Given function:
$$f(x) = 2x – 5$$
(i)
$$f(0) = 2 \cdot 0 – 5 = -5$$
(ii)
$$f(7) = 2 \cdot 7 – 5 = 14 – 5 = 9$$
(iii)
$$f(-3) = 2 \cdot (-3) – 5 = -6 – 5 = -11$$
Learn more about functions with easy examples and solutions — perfect study support from Anand Classes for school exams and competitive preparation!
NCERT Question 4 : The function $t$ which maps temperature in degree Celsius into temperature in degree Fahrenheit is defined by
$$t(C) = \frac{9C}{5} + 32$$
Find: (i) $t(0)$ (ii) $t(28)$ (iii) $t(-10)$
(iv) The value of $C$ when $t(C) = 212$
Solution:
Given:
$$t(C) = \frac{9C}{5} + 32$$
(i)
$$t(0) = \frac{9 \cdot 0}{5} + 32 = 32$$
(ii)
$$t(28) = \frac{9 \cdot 28}{5} + 32 = \frac{252}{5} + 32 = \frac{252 + 160}{5} = \frac{412}{5}$$
(iii)
$$t(-10) = \frac{9(-10)}{5} + 32 = -18 + 32 = 14$$
(iv) To find $C$ when $t(C) = 212$:
$$\frac{9C}{5} + 32 = 212$$
$$\frac{9C}{5} = 180$$
$$9C = 900$$
$$C = 100$$
✅ Therefore, the value of $C$ is $100^\circ$C.
For more clear concept-based solutions on Functions and Number Systems, keep learning with Anand Classes — Top-quality study material for Class 11 Maths and competitive exams!
NCERT Question 5 : Find the range of each of the following functions:
(i) $f(x) = 2 – 3x, x \in \mathbb{R}, x > 0$
(ii) $f(x) = x^2 + 2, x \in \mathbb{R}$
(iii) $f(x) = x, x \in \mathbb{R}$
Solution :
(i) Given:
$$f(x) = 2 – 3x,\quad x > 0$$
Since $$x > 0$$
Multiply both sides by $-3$ (inequality reverses):
$$-3x < 0$$
Add $2$ to both sides:
$$2 – 3x < 2$$
$$\Rightarrow f(x) < 2$$
📌 Range:
$$(-\infty, 2)$$
(ii)
Given:
$$f(x) = x^2 + 2$$
For all real $x$: $$x^2 \ge 0$$
$$x^2 + 2 \ge 2$$
$$\Rightarrow f(x) \ge 2$$
📌 Range:
$$[2, \infty)$$
(iii)
Given:
$$f(x) = x,\quad x \in \mathbb{R}$$
Here, $f(x)$ takes every real value taken by $x$.
📌 Range: $\mathbb{R}$
$$(-\infty, \infty)$$
✨ Keep practicing Functions with Anand Classes — best notes for CBSE & JEE aspirants, downloadable for free!
Summary
Exercise 2.3 deals with analyzing different types of functions and their properties. Key aspects include:
- Domain and Range: Identifying the set of possible input values (domain) and output values (range) for a function.
- Function Characteristics: Determining if a relation is a function using the vertical line test and understanding one-to-one (injective) and onto (surjective) functions.
