Anand Classes provides comprehensive NCERT Solutions for Class 11 Maths Chapter 2 – Relations and Functions (Miscellaneous Exercise) in an easy-to-understand PDF format for free download. This exercise combines all major concepts from the chapter, including Cartesian products, domain, range, codomain, relations, and functions, offering students a complete revision tool before exams. Each question is solved step by step according to the latest CBSE and NCERT syllabus, helping learners strengthen their mathematical reasoning and analytical skills. The detailed explanations are crafted by experienced teachers at Anand Classes, ensuring conceptual clarity for every type of question, from simple definitions to complex functional problems. These NCERT solutions are perfect for Class 11 students aiming to score high in board exams and competitive exams like JEE Main. Click the print button to download study material and notes.
NCERT Question 1: The relation $f$ is defined as:
$$
f(x) =
\begin{cases}
x^2, & 0 \le x \le 3 \ \\
3x, & 3 \le x \le 10
\end{cases}
$$
and the relation $g$ is defined as:
$$
g(x) =
\begin{cases}
x^2, & 0 \le x \le 2 \ \\
3x, & 2 \le x \le 10
\end{cases}
$$
Show that $f$ is a function and $g$ is not a function
Solution :
Given
The relation $f$ is defined as:
$$
f(x) =
\begin{cases}
x^2, & 0 \le x \le 3 \ \\
3x, & 3 \le x \le 10
\end{cases}
$$
The relation $g$ is defined as:
$$
g(x) =
\begin{cases}
x^2, & 0 \le x \le 2 \ \\
3x, & 2 \le x \le 10
\end{cases}
$$
For function $f(x)$
For the first condition $(0 \le x < 3)$ :
$$f(x) = x^2$$
For the second condition $(3 < x \le 10)$ :
$$f(x) = 3x$$
Now check at $(x = 3)$ :
- From $(x^2)$:
$$f(3) = 3^2 = 9$$ - From $(3x)$ :
$$f(3) = 3 \cdot 3 = 9$$
✅ Both conditions give the same image when $(x = 3)$.
So, for every value of $x$ in the domain $(0 \le x \le 10)$, the image of $f(x)$ is unique.
Since every point of the domain has one and only one image under $f$, this relation is a function. Thus, $f$ is a function.
For function $g(x)$
For the first condition $(0 \le x < 2)$ :
$$g(x) = x^2$$
For the second condition $(2 < x \le 10)$ :
$$g(x) = 3x$$
Now check at $(x = 2)$ :
- From $(x^2)$ :
$$g(2) = 2^2 = 4$$ - From $(3x)$ :
$$g(2) = 3 \cdot 2 = 6$$
❌ For (x = 2), the element corresponds to two different images (4) and (6).
Therefore, $g$ is not a function.
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NCERT Question 2 : If $f(x) = x^2$, find
$$
\frac{f(1.1) – f(1)}{1.1 – 1}
$$
Solution :
Given:
$$
f(x) = x^2
$$
We find the value of :
$$\frac{f(1.1) – f(1)}{1.1 – 1}$$
Substituting the values:
$$\frac{f(1.1) – f(1)}{1.1 – 1}=\frac{(1.1)^2 – (1)^2}{1.1 – 1}$$
$$
\frac{f(1.1) – f(1)}{1.1 – 1}= \frac{1.21 – 1}{0.1}
$$
$$
\frac{f(1.1) – f(1)}{1.1 – 1}= \frac{0.21}{0.1} = 2.1
$$
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NCERT Question 3 : Find the domain of the function
$$
f(x) = \frac{x^2 + 2x + 1}{x^2 – 8x + 12}
$$
Solution :
Given:
$$
f(x) = \frac{x^2 + 2x + 1}{x^2 – 8x + 12}
$$
Factorizing:
$$
x^2 + 2x + 1 = (x+1)^2
$$
$$
x^2 – 8x + 12 = (x-6)(x-2)
$$
So,
$$
f(x) = \frac{(x+1)^2}{(x-6)(x-2)}
$$
The function is undefined when the denominator is zero:
$$
(x – 6)(x – 2) = 0
$$
So,
$$
x = 6, x = 2
$$
✅ Domain of $f$
All real numbers except $x = 2$ and $x = 6$:
$$
\text{Domain} = \mathbb{R} – \{2, 6\}
$$
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NCERT Question 4 : Find the domain and range of the real function
$$
f(x) = \sqrt{x – 1}
$$
Solution :
Given the real function:
$$
f(x) = \sqrt{x – 1}
$$
For the square root to be defined,
$$
x – 1 \ge 0 \Rightarrow x \ge 1
$$
✅ Therefore, the domain of $f$ is:
$$
\text{Domain} = [1, \infty)
$$
Now,
if $x \ge 1$ then:
$$
x – 1 \ge 0 \Rightarrow \sqrt{x – 1} \ge 0
$$
✅ Therefore, the range of $f$ is:
$$
\text{Range} = [0, \infty)
$$
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NCERT Question 5 : Find the domain and the range of the real function
$$
f(x) = |x – 1|
$$
Solution :
Given:
$$
f(x) = |x – 1|
$$
The absolute value function is defined for all real numbers.
✅ So, the domain is:
$$
\text{Domain} = \mathbb{R}
$$
Also, $|x – 1| \ge 0$ for every real value of $x$, so the output is always non-negative.
✅ Therefore, the range is:
$$
\text{Range} = [0, \infty)
$$
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NCERT Question 6 : Let
$$
f = \{(x,\frac{x^2}{1+x^2}): x \in \mathbb{R}\}
$$
be a function from $\mathbb{R}$ into $\mathbb{R}$. Determine the range of $f$.
Solution :
Given function:
$$
f(x) = \frac{x^2}{1 + x^2}
$$
Since $x \in \mathbb{R}$, we know:
$$
x^2 \ge 0
$$
Thus,
$$
1 + x^2 > x^2
\quad \Rightarrow \quad
\frac{x^2}{1 + x^2} < 1
$$
Also, numerator $x^2 \ge 0$ implies:
$$
\frac{x^2}{1 + x^2} \ge 0
$$
Therefore:
$$
0 \le \frac{x^2}{1 + x^2} < 1
$$
✅ So, the range of $f$ is:
$$
[0, 1)
$$
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NCERT Question 7 : Let $f, g: \mathbb{R} \to \mathbb{R}$ be defined respectively by
$$
f(x) = x + 1, \quad g(x) = 2x – 3
$$
Find $f + g$, $f – g$, and $\dfrac{f}{g}$.
Solution :
Given:
$$
f(x) = x + 1, \quad g(x) = 2x – 3
$$
✅ Sum of two functions
$$
(f+g)(x) = f(x) + g(x)
$$
$$
(f+g)(x) = (x+1) + (2x – 3) = 3x – 2
$$
✅ Difference of two functions
$$
(f-g)(x) = f(x) – g(x)
$$
$$
(f-g)(x) = (x+1) – (2x – 3) = -x + 4
$$
✅ Division of two functions
Defined only when $g(x) \ne 0$:
$$
\left(\frac{f}{g}\right)(x) = \frac{f(x)}{g(x)} = \frac{x+1}{2x-3}
$$
Since $2x – 3 \ne 0$:
$$
x \ne \frac{3}{2}
$$
Final Answers
$$
(f+g)(x) = 3x – 2
$$
$$
(f-g)(x) = -x + 4
$$
$$
\left(\frac{f}{g}\right)(x) = \frac{x+1}{2x-3}, \quad x \ne \frac{3}{2}
$$
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NCERT Question 8 : Let
$$
f = {(1,1), (2,3), (0,-1), (-1,-3)}
$$
be a function from $\mathbb{Z}$ to $\mathbb{Z}$ defined by
$$
f(x) = ax + b
$$
for some integers $a, b$. Determine $a$ and $b$.
Solution :
Given:
$$
f(x) = ax + b
$$
Using the ordered pair $(1,1)$:
$$
f(1) = 1 \Rightarrow a(1) + b = 1
$$
$$
a + b = 1 \quad \text{(i)}
$$
Using the ordered pair $(0,-1)$:
$$
f(0) = -1 \Rightarrow a(0) + b = -1
$$
$$
b = -1
$$
Substitute $b = -1$ in equation (i):
$$
a + (-1) = 1
$$
$$
a = 2
$$
✅ Final Answer
$$
a = 2, \quad b = -1
$$
Thus,
$$
f(x) = 2x – 1
$$
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NCERT Question 9 : Let $R$ be a relation from $\mathbb{N}$ to $\mathbb{N}$ defined by
$$R = \{(a, b) : a, b \in \mathbb{N} \text{ and } a = b^2\}.$$
Are the following true?
(i) $(a,a) \in R$, for all $a \in \mathbb{N}$
(ii) $(a,b) \in R \Rightarrow (b,a) \in R$
(iii) $(a,b) \in R$ and $(b,c) \in R \Rightarrow (a,c) \in R$
Justify your answer in each case.
Solution:
Given:
$$R = {(a, b) : a = b^2, \ a, b \in \mathbb{N}}$$
(i)
Check if $(a, a) \in R$ for all $a \in \mathbb{N}$.
For $(a, a) \in R$, it must satisfy
$$a = a^2$$
which is not true for all $a \in \mathbb{N}$.
Example: $a = 2$
$$2 \neq 2^2 = 4$$
Hence, the statement is not true.
(ii)
Check if $(a, b) \in R \Rightarrow (b, a) \in R$.
Example: $(9, 3) \in R$ because
$$9 = 3^2$$
But we check $(3, 9)$:
$$3 \neq 9^2 = 81$$
So $(3, 9) \notin R$.
Hence, the statement is not true.
(iii)
Check if $(a, b) \in R$ and $(b, c) \in R \Rightarrow (a, c) \in R$.
Example:
$$(16, 4) \in R \quad \text{because } 16 = 4^2$$
$$(4, 2) \in R \quad \text{because } 4 = 2^2$$
Now check $(16, 2)$:
$$16 \neq 2^2 = 4$$
So $(16, 2) \notin R$.
Hence, the statement is not true.
✅ Final Conclusion:
All three statements (i), (ii), and (iii) are false for the given relation.
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NCERT Question 10 : Let
$A = \{1, 2, 3, 4\}$, $B = \{1, 5, 9, 11, 15, 16\}$ and
$f = \{(1, 5), (2, 9), (3, 1), (4, 5), (2, 11)\}.$
Are the following true?
(i) $f$ is a relation from $A$ to $B$
(ii) $f$ is a function from $A$ to $B$
Justify your answer in each case.
Solution:
Given:
$$A = \{1, 2, 3, 4\}, \quad B = \{1, 5, 9, 11, 15, 16\}$$
and
$$f = \{(1, 5), (2, 9), (3, 1), (4, 5), (2, 11)\}$$
(i)
A relation from $A$ to $B$ is any subset of $A \times B$.
All ordered pairs in $f$ have their first element from $A$ and second element from $B$.
Thus,
$$f \subseteq A \times B$$
Hence, $f$ is a relation from $A$ to $B$.
✅ True
(ii)
A function from $A$ to $B$ must assign each element of $A$ to exactly one element of $B$.
But in $f$,
$$2 \to 9 \quad \text{and} \quad 2 \to 11$$
The element $2 \in A$ has two different images, which violates the definition of function.
Hence, $f$ is not a function from $A$ to $B$.
❌ False
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NCERT Question 11 : Let $f$ be the subset of $\mathbb{Z} \times \mathbb{Z}$ defined by
$$f = \{(ab, a + b) : a, b \in \mathbb{Z}\}.$$
Is $f$ a function from $\mathbb{Z}$ to $\mathbb{Z}$? Justify your answer.
Solution:
Given:
$$f = \{(ab, a + b) : a, b \in \mathbb{Z}\}.$$
A relation $f$ from a set $A$ to a set $B$ is a function only if every element of $A$ has a unique image in $B$.
Take $a = 2, b = 6$ and $a = -2, b = -6$:
- For $(2, 6)$:
$$ab = 2 \cdot 6 = 12,\quad a + b = 2 + 6 = 8$$
So $(12, 8) \in f$. - For $(-2, -6)$:
$$ab = (-2) \cdot (-6) = 12,\quad a + b = -2 + (-6) = -8$$
So $(12, -8) \in f$.
Thus, the same first component $12$ has two different second components $8$ and $-8$.
Therefore, each input does not have a unique output.
Hence, $f$ is not a function from $\mathbb{Z}$ to $\mathbb{Z}$.
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NCERT Question 12 : Let $A = \{9, 10, 11, 12, 13\}$ and let $f: A \to \mathbb{N}$ be defined by
$$f(n) = \text{the highest prime factor of } n.$$
Solution:
Given: $A = \{9, 10, 11, 12, 13\}$
Function: $f(n) =$ highest prime factor of $n$
- So, Prime factor of 9 = 3
- Prime factors of 10 = 2, 5
- Prime factor of 11 = 11
- Prime factors of 12 = 2, 3
- Prime factor of 13 = 13
Hence, it can be expressed as:
Here,
- f(9) means the highest prime factor of 9 = 3
- f(10) means the highest prime factor of 10 = 5
- f(11) means the highest prime factor of 11 = 11
- f(12) means the highest prime factor of 12 = 3
- f(13) means the highest prime factor of 13 = 13
Now checking each element:
- $9 = 3 \cdot 3 \Rightarrow f(9) = 3$
- $10 = 2 \cdot 5 \Rightarrow f(10) = 5$
- $11$ is a prime number $\Rightarrow f(11) = 11$
- $12 = 2^2 \cdot 3 \Rightarrow f(12) = 3$
- $13$ is a prime number $\Rightarrow f(13) = 13$
Thus,
$$\text{Range}(f) = \{3, 5, 11, 13\}.$$
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Summary
Chapter 2 of the NCERT textbook covers the fundamental concepts of Relations and Functions. A relation is a collection of ordered pairs derived from two sets, where the first element belongs to one set and the second to another. A function is a special type of relation that associates each element in the domain with exactly one element in the codomain. This chapter’s miscellaneous exercise helps solidify these concepts by applying them to various problems, including finding domains and ranges, determining whether a given relation is a function, and working with composite functions.
