CBSE Class 12 Mathematics Relations and Functions VBQs Set A

Relations and Functions

Relation: If \(A\) and \(B\) are two non-empty sets, then any subset \(R\) of \(A \times B\) is called relation from set \(A\) to set \(B\).
i.e., \(R : A \rightarrow B \Leftrightarrow R \subseteq A \times B\)
For example: Let \(A = \{1, 2\}, B = \{3, 4\}\)
Then \(A \times B = \{(1, 3), (1, 4), (2, 3), (2, 4)\}\)
A subset \(R_1 = \{(1, 3), (2, 4)\} \subseteq A \times B\) is called relation from \(A\) to \(B\).
Similarly, other subsets of \(A \times B\) are also relation from \(A\) to \(B\).
If \((x, y) \in R\), then we write \(x R y\) (read as \(x\) is \(R\) related to \(y\)) and if \((x, y) \notin R\), then we write \(x \not{R} y\) (read as \(x\) is not \(R\) related to \(y\)).

Domain and Range of a Relation: If \(R\) is any relation from set \(A\) to set \(B\) then,
(a) Domain of \(R\) is the set of all first coordinates of elements of \(R\) and it is denoted by Dom (\(R\)).
(b) Range of \(R\) is the set of all second coordinates of \(R\) and it is denoted by Range (\(R\)).
A relation \(R\) on set \(A\) means, the relation from \(A\) to \(A\) i.e., \(R \subseteq A \times A\).

Some Standard Types of Relations:
Let \(A\) be a non-empty set. Then, a relation \(R\) on set \(A\) is said to be
(a) Reflexive: If \((x, x) \in R\) for each element \(x \in A\), i.e., if \(xRx\) for each element \(x \in A\).
(b) Symmetric: If \((x, y) \in R \Rightarrow (y, x) \in R\) for all \(x, y \in A\), i.e., if \(xRy \Rightarrow yRx\) for all \(x, y \in A\).
(c) Transitive: If \((x, y) \in R\) and \((y, z) \in R \Rightarrow (x, z) \in R\) for all \(x, y, z \in A\), i.e., if \(xRy\) and \(yRz \Rightarrow xRz\).

Equivalence Relation: Any relation \(R\) on a set \(A\) is said to be an equivalence relation if \(R\) is reflexive, symmetric and transitive.

Antisymmetric Relation: A relation \(R\) in a set \(A\) is antisymmetric
if \((a, b) \in R, (b, a) \in R \Rightarrow a = b \forall a, b \in R\), or \(aRb\) and \(bRa \Rightarrow a = b, \forall a, b \in R\).
For example, the relation “greater than or equal to”, “\(\ge\)” is antisymmetric relation as
\(a \ge b, b \ge a \Rightarrow a = b \forall a, b \in R\)
[Note: “Antisymmetric” is completely different from not symmetric.]

Equivalence Class: Let \(R\) be an equivalence relation on a non-empty set \(A\). For all \(a \in A\), the equivalence class of ‘\(a\)’ is defined as the set of all such elements of \(A\) which are related to ‘\(a\)’ under \(R\). It is denoted by \([a]\).
i.e., \([a]\) = equivalence class of ‘\(a\)’ = \(\{x \in A : (x, a) \in R\}\)
For example, Let \(A = \{1, 2, 3\}\) and \(R\) be the equivalence relation on \(A\) given by
\(R = \{(1, 1), (2, 2), (3, 3), (1, 2), (2, 1)\}\)
The equivalence classes are
\([1] = \text{equivalence class of } 1 = \{x \in A : (x, 1) \in R\} = \{1, 2\}\)
Similarly, \([2] = \{2, 1\}\) and \([3] = \{3\}\)

Function: Let \(X\) and \(Y\) be two non-empty sets. Then, a rule \(f\) which associates to each element \(x \in X\), a unique element, denoted by \(f(x)\) of \(Y\), is called a function from \(X\) to \(Y\) and written as \(f : X \rightarrow Y\) where, \(f(x)\) is called image of \(x\) and \(x\) is called the pre-image of \(f(x)\) and the set \(Y\) is called the co-domain of \(f\) and \(f(X) = \{f(x): x \in X\}\) is called the range of \(f\).

Types of Function:
(i) One-one function (injective function): A function \(f : X \rightarrow Y\) is defined to be one-one if the image of distinct element of \(X\) under rule \(f\) are distinct, i.e., for every \(x_1, x_2 \in X, f(x_1) = f(x_2)\) implies that \(x_1 = x_2\).
(ii) Onto function (Surjective function): A function \(f : X \rightarrow Y\) is said to be onto function if each element of \(Y\) is the image of some element of \(x\) i.e., for every \(y \in Y\), there exists some \(x \in X\), such that \(y = f(x)\). Thus \(f\) is onto if range of \(f\) = co-domain of \(f\).
(iii) One-one onto function (Bijective function): A function \(f : X \rightarrow Y\) is said to be one-one onto, if \(f\) is both one-one and onto.
(iv) Many-one function: A function \(f : X \rightarrow Y\) is said to be a many-one function if two or more elements of set \(X\) have the same image in \(Y\). i.e.,
\(f : X \rightarrow Y\) is a many-one function if there exist \(a, b \in X\) such that \(a \ne b\) but \(f(a) = f(b)\).

Identity Function: Let \(R\) be the set of real numbers. A function \(I : R \rightarrow R\) such that \(I(x) = x \forall x \in R\) is called identity function.
Obviously, identity function associates each real number to itself.

Invertible Function: For \(f : A \rightarrow B\), if there exists a function \(g : B \rightarrow A\) such that \(gof = I_A\) and \(fog = I_B\), where \(I_A\) and \(I_B\) are identity functions, then \(f\) is called an invertible function, and \(g\) is called the inverse of \(f\) and it is written as \(f^{-1} = g\).

Number of Functions: If \(X\) and \(Y\) are two finite sets having \(m\) and \(n\) elements respectively then the number of functions from \(X\) to \(Y\) is \(n^m\).

Question. Show that the relation \(R\) on the set \(\mathbb{R}\) of real numbers, defined as \(R = \{(a, b): a \le b^2\}\) is neither reflexive nor symmetric nor transitive. 
Answer: We have, \(R = \{(a, b): a \le b^2\}\), where \(a, b \in \mathbb{R}\)
Reflexivity: Obviously, \(\frac{1}{2}\) is a real number and \(\frac{1}{2} \le (\frac{1}{2})^2\) is not true. Therefore, \(R\) is not reflexive.
Symmetry: Consider the real numbers 1 and 2. Obviously, \(1 \le 2^2 \Rightarrow (1, 2) \in R\). But, \(2 \le (1)^2\) is not true and so, \((2, 1) \notin R\). Therefore, \(R\) is not symmetric.

Thus, \( (1, 2) \in R \) but \( (2, 1) \notin R \)
Hence, \( R \) is not symmetric.
Transitivity: By taking real numbers \( 2, -2 \) and \( 1 \), we have, \( 2 \le (-2)^2 \) and \( -2 \le (1)^2 \) but \( 2 \le (1)^2 \) is not true.
Thus, \( (2, -2) \in R \) and \( (-2, 1) \in R \), but \( (2, 1) \notin R \).
Hence, \( R \) is not transitive.

Question. Check whether the relation \( R \) in \( \mathbb{R} \) defined by \( R = \{(a, b) : a \le b^3\} \) is reflexive, symmetric or transitive.
Answer: \( R = \{(a, b) : a \le b^3, \forall a, b \in \mathbb{R}\} \)
Reflexivity: Here \( \frac{1}{3} \in \mathbb{R} \) (Real number)
and \( \frac{1}{3} > \frac{1}{27} \) or \( \frac{1}{3} > \left(\frac{1}{3}\right)^3 \) or \( \frac{1}{3} \not\le \left(\frac{1}{3}\right)^3 \)
So, \( \left(\frac{1}{3}, \frac{1}{3}\right) \notin R \)
\( \therefore R \) is not reflexive.
Symmetry: \( 1, 2 \in \mathbb{R} \) (Real number)
and \( 1 \le 8 \) or \( 1 \le 2^3 \)
So, \( (1, 2) \in R \) but \( (2, 1) \notin R \) [\( \because 2 > 1 \) or \( 2 > 1^3 \)]
\( \therefore R \) is not symmetric.
Transitivity: Here \( 10, 3, 2 \in \mathbb{R} \) (Real number)
and \( 10 \le 27 \) or \( 10 \le 3^3 \)
so, \( (10, 3) \in R \) and \( 3 \le 8 \) or \( 3 \le 2^3 \)
so, \( (3, 2) \in R \)
But \( 10 \ge 8 \) or \( 10 \ge 2^3 \) or \( 10 \not\le 2^3 \)
So, \( (10, 2) \notin R \)
So, here \( (10, 3) \in R \) and \( (3, 2) \in R \) but \( (10, 2) \notin R \)
\( \therefore R \) is not transitive.

Question. Show that the relation \( R \) in the set \( A = \{1, 2, 3, 4, 5\} \) given by \( R = \{(a, b) : |a - b| \text{ is even}\} \) is an equivalence relation. Show that all the elements of \( \{1, 3, 5\} \) are related to each other and all the elements of \( \{2, 4\} \) are related to each other. But no element of \( \{1, 3, 5\} \) is related to any element of \( \{2, 4\} \). 
Answer: For the given relation \( R \) on \( A \), we have
\( R = \{(1, 1), (1, 3), (1, 5), (2, 2), (2, 4), (3, 1), (3, 3), (3, 5), (4, 2), (4, 4), (5, 1), (5, 3), (5, 5)\} \)
For an equivalence relation, it must be reflexive, symmetric and transitive.
Reflexivity: Given that, \( A = \{1, 2, 3, 4, 5\} \) and \( R = \{(a, b) : |a - b| \text{ is even}\} \)
Here \( (a, a) \in R \) as \( |a - a| = 0 \) is even for \( a \in A \), so it is reflexive.
Symmetry: Let \( (a, b) \in R \) i.e., \( |a - b| \) is even \( \Rightarrow |b - a| \) is also even \( \Rightarrow (b, a) \in R \)
Thus, it is symmetric.
Transitivity: Now, if \( (a, b) \in R \) i.e., \( |a - b| \) is even \( \Rightarrow a - b = \pm 2m, m \in \mathbb{N} \)
and \( (b, c) \in R \) i.e., \( |b - c| \) is even \( \Rightarrow b - c = \pm 2n, n \in \mathbb{N} \)
Then, \( a - c = (a - b) + (b - c) = (\pm 2m) + (\pm 2n) = \pm 2(m + n) \)
\( \therefore |a - c| = 2(m + n) \)
Thus, \( |a - c| \) is even.
Hence, \( (a, c) \in R \Rightarrow R \) is transitive.
Hence, it is an equivalence relation.
In set \( R \) all the elements corresponding to \( \{1, 3, 5\} \) i.e., \( (1, 3), (3, 1), (1, 5), (5, 1), (3, 5), (5, 3) \) are related to each other because difference of these elements are even.
Again, all elements corresponding to \( \{2, 4\} \) are related to each other.
But no element of \( \{1, 3, 5\} \) is related to elements of \( \{2, 4\} \) because the difference of elements of the two sets are not even.

Question. Show that each of the relation \( R \) in the set \( A = \{x \in \mathbb{Z} : 0 \le x \le 12\} \), given by 
(i) \( R = \{(a, b) : |a - b| \text{ is a multiple of 4}\} \)
(ii) \( R = \{(a, b) : a = b\} \) is an equivalence relation.
Find the set of all elements related to 1 in each case.
Answer: \( A = \{x \in \mathbb{Z} : 0 \le x \le 12\} \)
(i) \( R = \{(a, b) : |a - b| \text{ is a multiple of 4}\} \)
Reflexive: Let \( x \in A \Rightarrow |x - x| = 0 \), which is a multiple of 4.
\( \Rightarrow (x, x) \in R, \forall x \in A \)
\( \therefore R \) is reflexive.
Symmetric: Let \( x, y \in A \) and \( (x, y) \in R \)
\( \Rightarrow |x - y| \) is a multiple of 4
or \( x - y = \pm 4p \) {\( p \) is any integer}
\( \Rightarrow y - x = \mp 4p \)
\( \Rightarrow |y - x| \) is a multiple of 4. \( \Rightarrow (y, x) \in R \)
\( \therefore R \) is symmetric.
Transitive: Let \( x, y, z \in A, (x, y) \in R \) and \( (y, z) \in R \)
\( \Rightarrow |x - y| \) is multiple of 4 and \( |y - z| \) is multiple of 4
\( \Rightarrow x - y \) is multiple of 4 and \( y - z \) is multiple of 4
\( \Rightarrow (x - y) + (y - z) \) is multiple of 4 \( \Rightarrow (x - z) \) is multiple of 4.
\( \Rightarrow |x - z| \) is multiple of 4.
\( \Rightarrow (x, z) \in R \Rightarrow R \) is transitive.
So, \( R \) is an equivalence relation.
Let \( B \) be the set of elements related to 1.
\( \therefore B = \{a \in A : |a - 1| \text{ is multiple of 4}\} \)
\( \Rightarrow B = \{1, 5, 9\} \) {as \( |1 - 1| = 0, |1 - 5| = 4, |1 - 9| = 8 \)}
(ii) \( R = \{(a, b) : a = b\} \)
Reflexive: Let \( x \in A \)
as \( x = x \Rightarrow (x, x) \in R \Rightarrow R \) is reflexive.
Symmetric: Let \( x, y \in A \) and \( (x, y) \in R \)
\( \Rightarrow x = y \Rightarrow y = x \)
\( \Rightarrow (y, x) \in R \)
\( \therefore R \) is symmetric.
Transitive: Let \( x, y, z \in A \)
and let \( (x, y) \in R \) and \( (y, z) \in R \)
\( \Rightarrow x = y \) and \( y = z \Rightarrow x = z \Rightarrow (x, z) \in R \Rightarrow R \) is transitive.
\( \therefore R \) is an equivalence relation.
Let \( C \) be the set of elements related to 1.
\( \therefore C = \{a \in A ; a = 1\} = \{1\} \).

Question. Prove that the greatest integer function \( f: \mathbb{R} \rightarrow \mathbb{R} \) given by \( f(x) = [x] \), is neither one-one nor onto, where \( [x] \) denotes the greatest integer less than or equal to \( x \).
Answer: \( f : \mathbb{R} \rightarrow \mathbb{R} \) given by \( f(x) = [x] \)
Injectivity: Let \( x_1 = 2.5 \) and \( x_2 = 2 \) be two elements of \( \mathbb{R} \).
\( f(x_1) = f(2.5) = [2.5] = 2 \)
\( f(x_2) = f(2) = [2] = 2 \)
\( \therefore f(x_1) = f(x_2) \) for \( x_1 \neq x_2 \)
\( \Rightarrow f(x) = [x] \) is not one-one i.e., not injective.
Surjectivity: Let \( y = 2.5 \in \mathbb{R} \) be any element.
\( \therefore f(x) = 2.5 \Rightarrow [x] = 2.5 \)
Which is not possible as \( [x] \) is always an integer.
\( \Rightarrow f(x) = [x] \) is not onto i.e., not surjective.

Question. Show that the modulus function \( f : \mathbb{R} \rightarrow \mathbb{R} \) given by \( f(x) = |x| \), is neither one-one nor onto, where \( |x| \) is \( x \), if \( x \) is positive or 0 and \( |x| \) is \( -x \), if \( x \) is negative.
Answer: \( f(x) = |x| = \begin{cases} x, & \text{if } x \ge 0 \\ -x, & \text{if } x < 0 \end{cases} \)
One-one: Let \( x_1 = 1, x_2 = -1 \) be two elements belonging to \( \mathbb{R} \)
\( f(x_1) = f(1) = |1| = 1 \) and \( f(x_2) = f(-1) = -(-1) = 1 \)
\( \Rightarrow f(x_1) = f(x_2) \) for \( x_1 \neq x_2 \)
\( \Rightarrow f(x) \) is not one-one.
Onto: Let \( f(x) = -1 \Rightarrow |x| = -1 \in \mathbb{R} \), which is not possible.
\( \Rightarrow f(x) \) is not onto.
Hence, \( f \) is neither one-one nor onto function.

Question. Let \( f : \mathbb{N} \rightarrow \mathbb{N} \) be defined by \( f(n) = \begin{cases} \frac{n+1}{2}, & \text{if } n \text{ is odd} \\ \frac{n}{2}, & \text{if } n \text{ is even} \end{cases} \). For all \( n \in \mathbb{N} \), state whether the function \( f \) is bijective. Justify your answer. 
Answer: Given, \( f(n) = \begin{cases} \frac{n+1}{2}, & \text{if } n \text{ is odd} \\ \frac{n}{2}, & \text{if } n \text{ is even} \end{cases} \)
Let \( x_1 = 1 \) and \( x_2 = 2 \) be two elements of \( \mathbb{N} \).
\( \therefore f(x_1) = f(1) = \frac{1+1}{2} = 1 \) and \( f(x_2) = f(2) = \frac{2}{2} = 1 \)
\( \therefore f(x_1) = f(x_2) \) for \( x_1 \neq x_2 \)
So, \( f \) is not one-one. Hence, \( f \) is not bijective.
\( f(x_1) = f(x_2) \) for \( x_1 \neq x_2 \)
\( f : N \rightarrow N \) is not one-one.
\( \Rightarrow \) As \( f \) is not one-one. \( f \) is not a bijective function.

Question. Consider \( f : R_+ \rightarrow [-5, \infty) \) given by \( f(x) = 9x^2 + 6x - 5 \). Show that \( f \) is invertible and \( f^{-1}(y) = \left( \frac{\sqrt{y+6}-1}{3} \right) \). 
Answer: Given function \( f : R_+ \rightarrow [-5, \infty) \) such that \( f(x) = 9x^2 + 6x - 5 \)
One-one: Let \( x_1, x_2 \in R_+ \) then
\( f(x_1) = f(x_2) \Rightarrow 9x_1^2 + 6x_1 - 5 = 9x_2^2 + 6x_2 - 5 \)
\( \Rightarrow 9(x_1^2 - x_2^2) + 6(x_1 - x_2) = 0 \Rightarrow 3\{3(x_1 + x_2) + 2\} \{x_1 - x_2\} = 0 \)
\( \Rightarrow x_1 - x_2 = 0 \) [\(\because x_1, x_2 \in R_+ \Rightarrow x_1 + x_2 \neq 0 \Rightarrow 3(x_1 + x_2) + 2 \neq 0\)]
\( \Rightarrow x_1 = x_2 \)
So, given function is one-one.
Onto: Let \( y \in [-5, \infty) \) then \( y = f(x) \)
\( \Rightarrow y = 9x^2 + 6x - 5 \)
i.e., \( 9x^2 + 6x - 5 = y \)
\( \Rightarrow (3x)^2 + 2 \cdot 3x \cdot 1 + (1)^2 - 1 - 5 = y \)
\( \Rightarrow (3x + 1)^2 = y + 6 \Rightarrow 3x + 1 = \sqrt{y + 6} \)
\( \Rightarrow x = \frac{-1 + \sqrt{y + 6}}{3} \) ...(i)
Clearly, \( x \in R \) for all \( y \in [-5, \infty) \)
Thus, for every \( y \in [-5, \infty) \) there exists \( x = \frac{-1 + \sqrt{y + 6}}{3} \in R \)
So, given function is onto.
Thus, \( f \) is both one-one and onto.
Hence, it is invertible.
Inverse:
From (i) we get,
\( x = \frac{-1 + \sqrt{y + 6}}{3} \) i.e., \( x = \frac{\sqrt{y+6}-1}{3} \)
\( \Rightarrow f^{-1}(y) = \frac{\sqrt{y+6}-1}{3} \)
\( \therefore f^{-1}(x) = \frac{\sqrt{x+6}-1}{3} \)

Question. Give examples of two functions \( f : N \rightarrow N \) and \( g : N \rightarrow N \) such that \( gof \) is onto but \( f \) is not onto.
Answer: Let \( f(x) = x + 1 \) and \( g(x) = \begin{cases} x - 1, & \text{if } x > 1 \\ 1, & \text{if } x = 1 \end{cases} \)
Let \( x \in N \) be any element.
\( \therefore x \geq 1 \Rightarrow x + 1 \geq 2 \)
\( \Rightarrow f(x) \geq 2 \forall x \in N \)
\( \therefore R_f \neq N \)
Hence, \( f \) is not onto.
Also, \( gof : N \rightarrow N \) is such that
\( (gof)(x) = g(f(x)) = g(x + 1) = (x + 1) - 1 = x \)
\( \Rightarrow (gof)(x) = x \forall x \in N \)
\( \therefore gof \) is an identity function.
Hence, \( gof \) is onto.

Question. Let \( A = \{-1, 0, 1, 2\}, B = \{-4, -2, 0, 2\} \) and \( f, g : A \rightarrow B \) be function defined by \( f(x) = x^2 - x, x \in A \) and \( g(x) = 2 \left| x - \frac{1}{2} \right| - 1, x \in A \). Are \( f \) and \( g \) equal? Justify your answer.
Answer: Given \( f : A \rightarrow B \) and \( g : A \rightarrow B \) defined as
\( f(x) = x^2 - x \) and \( g(x) = 2 \left| x - \frac{1}{2} \right| - 1 \forall x \in A \)
\( f(-1) = (-1)^2 - (-1) = 2 \); \( g(-1) = 2 \left| -1 - \frac{1}{2} \right| - 1 = 2 \left| \frac{-3}{2} \right| - 1 = 2 \left( \frac{3}{2} \right) - 1 = 2 \)
\( f(0) = 0^2 - 0 = 0 \); \( g(0) = 2 \left| 0 - \frac{1}{2} \right| - 1 = 2 \left| -\frac{1}{2} \right| - 1 = 1 - 1 = 0 \)
\( f(1) = 1^2 - 1 = 1 - 1 = 0 \); \( g(1) = 2 \left| 1 - \frac{1}{2} \right| - 1 = 2 \left( \frac{1}{2} \right) - 1 = 1 - 1 = 0 \)
\( f(2) = 2^2 - 2 = 2 \); \( g(2) = 2 \left| 2 - \frac{1}{2} \right| - 1 = 2 \left( \frac{3}{2} \right) - 1 = 3 - 1 = 2 \)
Clearly, \( f(-1) = g(-1) \); \( f(0) = g(0) \); \( f(1) = g(1) \) and \( f(2) = g(2) \)
\( \therefore f(x) = g(x) \forall x \in A \)

Question. Let \( f : R \rightarrow R \) be the Signum function defined as \( f(x) = \begin{cases} 1, & x > 0 \\ 0, & x = 0 \\ -1, & x < 0 \end{cases} \) and \( g : R \rightarrow R \) be the Greatest Integer Function given by \( g(x) = [x] \). Then, do \( fog \) and \( gof \) coincide in \( (0, 1] \)?
Answer: Given \( f : R \rightarrow R \) and \( g : R \rightarrow R \) defined as
\( f(x) = \begin{cases} 1, & x > 0 \\ 0, & x = 0 \\ -1, & x < 0 \end{cases} \) and \( g(x) = [x] \)
\( fog(x) = f(g(x)) = f([x]) \)
\( = \begin{cases} f(0), & \text{if } 0 < x < 1 \\ f(1), & \text{if } x = 1 \end{cases} \)
\( \therefore fog(x) = \begin{cases} 0, & \text{if } 0 < x < 1 \\ 1, & \text{if } x = 1 \end{cases} \)
and \( gof(x) = g(f(x)) \)
\( = g(1) \forall x \in (0, 1] \)
\( = [1] = 1 \)
\( \therefore gof(x) = 1 \forall x \in (0, 1] \)
Clearly, \( fog(x) \) and \( gof(x) \) do not coincide \( \forall x \in (0, 1] \).

Fill in the Blanks

Question. A relation from a set A to a set B is a ___________ of \( A \times B \).
Answer: Subset

Question. A relation R from set A to set B is said to be ___________ if \( R = A \times B \).
Answer: The universal relation

Question. If any set A contains \( n \) elements. Then, the total number of injective functions from A onto itself is ___________ .
Answer: \( n! \)

Question. The domain of the function \( f : R \rightarrow R \) defined by \( f(x) = \sqrt{x^2 - 3x + 2} \) is ___________ .
Answer: \( (-\infty, 1] \cup [2, \infty) \)

Question. If \( f(x) = \{4 - (x - 7)^3\} \), then \( f^{-1}(x) = \) ___________ .
Answer: \( 7 + (4 - x)^{\frac{1}{3}} \)

Solutions of Selected Fill in the Blanks

Question. For \( f(x) \) to be defined
\( x^2 - 3x + 2 \ge 0 \)
\( \Rightarrow x^2 - 2x - x + 2 \ge 0 \Rightarrow x(x - 2) - 1 (x - 2) \ge 0 \)
\( \Rightarrow (x - 2) (x - 1) \ge 0 \Rightarrow (x - 1) (x - 2) \ge 0 \)
\( \therefore x \le 1 \text{ or } x \ge 2 \)
Domain of \( f = (-\infty, 1] \cup [2, \infty) \).
Answer: \( (-\infty, 1] \cup [2, \infty) \)

Question. Let \( y = f(x) \Rightarrow y = \{4 - (x - 7)^3\} \)
\( \Rightarrow y - 4 = -(x - 7)^3 \)
\( \Rightarrow (x - 7)^3 = 4 - y \)
\( \Rightarrow (x - 7) = (4 - y)^{\frac{1}{3}} \Rightarrow x = 7 + (4 - y)^{\frac{1}{3}} \)
\( \Rightarrow f^{-1}(x) = 7 + (4 - x)^{\frac{1}{3}} \)
Answer: \( f^{-1}(x) = 7 + (4 - x)^{\frac{1}{3}} \)

Very Short Answer Questions

Question. Let \( A = \{1, 2, 3, 4\} \). Let \( R \) be the equivalence relation on \( A \times A \) defined by \( (a, b) R (c, d) \) iff \( a + d = b + c \). Find the equivalence class \( [(1, 3)] \).
Answer: \( [(1, 3)] = \{(x, y) \in A \times A : x + 3 = y + 1\} = \{(x, y) \in A \times A : y - x = 2\} \)
\( = \{(1, 3), (2, 4)\} \)

Question. If \( R = \{(x, y) : x + 2y = 8\} \) is a relation on \( N \), write the range of \( R \).
Answer: Given: \( R = \{(x, y) : x + 2y = 8\} \)
\( \because x + 2y = 8 \)
\( \Rightarrow y = \frac{8 - x}{2} \)
when \( x = 6, y = 1; x = 4, y = 2; x = 2, y = 3 \).
\( \therefore \text{Range} = \{1, 2, 3\} \)

Question. State the reason for the relation \( R \) in the set \( \{1, 2, 3\} \) given by \( R = \{(1, 2), (2, 1)\} \) not to be transitive.
Answer: \( R \) is not transitive as \( (1, 2) \in R \) and \( (2, 1) \in R \) but \( (1, 1) \notin R \)
[Note: A relation \( R \) in a set \( A \) is said to be transitive if \( (a, b) \in R \) and \( (b, c) \in R \Rightarrow (a, c) \in R, \forall a, b, c \in R \)]

Question. Let \( R = \{(a, a^3) : a \text{ is a prime number less than 5}\} \) be a relation. Find the range of \( R \).
Answer: Here \( R = \{(a, a^3) : a \text{ is a prime number less than 5}\} \)
\( \Rightarrow R = \{(2, 8), (3, 27)\} \)
Hence Range of \( R = \{8, 27\} \)

Question. If \( X \) and \( Y \) are two sets having 2 and 3 elements respectively, then find the number of functions from \( X \) to \( Y \).
Answer: Number of functions from \( X \) to \( Y = 3^2 = 9 \).

Question. If the mapping \( f \) and \( g \) are given by \( f = \{(1, 2), (3, 5), (4, 1)\} \) and \( g = \{(2, 3), (5, 1), (1, 3)\} \), then write \( fog \).
Answer: Obviously, domain of “\( fog \)” is domain of “\( g \)” i.e., \( \{2, 5, 1\} \).
Now, \( fog(2) = f(g(2)) = f(3) = 5 \),
\( fog(5) = f(g(5)) = f(1) = 2 \)
\( fog(1) = f(g(1)) = f(3) = 5 \)
\( \Rightarrow fog = \{(2, 5), (5, 2), (1, 5)\} \)

Question. If \( f : R \rightarrow R \) is given by \( f(x) = (3 - x^3)^{\frac{1}{3}} \), then determine \( f(f(x)) \).
Answer: We have, \( f(x) = (3 - x^3)^{\frac{1}{3}} \)
\( \therefore f(f(x)) = f[(3 - x^3)^{\frac{1}{3}}] = [3 - \{(3 - x^3)^{\frac{1}{3}}\}^3]^{\frac{1}{3}} \)
\( = [3 - (3 - x^3)]^{\frac{1}{3}} = (x^3)^{\frac{1}{3}} = x \)

Question. Find \( fog(x) \), if \( f(x) = |x| \) and \( g(x) = |5x - 2| \).
Answer: \( fog(x) = f(g(x)) = f(|5x - 2|) = ||5x - 2|| = |5x - 2| \)

Question. Write \( fog \), if \( f : R \rightarrow R \) and \( g : R \rightarrow R \) are given by \( f(x) = 8x^3 \) and \( g(x) = x^{\frac{1}{3}} \).
Answer: \( fog(x) = f(g(x)) = f(x^{\frac{1}{3}}) = 8(x^{\frac{1}{3}})^3 = 8x \)

Question. If \( f : R \rightarrow R \) is defined by \( f(x) = 3x + 2 \), define \( f[f(x)] \).
Answer: \( f(f(x)) = f(3x + 2) = 3(3x + 2) + 2 \)
\( = 9x + 6 + 2 = 9x + 8 \)

Question. If \( f : R \rightarrow R \) is given by \( f(x) = x^2 \), find the value of \( f^{-1} (25) \).
Answer: Let \( y = f(x) \Rightarrow y = x^2 \Rightarrow x = \sqrt{y} \)
\( \Rightarrow f^{-1}(x) = \sqrt{x} \)
\( \Rightarrow f^{-1}(25) = \sqrt{25} = \pm 5 \)
\( \therefore f^{-1}(25) = \{-5, 5\} \)

Short Answer Questions-I

Question. Write the inverse relation corresponding to the relation \( R \) given by \( R = \{(x, y): x \in N, x < 5, y = 3\} \). Also write the domain and range of inverse relation.
Answer: Given, \( R = \{(x, y) : x \in N, x < 5, y = 3\} \)
\( \Rightarrow R = \{(1, 3), (2, 3), (3, 3), (4, 3)\} \)
Hence, required inverse relation is
\( R^{-1} = \{(3, 1), (3, 2), (3, 3), (3, 4)\} \)
\( \therefore \text{Domain of } R^{-1} = \{3\} \) and
\( \text{Range of } R^{-1} = \{1, 2, 3, 4\} \)

Question. Check if the relation \( R \) in the set \( A = \{1, 2, 3, 4\} \) defined as \( R = \{(a, b) : a \text{ divides } b\} \) is (i) symmetric (ii) transitive.
Answer: In the set \( A = \{1, 2, 3, 4\} \)
Relation is defined as \( R = \{(a, b) : a \text{ divides } b\} \)
(i) Symmetric: Take \( a = 2, b = 4, a, b \in A \)
\( \because 2 \text{ divides } 4 \Rightarrow (a, b) \in R \)
but 4 does not divide 2 \( \Rightarrow (b, a) \notin R \)
\( \therefore \text{It is not symmetric} \)
(ii) Transitive: Let \( a = 1, b = 2 \text{ and } c = 4 \)
Here 1 divides 2 \( \Rightarrow (1, 2) \in R \)
and 2 divides 4 \( \Rightarrow (2, 4) \in R \)
\( \Rightarrow 1 \text{ divides } 4 \text{ also} \Rightarrow (1, 4) \in R \)
\( \therefore \text{It is transitive} \).

Question. If \( f \) is an invertible function, defined as \( f(x) = \frac{3x - 4}{5} \), write \( f^{-1}(x) \).
Answer: Since \( f^{-1} \) is inverse of \( f \).
\( \therefore fof^{-1} = I \Rightarrow fof^{-1} (x) = I (x) \)
\( \Rightarrow f(f^{-1}(x)) = (x) \)
\( \Rightarrow \frac{3(f^{-1}(x)) - 4}{5} = x \Rightarrow f^{-1}(x) = \frac{5x + 4}{3} \)

Question. What is the range of the function \( f(x) = \frac{|x - 1|}{(x - 1)} \)?
Answer: Given \( f(x) = \frac{|x - 1|}{(x - 1)} \)
Obviously, \( |x - 1| = \begin{cases} (x - 1) & \text{if } x - 1 > 0 \text{ or } x > 1 \\ -(x - 1) & \text{if } x - 1 < 0 \text{ or } x < 1 \end{cases} \)
Now, (i) \( \forall x > 1, f(x) = \frac{(x - 1)}{(x - 1)} = 1 \), (ii) \( \forall x < 1, f(x) = \frac{-(x - 1)}{(x - 1)} = -1 \),
i.e., \( f(x) = -1, 1 \)
\( \therefore \text{Range of } f(x) = \{-1, 1\} \).

Question. Let \( f : R \rightarrow R \) be the function defined by \( f(x) = \frac{1}{2 - \cos x} \), \( \forall x \in R \). Then, find the range of \( f \).
Answer: Given function, \( f(x) = \frac{1}{2 - \cos x} \), \( \forall x \in R \)
\( y = \frac{1}{2 - \cos x} \)
\( \Rightarrow 2y - y \cos x = 1 \Rightarrow y \cos x = 2y - 1 \)
\( \Rightarrow \cos x = \frac{2y - 1}{y} = 2 - \frac{1}{y} \)
\( \Rightarrow -1 \le \cos x \le 1 \Rightarrow -1 \le 2 - \frac{1}{y} \le 1 \)
\( \Rightarrow -3 \le -\frac{1}{y} \le -1 \Rightarrow 1 \le \frac{1}{y} \le 3 \)
\( \Rightarrow \frac{1}{3} \le y \le 1 \)
So, range of \( y \) is \( \left[\frac{1}{3}, 1\right] \).

Question. If \( f : R \rightarrow R \) is defined by \( f(x) = x^2 - 3x + 2 \), write \( f\{f(x)\} \).
Answer: Given that, \( f(x) = x^2 - 3x + 2 \)
\( f\{f(x)\} = f(x^2 - 3x + 2) \)
\( = (x^2 - 3x + 2)^2 - 3(x^2 - 3x + 2) + 2 \)
\( = x^4 + 9x^2 + 4 - 6x^3 - 12x + 4x^2 - 3x^2 + 9x - 6 + 2 \)
\( = x^4 - 6x^3 + 10x^2 - 3x \)