CBSE Class 12 Mathematics Relations And Functions Worksheet Set 02

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

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

Question. If \( f(x) = 27x^3 \) and \( g(x) = x^{1/3} \), find \( gof(x) \).
Answer: Given \( f(x) = 27x^3 \) and \( g(x) = x^{1/3} \)
\( (gof)(x) = g\{f(x)\} = g[27x^3] = [27x^3]^{1/3} = 3x \)

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

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

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. Check whether the relation \( R \) in the set \( \{1, 2, 3\} \) given by \( R = \{(1, 2), (2, 1)\} \) is transitive.
Answer: No, it is not transitive because \( 1R2, 2R1 \) but \( 1 \not R 1 \), i.e., \( (1, 1) \) does not lie in \( R \).

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 \( (1, 2) \in R, (2, 1) \in R \) But \( (1, 1) \notin R \)
[Note: A relation \( R \) in a set \( A \) is said to be transitive if \( (a, b) \in R, (b, c) \in R \)
\( \Rightarrow (a, c) \in R \forall a, b, c \in R \)]

Question. Let \( A = \{1, 2, 3\}, B = \{4, 5, 6, 7\} \) and let \( f = \{(1, 4), (2, 5), (3, 6)\} \) be a function from \( A \) to \( B \). State whether \( f \) is one-one or not.
Answer: \( f \) is one-one because
\( f(1) = 4; \)
\( f(2) = 5; \)
\( f(3) = 6 \)
i.e., no two elements of \( A \) have same \( f \) image.

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, \( fog(2) = f(g(2)) = f(3) = 5 \)
\( \Rightarrow fog(5) = f(g(5)) = f(1) = 2 \)
\( fog(1) = f(g(1)) = f(3) = 5 \)
\( \Rightarrow fog = \{(2, 5), (5, 2), (1, 5)\} \)

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

Short Answer Questions

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) \( \forall x > 1, f(x) = \frac{(x - 1)}{(x - 1)} = 1, \quad \) (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. 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 \( f \).
\( \therefore fof^{-1} = I \)
\( \Rightarrow fof^{-1}(x) = I(x) \)
\( \Rightarrow fof^{-1}(x) = x \)
\( \Rightarrow f(f^{-1}(x)) = (x) \)
\( \Rightarrow \frac{3(f^{-1}(x)) - 4}{5} = x \)
\( \Rightarrow f^{-1}(x) = \frac{5x + 4}{3} \)

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

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

Long Answer Questions

Question. Show that the relation \( R \) in the set \( \mathbb{N} \times \mathbb{N} \) defined by \( (a, b)R(c, d) \) iff \( a^2 + d^2 = b^2 + c^2 \forall a, b, c, d \in \mathbb{N} \), is an equivalence relation. 
Answer: Given, \( R \) is a relation in \( \mathbb{N} \times \mathbb{N} \) defined by \( (a, b) R (c, d) \)
\( \Rightarrow a^2 + d^2 = b^2 + c^2 \)
Reflexivity:
\( \because a^2 + b^2 = b^2 + a^2 \forall a, b \in \mathbb{N} \)
\( \Rightarrow (a, b) R (a, b) \)
\( \Rightarrow R \text{ is reflexive} \)
Symmetry: Let \( (a, b)R(c, d) \)
\( \Rightarrow a^2 + d^2 = b^2 + c^2 \)
\( \Rightarrow b^2 + c^2 = a^2 + d^2 \)
\( \Rightarrow c^2 + b^2 = d^2 + a^2 \)
\( \Rightarrow (c, d) R (a, b) \)
\( \Rightarrow R \text{ is symmetric} \)
Transitivity: Let \( (a, b) R (c, d) \text{ and } (c, d) R (e, f) \)
\( \Rightarrow a^2 + d^2 = b^2 + c^2 \text{ and } c^2 + f^2 = d^2 + e^2 \)
\( \Rightarrow a^2 + d^2 + c^2 + f^2 = b^2 + c^2 + d^2 + e^2 \)
\( \Rightarrow a^2 + f^2 = b^2 + e^2 \)
\( \Rightarrow (a, b) R (e, f) \)
\( \Rightarrow R \text{ is transitive.} \)
Hence, \( R \) is an equivalence relation.

Question. Let \( A = \{x \in \mathbb{R} : -1 \le x \le 1\} = B \). Show that \( f : A \rightarrow B \) given by \( f(x) = x |x| \) is a bijection. 
Answer: We have,
\( f(x) = x |x| = \begin{cases} x^2, & \text{if } x \ge 0 \\ -x^2, & \text{if } x < 0 \end{cases} \)
For \( x \ge 0, f(x) = x^2 \) represents a parabola opening upward and for \( x < 0, f(x) = -x^2 \) represents a parabola opening downward.
So, the graph of \( f(x) \) is as shown in figure.
Since any line parallel to x-axis, will cut the graph at only one point, so \( f \) is one-one. Also, any line parallel to y-axis will cut the graph, so \( f \) is onto.
Thus, it is evident from the graph of \( f(x) \) that \( f \) is one-one and onto.

Question. If \( f(x) = \sqrt{x} (x \ge 0) \) and \( g(x) = x^2 - 1 \) are two real functions, then find \( fog \) and \( gof \) and check whether \( fog = gof \). 
Answer: The given functions are \( f(x) = \sqrt{x}, x \ge 0 \) and \( g(x) = x^2 - 1 \)
We have, domain of \( f = [0, \infty) \) and range of \( f = [0, \infty) \)
domain of \( g = \mathbb{R} \) and range of \( g = [-1, \infty) \quad \left[ \because x^2 \ge 0 \text{ for all } x \in \mathbb{R} \right] \)
\( \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \left[ \therefore x^2 - 1 \ge -1 \text{ for all } x \in \mathbb{R} \right] \)
Computation of \( gof \): We observe that range of \( f = [0, \infty) \subseteq \) domain of \( g \)
\( \therefore gof \text{ exists and } gof : [0, \infty) \rightarrow \mathbb{R} \)
Also, \( gof(x) = g(f(x)) = g(\sqrt{x}) = (\sqrt{x})^2 - 1 = x - 1 \)
Thus, \( gof : [0, \infty) \rightarrow \mathbb{R} \text{ is defined as } gof(x) = x - 1 \)
Computation of \( fog \): We observe that range of \( g = [-1, \infty) \not\subseteq \) domain of \( f \).
\( \therefore \text{Domain of } fog = \{x : x \in \text{domain of } g \text{ and } g(x) \in \text{domain of } f\} \)
\( \Rightarrow \text{Domain of } fog = \{x : x \in \mathbb{R} \text{ and } g(x) \in [0, \infty)\} \)
\( \Rightarrow \text{Domain of } fog = \{x : x \in \mathbb{R} \text{ and } x^2 - 1 \in [0, \infty)\} \)
\( \Rightarrow \text{Domain of } fog = \{x : x \in \mathbb{R} \text{ and } x^2 - 1 \ge 0\} \)
Domain of \( fog = \{x : x \in \mathbb{R} \text{ and } x \le -1 \text{ or } x \ge 1\} \)
\( \therefore \text{Domain of } fog = x : x \in (-\infty, -1] \cup [1, \infty) \)
Also, \( fog(x) = f(g(x)) = f(x^2 - 1) = \sqrt{x^2 - 1} \)
Thus, \( fog : (-\infty, -1] \cup [1, \infty) \rightarrow \mathbb{R} \text{ is defined as } fog(x) = \sqrt{x^2 - 1} \).
We find that \( fog \) and \( gof \) have distinct domains. Also, their formulae are not same.
Hence, \( fog \neq gof \)

Question. Let \( A = \{-1, 0, 1, 2\}, B = \{-4, -2, 0, 2\} \) and \( f, g : A \rightarrow B \) be functions 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: For two functions \( f : A \rightarrow B \) and \( g : A \rightarrow B \) to be equal, \( f(a) = g(a) \forall a \in A \text{ and } R_f = R_g \).
Here, we have \( f(x) = x^2 - x \)
\( g(x) = 2 \left| x - \frac{1}{2} \right| - 1 \quad [x \in A = \{-1, 0, 1, 2\}] \)
We see that, \( f(-1) = (-1)^2 - (-1) = 2 \)
\( g(-1) = 2 \left| (-1) - \frac{1}{2} \right| - 1 = 2 \times \frac{3}{2} - 1 = 3 - 1 = 2 \)
So, \( f(-1) = g(-1) \)
Again, we check that, \( f(0) = g(0) = 0, f(1) = g(1) = 0 \text{ and } f(2) = g(2) = 2 \).
Hence, \( f \) and \( g \) are equal functions.

Question. Let \( g(x) = 1 + x - [x] \) and \( f(x) = \begin{cases} -1, & x < 0 \\ 0, & x = 0 \\ 1, & x > 0 \end{cases} \) then for all \( x \) find \( fog(x) \).
Answer: \( fog(x) = f(g(x)) = f(1 + x - [x]) = f(1 + \{x\}) = 1 \)
Here, \( \{x\} = x - [x] \)
Obviously, \( 0 \le \{x\} < 1 \)
\( \Rightarrow 0 \le \{x\} < 1 \)
\( \Rightarrow 1 + \{x\} \ge 1 \)
\( \therefore fog(x) = f(1 + \{x\}) = 1 \)
Note: Symbol \( \{x\} \) denotes the partial part or decimal part of \( x \).
For example, \( \{4.25\} = 0.25, \{4\} = 0, \{-3.45\} = 0.45 \)
In this way \( \{x\} = x - [x] \)
\( \Rightarrow 0 \le \{x\} < 1 \)

Question. Let \( A = \mathbb{R} - \{3\}, B = \mathbb{R} - \{1\} \). If \( f : A \rightarrow B \) be defined by \( f(x) = \frac{x - 2}{x - 3}, \forall x \in A \). Then, show that \( f \) is bijective. 
Answer: Given that, \( A = \mathbb{R} - \{3\}, B = \mathbb{R} - \{1\} \).
\( f : A \rightarrow B \text{ is defined by } f(x) = \frac{x - 2}{x - 3}, \forall x \in A \)
For injectivity
Let \( f(x_1) = f(x_2) \)
\( \Rightarrow \frac{x_1 - 2}{x_1 - 3} = \frac{x_2 - 2}{x_2 - 3} \)
\( \Rightarrow (x_1 - 2)(x_2 - 3) = (x_2 - 2)(x_1 - 3) \)
\( \Rightarrow x_1 x_2 - 3x_1 - 2x_2 + 6 = x_1 x_2 - 3x_2 - 2x_1 + 6 \)
\( \Rightarrow -3x_1 - 2x_2 = -3x_2 - 2x_1 \)
\( \Rightarrow -x_1 = -x_2 \)
\( \Rightarrow x_1 = x_2 \)
So, \( f(x) \) is an injective function.
For surjectivity
Let \( y = \frac{x - 2}{x - 3} \)
\( \Rightarrow x - 2 = xy - 3y \)
\( \Rightarrow x(1 - y) = 2 - 3y \)
\( \Rightarrow x = \frac{2 - 3y}{1 - y} \)
\( \Rightarrow x = \frac{3y - 2}{y - 1} \in A, \forall y \in B \quad [\text{codomain}] \)
So, \( f(x) \) is surjective function.
Hence, \( f(x) \) is a bijective function.

Question. Let \( A = \{1, 2, 3, \dots, 9\} \) and \( R \) be the relation in \( A \times A \) defined by \( (a, b) R (c, d) \) if \( a + d = b + c \) for \( (a, b), (c, d) \text{ in } A \times A \). Prove that \( R \) is an equivalence relation and also obtain the equivalent class \([(2, 5)]\). 
Answer: Given that, \( A = \{1, 2, 3, \dots, 9\} \) and \( (a, b) R (c, d) \) if \( a + d = b + c \) for \( (a, b) \in A \times A \) and \( (c, d) \in A \times A \).
Let \( (a, b) R (a, b) \) then
\( a + b = b + a, \forall a, b \in A \)
Hence, \( R \) is reflexive.
Let \( (a, b) R (c, d) \) then
\( a + d = b + c \)
\( \Rightarrow c + b = d + a \)
\( \Rightarrow (c, d) R (a, b) \)
Hence, \( R \) is symmetric.
Let \( (a, b) R (c, d) \text{ and } (c, d) R (e, f) \text{ then} \)
\( a + d = b + c \text{ and } c + f = d + e \)
\( \Rightarrow a + d = b + c \text{ and } d + e = c + f \)
\( \therefore (a + d) - (d + e) = (b + c) - (c + f) \)
\( \Rightarrow (a - e) = b - f \)
\( \Rightarrow a + f = b + e \)
\( \Rightarrow (a, b) R (e, f) \)
So, \( R \) is transitive.
Hence, \( R \) is an equivalence relation.
Now, equivalence class containing [(2, 5)] is \{(1, 4), (2, 5), (3, 6), (4, 7), (5, 8), (6, 9)\}.

Long Answer Questions-II

Question. Consider \( f : \mathbb{R}_+ \rightarrow [-9, \infty) \) given by \( f(x) = 5x^2 + 6x - 9 \). Prove that \( f \) is invertible with \( f^{-1}(y) = \left( \frac{\sqrt{54 + 5y} - 3}{5} \right) \). 
Answer: To prove \( f \) is invertible, it is sufficient to prove \( f \) is one–one onto
Here, \( f(x) = 5x^2 + 6x - 9 \)
One-one: Let \( x_1, x_2 \in \mathbb{R}_+ \), then
\( f(x_1) = f(x_2) \)
\( \Rightarrow 5x_1^2 + 6x_1 - 9 = 5x_2^2 + 6x_2 - 9 \)
\( \Rightarrow 5x_1^2 + 6x_1 - 5x_2^2 - 6x_2 = 0 \)
\( \Rightarrow 5(x_1^2 - x_2^2) + 6(x_1 - x_2) = 0 \)
\( \Rightarrow 5(x_1 - x_2)(x_1 + x_2) + 6(x_1 - x_2) = 0 \)
\( \Rightarrow (x_1 - x_2)(5x_1 + 5x_2 + 6) = 0 \)
\( \Rightarrow x_1 - x_2 = 0 \quad [\because 5x_1 + 5x_2 + 6 \neq 0] \)
\( \Rightarrow x_1 = x_2 \)
i.e., \( f \) is one-one function.
Onto: Let \( f(x) = y \)
\( \therefore y = 5x^2 + 6x - 9 \)
\( \Rightarrow 5x^2 + 6x - (9 + y) = 0 \)
\( \Rightarrow x = \frac{-6 \pm \sqrt{36 + 4 \times 5(9 + y)}}{10} \)
\( \Rightarrow x = \frac{-6 \pm \sqrt{216 + 20y}}{10} \)
\( \Rightarrow x = \frac{\pm \sqrt{54 + 5y} - 3}{5} \)
\( \Rightarrow x = \frac{\sqrt{54 + 5y} - 3}{5} \quad [\because x \in \mathbb{R}_+] \)
Obviously, \( \forall y \in [-9, \infty] \) the value of \( x \in \mathbb{R}_+ \).
\( \Rightarrow f \) is onto function.
Hence, \( f \) is one-one onto function, i.e., invertible.
Also, \( f \) is invertible with \( f^{-1}(y) = \frac{\sqrt{54 + 5y} - 3}{5} \).

Question. Let \( A = \mathbb{R} - \{3\} \) and \( B = \mathbb{R} - \{1\} \). Consider the function \( f : A \rightarrow B \) defined by \( f(x) = \left( \frac{x - 2}{x - 3} \right) \). Show that \( f \) is one-one and onto and hence find \( f^{-1} \).
OR
Let \( A = \mathbb{R} - \{2\} \) and \( B = \mathbb{R} - \{1\} \). If \( f : A \rightarrow B \) is a function defined by \( f(x) = \left( \frac{x - 1}{x - 2} \right) \), show that \( f \) is one-one and onto. Hence, find \( f^{-1} \).
Answer: One-one:
Let \( x_1, x_2 \in A \)
Now, \( f(x_1) = f(x_2) \)
\( \Rightarrow \frac{x_1 - 2}{x_1 - 3} = \frac{x_2 - 2}{x_2 - 3} \)
\( \Rightarrow (x_1 - 2)(x_2 - 3) = (x_1 - 3)(x_2 - 2) \)
\( \Rightarrow x_1 x_2 - 3x_1 - 2x_2 + 6 = x_1 x_2 - 2x_1 - 3x_2 + 6 \)
\( \Rightarrow -3x_1 - 2x_2 = -2x_1 - 3x_2 \)
\( \Rightarrow -x_1 = -x_2 \)
\( \Rightarrow x_1 = x_2 \)
Hence, \( f \) is one-one function.
Onto:
Let \( y = \frac{x - 2}{x - 3} \)
\( \Rightarrow xy - 3y = x - 2 \)
\( \Rightarrow xy - x = 3y - 2 \)
\( \Rightarrow x(y - 1) = 3y - 2 \)
\( \Rightarrow x = \frac{3y - 2}{y - 1} \quad \dots(i) \)
From above it is obvious that \( \forall y \) except 1, i.e., \( \forall y \in B = \mathbb{R} - \{1\} \exists x \in A \)
Hence, \( f \) is onto function.
Thus, \( f \) is one-one onto function.
If \( f^{-1} \) is inverse function of \( f \) then \( f^{-1}(y) = \frac{3y - 2}{y - 1} \). [from (i)]
OR
Solution is similar only values are changed.
Ans. \( \frac{2x - 1}{x - 1} \)

Question. If \( f, g : \mathbb{R} \rightarrow \mathbb{R} \) be two functions defined as \( f(x) = |x| + x \) and \( g(x) = |x| - x, \forall x \in \mathbb{R} \). Then find \( fog \) and \( gof \). Hence find \( fog(-3), fog(5) \) and \( gof(-2) \). 
Answer: Here, \( f(x) = |x| + x \) can be written as
\( f(x) = \begin{cases} 2x & \text{if } x \ge 0 \\ 0 & \text{if } x < 0 \end{cases} \)
And \( g(x) = |x| - x \), can be written as
\( g(x) = \begin{cases} 0 & \text{if } x \ge 0 \\ -2x & \text{if } x < 0 \end{cases} \)
Therefore, \( gof \) is defined as
For \( x \ge 0, gof(x) = g(f(x)) \)
\( \Rightarrow gof(x) = g(2x) = 0 \)
and for \( x < 0, gof(x) = g(f(x)) = g(0) = 0 \)
Hence, \( gof(x) = 0 \forall x \in \mathbb{R} \).
Again, \( fog \) is defined as
For \( x \ge 0, fog(x) = f(g(x)) = f(0) = 0 \)
and for \( x < 0, fog(x) = f(g(x)) = f(-2x) = 2(-2x) = -4x \)
Hence, \( fog(x) = \begin{cases} 0, & x \ge 0 \\ -4x, & x < 0 \end{cases} \)
2nd part
\( fog(5) = 0 \quad [\because 5 \ge 0] \)
\( fog(-3) = -4 \times (-3) = 12 \quad [\because -3 < 0] \)
\( gof(-2) = 0 \)

Question. Let \( \mathbb{N} \) denote the set of all natural numbers and \( R \) be the relation on \( \mathbb{N} \times \mathbb{N} \) defined by \( (a, b) R (c, d) \) if \( ad(b + c) = bc(a + d) \). Show that \( R \) is an equivalence relation.
Answer: Here \( R \) is a relation defined as
\( R = \{[(a, b), (c, d)] : ad(b + c) = bc(a + d)\} \)
Reflexivity: By commutative law under addition and multiplication
\( b + a = a + b \quad \forall a, b \in \mathbb{N} \)
\( ab = ba \quad \forall a, b \in \mathbb{N} \)
\( \therefore ab(b + a) = ba(a + b) \quad \forall a, b \in \mathbb{N} \)
\( \Rightarrow (a, b) R (a, b) \)
Hence, \( R \) is reflexive.
Symmetry: Let \( (a, b) R (c, d) \)
\( (a, b) R (c, d) \)
\( \Rightarrow ad(b + c) = bc(a + d) \)
\( \Rightarrow bc(a + d) = ad(b + c) \)
\( \Rightarrow cb(d + a) = da(c + b) \quad [\text{By commutative law under addition and multiplication}] \)
\( \Rightarrow (c, d) R (a, b) \)
Hence, \( R \) is symmetric.
Transitivity: Let \( (a, b) R (c, d) \text{ and } (c, d) R (e, f) \)
Now, \( (a, b) R (c, d) \text{ and } (c, d) R (e, f) \)
\( \Rightarrow ad(b + c) = bc(a + d) \text{ and } cf(d + e) = de(c + f) \)
\( \Rightarrow \frac{b + c}{bc} = \frac{a + d}{ad} \text{ and } \frac{d + e}{de} = \frac{c + f}{cf} \)
\( \Rightarrow \frac{1}{c} + \frac{1}{b} = \frac{1}{d} + \frac{1}{a} \text{ and } \frac{1}{e} + \frac{1}{d} = \frac{1}{f} + \frac{1}{c} \)
Adding both, we get
\( \Rightarrow \frac{1}{c} + \frac{1}{b} + \frac{1}{e} + \frac{1}{d} = \frac{1}{d} + \frac{1}{a} + \frac{1}{f} + \frac{1}{c} \)
\( \Rightarrow \frac{1}{b} + \frac{1}{e} = \frac{1}{a} + \frac{1}{f} \)
\( \Rightarrow \frac{e + b}{be} = \frac{f + a}{af} \)
\( \Rightarrow af(b + e) = be(a + f) \)
\( \Rightarrow (a, b) R (e, f) \quad [c, d \neq 0] \)
Hence, \( R \) is transitive.
In this way, \( R \) is reflexive, symmetric and transitive.
Therefore, \( R \) is an equivalence relation.

Question. Consider \( f : \mathbb{R}_+ \rightarrow [4, \infty) \) given by \( f(x) = x^2 + 4 \). Show that \( f \) is invertible with the inverse (\( f^{-1} \)) of \( f \) given by \( f^{-1}(y) = \sqrt{y - 4} \), where \( \mathbb{R}_+ \) is the set of all non-negative real numbers. 
Answer: One-one: Let \( x_1, x_2 \in \mathbb{R} \) (Domain)
\( f(x_1) = f(x_2) \)
\( \Rightarrow x_1^2 + 4 = x_2^2 + 4 \)
\( \Rightarrow x_1^2 = x_2^2 \)
\( \Rightarrow x_1 = x_2 \quad [\because x_1, x_2 \text{ are +ve real number}] \)
Hence, \( f \) is one-one function.
Onto: Let \( y \in [4, \infty) \) such that
\( y = f(x) \forall x \in \mathbb{R}_+ \quad [\text{set of non-negative reals}] \)
\( \Rightarrow y = x^2 + 4 \)
\( \Rightarrow x = \sqrt{y - 4} \quad [\because x \text{ is +ve real number}] \)
Obviously, \( \forall y \in [4, \infty) \), \( x \) is real number \( \in \mathbb{R} \) (domain)
i.e., all elements of codomain have pre image in domain.
\( \Rightarrow f \) is onto.
Hence, \( f \) is invertible being one-one onto.
Inverse function: If \( f^{-1} \) is inverse of \( f \), then
\( fof^{-1} = I \quad [\text{Identity function}] \)
\( \Rightarrow fof^{-1}(y) = y \forall y \in [4, \infty) \)
\( \Rightarrow f(f^{-1}(y)) = y \)
\( \Rightarrow (f^{-1}(y))^2 + 4 = y \quad [\because f(x) = x^2 + 4] \)
\( \Rightarrow f^{-1}(y) = \sqrt{y - 4} \)
Therefore, required inverse function is \( f^{-1} : [4, \infty) \rightarrow \mathbb{R} \) defined by
\( f^{-1}(y) = \sqrt{y - 4} \quad \forall y \in [4, \infty) \)

Question. Determine whether the relation \( R \) defined on the set \( \mathbb{R} \) of all real numbers as \( R = \{(a, b) : a, b \in \mathbb{R} \text{ and } a - b + \sqrt{3} \in S \} \), where \( S \) is the set of all irrational numbers}, is reflexive, symmetric and transitive. 
Answer: Here, relation \( R \) defined on the set \( \mathbb{R} \) is given as
\( R = \{(a, b) : a, b \in \mathbb{R} \text{ and } a - b + \sqrt{3} \in S\} \)
Reflexivity: Let \( a \in \mathbb{R} \) (set of real numbers)
Now, \( (a, a) \in R \) as \( a - a + \sqrt{3} = \sqrt{3} \in S \)
i.e., \( R \) is reflexive.
Symmetry: Taking \( a = \sqrt{3} \) and \( b = 1 \), we have
\( (a, b) \in R \)
\( \Rightarrow a - b + \sqrt{3} = \sqrt{3} - 1 + \sqrt{3} = 2\sqrt{3} - 1 \in S \)
\( \Rightarrow (a, b) \in R \)
But \( b - a + \sqrt{3} = 1 - \sqrt{3} + \sqrt{3} = 1 \notin S \)
\( \Rightarrow (b, a) \notin R \)
\( \therefore R \) is not symmetric.
Transitivity: Taking \( a = 1, b = \sqrt{2} \) and \( c = \sqrt{3} \)
\( (a, b) \in R \)
\( \Rightarrow a - b + \sqrt{3} = 1 - \sqrt{2} + \sqrt{3} \in S \)
\( \Rightarrow (a, b) \in R \)
and \( b - c + \sqrt{3} = \sqrt{2} - \sqrt{3} + \sqrt{3} = \sqrt{2} \in S \)
\( \Rightarrow (b, c) \in R \)
But \( a - c + \sqrt{3} = 1 - \sqrt{3} + \sqrt{3} = 1 \notin S \)
\( \Rightarrow (a, c) \notin R \)
\( \therefore R \) is not transitive.
Hence, \( R \) is reflexive but neither symmetric nor transitive.

Question. Show that the function \( f \) in \( A = \mathbb{R} - \left\{\frac{2}{3}\right\} \) defined as \( f(x) = \frac{4x + 3}{6x - 4} \) is one-one and onto. Hence, find \( f^{-1} \). 
Answer: One-one: Let \( x_1, x_2 \in A \)
Now, \( f(x_1) = f(x_2) \)
\( \Rightarrow \frac{4x_1 + 3}{6x_1 - 4} = \frac{4x_2 + 3}{6x_2 - 4} \)
\( \Rightarrow 24x_1 x_2 + 18x_2 - 16x_1 - 12 = 24x_1 x_2 + 18x_1 - 16x_2 - 12 \)
\( \Rightarrow -34x_1 = -34x_2 \)
\( \Rightarrow x_1 = x_2 \)
Hence, \( f \) is one-one function.
Onto:
Let \( y = \frac{4x + 3}{6x - 4} \)
\( \Rightarrow 6xy - 4y = 4x + 3 \)
\( \Rightarrow 6xy - 4x = 4y + 3 \)
\( \Rightarrow x(6y - 4) = 4y + 3 \)
\( \Rightarrow x = \frac{4y + 3}{6y - 4} \)
\( \Rightarrow \forall y \in \text{codomain } \exists x \in \text{domain} \left[ \because x \neq \frac{2}{3} \right] \)
\( \Rightarrow f \text{ is onto function.} \)
Thus, \( f \) is one-one onto function.
Also, \( f^{-1}(x) = \frac{4x + 3}{6x - 4} \)

Question. Let \( f : W \rightarrow W \), be defined as \( f(x) = x - 1 \), if \( x \) is odd and \( f(x) = x + 1 \), if \( x \) is even. Show that \( f \) is invertible. Find the inverse of \( f \), where \( W \) is the set of all whole numbers. 
Answer: One-one:
Case I : When \( x_1, x_2 \) are even number
Now, \( f(x_1) = f(x_2) \)
\( \Rightarrow x_1 + 1 = x_2 + 1 \)
\( \Rightarrow x_1 = x_2 \)
i.e., \( f \) is one-one.
Case II : When \( x_1, x_2 \) are odd number
Now, \( f(x_1) = f(x_2) \)
\( \Rightarrow x_1 - 1 = x_2 - 1 \)
\( \Rightarrow x_1 = x_2 \)
i.e., \( f \) is one-one.
Case III : When \( x_1 \) is odd and \( x_2 \) is even number
Then, \( x_1 \neq x_2 \). Also, in this case \( f(x_1) \) is even and \( f(x_2) \) is odd and so
\( f(x_1) \neq f(x_2) \)
i.e. \( x_1 \neq x_2 \)
\( \Rightarrow f(x_1) \neq f(x_2) \)
i.e., \( f \) is one-one.
Case IV : When \( x_1 \) is even and \( x_2 \) is odd number
Similar as Case III, we can prove \( f \) is one-one.
Onto:
Given, \( f(x) = \begin{cases} x - 1, & \text{if } x \text{ is odd} \\ x + 1, & \text{if } x \text{ is even} \end{cases} \)
\( \Rightarrow \text{For every even number `y' of codomain } \exists \text{ odd number } y + 1 \text{ in domain and for every odd number } y \text{ of codomain there exists even number } y - 1 \text{ in domain i.e., } f \text{ is onto function.} \)
Hence, \( f \) is one-one onto i.e., invertible function.
Inverse:
Let \( f(x) = y \)
Now, \( y = x + 1 \)
\( \Rightarrow x = y - 1 \)
And, \( y = x - 1 \)
\( \Rightarrow x = y + 1 \)
Therefore, required inverse function is given by
\( f^{-1}(x) = \begin{cases} x + 1, & \text{if } x \text{ is odd} \\ x - 1, & \text{if } x \text{ is even} \end{cases} \)

Question. If the function \( f : \mathbb{R} \rightarrow \mathbb{R} \) be defined by \( f(x) = 2x - 3 \) and \( g : \mathbb{R} \rightarrow \mathbb{R} \) by \( g(x) = x^3 + 5 \), then find the value of \( (fog)^{-1} (x) \). 
Answer: Here \( f : \mathbb{R} \rightarrow \mathbb{R} \) and \( g : \mathbb{R} \rightarrow \mathbb{R} \) be two functions such that
\( f(x) = 2x - 3 \) and \( g(x) = x^3 + 5 \)
\( \because f \text{ and } g \text{ both are bijective (one-one onto) function.} \)
\( \Rightarrow fog \text{ is also bijective function.} \)
\( \Rightarrow fog \text{ is invertible function.} \)
Now, \( fog(x) = f\{g(x)\} \)
\( \Rightarrow fog(x) = f(x^3 + 5) \)
\( \Rightarrow fog(x) = 2(x^3 + 5) - 3 \)
\( \Rightarrow fog(x) = 2x^3 + 10 - 3 \)
\( \Rightarrow fog(x) = 2x^3 + 7 \quad \dots(i) \)
For inverse of \( fog(x) \)
Let \( fog(x) = y \)
\( \Rightarrow x = fog^{-1}(y) \)
\( (i) \)
\( \Rightarrow y = 2x^3 + 7 \)
\( \Rightarrow 2x^3 = y - 7 \)
\( \Rightarrow x^3 = \frac{y - 7}{2} \)
\( \Rightarrow x = \left( \frac{y - 7}{2} \right)^{\frac{1}{3}} \)
\( \Rightarrow fog^{-1}(y) = \left( \frac{y - 7}{2} \right)^{\frac{1}{3}} \)
\( \Rightarrow fog^{-1}(x) = \left( \frac{x - 7}{2} \right)^{\frac{1}{3}} \)

Question. Let \( f : \mathbb{N} \rightarrow \mathbb{R} \) be a function defined as \( f(x) = 4x^2 + 12x + 15 \). Show that \( f : \mathbb{N} \rightarrow S \) is invertible, where \( S \) is the range of \( f \). Hence, find inverse of \( f \). 
Answer: Let \( y \in S \), then \( y = 4x^2 + 12x + 15 \), for some \( x \in \mathbb{N} \)
\( \Rightarrow y = (2x + 3)^2 + 6 \)
\( \Rightarrow x = \frac{(\sqrt{y - 6}) - 3}{2}, \text{ as } y > 6 \)
Let \( g : S \rightarrow \mathbb{N} \) is defined by \( g(y) = \frac{(\sqrt{y - 6}) - 3}{2} \)
\( \therefore gof(x) = g(4x^2 + 12x + 15) = g((2x + 3)^2 + 6) = \frac{\sqrt{(2x + 3)^2} - 3}{2} = x \)
and \( fog(y) = f\left( \frac{(\sqrt{y - 6}) - 3}{2} \right) = \left[ 2 \left( \frac{(\sqrt{y - 6}) - 3}{2} \right) + 3 \right]^2 + 6 = y \)
Hence, \( fog(y) = I_S \) and \( gof(x) = I_N \)
\( f \text{ is invertible, } f^{-1} = g. \)

Question. Let \( \mathbb{Z} \) be the set of all integers and \( R \) be relation on \( \mathbb{Z} \) defined as \( R = \{(a, b) : a, b \in \mathbb{Z} \text{ and } (a - b) \text{ is divisible by 5}\} \). Prove that \( R \) is an equivalence relation. 
OR
Show that the relation \( S \) in the set \( A = \{x \in \mathbb{Z} : 0 \le x \le 12\} \) given by \( S = \{(a, b) : a, b \in \mathbb{Z}, |a - b| \text{ is divisible by 3}\} \) is an equivalence relation.
Answer: Given \( R = \{(a, b) : a, b \in \mathbb{Z} \text{ and } (a - b) \text{ is divisible by 5}\} \)
Reflexivity: \( \forall a \in \mathbb{Z} \)
\( a - a = 0 \text{ is divisible by 5} \)
\( \Rightarrow (a, a) \in R \forall a \in \mathbb{Z} \)
Hence, \( R \) is reflexive.
Symmetry: Let \( (a, b) \in R \)
\( \Rightarrow a - b \text{ is divisible by 5} \)
\( \Rightarrow -(b - a) \text{ is divisible by 5} \)
\( \Rightarrow (b - a) \text{ is divisible by 5} \)
\( \Rightarrow (b, a) \in R \)
Hence, \( R \) is symmetric.
Transitivity: Let \( (a, b), (b, c) \in R \)
\( \Rightarrow (a - b) \text{ and } (b - c) \text{ are divisible by 5} \)
\( \Rightarrow (a - b + b - c) \text{ is divisible by 5} \)
\( \Rightarrow a - c \text{ is divisible by 5} \)
\( \Rightarrow (a, c) \in R \)
Hence, \( R \) is transitive.
Thus, \( R \) is an equivalence relation.
OR
Solution is similar only values are changed.

Question. Let \( f : \mathbb{R} - \left\{-\frac{4}{3}\right\} \rightarrow \mathbb{R} \) be a function defined as \( f(x) = \frac{4x}{3x + 4} \). Show that, in \( f : \mathbb{R} - \left\{-\frac{4}{3}\right\} \rightarrow \text{Range of } f \), \( f \) is one-one and onto. Hence find \( f^{-1} : \text{Range } f \rightarrow \mathbb{R} - \left\{-\frac{4}{3}\right\} \). 
Answer: Let \( x_1, x_2 \in \mathbb{R} - \left\{-\frac{4}{3}\right\} \)
Now \( f(x_1) = f(x_2) \)
\( \Rightarrow \frac{4x_1}{3x_1 + 4} = \frac{4x_2}{3x_2 + 4} \)
\( \Rightarrow 12 x_1 x_2 + 16 x_1 = 12 x_1 x_2 + 16 x_2 \)
\( \Rightarrow 16 x_1 = 16 x_2 \)
\( \Rightarrow x_1 = x_2 \)
Hence \( f \) is one-one function.
Since, co-domain \( f \) is range of \( f \)
So, \( f : \mathbb{R} - \left\{-\frac{4}{3}\right\} \rightarrow \mathbb{R} \) is one-one onto function.
For inverse
Let \( f(x) = y \)
\( \Rightarrow \frac{4x}{3x + 4} = y \)
\( \Rightarrow 3xy + 4y = 4x \)
\( \Rightarrow 4x - 3xy = 4y \)
\( \Rightarrow x(4 - 3y) = 4y \)
\( \Rightarrow x = \frac{4y}{4 - 3y} \)
Therefore, \( f^{-1} : \text{Range of } f \rightarrow \mathbb{R} - \{-4/3\} \) is \( f^{-1}(y) = \frac{4y}{4 - 3y} \).

Selected NCERT Questions

Question. Show that the relation \( R \) on the set \( R \) 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 \( a, b \in R \)
Reflexivity: Obviously, \( \frac{1}{2} \) is a real number and \( \frac{1}{2} \le \left(\frac{1}{2}\right)^2 \) is not true.
Therefore, \( R \) is not reflexive.
Symmetry: Consider the real numbers 1 and 2.
Obviously, \( 1 \le 2^2 \)
\( \implies \) \( (1, 2) \in R \)
But, \( 2 \le (1)^2 \) is not true and so, \( (2, 1) \notin R \).
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 \( (2, 1) \notin R \quad [\because 2 \ge 1 \text{ or } 2 \ge 1^3] \)
\( \therefore R \) 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
\( \implies \) \( |b - a| \) is also even
\( \implies \) \( (b, a) \in R \)
Thus, it is symmetric.
Transitivity: Now, if \( (a, b) \in R \) i.e., \( |a - b| \) is even
\( \implies \) \( a - b = \pm 2m \quad m \in N \)
and \( (b, c) \in R \) i.e., \( |b - c| \) is even
\( \implies \) \( b - c = \pm 2n, \quad n \in N \)
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, \( (a, c) \in R \)
\( \implies \) \( R \) 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 \( A = \{x \in Z : 0 \le x \le 12\} \), 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: \( A = \{x \in Z : 0 \le x \le 12\} \)
(i) \( R = \{(a, b) : |a - b| \text{ is a multiple of } 4\} \)
Reflexive: Let \( x \in A \)
\( \implies \) \( |x - x| = 0 \), which is a multiple of 4.
\( \implies \) \( (x, x) \in R \; \forall \; x \in A \)
\( \therefore R \) is reflexive.
Symmetric: Let \( x, y \in A \) and \( (x, y) \in R \)
\( \implies \) \( |x - y| \) is a multiple of 4
or \( x - y = \pm 4p \quad \{p \text{ is any integer}\} \)
\( \implies \) \( y - x = \mp 4p \)
\( \implies \) \( |y - x| \) is a multiple of 4.
\( \implies \) \( (y, x) \in R \)
\( \implies \) \( R \) is symmetric.
Transitive: Let \( x, y, z \in A \), \( (x, y) \in R \) and \( (y, z) \in R \)
\( \implies \) \( |x - y| \) is multiple of 4 and \( |y - z| \) is multiple of 4
\( \implies \) \( x - y \) is multiple of 4 and \( y - z \) is multiple of 4
\( \implies \) \( (x - y) + (y - z) \) is multiple of 4
\( \implies \) \( (x - z) \) is multiple of 4.
\( \implies \) \( |x - z| \) is multiple of 4.
\( \implies \) \( (x, z) \in R \)
\( \implies \) \( R \) is transitive.
So, \( R \) is an equivalence relation.
Let \( B \) be the set of elements related to 1.
\( \therefore B = \{a \in A : |a - 1| \text{ is multiple of } 4\} \)
\( \implies \) \( B = \{1, 5, 9\} \quad \{\text{as } |1 - 1| = 0, \; |1 - 5| = 4, \; |1 - 9| = 8\} \)
(ii) \( R = \{(a, b) : a = b\} \)
Reflexive: Let \( x \in A \)
as \( x = x \)
\( \implies \) \( (x, x) \in R \)
\( \implies \) \( R \) is reflexive.
Symmetric: Let \( x, y \in A \) and \( (x, y) \in R \)
\( \implies \) \( x = y \)
\( \implies \) \( y = x \)
\( \implies \) \( (y, x) \in R \)
\( \therefore R \) is symmetric.
Transitive: Let \( x, y, z \in A \)
and let \( (x, y) \in R \) and \( (y, z) \in R \)
\( \implies \) \( x = y \) and \( y = z \)
\( \implies \) \( x = z \)
\( \implies \) \( (x, z) \in R \)
\( \implies \) \( R \) 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 \( f: \mathbb{R} \to \mathbb{R} \) given by \( f(x) = [x] \), is neither one-one nor onto, where \( [x] \) denotes the greatest integer less than or equal to \( x \).
Answer: \( f: \mathbb{R} \to \mathbb{R} \) 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 \( x_1 \ne x_2 \)
\( \implies \) \( f(x) = [x] \) is not one-one i.e., not injective.
Surjectivity: Let \( y = 2.5 \in \mathbb{R} \) be any element.
\( \therefore f(x) = 2.5 \)
\( \implies \) \( [x] = 2.5 \)
Which is not possible as \( [x] \) is always an integer.
\( \implies \) \( f(x) = [x] \) is not onto i.e., not surjective.

Question. Show that the modulus function \( f : \mathbb{R} \to \mathbb{R} \) 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: \( f(x) = |x| \)
Let \( x_1, x_2 \in \mathbb{R} \) and \( f(x_1) = f(x_2) \)
\( \implies \) \( |x_1| = |x_2| \)
\( \implies \) \( x_1 = \pm x_2 \)
\( \implies \) \( f \) is not one-one when \( x_1 = -x_2 \).
\( \because \) Range of \( |x| \) is all non-negative real numbers. i.e., \( [0, \infty) \) and \( [0, \infty) \ne \mathbb{R} = \text{Co-domain of } f \).
\( \therefore f \) is not onto.

Question. Let \( f : N \to N \) 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 \( n \in N \), state whether the function \( f \) is bijective. Justify your answer. 
Answer: Given, \( f(x) = \begin{cases} \frac{n+1}{2}, & \text{if } n \text{ is odd} \\ \frac{n}{2}, & \text{if } n \text{ is even} \end{cases} \)
Let \( x_1 = 1 \) and \( x_2 = 2 \) be two elements of \( N \).
\( \therefore f(x_1) = f(1) = \frac{1+1}{2} = 1 \)
and \( f(x_2) = f(2) = \frac{2}{2} = 1 \)
\( f(x_1) = f(x_2) \) for \( x_1 \ne x_2 \)
\( f: N \to N \) is not one-one.
\( \implies \) \( f \) is not one-one. \( f \) is not a bijective function.

Question. Consider \( f : R_+ \to [-5, \infty) \) 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 \( f : R_+ \to [-5, \infty) \) such that \( f(x) = 9x^2 + 6x - 5 \)
One-one: Let \( x_1, x_2 \in R_+ \) then
\( f(x_1) = f(x_2) \)
\( \implies \) \( 9x_1^2 + 6x_1 - 5 = 9x_2^2 + 6x_2 - 5 \)
\( \implies \) \( 9(x_1^2 - x_2^2) + 6(x_1 - x_2) = 0 \)
\( \implies \) \( 3[3(x_1 + x_2) + 2] (x_1 - x_2) = 0 \)
\( \implies \) \( x_1 - x_2 = 0 \quad [\because x_1, x_2 \in R_+ \)
\( \implies \) \( x_1 + x_2 \ne 0 \)
\( \implies \) \( 3(x_1 + x_2) + 2 \ne 0] \)
\( \implies \) \( x_1 = x_2 \)
So, given function is one-one.
Onto: Let \( y \in [-5, \infty) \) then \( y = f(x) \)
\( \implies \) \( y = 9x^2 + 6x - 5 \)
i.e., \( 9x^2 + 6x - 5 = y \)
\( \implies \) \( (3x)^2 + 2.3x.1 + (1)^2 - 1 - 5 = y \)
\( \implies \) \( (3x + 1)^2 = y + 6 \)
\( \implies \) \( 3x + 1 = \sqrt{y + 6} \)
\( \implies \) \( x = \frac{-1 + \sqrt{y + 6}}{3} \quad ...(i) \)
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,
\( x = \frac{-1 + \sqrt{y + 6}}{3} \quad i.e., \quad x = \frac{\sqrt{y + 6} - 1}{3} \)
\( \implies \) \( f^{-1}(y) = \frac{\sqrt{y + 6} - 1}{3} \)
\( \therefore f^{-1}(x) = \frac{\sqrt{x + 6} - 1}{3} \)

Question. Give examples of two functions \( f : N \to N \) and \( g : N \to N \) 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.
\( \therefore x \ge 1 \)
\( \implies \) \( x + 1 \ge 2 \)
\( \implies \) \( f(x) \ge 2 \; \forall \; x \in N \)
\( \therefore R_f \ne N \)
Hence, \( f \) is not onto.
Also, \( gof : N \to N \) is such that
\( (gof)(x) = g(f(x)) = g(x + 1) = (x + 1) - 1 = x \)
\( \implies \) \( (gof)(x) = x \; \forall \; x \in N \)
\( \therefore gof \) is an identity function.
Hence, \( gof \) is onto.

Question. Let \( A = \{-1, 0, 1, 2\} \), \( B = \{-4, -2, 0, 2\} \) and \( f, g : A \to B \) 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 \( f : A \to B \) and \( g : A \to B \) defined as
\( f(x) = x^2 - x \text{ and } g(x) = 2 \left|x - \frac{1}{2}\right| \quad \forall \; x \in A \)
\( f(-1) = (-1)^2 - (-1) = 2; \quad g(-1) = 2\left|-1 - \frac{1}{2}\right| - 1 = 2 \left|\frac{-3}{2}\right| - 1 = 2 \)
\( f(0) = 0^2 - 0 = 0; \quad g(0) = 2\left|0 - \frac{1}{2}\right| - 1 = 1 - 1 = 0 \)
\( f(1) = 1^2 - 1 = 1 - 1 = 0; \quad g(1) = 2\left|1 - \frac{1}{2}\right| - 1 = 2 \left|\frac{1}{2}\right| - 1 = 1 - 1 = 0 \)
\( f(2) = 2^2 - 2 = 2; \quad g(2) = 2\left|2 - \frac{1}{2}\right| - 1 = 2 \left|\frac{3}{2}\right| - 1 = 2 \)
Clearly, \( f(-1) = g(-1); \; f(0) = g(0); \; f(1) = g(1) \text{ and } f(2) = g(2) \)
\( \therefore f(x) = g(x) \; \forall \; x \in A \)

Question. Let \( f : \mathbb{R} \to \mathbb{R} \) be the Signum function defined as \( f(x) = \begin{cases} 1, & x > 0 \\ 0, & x = 0 \\ -1, & x < 0 \end{cases} \) and \( g : \mathbb{R} \to \mathbb{R} \) be the Greatest Integer Function given by \( g(x) = [x] \). Then, do \( fog \) and \( gof \) coincide in \( (0, 1] \)?
Answer: Given \( f : \mathbb{R} \to \mathbb{R} \) and \( g : \mathbb{R} \to \mathbb{R} \) defined as
\( f(x) = \begin{cases} 1, & x > 0 \\ 0, & x = 0 \\ -1, & x < 0 \end{cases} \text{ and } g(x) = [x] \)
\( fog(x) = f(g(x)) = f([x]) \)
\( = \begin{cases} f(0), & \text{if } 0 < x < 1 \\ f(1), & \text{if } x = 1 \end{cases} \)
\( \therefore fog(x) = \begin{cases} 0 & \text{if } 0 < x < 1 \\ 1 & \text{if } x = 1 \end{cases} \)
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 \( \forall \; x \in (0, 1] \).in the graph, intersects the graph at only one point then the relation is a function, otherwise it is not a function.

(a) It is graph of function i.e., \( f(x) = x^2 \) is a function.
(b) It is not the graph of function i.e., \( f(x) = \pm\sqrt{x} \) is not a function