CBSE Class 12 Mathematics HOTs Relations and Functions Set 03

Composition of Functions and Invertible Function

SA (4 marks)

Question. If the function \( f : \mathbb{R} \to \mathbb{R} \) be given by \( f(x) = x^2 + 2 \) and \( g : \mathbb{R} \to \mathbb{R} \) be given by \( g(x) = \frac{x}{x - 1} \), \( x \neq 1 \), find \( f \circ g \) and \( g \circ f \) and hence find \( f \circ g(2) \) and \( g \circ f(-3) \).
Answer: We have \( f(x) = x^2 + 2 \) and \( g(x) = \frac{x}{x - 1} \).
\[ (f \circ g)(x) = f(g(x)) = \left(\frac{x}{x-1}\right)^2 + 2, \quad x \neq 1 \]
\[ (g \circ f)(x) = g(f(x)) = \frac{x^2 + 2}{x^2 + 2 - 1} = \frac{x^2 + 2}{x^2 + 1} \]
Now, we calculate:
\[ (f \circ g)(2) = \left(\frac{2}{2-1}\right)^2 + 2 = 4 + 2 = 6 \]
\[ (g \circ f)(-3) = \frac{(-3)^2 + 2}{(-3)^2 + 1} = \frac{11}{10} \].

Question. Let \( f : W \to W \), 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: Since \( f \) is bijective, it is invertible.
Let \( f(x) = y \).
If \( x \) is even, \( y = x + 1 \) (where \( y \) is odd) \( \implies x = y - 1 \).
If \( x \) is odd, \( y = x - 1 \) (where \( y \) is even) \( \implies x = y + 1 \).
Thus, \( f^{-1}(y) = \begin{cases} y - 1, & \text{if } y \text{ is odd} \\ y + 1, & \text{if } y \text{ is even} \end{cases} \).
Hence, \( f^{-1}(x) = \begin{cases} x - 1, & \text{if } x \text{ is odd} \\ x + 1, & \text{if } x \text{ is even} \end{cases} \).

Question. Let \( A = \mathbb{R} - \{3\} \), \( B = \mathbb{R} - \{1\} \). Let \( f : A \to B \) be defined by \( f(x) = \frac{x - 2}{x - 3} \), for all \( x \in A \). Then show that \( f \) is bijective. Hence find \( f^{-1}(x) \).
Answer: We have \( f(x) = \frac{x - 2}{x - 3} \).
(i) One-one: Let \( f(x_1) = f(x_2) \implies \frac{x_1 - 2}{x_1 - 3} = \frac{x_2 - 2}{x_2 - 3} \implies (x_1 - 2)(x_2 - 3) = (x_1 - 3)(x_2 - 2) \implies x_1x_2 - 3x_1 - 2x_2 + 6 = x_1x_2 - 2x_1 - 3x_2 + 6 \implies -3x_1 - 2x_2 = -2x_1 - 3x_2 \implies x_1 = x_2 \). Thus, \( f \) is one-one.
(ii) Onto: Let \( y \in B \). We set \( y = \frac{x - 2}{x - 3} \implies y(x - 3) = x - 2 \implies xy - 3y = x - 2 \implies x(y - 1) = 3y - 2 \implies x = \frac{3y - 2}{y - 1} \). Since \( y \neq 1 \), \( x \) is a well-defined real number. Moreover, \( x \neq 3 \) because \( \frac{3y - 2}{y - 1} = 3 \implies 3y - 2 = 3y - 3 \implies -2 = -3 \), which is impossible. Thus \( x \in A \), and \( f \) is onto.
Hence, \( f \) is bijective.
The inverse function is given by \( f^{-1}(x) = \frac{3x - 2}{x - 1} \).

Question. Let \( f, g : \mathbb{R} \to \mathbb{R} \) be two functions defined as \( f(x) = |x| + x \) and \( g(x) = |x| - x \), for all \( x \in \mathbb{R} \). Then find \( f \circ g \) and \( g \circ f \).
Answer: We can rewrite \( f(x) \) and \( g(x) \) as:
\[ f(x) = \begin{cases} 2x, & x \ge 0 \\ 0, & x < 0 \end{cases} \quad \text{and} \quad g(x) = \begin{cases} 0, & x \ge 0 \\ -2x, & x < 0 \end{cases} \]
Now, \( (f \circ g)(x) = f(g(x)) \). Since \( g(x) \ge 0 \) for all \( x \in \mathbb{R} \):
If \( x \ge 0 \), \( g(x) = 0 \implies f(g(x)) = f(0) = 0 \).
If \( x < 0 \), \( g(x) = -2x \ge 0 \implies f(g(x)) = 2(-2x) = -4x \).
Therefore, \( (f \circ g)(x) = \begin{cases} 0, & x \ge 0 \\ -4x, & x < 0 \end{cases} \).

Now, \( (g \circ f)(x) = g(f(x)) \). Since \( f(x) \ge 0 \) for all \( x \in \mathbb{R} \):
For any \( x \in \mathbb{R} \), \( f(x) \ge 0 \implies g(f(x)) = 0 \) (since \( g(t) = 0 \) for any \( t \ge 0 \)).
Therefore, \( (g \circ f)(x) = 0 \) for all \( x \in \mathbb{R} \).

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: (i) One-one: Let \( f(x_1) = f(x_2) \implies \frac{4x_1 + 3}{6x_1 - 4} = \frac{4x_2 + 3}{6x_2 - 4} \implies (4x_1 + 3)(6x_2 - 4) = (6x_1 - 4)(4x_2 + 3) \implies -16x_1 + 18x_2 = 18x_1 - 16x_2 \implies 34x_1 = 34x_2 \implies x_1 = x_2 \). Thus, \( f \) is one-one.
(ii) Onto: Let \( y \in A \). We set \( y = \frac{4x + 3}{6x - 4} \implies y(6x - 4) = 4x + 3 \implies 6xy - 4y = 4x + 3 \implies x(6y - 4) = 4y + 3 \implies x = \frac{4y + 3}{6y - 4} \). Since \( y \neq \frac{2}{3} \), \( x \) is a well-defined real number and \( x \neq \frac{2}{3} \). Thus, \( f \) is onto.
Hence, \( f^{-1}(x) = \frac{4x + 3}{6x - 4} \).

Question. Consider \( f : \mathbb{R}_+ \to [4, \infty) \) 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: (i) **One-one:** Let \( x_1, x_2 \in \mathbb{R}_+ \) such that \( f(x_1) = f(x_2) \implies x_1^2 + 4 = x_2^2 + 4 \implies x_1^2 = x_2^2 \implies x_1 = x_2 \) (since \( x_1, x_2 \ge 0 \)). Thus, \( f \) is one-one.
(ii) **Onto:** Let \( y \in [4, \infty) \). We set \( y = x^2 + 4 \implies x^2 = y - 4 \implies x = \sqrt{y - 4} \) (since \( x \ge 0 \)). Since \( y \ge 4 \), \( x \) is a real non-negative number, so \( x \in \mathbb{R}_+ \). Also, \( f(x) = (\sqrt{y-4})^2 + 4 = y \). Thus, \( f \) is onto.
Hence, \( f \) is invertible and the inverse is \( f^{-1}(y) = \sqrt{y - 4} \).

Question. Let \( A = \mathbb{R} - \{2\} \) and \( B = \mathbb{R} - \{1\} \). If \( f : A \to B \) is a function defined by \( f(x) = \frac{x - 1}{x - 2} \), show that \( f \) is one-one and onto. Hence find \( f^{-1} \).
Answer: (i) One-one: Let \( f(x_1) = f(x_2) \implies \frac{x_1 - 1}{x_1 - 2} = \frac{x_2 - 1}{x_2 - 2} \implies (x_1 - 1)(x_2 - 2) = (x_1 - 2)(x_2 - 1) \implies -2x_1 - x_2 = -x_1 - 2x_2 \implies x_1 = x_2 \). Thus, \( f \) is one-one.
(ii) Onto: Let \( y \in B \). We set \( y = \frac{x - 1}{x - 2} \implies xy - 2y = x - 1 \implies x(y - 1) = 2y - 1 \implies x = \frac{2y - 1}{y - 1} \). Since \( y \neq 1 \), \( x \) is a well-defined real number and \( x \neq 2 \). Thus \( f \) is onto.
Hence, \( f^{-1}(x) = \frac{2x - 1}{x - 1} \).

Question. Let \( A = \mathbb{R} - \{3\} \) and \( B = \mathbb{R} - \{1\} \). Consider the function \( f : A \to B \) defined by \( f(x) = \frac{x - 2}{x - 3} \). Is \( f \) one-one and onto? Justify your answer.
Answer: Yes, \( f \) is both one-one and onto. Refer to Question 34 for the detailed justification.

Question. Let \( f : \mathbb{R} \to \mathbb{R} \) be defined as \( f(x) = 10x + 7 \). Find the function \( g : \mathbb{R} \to \mathbb{R} \) such that \( g \circ f = f \circ g = I_{\mathbb{R}} \).
Answer: We set \( y = 10x + 7 \implies x = \frac{y - 7}{10} \).
So, we define \( g : \mathbb{R} \to \mathbb{R} \) by \( g(y) = \frac{y - 7}{10} \).
We verify:
\( (g \circ f)(x) = g(f(x)) = g(10x + 7) = \frac{10x + 7 - 7}{10} = x \)
\( (f \circ g)(y) = f(g(y)) = f\left(\frac{y - 7}{10}\right) = 10\left(\frac{y - 7}{10}\right) + 7 = y \).
Thus, \( g \circ f = f \circ g = I_{\mathbb{R}} \), and the function is \( g(x) = \frac{x - 7}{10} \).

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Question. If the function \( f : \mathbb{R} \to \mathbb{R} \) is given by \( f(x) = \frac{x + 3}{2} \) and \( g : \mathbb{R} \to \mathbb{R} \) is given by \( g(x) = 2x - 3 \). Find (i) \( fog \) and (ii) \( gof \). Is \( f^{-1} = g \)?
Answer:
(i) \( (fog)(x) = f(g(x)) = \frac{2x - 3 + 3}{2} = \frac{2x}{2} = x \)
(ii) \( (gof)(x) = g(f(x)) = 2\left(\frac{x + 3}{2}\right) - 3 = x + 3 - 3 = x \)
So, \( fog = gof = I_{\mathbb{R}} \). Hence, \( f^{-1} = g \).

Question. If \( f : \mathbb{R} \to \mathbb{R} \) and \( g : \mathbb{R} \to \mathbb{R} \) are defined respectively as \( f(x) = x^2 + 3x + 1 \) and \( g(x) = 2x - 3 \). Find (a) \( fog \) (b) \( gof \).
Answer:
(a) \( (fog)(x) = f(g(x)) = (2x - 3)^2 + 3(2x - 3) + 1 = 4x^2 - 12x + 9 + 6x - 9 + 1 = 4x^2 - 6x + 1 \)
(b) \( (gof)(x) = g(f(x)) = 2(x^2 + 3x + 1) - 3 = 2x^2 + 6x + 2 - 3 = 2x^2 + 6x - 1 \).

Question. If \( f \) be a greatest integer function and \( g \) be an absolute value function, find the value of \( (fog)\left(-\frac{3}{2}\right) + (gof)\left(\frac{4}{3}\right) \).
Answer:
Here \( f(x) = [x] \) and \( g(x) = |x| \).
\( (fog)(x) = f(g(x)) = f(|x|) = [|x|] \)
\( \therefore (fog)\left(-\frac{3}{2}\right) = \left[\left|-\frac{3}{2}\right|\right] = \left[\frac{3}{2}\right] = 1 \)
\( (gof)(x) = g(f(x)) = g([x]) = |[x]| \)
\( \therefore (gof)\left(\frac{4}{3}\right) = \left|\left[\frac{4}{3}\right]\right| = |1| = 1 \)
\( \therefore (fog)\left(-\frac{3}{2}\right) + (gof)\left(\frac{4}{3}\right) = 1 + 1 = 2 \).

Question. Let \( f : \mathbb{N} \to \mathbb{N} \) be a function defined as \( f(x) = 9x^2 + 6x - 5 \). Show that \( f : \mathbb{N} \to S \), where \( S \) is the range of \( f \), is invertible. Find the inverse of \( f \) and hence find \( f^{-1}(43) \) and \( f^{-1}(163) \).
Answer:
Let \( f : \mathbb{N} \to S \), \( f(x) = 9x^2 + 6x - 5 \).
Consider \( f(x_1) = f(x_2) \)
\( \Rightarrow 9x_1^2 + 6x_1 - 5 = 9x_2^2 + 6x_2 - 5 \)
\( \Rightarrow 9(x_1^2 - x_2^2) + 6(x_1 - x_2) = 0 \)
\( \Rightarrow (x_1 - x_2)[9x_1 + 9x_2 + 6] = 0 \)
\( \Rightarrow x_1 = x_2 \) [since \( x_1, x_2 \in \mathbb{N} \) and \( 9x_1 + 9x_2 + 6 \neq 0 \)]
\( \Rightarrow f \) is one-one.
Since, \( S \) is the range of \( f \), therefore \( f \) is onto.
Since, \( f \) is one-one and onto, \( f \) is invertible.
Let \( y \in S \) be an arbitrary number.
Consider, \( y = f(x) \Rightarrow x = f^{-1}(y) \)
\( \Rightarrow y = 9x^2 + 6x - 5 \Rightarrow y = (3x + 1)^2 - 6 \)
\( \Rightarrow 3x + 1 = \sqrt{y + 6} \Rightarrow x = \frac{\sqrt{y + 6} - 1}{3} \)
Also, \( f^{-1}(y) = \frac{\sqrt{y + 6} - 1}{3} \) or \( f^{-1}(x) = \frac{\sqrt{x + 6} - 1}{3} \)
Now, \( f^{-1}(43) = \frac{\sqrt{43 + 6} - 1}{3} = \frac{7 - 1}{3} = 2 \)
and \( f^{-1}(163) = \frac{\sqrt{163 + 6} - 1}{3} = \frac{13 - 1}{3} = 4 \).

Question. If \( f, g : \mathbb{R} \to \mathbb{R} \) 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:
We can rewrite the given functions as:
\( f(x) = \begin{cases} 2x, & x \ge 0 \\ 0, & x < 0 \end{cases} \)
and \( g(x) = \begin{cases} 0, & x \ge 0 \\ -2x, & x < 0 \end{cases} \)
Now, let us find \( fog(x) \):
If \( x \ge 0 \), then \( g(x) = 0 \), so \( fog(x) = f(g(x)) = f(0) = 0 \).
If \( x < 0 \), then \( g(x) = -2x > 0 \), so \( fog(x) = f(g(x)) = f(-2x) = 2(-2x) = -4x \).
Thus, \( fog(x) = \begin{cases} 0, & x \ge 0 \\ -4x, & x < 0 \end{cases} \)
Now, let us find \( gof(x) \):
If \( x \ge 0 \), then \( f(x) = 2x \ge 0 \), so \( gof(x) = g(f(x)) = g(2x) = 0 \).
If \( x < 0 \), then \( f(x) = 0 \), so \( gof(x) = g(f(x)) = g(0) = 0 \).
Thus, \( gof(x) = 0 \) for all \( x \in \mathbb{R} \).
Now, we evaluate the required values:
\( fog(-3) = -4(-3) = 12 \)
\( fog(5) = 0 \)
\( gof(-2) = 0 \).

Question. Consider \( f : \mathbb{R}_+ \to [-9, \infty[ \) given by \( f(x) = 5x^2 + 6x - 9 \). Prove that \( f \) is invertible with \( f^{-1}(y) = \frac{\sqrt{54 + 5y} - 3}{5} \).
Answer:
First we shall show that \( f \) is one-one.
Let \( f(x) = f(y) \), for \( x, y \in \mathbb{R}_+ \)
\( \Rightarrow 5x^2 + 6x - 9 = 5y^2 + 6y - 9 \)
\( \Rightarrow 5(x^2 - y^2) + 6(x - y) = 0 \)
\( \Rightarrow (x - y)[5(x + y) + 6] = 0 \)
\( \Rightarrow x = y \) [since for \( x, y \in \mathbb{R}_+ \), \( 5(x + y) + 6 \neq 0 \)]
\( \Rightarrow f \) is one-one.
Let \( y \in [-9, \infty[ \) be such that \( f(x) = y \)
\( \Leftrightarrow 5x^2 + 6x - 9 = y \Rightarrow 5x^2 + 6x - (9 + y) = 0 \)
\( \Rightarrow x = \frac{-6 \pm \sqrt{6^2 + 4 \cdot 5 \cdot (9 + y)}}{2 \cdot 5} = \frac{-6 \pm \sqrt{36 + 180 + 20y}}{10} = \frac{-6 \pm \sqrt{216 + 20y}}{10} = \frac{-3 \pm \sqrt{54 + 5y}}{5} \)
Taking only the positive sign (since for the negative sign, \( x \notin \mathbb{R}_+ \)), we get:
\( x = \frac{-3 + \sqrt{54 + 5y}}{5} \in \mathbb{R}_+ \) for which \( f(x) = y \).
\( \Rightarrow f \) is onto.
Since \( f \) is both one-one and onto, \( f \) is invertible and \( f^{-1} \) is given by:
\( f^{-1}(y) = x = \frac{-3 + \sqrt{54 + 5y}}{5} \).

Question. Let \( f : \mathbb{N} \to \mathbb{R} \) be a function defined as \( f(x) = 4x^2 + 12x + 15 \). Show that \( f : \mathbb{N} \to S \), where \( S \) is the range of \( f \), is invertible. Also find the inverse of \( f \).
Answer:
Consider, \( f(x_1) = f(x_2) \)
\( \Rightarrow 4x_1^2 + 12x_1 + 15 = 4x_2^2 + 12x_2 + 15 \)
\( \Rightarrow 4(x_1^2 - x_2^2) + 12(x_1 - x_2) = 0 \)
\( \Rightarrow (x_1 - x_2)[4x_1 + 4x_2 + 12] = 0 \)
\( \Rightarrow x_1 = x_2 \) [since \( x_1, x_2 \in \mathbb{N} \) and \( 4x_1 + 4x_2 + 12 \neq 0 \)]
\( \therefore f \) is one-one.
Since, \( S \) is the range of \( f \), \( f \) is onto.
Since, \( f \) is one-one and onto, \( f \) is invertible.
Let \( y \in S \) be an arbitrary number such that \( f(x) = y \).
\( \Rightarrow y = 4x^2 + 12x + 15 \)
\( \Rightarrow y = (2x + 3)^2 + 6 \)
\( \Rightarrow \sqrt{y - 6} = 2x + 3 \Rightarrow x = \frac{\sqrt{y - 6} - 3}{2} \)
Also, \( f(x) = y \Rightarrow x = f^{-1}(y) \)
\( \Rightarrow f^{-1}(y) = \frac{\sqrt{y - 6} - 3}{2} \) or \( f^{-1}(x) = \frac{\sqrt{x - 6} - 3}{2} \).

Binary Operations

Question. Let \(*\) be a binary operation on the set of all non-zero real numbers, given by \( a * b = \frac{ab}{5} \) for all \( a, b \in \mathbb{R} - \{0\} \). Find the value of \( x \text{, given that } 2 * (x * 5) = 10 \).
Answer:
We have, \( 2 * (x * 5) = 10 \Rightarrow 2 * \left(\frac{5x}{5}\right) = 10 \)
\( \Rightarrow 2 * x = 10 \Rightarrow \frac{2x}{5} = 10 \Rightarrow x = 25 \).

Question. Let \( * : \mathbb{R} \times \mathbb{R} \to \mathbb{R} \), given by \( (a, b) \to a + 4b^2 \) is a binary operation. Compute \( (-5) * (2 * 0) \).
Answer:
Here, \( a * b = a + 4b^2 \).
\( (-5) * (2 * 0) = (-5) * (2 + 4 \cdot 0^2) = -5 * 2 \)
\( = -5 + 4 \cdot 2^2 = -5 + 16 = 11 \).

Question. Let \(*\) be a binary operation on \(\mathbb{N}\) given by \( a * b = \text{L.C.M. }(a, b) \) for all \( a, b \in \mathbb{N} \). Find \( 5 * 7 \).
Answer:
\( 5 * 7 = \text{L.C.M. }(5, 7) = 35 \).

Question. The binary operation \( * : \mathbb{R} \times \mathbb{R} \to \mathbb{R} \), is defined as \( a * b = 2a + b \). Find \( (2 * 3) * 4 \).
Answer:
\( (2 * 3) * 4 = (2 \times 2 + 3) * 4 = 7 * 4 \)
\( = 2 \times 7 + 4 = 18 \).

Question. If the binary operation \(*\) on the set of integers \(\mathbb{Z}\), is defined by \( a * b = a + 3b^2 \), then find the value of \( 8 * 3 \).
Answer:
Here, \( a * b = a + 3b^2 \) for all \( a, b \in \mathbb{Z} \).
\( \Rightarrow 8 * 3 = 8 + 3 \cdot 3^2 = 8 + 27 = 35 \).

Question. Let \(*\) be a binary operation defined on the set of integers by \( a * b = 2a + b - 3 \). Find \( 3 * 4 \).
Answer:
Here, \( a * b = 2a + b - 3 \).
\( \therefore 3 * 4 = 2(3) + 4 - 3 = 7 \).

Question. Let \(*\) be a binary operation defined by \( a * b = 3a + 4b - 2 \). Find \( 4 * 5 \).
Answer:
Here, \( a * b = 3a + 4b - 2 \).
\( \dots 4 * 5 = 3(4) + 4(5) - 2 = 12 + 20 - 2 = 30 \).

Question. If the binary operation \(*\) on the set of integers \(\mathbb{Z}\) is defined by \( a * b = a + 3b^2 \), then find the value \( 2 * 4 \).
Answer:
Here, \( a * b = a + 3b^2 \).
\( \therefore 2 * 4 = 2 + 3(4)^2 = 2 + 3 \times 16 = 50 \).

Question. Let \(*\) be a binary operation on \(\mathbb{N}\) given by \( a * b = \text{H.C.F. }(a, b) \); \( a, b \in \mathbb{N} \). Write the value of \( 22 * 4 \).
Answer:
Here, \( a * b = \text{H.C.F. }(a, b) \).
\( \therefore 22 * 4 = \text{H.C.F. }(22, 4) = 2 \).

Question. Let \(*\) be a binary operation on set \(\mathbb{Q}\) of rational numbers defined as \( a * b = \frac{ab}{5} \), write the identity for \(*\), if any.
Answer:
Here, \( a * b = \frac{ab}{5} \).
For identity, \( a * e = a = e * a \)
\( \Rightarrow \frac{ae}{5} = a \Rightarrow e = 5 \)
\( \therefore \) Identity element for \(*\) is \( 5 \).

Question. Let \(S\) be the set of all rational numbers except 1 and \(*\) be defined on \(S\) by \( a * b = a + b - ab \), for all \( a, b \in S \). Prove that (i) \(*\) is a binary operation on \(S\). (ii) \(*\) is commutative as well as associative.
Answer:
We have, \( S = \mathbb{Q} - \{1\} \).
\( a * b = a + b - ab \), for all \( a, b \in S \).
(i) As \( a, b \in S \Rightarrow a, b \in \mathbb{Q} \) and \( a \neq 1 \), \( b \neq 1 \) ...(1)
\( \therefore a + b - ab \in \mathbb{Q} \)
We check: \( a + b - ab \neq 1 \).
Suppose \( a + b - ab = 1 \)
\( \Rightarrow a + b - ab - 1 = 0 \)
\( \Rightarrow a - 1 + b(1 - a) = 0 \)
\( \Rightarrow -(1 - a) + b(1 - a) = 0 \)
\( \Rightarrow (1 - a)(-1 + b) = 0 \)
\( \Rightarrow \) Either \( 1 - a = 0 \) or \( -1 + b = 0 \)
\( \Rightarrow a = 1 \) or \( b = 1 \).
This contradicts (1).
\( \therefore a + b - ab \neq 1 \).
\( \Rightarrow a + b - ab \in \mathbb{Q} - \{1\} = S \).
\( \Rightarrow * \) is a binary operation on \( S \).
(ii) Let \( a, b \in S \).
\( a * b = a + b - ab = b + a - ba = b * a \), hence \(*\) is commutative in \(S\).
Let \( a, b, c \in S \).
Then, \( a * (b * c) = a * (b + c - bc) \)
\( = a + b + c - bc - a(b + c - bc) \)
\( = a + b + c - ab - bc - ca + abc \)
\( = a + b - ab + c - (a + b - ab)c \)
\( = (a * b) * c \).
\( \therefore * \) is associative.

Question. Consider the binary operations \( * : \mathbb{R} \times \mathbb{R} \to \mathbb{R} \) and \( o : \mathbb{R} \times \mathbb{R} \to \mathbb{R} \) defined as \( a * b = |a - b| \) and \( a \ o \ b = a \) for all \( a, b \in \mathbb{R} \). Show that \(*\) is commutative but not associative, \(o\) is associative but not commutative.
Answer:
\( b * a = |b - a| = |a - b| = a * b \) for all \( a, b \in \mathbb{R} \).
\( \Rightarrow * \) is commutative on \(\mathbb{R}\).
Also, for \( a = 2 \), \( b = 4 \), \( c = 5 \):
\( (a * b) * c = (2 * 4) * 5 = |2 - 4| * 5 = 2 * 5 = |2 - 5| = 3 \).
And \( a * (b * c) = 2 * (4 * 5) = 2 * |4 - 5| = 2 * 1 = |2 - 1| = 1 \).
\( \therefore (a * b) * c \neq a * (b * c) \).
\( \Rightarrow * \) is not associative on \(\mathbb{R}\).
Also, \( (a \ o \ b) \ o \ c = a \ o \ c = a \) and \( a \ o \ (b \ o \ c) = a \ o \ b = a \).
\( \Rightarrow (a \ o \ b) \ o \ c = a \ o \ (b \ o \ c) \) for all \( a, b, c \in \mathbb{R} \).
\( \Rightarrow o \) is associative on \(\mathbb{R}\).
Also, for \( a = 3 \), \( b = 2 \):
\( a \ o \ b = 3 \ o \ 2 = 3 \)
\( b \ o \ a = 2 \ o \ 3 = 2 \)
\( \Rightarrow a \ o \ b \neq b \ o \ a \).
\( \Rightarrow o \) is not commutative on \(\mathbb{R}\).

Question. Consider the binary operation \(*\) on the set \(\{1, 2, 3, 4, 5\}\) defined by \( a * b = \min(a, b) \). Write the operation table of the operation \(*\).
Answer:
Let \( A = \{1, 2, 3, 4, 5\} \).
\( a * b = \min(a, b) \).
The operation table is as follows:

Question. A binary operation \(*\) on the set \(\{0, 1, 2, 3, 4, 5\}\) is defined as:
\( a * b = \begin{cases} a + b, & \text{if } a + b < 6 \\ a + b - 6, & \text{if } a + b \ge 6 \end{cases} \)
Show that zero is the identity for this operation and each element \( a \neq 0 \) of the set is invertible with \( 6 - a \) being the inverse of \( a \).
Answer:
The operation table is given by:

Identity: Let \( e \) be the identity element, then \( a * e = a = e * a \).
Now, \( a * 0 = a + 0 = a \) and \( 0 * a = 0 + a = a \).
Thus, \( a * 0 = a = 0 * a \). Hence, 0 is the identity element of the operation.
Inverse: Since each row or column contains the identity element i.e., 0, each element is invertible.
Now, \( a * (6 - a) = a + (6 - a) - 6 = 0 \) and \( (6 - a) * a = (6 - a) + a - 6 = 0 \).
Hence, each element \( a \) of the set is invertible with inverse \( 6 - a \).

Question. Let \(*\) be a binary operation on \(\mathbb{Q}\) defined by \( a * b = \frac{3ab}{5} \). Show that \(*\) is commutative as well as associative. Also, find its identity element, if it exists.
Answer:
Commutativity: \( a * b = \frac{3ab}{5} = \frac{3ba}{5} = b * a \).
\( \therefore * \) is commutative.
Associativity: \( (a * b) * c = \left(\frac{3ab}{5}\right) * c = \frac{9abc}{25} \)
and \( a * (b * c) = a * \left(\frac{3bc}{5}\right) = \frac{9abc}{25} \).
\( \Rightarrow (a * b) * c = a * (b * c) \).
\( \therefore * \) is associative.
Identity: Let \( e \in \mathbb{Q} \) be the identity element. Then, \( a * e = a = e * a \)
\( \Rightarrow \frac{3ae}{5} = a \Rightarrow e = \frac{5}{3} \in \mathbb{Q} \).
\( \therefore \frac{5}{3} \) is the identity element.

Question. Let \(*\) be a binary operation on the set of rational numbers given as \( a * b = (2a - b)^2 \), \( a, b \in \mathbb{Q} \). Find \( 3 * 5 \) and \( 5 * 3 \). Is \( 3 * 5 = 5 * 3 \)?
Answer:
We have, \( a * b = (2a - b)^2 \).
\( \therefore 3 * 5 = (2 \times 3 - 5)^2 = (6 - 5)^2 = 1 \).
\( 5 * 3 = (2 \times 5 - 3)^2 = (10 - 3)^2 = 49 \).
Thus, \( 3 * 5 \neq 5 * 3 \).

Question. Let \(*\) be the binary operation on \(\mathbb{N}\) given by \( a * b = \text{L.C.M. of } a \text{ and } b \). Find the value of \( 20 * 16 \). Is \(*\) (i) commutative, (ii) associative?
Answer:
We have, \( a * b = \text{L.C.M. of } a \text{ and } b \).
\( \therefore 20 * 16 = \text{L.C.M. of 20 and 16} = 80 \).
(i) Commutativity: \( a * b = \text{L.C.M. of } a \text{ and } b = \text{L.C.M. of } b \text{ and } a = b * a \).
\( \Rightarrow a * b = b * a \). So, \(*\) is commutative.
(ii) Associativity: \( (a * b) * c = [\text{L.C.M. }(a, b)] * c = \text{L.C.M. }[\text{L.C.M. }(a, b), c] = \text{L.C.M. }(a, b, c) \)
and \( a * (b * c) = a * [\text{L.C.M. }(b, c)] = \text{L.C.M. }[a, \text{L.C.M. }(b, c)] = \text{L.C.M. }(a, b, c) \).
\( \Rightarrow a * (b * c) = (a * b) * c \). So, \(*\) is associative.

Question. Show that the binary operation \(*\) on \( A = \mathbb{R} - \{-1\} \) defined as \( a * b = a + b + ab \) for all \( a, b \in A \) is commutative and associative on \( A \). Also find the identity element of \(*\) in \( A \) and prove that every element of \( A \) is invertible.
Answer:
We have, \( a * b = a + b + ab \) for all \( a, b \in A \), where \( A = \mathbb{R} - \{-1\} \).
Commutativity: Let \( a, b \in \mathbb{R} - \{-1\} \).
We have, \( a * b = a + b + ab = b + a + ba = b * a \).
Hence, \(*\) is commutative.
Associativity: Let \( a, b, c \in \mathbb{R} - \{-1\} \).
We have, \( a * (b * c) = a * (b + c + bc) \)
\( = a + (b + c + bc) + a(b + c + bc) \)
\( = a + b + c + bc + ab + ac + abc \)
\( = a + b + ab + c + (a + b + ab)c \)
\( = (a + b + ab) * c = (a * b) * c \).
Hence, \(*\) is associative.
Identity: Let \( e \in A \) be the identity element. Then, \( a * e = a = e * a \)
\( \Rightarrow a + e + ae = a \Rightarrow e(1 + a) = 0 \Rightarrow e = 0 \) [since \( a \neq -1 \)].
Hence, the identity element for \(*\) is \( e = 0 \).
Existence of inverse: Let \( a \in \mathbb{R} - \{-1\} \) and \( b \) be the inverse of \( a \).
Then, \( a * b = e \Rightarrow a + b + ab = 0 \)
\( \Rightarrow b(1 + a) = -a \Rightarrow b = \frac{-a}{a + 1} \).
Since \( a \in \mathbb{R} - \{-1\} \), \( a \neq -1 \Rightarrow a + 1 \neq 0 \Rightarrow b = \frac{-a}{a + 1} \in \mathbb{R} \).
Also, if \( \frac{-a}{a + 1} = -1 \Rightarrow -a = -a - 1 \Rightarrow 0 = -1 \), which is not possible.
Hence, \( \frac{-a}{a + 1} \in \mathbb{R} - \{-1\} \).
So, every element of \( \mathbb{R} - \{-1\} \) is invertible and the inverse of an element \( a \) is \( \frac{-a}{a + 1} \).

Question. Let \( A = \mathbb{R} \times \mathbb{R} \) and \(*\) be the binary operation on \( A \) defined by \( (a, b) * (c, d) = (a + c, b + d) \). Show that \(*\) is commutative and associative. Find the identity element for \(*\) on \( A \).
Answer:
Here \( A = \mathbb{R} \times \mathbb{R} \) and \(*\) on \( A \) is defined as \( (a, b) * (c, d) = (a + c, b + d) \) for all \( (a, b), (c, d) \in \mathbb{R} \times \mathbb{R} \).
Now, \( (c, d) * (a, b) = (c + a, d + b) = (a + c, b + d) = (a, b) * (c, d) \).
\( \Rightarrow * \) is commutative on \( A \).
Again, \( [(a, b) * (c, d)] * (e, f) = (a + c, b + d) * (e, f) \)
\( = ((a + c) + e, (b + d) + f) = (a + (c + e), b + (d + f)) \)
\( = (a, b) * (c + e, d + f) = (a, b) * [(c, d) * (e, f)] \).
\( \Rightarrow * \) is associative on \( A \).
Also, \( 0 \in \mathbb{R} \) and \( (0, 0) \in A \).
\( \forall (a, b) \in A \), \( (a, b) * (0, 0) = (a + 0, b + 0) = (a, b) \) and \( (0, 0) * (a, b) = (0 + a, 0 + b) = (a, b) \).
\( \Rightarrow (0, 0) \) acts as an identity element in \( A \) w.r.t. \(*\).