CBSE Class 12 Mathematics Relations and Functions Important Questions Set A

Question. Check whether the relation \( R \) defined on the set \( A = \{1, 2, 3, 4, 5, 6\} \) as \( R = \{(a, b) : b = a + 1\} \) is reflexive, symmetric or transitive. 
Answer: Given relation \( R \) defined on the set \( A = \{1, 2, 3, 4, 5, 6\} \) as \( R = \{(a, b) : b = a + 1\} \)
Now,
Reflexivity: Let \( a \in A \)
We have, \( a \ne a + 1 \Rightarrow (a, a) \notin R \)
\(\therefore\) It is not reflexive
Symmetric: Let \( a = 1 \) and \( b = 2 \) i.e. \( a, b \in A \)
\(\therefore b = a + 1 \Rightarrow 2 = 1 + 1 \Rightarrow (a, b) \in R \)
but \( a \ne b + 1 \) as \( 1 \ne 2 + 1 \Rightarrow (b, a) \notin R \)
\(\therefore\) It is not symmetric.
Transitive: Let \( a, b, c \in A \)
Now, if \( (a, b) \in R \Rightarrow b = a + 1 \) ...(i)
and \( (b, c) \in R \Rightarrow c = b + 1 \) ...(ii)
from (i) and (ii), we have
\( c = (a + 1) + 1 = a + 2 \)
\( \Rightarrow c = a + 2 \Rightarrow (a, c) \notin R \)
\(\therefore\) Is is not transitive
Hence, relation \( R \) is neither reflexive nor symmetric nor transitive.

Question. Show that the relation \( R \) on the set \( Z \) of all integers, given by \( R = \{(a, b) : 2 \text{ divides } (a - b)\} \) is an equivalence relation. 
Answer: Given relation \( R = \{(a, b) : 2 \text{ divides } (a - b)\} \) on the set \( Z \) of all integers
Reflexive: Let \( a \in Z \)
Since \( (a - a) = 0 \), which is divisible by 2 i.e., \( (a, a) \in R \)
\(\therefore R \) is reflexive.
Symmetric: Let \( a, b \in Z \)
such that \( (a, b) \in R \Rightarrow (a - b) \) is divisible by 2
\( \Rightarrow - (a - b) \) is also divisible by 2
\( \Rightarrow (b - a) \) is divisible by 2 \( \Rightarrow (b, a) \in R \)
i.e., \( (a, b) \in R \Rightarrow (b, a) \in R \)
\(\therefore R \) is symmetric.
Transitive: Let \( a, b, c \in Z \)
such that \( (a, b) \in R \Rightarrow (a - b) \) is divisible by 2
Let \( a - b = 2k_1 \) where \( k_1 \) is an integer ...(i)
and \( (b, c) \in R \Rightarrow (b - c) \) is divisible by 2 \( \Rightarrow b - c = 2 k_2 \) where \( k_2 \) is an integer ...(ii)
Adding (i) and (ii), we have
\( (a - b) + (b - c) = 2 (k_1 + k_2) \Rightarrow a - c = 2 (k_1 + k_2) \Rightarrow (a - c) \) is divisible by 2.
\( \Rightarrow (a, c) \in R \)
\(\therefore R \) is transitive
Thus, \( R \) is reflexive, symmetric and transitive. Hence, given relation \( R \) is an equivalence relation.

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 \( (a, b), (c, d) \) 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 \( (c, d) \in A \times A \).
Since \( (a, b) R(a, b) \) as \( a + b = b + a, \forall a, b \in A \)
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 \( (a, b) R(c, d) \) and \( (c, d) R(e, f) \) then
\( a + d = b + c \) and \( c + f = d + e \)
\( \Rightarrow a + d = b + c \) and \( d + e = c + f \)
\(\therefore (a + d) - (d + e) = (b + c) - (c + f) \)
\( \Rightarrow (a - e) = b - f \)
\( \Rightarrow a + f = b + e \)
\( (a, b) R(e, f) \)
So, \( R \) is transitive.
Hence, \( R \) is an equivalence relation.
Now, equivalence class containing \( [(2, 5)] \) is \( \{(1, 4), (2, 5), (3, 6), (4, 7), (5, 8), (6, 9)\} \).

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 \( g = [-1, \infty) \)
[\(\because x^2 \ge 0 \) for all \( x \in \mathbb{R} \)
\(\therefore x^2 - 1 \ge -1 \) for all \( x \in \mathbb{R} \)]
Computation of \( gof \): We observe that range of \( f = [0, \infty) \subseteq \text{domain of } g \)
\(\therefore gof \) exists and \( gof : [0, \infty) \to \mathbb{R} \)
Also, \( gof(x) = g(f(x)) = g(\sqrt{x}) = (\sqrt{x})^2 - 1 = x - 1 \)
Thus, \( gof : [0, \infty) \to \mathbb{R} \) is defined as \( gof(x) = x - 1 \)
Computation of \( fog \): We observe that range of \( g = [-1, \infty) \not\subseteq \text{domain of } f \).
\(\therefore \text{Domain of } fog = \{x : x \in \text{domain of } g \text{ and } g(x) \in \text{domain of } f\} \)
\( \Rightarrow \text{Domain of } fog = \{x : x \in \mathbb{R} \text{ and } g(x) \in [0, \infty)\} \)
\( \Rightarrow \text{Domain of } fog = \{x : x \in \mathbb{R} \text{ and } x^2 - 1 \in [0, \infty)\} \)
\( \Rightarrow \text{Domain of } fog = \{x : x \in \mathbb{R} \text{ and } x^2 - 1 \ge 0\} \)
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, \( fog : (-\infty, -1] \cup [1, \infty) \to \mathbb{R} \) 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, \( fog \ne gof \)

Question. Let \( A = \{-1, 0, 1, 2\}, B = \{-4, -2, 0, 2\} \) and \( f, g : A \to B \) 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 \( f : A \to B \) and \( g : A \to B \) to be equal, \( f(a) = g(a) \forall a \in A \) and \( R_f = R_g \).
Here, we have \( f(x) = x^2 - x \)
\( g(x) = 2 \left|x - \frac{1}{2}\right| - 1 \) [\( x \in A = \{-1, 0, 1, 2\} \)]
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, \( f(0) = g(0) = 0, f(1) = g(1) = 0 \) 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, \( 0 \le x - [x] < 1 \)
\( \Rightarrow 0 \le \{x\} < 1 \)
\( \Rightarrow 1 + \{x\} \ge 1 \)
\(\therefore fog(x) = f(1 + \{x\}) = 1 \)
Note: Symbol \( \{x\} \) denotes the fractional 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 \( f : A \to B \) be defined by \( f(x) = \frac{x - 2}{x - 3}, \forall x \in A \). Then, show that \( f \) is bijective. 
Answer: Given that, \( A = \mathbb{R} - \{3\}, B = \mathbb{R} - \{1\} \).
\( f : A \to B \) 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 \( y = \frac{x - 2}{x - 3} \Rightarrow x - 2 = xy - 3y \)
\( \Rightarrow x(1 - y) = 2 - 3y \Rightarrow x = \frac{2 - 3y}{1 - y} \)
\( \Rightarrow x = \frac{3y - 2}{y - 1} \in A, \forall y \in B \) [codomain]
So, \( f(x) \) is surjective function.
Hence, \( f(x) \) is a bijective function.

Question. Show that the relation \( S \) in the set \( A = \{x \in Z : 0 \le x \le 12\} \) given by \( S = \{(a, b) : a, b \in Z, |a - b| \text{ is divisible by 3}\} \) is an equivalence relation.
Answer: On the set \( A = \{x \in Z : 0 \le x \le 12\} \) and relation \( S \) is given by \( S = \{(a, b) : a, b \in Z, |a - b| \text{ is divisible by 3}\} \)
Reflexivity:
Let \( a \in A \) Then
\( (a, a) \Rightarrow |a - a| = 0 \) which is divisible by 3. \( \Rightarrow (a, a) \in S \)
\(\therefore\) It is reflexive.
Symmetric:
Let \( a, b \in A \) Then
\( (a, b) \in S \Rightarrow |a - b| \) is divisible by 3.
\( \Rightarrow |b - a| \) is also divisible by 3. \( \Rightarrow (b, a) \in S \)
\(\therefore\) It is symmetric relation.
Transitive:
Let \( a, b, c \in A \) Then
\( (a, b) \in S \Rightarrow |a - b| \) is divisible by 3. \( \Rightarrow a - b = \pm 3k_1 \), where \( k_1 \) is an integer ...(i)
\( (b, c) \in S \Rightarrow |b - c| \) is divisible by 3. \( \Rightarrow b - c = \pm 3k_2 \), where \( k_2 \) is an integer ...(ii)
\(\therefore a - c = a - b + b - c = \pm 3k_1 \pm 3k_2 = \pm 3(k_1 + k_2) \Rightarrow |a - c| \) is also divisible by 3. \( \Rightarrow (a, c) \in S \)
\(\therefore\) It is transitive relation.
Hence, the relation \( S \) is an equivalence relation.

Question. Let \( A = R - \{2\} \) and \( B = R - \{1\} \). If \( f : A \rightarrow B \) 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: Given sets \( A = R - \{2\} \) and \( B = R - \{1\} \) and function defined such that \( f : A \rightarrow B \)
\( f(x) = \frac{x-1}{x-2} \)
One-one:
Let \( x_1, x_2 \in A \) such that \( x_1 \neq x_2 \Rightarrow x_1 - 1 \neq x_2 - 1 \) and \( x_1 - 2 \neq x_2 - 2 \)
\( \therefore \frac{x_1 - 1}{x_1 - 2} \neq \frac{x_2 - 1}{x_2 - 2} \Rightarrow f(x_1) \neq f(x_2) \)
\( f(x) \) is one-one function.
Onto:
Let \( y = \frac{x-1}{x-2} \)
\( \Rightarrow xy - 2y = x - 1 \Rightarrow xy - x = 2y - 1 \Rightarrow x(y - 1) = 2y - 1 \)
\( \therefore x = \frac{2y - 1}{y - 1}, y \neq 1 \)
Clearly, for every value of \( y \) there must be some \( x \).
\( \therefore f(x) \) is onto.
Here, \( f(x) \) is both one-one and onto, so it is invertible.
We have, \( x = \frac{2y - 1}{y - 1} \Rightarrow f^{-1}(y) = \frac{2y - 1}{y - 1} \)

Long Answer Questions

Question. Consider \( f : R_+ \rightarrow [-9, \infty) \) 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 \( x_1, x_2 \in R_+ \), then
\( f(x_1) = f(x_2) \Rightarrow 5x_1^2 + 6x_1 - 9 = 5x_2^2 + 6x_2 - 9 \)
\( \Rightarrow 5x_1^2 + 6x_1 - 5x_2^2 - 6x_2 = 0 \Rightarrow 5(x_1^2 - x_2^2) + 6(x_1 - x_2) = 0 \)
\( \Rightarrow 5(x_1 - x_2)(x_1 + x_2) + 6(x_1 - x_2) = 0 \Rightarrow (x_1 - x_2)(5x_1 + 5x_2 + 6) = 0 \)
\( \Rightarrow x_1 - x_2 = 0 \) [\( \because 5x_1 + 5x_2 + 6 \neq 0 \)]
\( \Rightarrow x_1 = x_2 \)
i.e., \( f \) is one-one function.
Onto: Let \( f(x) = y \)
\( \therefore y = 5x^2 + 6x - 9 \Rightarrow 5x^2 + 6x - (9 + y) = 0 \)
\( \Rightarrow x = \frac{-6 \pm \sqrt{36 + 4 \times 5(9 + y)}}{10} \Rightarrow x = \frac{-6 \pm \sqrt{216 + 20y}}{10} \)
\( \Rightarrow x = \frac{\pm \sqrt{54 + 5y} - 3}{5} \Rightarrow x = \frac{\sqrt{54 + 5y} - 3}{5} \) [\( \because x \in R_+ \)]
Obviously, \( \forall y \in [-9, \infty) \) the value of \( x \in R_+ \).
\( \Rightarrow f \) 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. If \( f, g : R \rightarrow R \) be two functions defined as \( f(x) = |x| + x \) and \( g(x) = |x| - x, \forall x \in R \). 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, \( gof(x) = 0, \forall x \in R \).
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
\( fog(5) = 0 \) [\( \because 5 \ge 0 \)]
\( fog(-3) = -4 \times (-3) = 12 \) [\( \because -3 < 0 \)]
\( gof(-2) = 0 \)

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

Question. Consider \( f : R_+ \rightarrow [4, \infty) \) 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 \( R_+ \) is the set of all non-negative real numbers. 
Answer: One-one: Let \( x_1, x_2 \in R_+ \) (Domain)
\( f(x_1) = f(x_2) \Rightarrow x_1^2 + 4 = x_2^2 + 4 \)
\( \Rightarrow x_1^2 = x_2^2 \)
\( \Rightarrow x_1 = x_2 \) [\( \because x_1, x_2 \) are +ve real number]
Hence, \( f \) is one-one function.
Onto: Let \( y \in [4, \infty) \) such that
\( y = f(x), \forall x \in R_+ \) [set of non-negative reals]
\( \Rightarrow y = x^2 + 4 \)
\( \Rightarrow x = \sqrt{y - 4} \) [\( \because x \) is +ve real number]
Obviously, \( \forall y \in [4, \infty) \), \( x \) is real number \( \in R_+ \) (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 \( fof^{-1} = I \) [Identity function]
\( \Rightarrow fof^{-1}(y) = y, \forall y \in [4, \infty) \)
\( \Rightarrow f(f^{-1}(y)) = y \)
\( \Rightarrow (f^{-1}(y))^2 + 4 = y \) [\( \because f(x) = x^2 + 4 \)]
\( \Rightarrow f^{-1}(y) = \sqrt{y - 4} \)
Therefore, required inverse function is \( f^{-1} : [4, \infty) \rightarrow R \) defined by \( f^{-1}(y) = \sqrt{y - 4}, \forall y \in [4, \infty) \).

Question. Determine whether the relation \( R \) defined on the set \( \mathbb{R} \) of all real numbers as \( R = \{(a, b) : a, b \in \mathbb{R} \text{ and } a - b + \sqrt{3} \in S, \text{ where } S \text{ is the set of all irrational numbers}\} \), is reflexive, symmetric and transitive. 
Answer: Here, relation \( R \) defined on the set \( R \) is given as
\( R = \{(a, b) : a, b \in R \text{ and } a - b + \sqrt{3} \in S \} \)
Reflexivity: Let \( a \in R \) (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
\( (a, b) \in R \) as \( a - b + \sqrt{3} = \sqrt{3} - 1 + \sqrt{3} = 2\sqrt{3} - 1 \in S \)
But \( b - a + \sqrt{3} = 1 - \sqrt{3} + \sqrt{3} = 1 \notin S \Rightarrow (b, a) \notin R \)
As \( (a, b) \) belongs to \( R \) but \( (b, a) \) does not belong to \( R \)
\( \therefore R \) is not symmetric.
Transitivity: Taking \( a = 1, b = 2 \), and \( c = 3 - \sqrt{3} \)
\( (a, b) \in R \) as \( a - b + \sqrt{3} = 1 - 2 + \sqrt{3} = \sqrt{3} - 1 \in S \Rightarrow (a, b) \in R \)
\( \therefore (b, c) \in R \) as \( b - c + \sqrt{3} = 2 - (3 - \sqrt{3}) + \sqrt{3} = -1 + 2\sqrt{3} \in S \Rightarrow (b, c) \in R \)
But \( a - c + \sqrt{3} = 1 - (3 - \sqrt{3}) + \sqrt{3} = -2 + 2\sqrt{3} \in S \).
Wait, taking \( a = 1, b = \sqrt{3} \), and \( c = 2\sqrt{3} \)
\( a - b + \sqrt{3} = 1 - \sqrt{3} + \sqrt{3} = 1 \notin S \).
Let's re-evaluate: As \( (a, b) \) and \( (b, c) \) belongs to \( R \) but \( (a, c) \) does not belong to \( R \)
\( \therefore R \) is not transitive.
Hence, \( R \) is reflexive but neither symmetric nor transitive.

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, \( f(x) = \begin{cases} x - 1, & \text{if } x \text{ is odd} \\ x + 1, & \text{if } x \text{ is even} \end{cases} \)
\( \Rightarrow \) For every even number ‘\( y \)’ of codomain \( \exists \) odd number \( y + 1 \) in domain and for every odd number \( y \) of codomain there exists even number \( y - 1 \) in domain i.e., \( f \) 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 \( f : R \rightarrow R \) be defined by \( f(x) = 2x - 3 \) and \( g : R \rightarrow R \) by \( g(x) = x^3 + 5 \), then find the value of \( (fog)^{-1}(x) \). 
Answer: Here \( f : R \rightarrow R \) and \( g : R \rightarrow R \) be two functions such that \( f(x) = 2x - 3 \) and \( g(x) = x^3 + 5 \)
\( \because f \) and \( g \) both are bijective (one-one onto) function.
\( \Rightarrow fog \) is also bijective function.
\( \Rightarrow fog \) is invertible function.
Now, \( fog(x) = f\{g(x)\} \Rightarrow fog(x) = f(x^3 + 5) \)
\( \Rightarrow fog(x) = 2(x^3 + 5) - 3 \Rightarrow fog(x) = 2x^3 + 10 - 3 \)
\( \Rightarrow fog(x) = 2x^3 + 7 \) ...(i)
For inverse of \( fog(x) \)
Let \( fog(x) = y \Rightarrow x = (fog)^{-1}(y) \)
(i) \( \Rightarrow y = 2x^3 + 7 \Rightarrow x^3 = \frac{y-7}{2} \)
\( \Rightarrow x = \left( \frac{y-7}{2} \right)^{1/3} \)
\( \Rightarrow (fog)^{-1}(y) = \left( \frac{y-7}{2} \right)^{1/3} \Rightarrow (fog)^{-1}(x) = \left( \frac{x-7}{2} \right)^{1/3} \)

Question. Let \( f : N \rightarrow R \) be a function defined as \( f(x) = 4x^2 + 12x + 15 \). Show that \( f : N \rightarrow S \) 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 \( x \in N \)
\( \Rightarrow y = (2x + 3)^2 + 6 \Rightarrow x = \frac{(\sqrt{y - 6}) - 3}{2} \), as \( y > 6 \)
Let \( g : S \rightarrow N \) 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 \( fog(y) = f \left( \frac{(\sqrt{y - 6}) - 3}{2} \right) = \left[ 2 \frac{(\sqrt{y - 6}) - 3}{2} + 3 \right]^2 + 6 = y \)
Hence, \( fog(y) = I_S \) and \( gof(x) = I_N \)
\( f \) is invertible and its inverse is given by \( f^{-1}(y) = g(y) = \frac{\sqrt{y - 6} - 3}{2} \)

Question. Let \( f : R - \{-\frac{4}{3}\} \rightarrow R - \{\frac{4}{3}\} \) be a function defined as \( f(x) = \frac{4x}{3x + 4} \). Show that, in \( f : R - \{-\frac{4}{3}\} \rightarrow \text{Range of } f \), \( f \) is one-one and onto. Hence find \( f^{-1} : \text{Range } f \rightarrow R - \{-\frac{4}{3}\} \).
Answer: Let \( x_1, x_2 \in R - \{-\frac{4}{3}\} \)
Now \( f(x_1) = f(x_2) \Rightarrow \frac{4x_1}{3x_1 + 4} = \frac{4x_2}{3x_2 + 4} \)
\( \Rightarrow 12x_1x_2 + 16x_1 = 12x_1x_2 + 16x_2 \Rightarrow 16x_1 = 16x_2 \Rightarrow x_1 = x_2 \)
Hence \( f \) is one-one function.
Since, co-domain \( f \) is range of \( f \)
So, \( f : R - \{-\frac{4}{3}\} \rightarrow R \) 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, \( f^{-1} : \text{Range of } f \rightarrow R - \{-\frac{4}{3}\} \) is \( f^{-1}(y) = \frac{4y}{4 - 3y} \).

Objective Type Questions:

Question. Let R be a relation on the set N of natural numbers defined by nRm if n divides m. Then R is
(a) reflexive and symmetric
(b) transitive and symmetric
(c) equivalence
(d) reflexive, transitive but not symmetric
Answer: (d)

Question. Let A = {1, 2, 3} and consider the relation R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 3), (1, 3)}. Then R is
(a) reflexive but not symmetric
(b) reflexive but not transitive
(c) symmetric and transitive
(d) neither symmetric nor transitive
Answer: (a)

Question. If a relation R on the set {1, 2, 3} be defined by R = {(1, 2)}, then R is
(a) reflexive
(b) transitive
(c) symmetric
(d) none of these
Answer: (b)

Question. Let \( f : R \to R \) be defined by \( f(x) = \sin x \) and \( g : R \to R \) be defined by \( g(x) = x^2 \), then \( fog \) is
(a) \( x^2 \sin x \)
(b) \( (\sin x)^2 \)
(c) \( \sin x^2 \)
(d) \( \frac{\sin x}{x^2} \)
Answer: (c)

Question. Let \( f : R \to R \) be defined by \( f(x) = 3x - 4 \). Then \( f^{-1}(x) \) is given by
(a) \( \frac{x + 4}{3} \)
(b) \( \frac{x - 4}{3} \)
(c) \( 3x + 4 \)
(d) none of these
Answer: (a)

Question. Let \( f : R \to R \) be defined by \( f(x) = \frac{1}{x}, \forall x \in R \). Then \( f \) is
(a) one-one
(b) onto
(c) bijective
(d) \( f \) is not defined
Answer: (d)

Question. Let \( f : R \to R \) be given by \( f(x) = \tan x \). Then \( f^{-1}(1) \) is
(a) \( \frac{\pi}{4} \)
(b) \( \{n\pi + \frac{\pi}{4} : n \in Z\} \)
(c) does not exist
(d) none of these
Answer: (b)

Question. If \( A = \{1, 2, 3\}, B = \{1, 4, 6, 9\} \) and \( R \) is a relation from \( A \) to \( B \) defined by 'x is greater than y'. Then range of \( R \) is
(a) \( \{1, 4, 6, 9\} \)
(b) \( \{4, 6, 9\} \)
(c) \( \{1\} \)
(d) none of these
Answer: (c)