GSEB Class 12 Maths Solutions Chapter 1 Relations and Functions Exercise 1.3

Question 1. Let \( f : \{1, 3, 4\} \rightarrow \{1, 2, 5\} \) and \( g : \{1, 2, 5\} \rightarrow \{1, 3\} \) be given by \( f = \{(1, 2), (3, 5), (4,1)\} \) and \( g = \{(1, 3), (2, 3), (5,1)\} \). Write down \( gof \).
Answer: We have \( f = \{(1, 2), (3, 5), (4,1)\} \) and \( g = \{(1, 3), (2, 3), (5, 1)\} \).
To find \( gof \), we need to compute \( g(f(x)) \) for each \( x \) in the domain of \( f \).
For \( x = 1 \): \( f(1) = 2 \). Then \( g(f(1)) = g(2) = 3 \). So, \( (1, 3) \) is a pair in \( gof \).
For \( x = 3 \): \( f(3) = 5 \). Then \( g(f(3)) = g(5) = 1 \). So, \( (3, 1) \) is a pair in \( gof \).
For \( x = 4 \): \( f(4) = 1 \). Then \( g(f(4)) = g(1) = 3 \). So, \( (4, 3) \) is a pair in \( gof \).
Therefore, \( gof = \{(1, 3), (3, 1), (4, 3)\} \).
In simple words: To calculate \( gof \), you first apply function \( f \) to an input, and then apply function \( g \) to the output of \( f \). We found the result for each number.

Exam Tip: When finding a composite function like \( gof \), always apply the inner function (here, \( f \)) first, then the outer function (here, \( g \)) to its result, mapping the original domain to the final codomain.

Question 2. Let f, g and h be functions from R to R. Show that
(i) \( (f + g)oh = foh + goh \).
(ii) \( (f \cdot g)oh = (foh) \cdot (goh) \).
Answer: Let \( x \in R \) be any real number.
(i) To show \( (f + g)oh = foh + goh \):
Consider the left-hand side (LHS):
\( (f + g)oh (x) = (f + g)[h(x)] \)
By the definition of function addition, \( (f + g)(y) = f(y) + g(y) \). So, let \( y = h(x) \).
\( (f + g)[h(x)] = f[h(x)] + g[h(x)] \)
By the definition of composite functions, \( f[h(x)] = (foh)(x) \) and \( g[h(x)] = (goh)(x) \).
\( f[h(x)] + g[h(x)] = (foh)(x) + (goh)(x) \)
This is the right-hand side (RHS).

\( \implies \) Therefore, \( (f + g)oh = foh + goh \).
(ii) To show \( (f \cdot g)oh = (foh) \cdot (goh) \):
Consider the left-hand side (LHS):
\( (f \cdot g)oh (x) = (f \cdot g)[h(x)] \)
By the definition of function multiplication, \( (f \cdot g)(y) = f(y) \cdot g(y) \). So, let \( y = h(x) \).
\( (f \cdot g)[h(x)] = f[h(x)] \cdot g[h(x)] \)
By the definition of composite functions, \( f[h(x)] = (foh)(x) \) and \( g[h(x)] = (goh)(x) \).
\( f[h(x)] \cdot g[h(x)] = (foh)(x) \cdot (goh)(x) \)
This is the right-hand side (RHS).

\( \implies \) Therefore, \( (f \cdot g)oh = (foh) \cdot (goh) \).
In simple words: This proves that for functions, composition distributes over addition and multiplication. It means you can apply the inner function \( h \) to each of \( f \) and \( g \) separately, and then add or multiply their results, which is the same as adding or multiplying \( f \) and \( g \) first, then applying \( h \).

Exam Tip: Remember that function composition is distributive over addition and multiplication. Clearly state the definitions of function addition, multiplication, and composition in your steps.

Question 3. Find \( gof \) and \( fog \), if
(i) \( f(x) = |x| \), and \( g(x) = |5x - 2| \).
(ii) \( f(x) = 8x^3 \) and \( g(x) = x^{1/3} \).
Answer:
(i) Given \( f(x) = |x| \) and \( g(x) = |5x - 2| \).
To find \( gof(x) \):
\( gof(x) = g[f(x)] \)
\( = g[|x|] \)
Substitute \( |x| \) into \( g(x) \): \( g(y) = |5y - 2| \). So, \( g[|x|] = |5|x| - 2| \).
Thus, \( gof(x) = |5|x| - 2| \).
To find \( fog(x) \):
\( fog(x) = f[g(x)] \)
\( = f[|5x - 2|] \)
Substitute \( |5x - 2| \) into \( f(x) \): \( f(y) = |y| \). So, \( f[|5x - 2|] = ||5x - 2|| \).
The absolute value of an absolute value is simply the absolute value itself, so \( ||5x - 2|| = |5x - 2| \).
Thus, \( fog(x) = |5x - 2| \).
(ii) Given \( f(x) = 8x^3 \) and \( g(x) = x^{1/3} \).
To find \( gof(x) \):
\( gof(x) = g[f(x)] \)
\( = g[8x^3] \)
Substitute \( 8x^3 \) into \( g(x) \): \( g(y) = y^{1/3} \). So, \( g[8x^3] = (8x^3)^{1/3} \).
Using power rules, \( (8x^3)^{1/3} = 8^{1/3} \cdot (x^3)^{1/3} = 2 \cdot x^1 = 2x \).
Thus, \( gof(x) = 2x \).
To find \( fog(x) \):
\( fog(x) = f[g(x)] \)
\( = f[x^{1/3}] \)
Substitute \( x^{1/3} \) into \( f(x) \): \( f(y) = 8y^3 \). So, \( f[x^{1/3}] = 8(x^{1/3})^3 \).
Using power rules, \( 8(x^{1/3})^3 = 8 \cdot x^{(1/3) \cdot 3} = 8 \cdot x^1 = 8x \).
Thus, \( fog(x) = 8x \).
In simple words: For \( gof \), you put the whole function \( f(x) \) inside \( g(x) \). For \( fog \), you put the whole function \( g(x) \) inside \( f(x) \). Then, you simplify the expression using the rules of exponents and absolute values.

Exam Tip: Remember to substitute the entire function expression when finding composite functions. Pay close attention to the order of operations and properties of absolute values and exponents.

Question 4. If \( f(x) = \frac{4x+3}{6x-4} \), \( x \neq \frac{2}{3} \), show that \( fof(x) = x \) for all \( x \neq \frac{2}{3} \). What is the inverse of \( f \)?
Answer: Given \( f(x) = \frac{4x+3}{6x-4} \), where \( x \neq \frac{2}{3} \).
(a) To show \( fof(x) = x \):
\( fof(x) = f[f(x)] \)
Substitute \( f(x) \) into itself:
\( f\left(\frac{4x+3}{6x-4}\right) = \frac{4\left(\frac{4x+3}{6x-4}\right)+3}{6\left(\frac{4x+3}{6x-4}\right)-4} \)
To simplify, find a common denominator for the numerator and the denominator:
Numerator: \( 4\left(\frac{4x+3}{6x-4}\right)+3 = \frac{4(4x+3) + 3(6x-4)}{6x-4} = \frac{16x+12 + 18x-12}{6x-4} = \frac{34x}{6x-4} \)
Denominator: \( 6\left(\frac{4x+3}{6x-4}\right)-4 = \frac{6(4x+3) - 4(6x-4)}{6x-4} = \frac{24x+18 - 24x+16}{6x-4} = \frac{34}{6x-4} \)
Now, divide the numerator by the denominator:
\( fof(x) = \frac{\frac{34x}{6x-4}}{\frac{34}{6x-4}} \)
Multiply by the reciprocal of the denominator:
\( = \frac{34x}{6x-4} \cdot \frac{6x-4}{34} \)
Cancel out \( (6x-4) \) and \( 34 \):
\( = x \)
Thus, \( fof(x) = x \).
(b) To find the inverse of \( f \):
Let \( y = f(x) \). So, \( y = \frac{4x+3}{6x-4} \).
We need to solve for \( x \) in terms of \( y \).
Multiply both sides by \( (6x-4) \):
\( y(6x-4) = 4x+3 \)
Distribute \( y \):
\( 6xy - 4y = 4x+3 \)
Move all terms with \( x \) to one side and terms without \( x \) to the other side:
\( 6xy - 4x = 4y+3 \)
Factor out \( x \):
\( x(6y - 4) = 4y+3 \)
Divide by \( (6y-4) \) to isolate \( x \):
\( x = \frac{4y+3}{6y-4} \)
This expression for \( x \) in terms of \( y \) is the inverse function, \( f^{-1}(y) \).
So, \( f^{-1}(y) = \frac{4y+3}{6y-4} \).
Notice that this form is identical to the original function \( f(x) \). This means that the inverse of \( f \) is \( f \) itself.
In simple words: We put the function into itself and simplified, which gave us \( x \). This shows that \( fof(x) = x \). To find the inverse, we switched \( y \) and \( x \) and solved for \( x \). The result was the same as the original function, meaning \( f \) is its own inverse.

Exam Tip: When proving \( fof(x) = x \), make sure to perform algebraic simplifications carefully, especially with fractions. To find the inverse, set \( y = f(x) \), then swap \( x \) and \( y \), and solve for \( y \).

Question 5. State with reasons whether each of the following functions has an inverse or not:
(i) \( f : \{1, 2, 3, 4\} \rightarrow \{10\} \) with \( f = \{(1, 10), (2, 10), (3, 10), (4, 10)\} \).
(ii) \( g : \{5, 6, 7, 8\} \rightarrow \{1, 2, 3, 4\} \) with \( g = \{(5, 4), (6, 3), (7, 4), (8, 2)\} \).
(iii) \( h : \{2, 3, 4, 5\} \rightarrow \{7, 9, 11, 13\} \) with \( h = \{(2, 7), (3, 9), (4, 11), (5, 13)\} \).
Answer: A function has an inverse if and only if it is both one-to-one (injective) and onto (surjective).
(i) Given \( f : \{1, 2, 3, 4\} \rightarrow \{10\} \) with \( f = \{(1, 10), (2, 10), (3, 10), (4, 10)\} \).
For this function, all elements in the domain \( \{1, 2, 3, 4\} \) map to the single element \( 10 \) in the codomain. Since multiple distinct elements (1, 2, 3, 4) have the same image (10), the function \( f \) is not one-to-one.

\( \implies \) Therefore, \( f \) does not have an inverse.
(ii) Given \( g : \{5, 6, 7, 8\} \rightarrow \{1, 2, 3, 4\} \) with \( g = \{(5, 4), (6, 3), (7, 4), (8, 2)\} \).
In this function, we observe that \( g(5) = 4 \) and \( g(7) = 4 \). Since two distinct elements (5 and 7) from the domain map to the same image (4), the function \( g \) is not one-to-one.

\( \implies \) Therefore, \( g \) does not have an inverse (it is not invertible).
(iii) Given \( h : \{2, 3, 4, 5\} \rightarrow \{7, 9, 11, 13\} \) with \( h = \{(2, 7), (3, 9), (4, 11), (5, 13)\} \).
To check if \( h \) is one-to-one: Each distinct element in the domain \( \{2, 3, 4, 5\} \) maps to a distinct element in the codomain \( \{7, 9, 11, 13\} \). For example, if \( h(x_1) = h(x_2) \), then \( x_1 = x_2 \). So, \( h \) is one-to-one.
To check if \( h \) is onto: Every element in the codomain \( \{7, 9, 11, 13\} \) has a pre-image in the domain. For example, \( 7 = h(2) \), \( 9 = h(3) \), \( 11 = h(4) \), and \( 13 = h(5) \). So, \( h \) is onto.
Since \( h \) is both one-to-one and onto, it is a bijective function.

\( \implies \) Therefore, \( h \) is invertible and has an inverse.
In simple words: A function needs to be "one-to-one" (each input gives a different output) and "onto" (every possible output is actually produced) to have an inverse. Function (i) and (ii) fail the one-to-one test, so they do not have inverses. Function (iii) passes both tests, so it does have an inverse.

Exam Tip: A function has an inverse if and only if it is a bijection (both injective and surjective). To quickly check injectivity, look for distinct elements in the domain that map to the same image. To check surjectivity, ensure every element in the codomain has at least one pre-image.

Question 6. Show that \( f: [-1, 1] \rightarrow R \), given by \( f(x) = \frac{x}{x+2} \), \( x \neq -2 \) is one-one. Find the inverse of the function \( f: [-1, 1] \rightarrow \text{Range } f \).
Answer: Given the function \( f: [-1, 1] \rightarrow R \) defined by \( f(x) = \frac{x}{x+2} \), where \( x \neq -2 \).
To show that \( f \) is one-one (injective):
Assume \( f(x_1) = f(x_2) \) for some \( x_1, x_2 \in [-1, 1] \).
\( \frac{x_1}{x_1+2} = \frac{x_2}{x_2+2} \)
Cross-multiply:
\( x_1(x_2+2) = x_2(x_1+2) \)
Expand both sides:
\( x_1x_2 + 2x_1 = x_1x_2 + 2x_2 \)
Subtract \( x_1x_2 \) from both sides:
\( 2x_1 = 2x_2 \)
Divide by 2:
\( x_1 = x_2 \)
Since \( f(x_1) = f(x_2) \) implies \( x_1 = x_2 \), the function \( f \) is one-one.
To find the inverse of \( f: [-1, 1] \rightarrow \text{Range } f \):
Let \( y = f(x) \). So, \( y = \frac{x}{x+2} \).
We need to solve for \( x \) in terms of \( y \).
Multiply both sides by \( (x+2) \):
\( y(x+2) = x \)
Distribute \( y \):
\( yx + 2y = x \)
Move all terms with \( x \) to one side and terms without \( x \) to the other side:
\( 2y = x - yx \)
Factor out \( x \):
\( 2y = x(1 - y) \)
Divide by \( (1-y) \) to isolate \( x \):
\( x = \frac{2y}{1-y} \)
This expression for \( x \) in terms of \( y \) is the inverse function, \( f^{-1}(y) \).
So, \( f^{-1}(y) = \frac{2y}{1-y} \).
The domain of \( f^{-1} \) is the range of \( f \). We need to determine the range of \( f \).
If \( x \in [-1, 1] \), then \( -1 \le x \le 1 \).
Since \( f(x) = 1 - \frac{2}{x+2} \), as \( x \) increases from \( -1 \) to \( 1 \):
When \( x = -1 \), \( f(-1) = \frac{-1}{-1+2} = \frac{-1}{1} = -1 \).
When \( x = 1 \), \( f(1) = \frac{1}{1+2} = \frac{1}{3} \).
The function is increasing on its domain \( [-1, 1] \). Therefore, the range of \( f \) is \( [-1, \frac{1}{3}] \).
Thus, the inverse function is \( f^{-1}: [-1, \frac{1}{3}] \rightarrow [-1, 1] \) given by \( f^{-1}(y) = \frac{2y}{1-y} \).
In simple words: To prove it's "one-one," we assumed two inputs gave the same output and showed that the inputs must be identical. To find the inverse, we set \( y \) equal to the function, then swapped \( x \) and \( y \) and solved for \( x \). We then found the range of the original function to define the domain of the inverse.

Exam Tip: For showing a function is one-one, always start with \( f(x_1) = f(x_2) \) and algebraically simplify to \( x_1 = x_2 \). When finding the inverse, make sure to state its domain, which is the range of the original function.

Question 7. Consider \( f : R \rightarrow R \) given by \( f(x) = 4x + 3 \). Show that \( f \) is invertible. Find the inverse of \( f \).
Answer: Given the function \( f : R \rightarrow R \) defined by \( f(x) = 4x + 3 \).
To show that \( f \) is invertible, we must show it is both one-one (injective) and onto (surjective).
To prove \( f \) is one-one:
Assume \( f(x_1) = f(x_2) \) for any \( x_1, x_2 \in R \).
\( 4x_1 + 3 = 4x_2 + 3 \)
Subtract 3 from both sides:
\( 4x_1 = 4x_2 \)
Divide by 4:
\( x_1 = x_2 \)
Since \( f(x_1) = f(x_2) \) implies \( x_1 = x_2 \), \( f \) is a one-one function.
To prove \( f \) is onto:
Let \( y \) be an arbitrary element in the codomain \( R \). We need to find an \( x \in R \) such that \( f(x) = y \).
\( y = 4x + 3 \)
Subtract 3 from both sides:
\( y - 3 = 4x \)
Divide by 4:
\( x = \frac{y-3}{4} \)
Since \( y \in R \), \( (y-3)/4 \) is always a real number. Thus, for every \( y \) in the codomain, there exists an \( x \) in the domain such that \( f(x) = y \). So, \( f \) is an onto function.
Since \( f \) is both one-one and onto, it is invertible.
To find the inverse of \( f \):
The expression we found for \( x \) in terms of \( y \) is the inverse function. So, \( f^{-1}(y) = \frac{y-3}{4} \).
We can also write it using \( x \) as the variable: \( f^{-1}(x) = \frac{x-3}{4} \).
In simple words: A function is invertible if it's both "one-to-one" (each output comes from only one input) and "onto" (all possible outputs are reached). We proved both conditions. Then, to get the inverse, we set \( y \) equal to the function and solved for \( x \), which gives us the inverse formula.

Exam Tip: To show invertibility, you must explicitly prove both one-one and onto conditions. The steps for finding the inverse often follow directly from proving the onto condition.

Question 8. Consider \( f : R_+ \rightarrow [4, \infty) \) given by \( f(x) = x^2 + 4 \). Show that \( f \) is invertible with the inverse of \( f^{-1} \) of \( f \) given by \( f^{-1}(y) = \sqrt{y-4} \), where \( R_+ \) is the set of all non-negative real numbers.
Answer: Given the function \( f : R_+ \rightarrow [4, \infty) \) defined by \( f(x) = x^2 + 4 \), where \( R_+ = [0, \infty) \) is the set of all non-negative real numbers.
To show that \( f \) is invertible, we must show it is both one-one (injective) and onto (surjective).
To prove \( f \) is one-one:
Assume \( f(x_1) = f(x_2) \) for any \( x_1, x_2 \in R_+ \).
\( x_1^2 + 4 = x_2^2 + 4 \)
Subtract 4 from both sides:
\( x_1^2 = x_2^2 \)
Taking the square root of both sides, we get \( x_1 = \pm x_2 \).
However, since the domain of \( f \) is \( R_+ = [0, \infty) \), both \( x_1 \) and \( x_2 \) must be non-negative. Therefore, we must have \( x_1 = x_2 \).
Since \( f(x_1) = f(x_2) \) implies \( x_1 = x_2 \), \( f \) is a one-one function.
To prove \( f \) is onto:
Let \( y \) be an arbitrary element in the codomain \( [4, \infty) \). We need to find an \( x \in R_+ \) such that \( f(x) = y \).
\( y = x^2 + 4 \)
Subtract 4 from both sides:
\( y - 4 = x^2 \)
Take the square root of both sides:
\( x = \sqrt{y-4} \) (We take the positive square root because \( x \in R_+ \)).
For \( y \in [4, \infty) \), we know that \( y-4 \ge 0 \), so \( \sqrt{y-4} \) is always a real and non-negative number. Thus, for every \( y \) in the codomain, there exists an \( x \) in the domain \( R_+ \) such that \( f(x) = y \). So, \( f \) is an onto function.
Since \( f \) is both one-one and onto, it is invertible.
To find the inverse of \( f \):
The expression we found for \( x \) in terms of \( y \) is the inverse function. So, \( f^{-1}(y) = \sqrt{y-4} \).
In simple words: We showed that the function is "one-to-one" because positive numbers squared lead to unique results. We also showed it's "onto" because for any output \( y \) in the given range, we can find a matching input \( x \). This means it is invertible, and its inverse is found by solving for \( x \) in terms of \( y \), which resulted in \( \sqrt{y-4} \).

Exam Tip: Pay close attention to the domain and codomain when dealing with functions involving squares or square roots, as they affect whether a function is one-one or onto and the form of its inverse.

Question 9. Consider \( f : R_+ \rightarrow [-5, \infty) \) given by \( f(x) = 9x^2 + 6x - 5 \). Show that \( f \) is invertible with \( f^{-1}(y) = \frac{\sqrt{y+6}-1}{3} \).
Answer: Given the function \( f : R_+ \rightarrow [-5, \infty) \) defined by \( f(x) = 9x^2 + 6x - 5 \), where \( R_+ = [0, \infty) \).
To show \( f \) is invertible, we must show it is one-one and onto.
Let's first rewrite \( f(x) \) by completing the square:
\( f(x) = 9x^2 + 6x - 5 \)
\( = (9x^2 + 6x + 1) - 1 - 5 \)
\( = (3x+1)^2 - 6 \)
To prove \( f \) is one-one:
Assume \( f(x_1) = f(x_2) \) for \( x_1, x_2 \in R_+ \).
\( (3x_1+1)^2 - 6 = (3x_2+1)^2 - 6 \)
\( (3x_1+1)^2 = (3x_2+1)^2 \)
Taking the square root of both sides:
\( 3x_1+1 = \pm (3x_2+1) \)
Since \( x_1, x_2 \in R_+ = [0, \infty) \), it means \( 3x_1+1 \ge 1 \) and \( 3x_2+1 \ge 1 \). Therefore, both terms are positive, and we must take the positive root:
\( 3x_1+1 = 3x_2+1 \)
\( 3x_1 = 3x_2 \)
\( x_1 = x_2 \)
So, \( f \) is one-one.
To prove \( f \) is onto:
Let \( y \) be an arbitrary element in the codomain \( [-5, \infty) \). We need to find an \( x \in R_+ \) such that \( f(x) = y \).
\( y = (3x+1)^2 - 6 \)
Add 6 to both sides:
\( y+6 = (3x+1)^2 \)
Take the square root of both sides:
\( \sqrt{y+6} = 3x+1 \) (Since \( 3x+1 \ge 1 \), we take the positive root).
Subtract 1:
\( \sqrt{y+6} - 1 = 3x \)
Divide by 3:
\( x = \frac{\sqrt{y+6}-1}{3} \)
For \( y \in [-5, \infty) \), \( y+6 \ge 1 \), so \( \sqrt{y+6} \ge 1 \). This means \( \sqrt{y+6}-1 \ge 0 \), and thus \( x = \frac{\sqrt{y+6}-1}{3} \ge 0 \). This value of \( x \) is in \( R_+ \).
Therefore, for every \( y \) in the codomain, there exists an \( x \) in the domain \( R_+ \) such that \( f(x) = y \). So, \( f \) is onto.
Since \( f \) is both one-one and onto, it is invertible.
The inverse function is \( f^{-1}(y) = \frac{\sqrt{y+6}-1}{3} \).
Let's verify \( gof(x) \) and \( fog(y) \). Let \( g(y) = \frac{\sqrt{y+6}-1}{3} \).
\( gof(x) = g[f(x)] = g[(3x+1)^2 - 6] \)
\( = \frac{\sqrt{((3x+1)^2 - 6) + 6}-1}{3} \)
\( = \frac{\sqrt{(3x+1)^2}-1}{3} \)
Since \( x \in R_+ \), \( 3x+1 \) is positive, so \( \sqrt{(3x+1)^2} = 3x+1 \).
\( = \frac{(3x+1)-1}{3} = \frac{3x}{3} = x \).

\( \implies \) So, \( gof(x) = x \).
\( fog(y) = f[g(y)] = f\left(\frac{\sqrt{y+6}-1}{3}\right) \)
\( = \left(3\left(\frac{\sqrt{y+6}-1}{3}\right)+1\right)^2 - 6 \)
\( = \left((\sqrt{y+6}-1)+1\right)^2 - 6 \)
\( = (\sqrt{y+6})^2 - 6 \)
\( = (y+6) - 6 = y \).

\( \implies \) So, \( fog(y) = y \).
Since \( gof(x) = x \) and \( fog(y) = y \), \( f \) is invertible and \( f^{-1}(y) = g(y) = \frac{\sqrt{y+6}-1}{3} \).
In simple words: First, we completed the square for \( f(x) \). Then, we showed the function is "one-to-one" by proving that if outputs are the same, inputs must be the same. We proved it's "onto" by finding an \( x \) for every \( y \) in the codomain. This confirms it is invertible. The inverse function was derived by solving \( y = f(x) \) for \( x \).

Exam Tip: For quadratic functions, completing the square is a crucial step to easily determine injectivity and surjectivity, and to derive the inverse function accurately.

Question 10. Let \( f : X \rightarrow Y \) be an invertible function. Show that \( f \) has a unique inverse.
Answer: Let \( f : X \rightarrow Y \) be an invertible function. By definition, this means \( f \) is both one-to-one and onto.
Assume, for contradiction, that \( f \) has two inverses, say \( g_1: Y \rightarrow X \) and \( g_2: Y \rightarrow X \).
Since \( g_1 \) is an inverse of \( f \), we have:
\( fog_1(y) = I_Y \) (the identity function on Y) for all \( y \in Y \).
\( g_1of(x) = I_X \) (the identity function on X) for all \( x \in X \).
Similarly, since \( g_2 \) is an inverse of \( f \), we have:
\( fog_2(y) = I_Y \) for all \( y \in Y \).
\( g_2of(x) = I_X \) for all \( x \in X \).
Now, consider \( g_1(y) \). We know that \( I_X(g_1(y)) = g_1(y) \).
We can write \( I_X \) as \( g_2of \). So, \( g_1(y) = (g_2of)(g_1(y)) \).
Using the definition of composition, \( g_1(y) = g_2(f(g_1(y))) \).
Since \( fog_1(y) = I_Y(y) = y \), we have \( f(g_1(y)) = y \).
Substitute this back:
\( g_1(y) = g_2(y) \)
Since this holds for all \( y \in Y \), it means \( g_1 = g_2 \).
Therefore, an invertible function has a unique inverse.
In simple words: We started by assuming a function could have two different inverses. Then, using the properties of identity functions and composition, we showed that these two inverses must actually be the same. This proves that a function can only have one unique inverse.

Exam Tip: The proof for the uniqueness of an inverse function is a standard method: assume two inverses exist and then show they must be identical by using the definition of inverse functions and identity functions.

Question 11. Consider \( f : \{1, 2, 3\} \rightarrow \{a, b, c\} \) given by \( f(1) = a, f(2) = b, f(3) = c \). Find \( f^{-1} \) and show that \( (f^{-1})^{-1} = f \).
Answer: Given the function \( f : \{1, 2, 3\} \rightarrow \{a, b, c\} \) defined by \( f(1) = a, f(2) = b, f(3) = c \).
To find \( f^{-1} \):
The inverse function \( f^{-1} \) maps elements from the codomain of \( f \) back to its domain. If \( f(x) = y \), then \( f^{-1}(y) = x \).
From \( f(1) = a \), we get \( f^{-1}(a) = 1 \).
From \( f(2) = b \), we get \( f^{-1}(b) = 2 \).
From \( f(3) = c \), we get \( f^{-1}(c) = 3 \).
So, \( f^{-1} : \{a, b, c\} \rightarrow \{1, 2, 3\} \) is given by \( f^{-1} = \{(a, 1), (b, 2), (c, 3)\} \).
To show that \( (f^{-1})^{-1} = f \):
Now, let's find the inverse of \( f^{-1} \). We apply the same principle: if \( f^{-1}(y) = x \), then \( (f^{-1})^{-1}(x) = y \).
From \( f^{-1}(a) = 1 \), we get \( (f^{-1})^{-1}(1) = a \).
From \( f^{-1}(b) = 2 \), we get \( (f^{-1})^{-1}(2) = b \).
From \( f^{-1}(c) = 3 \), we get \( (f^{-1})^{-1}(3) = c \).
So, \( (f^{-1})^{-1} : \{1, 2, 3\} \rightarrow \{a, b, c\} \) is given by \( (f^{-1})^{-1} = \{(1, a), (2, b), (3, c)\} \).
By comparing this with the original function \( f \), we can see that \( (f^{-1})^{-1} \) is exactly the same as \( f \).
Thus, \( (f^{-1})^{-1} = f \).
In simple words: First, we found the inverse function by reversing the input-output pairs. Then, we found the inverse of that inverse function, which essentially reversed the pairs back again. We found that reversing the inverse brings us back to the original function.

Exam Tip: The concept that the inverse of an inverse is the original function, i.e., \( (f^{-1})^{-1} = f \), is fundamental in function theory. Make sure to clearly list the mapping for \( f \), then \( f^{-1} \), and finally \( (f^{-1})^{-1} \) to demonstrate this property.

Question 12. Let \( f : X \rightarrow Y \) be an invertible function. Show that the inverse of \( f^{-1} \) is \( f \), i.e., \( (f^{-1})^{-1} = f \).
Answer: Let \( f : X \rightarrow Y \) be an invertible function. This means there exists a function \( g : Y \rightarrow X \) such that \( gof = I_X \) and \( fog = I_Y \). By definition, \( g = f^{-1} \).
Now, we want to find the inverse of \( f^{-1} \), which is \( (f^{-1})^{-1} \). Let's call this function \( h \). So, \( h = (f^{-1})^{-1} \).
By the definition of an inverse, \( h \) is the inverse of \( f^{-1} \) if:
1. \( h \circ f^{-1} = I_Y \) (identity on Y)
2. \( f^{-1} \circ h = I_X \) (identity on X)
We know that \( f \circ f^{-1} = I_Y \).
Comparing this with condition 1, we can see that \( h \) must be \( f \). So, \( (f^{-1})^{-1} = f \).
Let's also check condition 2. We know that \( f^{-1} \circ f = I_X \).
Comparing this with condition 2, we can also see that \( h \) must be \( f \).
Thus, the inverse of \( f^{-1} \) is \( f \), i.e., \( (f^{-1})^{-1} = f \).
In simple words: If a function \( f \) has an inverse \( f^{-1} \), then applying the inverse function and then taking its inverse again leads you back to the original function \( f \). This is because the inverse operation essentially undoes the original function, and undoing the undo operation brings you back to the start.

Exam Tip: This proof relies on the definitions of invertible functions and identity functions. Clearly state what it means for \( g \) to be the inverse of \( f \), and then apply the same definition to find the inverse of \( g \).

Question 13. If \( f : R \rightarrow R \) be given by \( f(x) = (3-x^3)^{1/3} \), when \( fof(x) \) is
(A) \( x^{1/3} \)
(B) \( x^3 \)
(C) \( x \)
(D) \( 3-x^3 \)
Answer: (C) x
Given \( f(x) = (3-x^3)^{1/3} \).
We need to find \( fof(x) \), which is \( f[f(x)] \).
Substitute \( f(x) \) into itself:
\( fof(x) = f[(3-x^3)^{1/3}] \)
Now replace \( x \) in the original function \( f(x) \) with \( (3-x^3)^{1/3} \):
\( = \left(3 - \left((3-x^3)^{1/3}\right)^3\right)^{1/3} \)
First, simplify the inner cube: \( \left((3-x^3)^{1/3}\right)^3 = (3-x^3)^{(1/3) \cdot 3} = (3-x^3)^1 = 3-x^3 \).
Substitute this back:
\( = (3 - (3-x^3))^{1/3} \)
Distribute the negative sign:
\( = (3 - 3 + x^3)^{1/3} \)
Simplify:
\( = (x^3)^{1/3} \)
Using power rules, \( (x^3)^{1/3} = x^{(3 \cdot 1/3)} = x^1 = x \).
So, \( fof(x) = x \).
The correct option is (C).
In simple words: To solve for \( fof(x) \), we took the function \( f(x) \) and substituted it back into itself. After carefully simplifying the powers and terms, the expression simplifies down to just \( x \).

Exam Tip: When evaluating composite functions like \( fof(x) \), substitute the entire function for \( x \) in the original expression. Be careful with exponent rules, especially when cubing a term raised to the power of one-third.

Question 14. Let \( f: R - \left\{-\frac{4}{3}\right\} \rightarrow R \) be a function, defined as \( f(x) = \frac{4x}{3x+4} \), \( x \neq -\frac{4}{5} \). The inverse of \( f \) is map \( g: \text{Range } f \rightarrow R \) is given by
(A) \( g(y) = \frac{3y}{3-4y} \)
(B) \( g(y) = \frac{4y}{4-3y} \)
(C) \( g(y) = \frac{4y}{3-4y} \)
(D) \( g(y) = \frac{3y}{4-3y} \)
Answer: (B) \( g(y) = \frac{4y}{4-3y} \)
Given the function \( f(x) = \frac{4x}{3x+4} \).
To find the inverse function, we let \( y = f(x) \) and solve for \( x \) in terms of \( y \).
\( y = \frac{4x}{3x+4} \)
Multiply both sides by \( (3x+4) \):
\( y(3x+4) = 4x \)
Distribute \( y \):
\( 3xy + 4y = 4x \)
Move all terms containing \( x \) to one side and terms not containing \( x \) to the other side:
\( 4y = 4x - 3xy \)
Factor out \( x \) from the right side:
\( 4y = x(4 - 3y) \)
Divide by \( (4-3y) \) to isolate \( x \):
\( x = \frac{4y}{4-3y} \)
This expression for \( x \) in terms of \( y \) is the inverse function, \( f^{-1}(y) \), which is also denoted as \( g(y) \).
So, \( g(y) = \frac{4y}{4-3y} \).
This matches option (B).
In simple words: To find the inverse function, we set \( y \) equal to \( f(x) \) and then rearranged the equation to solve for \( x \) in terms of \( y \). This new expression gives us the formula for the inverse function, \( g(y) \).

Exam Tip: Finding the inverse of a rational function involves cross-multiplication and careful algebraic manipulation to isolate \( x \). Remember to factor out \( x \) correctly when it appears in multiple terms.