OP Malhotra Class 12 Maths Solutions Chapter 2 Functions Exercise 2 (C)

S Chand Class 12 ICSE Maths Solutions Chapter 2 Functions Ex 2(c)

 

Question 1.
(i) If \( f : R \rightarrow R \) is defined by \( f(x) = 2x + 3 \), then find \( f^{-1}(x) \).
(ii) If the function \( f : R \rightarrow R \), defined by \( f(x) = 3x - 4 \) is invertible, then find \( f^{-1} \).
(iii) If \( f : R \rightarrow R \) defined by \( f(x) = \frac { 3x+5 }{ 2 } \) is an invertible function, then find \( f^{-1} \).
Answer:
(i) Given \( f : R \rightarrow R \) defined by \( f(x) = 2x + 3 \).
For Injectivity (one-one function):
Assume \( f(x) = f(y) \) for any \( x, y \in R \).
\( \implies 2x + 3 = 2y + 3 \)
\( \implies 2x = 2y \)
\( \implies x = y \)
So, \( f \) is an injective (one-one) function.

For Surjectivity (onto function):
Let \( y \in R \) be any arbitrary element in the codomain. We need to find \( x \in R \) such that \( f(x) = y \).
\( 2x + 3 = y \)
\( \implies 2x = y - 3 \)
\( \implies x = \frac{y-3}{2} \)
Since \( y \in R \), \( \frac{y-3}{2} \) will also be in \( R \). So, for every \( y \) in the codomain, there is an \( x \) in the domain.
Thus, \( f \) is a surjective (onto) function.

Since \( f \) is both one-one and onto, it is a bijective function, which means its inverse \( f^{-1} \) exists.

To find \( f^{-1}(x) \):
Let \( y = f(x) \). Then \( x = f^{-1}(y) \).
\( y = 2x + 3 \)
\( \implies 2x = y - 3 \)
\( \implies x = \frac{y-3}{2} \)
So, \( f^{-1}(y) = \frac{y-3}{2} \).
Replacing \( y \) with \( x \), we get \( f^{-1}(x) = \frac{x-3}{2} \).

(ii) Given \( f : R \rightarrow R \) defined by \( f(x) = 3x - 4 \).
For Injectivity (one-one function):
Assume \( f(x) = f(y) \) for any \( x, y \in R \).
\( \implies 3x - 4 = 3y - 4 \)
\( \implies 3x = 3y \)
\( \implies x = y \)
So, \( f \) is an injective (one-one) function.

For Surjectivity (onto function):
Let \( y \in R \) be any arbitrary element in the codomain. We need to find \( x \in R \) such that \( f(x) = y \).
\( 3x - 4 = y \)
\( \implies 3x = y + 4 \)
\( \implies x = \frac{y+4}{3} \)
Since \( y \in R \), \( \frac{y+4}{3} \) will also be in \( R \). So, for every \( y \) in the codomain, there is an \( x \) in the domain.
Thus, \( f \) is a surjective (onto) function.

Since \( f \) is both one-one and onto, its inverse \( f^{-1} \) exists.

To find \( f^{-1}(x) \):
Let \( y = f(x) \). Then \( x = f^{-1}(y) \).
\( y = 3x - 4 \)
\( \implies 3x = y + 4 \)
\( \implies x = \frac{y+4}{3} \)
So, \( f^{-1}(y) = \frac{y+4}{3} \).
Replacing \( y \) with \( x \), we get \( f^{-1}(x) = \frac{x+4}{3} \).

(iii) Given \( f : R \rightarrow R \) defined by \( f(x) = \frac{3x+5}{2} \).
For Injectivity (one-one function):
Assume \( f(x) = f(y) \) for any \( x, y \in R \).
\( \implies \frac{3x+5}{2} = \frac{3y+5}{2} \)
\( \implies 3x + 5 = 3y + 5 \)
\( \implies 3x = 3y \)
\( \implies x = y \)
So, \( f \) is an injective (one-one) function.

For Surjectivity (onto function):
Let \( y \in R \) be any arbitrary element in the codomain. We need to find \( x \in R \) such that \( f(x) = y \).
\( \frac{3x+5}{2} = y \)
\( \implies 3x + 5 = 2y \)
\( \implies 3x = 2y - 5 \)
\( \implies x = \frac{2y-5}{3} \)
Since \( y \in R \), \( \frac{2y-5}{3} \) will also be in \( R \). So, for every \( y \) in the codomain, there is an \( x \) in the domain.
Thus, \( f \) is a surjective (onto) function.

Since \( f \) is both one-one and onto, its inverse \( f^{-1} \) exists.

To find \( f^{-1}(x) \):
Let \( y = f(x) \). Then \( x = f^{-1}(y) \).
\( y = \frac{3x+5}{2} \)
\( \implies 3x + 5 = 2y \)
\( \implies 3x = 2y - 5 \)
\( \implies x = \frac{2y-5}{3} \)
So, \( f^{-1}(y) = \frac{2y-5}{3} \).
Replacing \( y \) with \( x \), we get \( f^{-1}(x) = \frac{2x-5}{3} \).
In simple words: For each function, we first check if it's one-to-one (different inputs give different outputs) and onto (every output value is reached). If it is, an inverse function exists. To find the inverse, we swap \( x \) and \( y \) in the function's equation and then solve for \( y \).

🎯 Exam Tip: Always verify both injectivity (one-one) and surjectivity (onto) to prove a function is invertible before attempting to find its inverse. This ensures the inverse function is well-defined.

 

Question 2. Let the function \( f \) which is invertible be defined by \( f(x) = \frac{2x+1}{1-3x} \), then show that \( f^{-1}(x) = \frac{x-1}{3x+2} \).
Answer:
Given that \( f \) is an invertible function defined by \( f(x) = \frac{2x+1}{1-3x} \). Since it's invertible, it must be both one-one and onto.

To find the inverse function \( f^{-1}(x) \), we follow these steps:
1. Let \( y = f(x) \).
\( y = \frac{2x+1}{1-3x} \)
2. Solve for \( x \) in terms of \( y \).
\( y(1-3x) = 2x+1 \)
\( \implies y - 3xy = 2x + 1 \)
\( \implies y - 1 = 2x + 3xy \)
\( \implies y - 1 = x(2 + 3y) \)
\( \implies x = \frac{y-1}{2+3y} \)
3. Replace \( x \) with \( f^{-1}(y) \).
\( f^{-1}(y) = \frac{y-1}{3y+2} \)
4. Replace \( y \) with \( x \) to get \( f^{-1}(x) \).
\( f^{-1}(x) = \frac{x-1}{3x+2} \)
This matches the expression given in the question, thus it is shown. This process helps us reverse the function's mapping.
In simple words: To find the inverse of a function, set the function equal to \( y \). Then, switch \( x \) and \( y \) around and solve for \( y \). The new \( y \) value is the inverse function.

🎯 Exam Tip: Always double-check your algebraic manipulation steps when solving for \( x \) to avoid errors. It's helpful to clearly state each step in the derivation of the inverse function.

 

Question 3. If \( f : R \rightarrow R \) is invertible and is defined as \( f(x) = (1 - x)^{1/3} \), then find \( f^{-1}(x) \).
Answer:
Given \( f : R \rightarrow R \) is an invertible function defined by \( f(x) = (1 - x)^{1/3} \).
Since \( f \) is invertible, its inverse \( f^{-1} \) exists.

To find \( f^{-1}(x) \):
1. Let \( y = f(x) \).
\( y = (1 - x)^{1/3} \)
2. Solve for \( x \) in terms of \( y \).
To remove the cube root, cube both sides:
\( y^3 = ((1 - x)^{1/3})^3 \)
\( \implies y^3 = 1 - x \)
\( \implies x = 1 - y^3 \)
3. Replace \( x \) with \( f^{-1}(y) \).
\( f^{-1}(y) = 1 - y^3 \)
4. Replace \( y \) with \( x \) to get \( f^{-1}(x) \).
\( f^{-1}(x) = 1 - x^3 \)
The inverse function essentially "undoes" what the original function did, mapping outputs back to their original inputs.
In simple words: When you have a function with a power like \( 1/3 \) (which is a cube root), to find its inverse, set the function equal to \( y \), then cube both sides to get rid of the \( 1/3 \) power. After that, rearrange to find \( x \) in terms of \( y \), and that will be your inverse.

🎯 Exam Tip: Remember that raising an expression to the power of \( n \) is the inverse operation of taking the \( n \)-th root. Use this property to simplify and solve for \( x \) in inverse function problems.

 

Question 4. If \( f(x) = [4 - (x - 7)^3]^{1/5} \), is a real invertible function, then find \( f^{-1}(x) \).
Answer:
Given that \( f(x) = [4 - (x - 7)^3]^{1/5} \) is a real invertible function. This means \( f \) is one-one and onto.

To find \( f^{-1}(x) \):
1. Let \( y = f(x) \).
\( y = [4 - (x - 7)^3]^{1/5} \)
2. Solve for \( x \) in terms of \( y \).
Raise both sides to the power of 5 to remove the \( 1/5 \) exponent:
\( y^5 = ([4 - (x - 7)^3]^{1/5})^5 \)
\( \implies y^5 = 4 - (x - 7)^3 \)
Rearrange the terms to isolate \( (x - 7)^3 \):
\( \implies (x - 7)^3 = 4 - y^5 \)
Take the cube root of both sides:
\( \implies x - 7 = \sqrt[3]{4 - y^5} \)
\( \implies x = 7 + \sqrt[3]{4 - y^5} \)
3. Replace \( x \) with \( f^{-1}(y) \).
\( f^{-1}(y) = 7 + \sqrt[3]{4 - y^5} \)
4. Replace \( y \) with \( x \) to get \( f^{-1}(x) \).
\( f^{-1}(x) = 7 + \sqrt[3]{4 - x^5} \)
This systematic approach ensures each step correctly reverses the original function's operations.
In simple words: To find the inverse, first set the function equal to \( y \). Then, do the opposite operations in reverse order. If there's a power of \( 1/5 \), raise both sides to the power of 5. If there's a cube, take the cube root. Keep rearranging until \( x \) is alone on one side, then swap \( x \) and \( y \).

🎯 Exam Tip: When dealing with multiple nested operations (like powers, subtractions, and roots), carefully apply inverse operations in the correct order, moving from the outermost operation inwards.

 

Question 5. If \( f : N \rightarrow Y \) is a function defined by \( f(x) = 4x + 3 \) where \( Y = \{y \in N : y = 4x + 3 \text{ for some } x \in N\} \), then, find the inverse of \( f \).
Answer:
Given \( f : N \rightarrow Y \) is defined by \( f(x) = 4x + 3 \), where \( Y = \{y \in N : y = 4x + 3 \text{ for some } x \in N\} \). This means \( Y \) is the range of \( f \).
The domain is \( N \) (natural numbers: 1, 2, 3, ...). The range \( Y \) would be \( \{4(1)+3, 4(2)+3, 4(3)+3, \dots\} = \{7, 11, 15, 19, \dots\} \).

For Injectivity (one-one function):
Assume \( f(x) = f(y) \) for any \( x, y \in N \).
\( \implies 4x + 3 = 4y + 3 \)
\( \implies 4x = 4y \)
\( \implies x = y \)
So, \( f \) is an injective (one-one) function.

For Surjectivity (onto function):
Since \( Y \) is defined as the set of all values \( 4x+3 \) for \( x \in N \), the codomain is equal to the range. By definition, if the codomain equals the range, the function is surjective (onto).
Thus, \( f \) is a surjective (onto) function.

Since \( f \) is both one-one and onto, its inverse \( f^{-1} \) exists.

To find \( f^{-1}(x) \):
1. Let \( y = f(x) \).
\( y = 4x + 3 \)
2. Solve for \( x \) in terms of \( y \).
\( y - 3 = 4x \)
\( \implies x = \frac{y-3}{4} \)
3. Replace \( x \) with \( f^{-1}(y) \).
\( f^{-1}(y) = \frac{y-3}{4} \)
4. Replace \( y \) with \( x \) to get \( f^{-1}(x) \).
\( f^{-1}(x) = \frac{x-3}{4} \)
This inverse function maps elements from the range \( Y \) back to the natural numbers \( N \).
In simple words: This function takes a natural number, multiplies it by 4, and adds 3. To reverse this, we subtract 3 and then divide by 4. This new rule is the inverse function.

🎯 Exam Tip: When the codomain is explicitly defined as the range, you can directly conclude that the function is surjective without further proof. This simplifies the checking process.

 

Question 6. Show that the inverse function of \( f : R \rightarrow R \) defined by \( f(x) = x^3 \) is \( f^{-1}(x) = \sqrt[3]{x} \).
Answer:
Given \( f : R \rightarrow R \) defined by \( f(x) = x^3 \).

For Injectivity (one-one function):
Assume \( f(x) = f(y) \) for any \( x, y \in R \).
\( \implies x^3 = y^3 \)
\( \implies x^3 - y^3 = 0 \)
Using the difference of cubes formula \( a^3 - b^3 = (a-b)(a^2 + ab + b^2) \):
\( \implies (x-y)(x^2 + xy + y^2) = 0 \)
For real numbers, \( x^2 + xy + y^2 = (x + \frac{y}{2})^2 + \frac{3y^2}{4} \), which is always greater than or equal to 0. It is 0 only if \( x=0 \) and \( y=0 \).
If \( x^2 + xy + y^2 = 0 \), then \( x=y=0 \). If \( x \) and \( y \) are not both zero, then \( x^2 + xy + y^2 > 0 \).
Therefore, for the product to be zero, we must have \( x-y = 0 \).
\( \implies x = y \)
So, \( f \) is an injective (one-one) function.

For Surjectivity (onto function):
Let \( y \in R \) be any arbitrary element in the codomain. We need to find \( x \in R \) such that \( f(x) = y \).
\( x^3 = y \)
\( \implies x = \sqrt[3]{y} \)
Since for every real number \( y \), there exists a real cube root \( \sqrt[3]{y} \), for every \( y \in R \) in the codomain, there is an \( x = \sqrt[3]{y} \in R \) in the domain.
Thus, \( f \) is a surjective (onto) function.

Since \( f \) is both one-one and onto, its inverse \( f^{-1} \) exists.

To find \( f^{-1}(x) \):
1. Let \( y = f(x) \).
\( y = x^3 \)
2. Solve for \( x \) in terms of \( y \).
Take the cube root of both sides:
\( \implies x = \sqrt[3]{y} \)
3. Replace \( x \) with \( f^{-1}(y) \).
\( f^{-1}(y) = \sqrt[3]{y} \)
4. Replace \( y \) with \( x \) to get \( f^{-1}(x) \).
\( f^{-1}(x) = \sqrt[3]{x} \)
This confirms that the inverse of cubing a number is taking its cube root. For example, \( f(2)=8 \), and \( f^{-1}(8)=\sqrt[3]{8}=2 \).
In simple words: The function \( f(x) = x^3 \) means it cubes any number you give it. To find the inverse, we need a function that "uncubes" a number. That function is the cube root, written as \( \sqrt[3]{x} \). We showed this by proving the original function is one-to-one and onto, and then by swapping \( x \) and \( y \) and solving.

🎯 Exam Tip: When proving injectivity for \( x^3 \), ensure you explain why \( x^2 + xy + y^2 \) cannot be zero unless \( x=y=0 \), or simply state that for real numbers, \( x^3 = y^3 \implies x=y \).

 

Question 7. If \( f : R \rightarrow R \) is defined by \( f(x) = |x| \), then
(a) \( f^{-1}(x) = -x \)
(b) \( f^{-1}(x) = \frac{1}{|x|} \)
(c) \( f^{-1}(x) \) does not exist
(d) \( f^{-1}(x) = \frac{1}{x} \)
Answer: (c) \( f^{-1}(x) \) does not exist
In simple words: The function \( f(x) = |x| \) turns both positive and negative numbers into their positive version. Because two different input numbers can give the same output (like \( |1| = 1 \) and \( |-1| = 1 \)), it's not a one-to-one function. For an inverse to exist, the function must be one-to-one, so no inverse can be found here.

🎯 Exam Tip: For an inverse function to exist, the original function must be both one-to-one (injective) and onto (surjective). If either condition fails, the inverse does not exist.

 

Question 8. Consider \( f : R \rightarrow R \) given by \( f(x) = 4x + 3 \). Show that \( f \) is invertible. Find the inverse of \( f \).
Answer:
Given \( f : R \rightarrow R \) defined by \( f(x) = 4x + 3 \).

For Injectivity (one-one function):
Assume \( f(x) = f(y) \) for any \( x, y \in R \).
\( \implies 4x + 3 = 4y + 3 \)
\( \implies 4x = 4y \)
\( \implies x = y \)
So, \( f \) is an injective (one-one) function.

For Surjectivity (onto function):
Let \( y \in R \) be any arbitrary element in the codomain. We need to find \( x \in R \) such that \( f(x) = y \).
\( 4x + 3 = y \)
\( \implies 4x = y - 3 \)
\( \implies x = \frac{y-3}{4} \)
Since \( y \in R \), \( \frac{y-3}{4} \) will also be in \( R \). So, for every \( y \) in the codomain, there is an \( x \) in the domain.
Thus, \( f \) is a surjective (onto) function.

Since \( f \) is both one-one and onto, \( f \) is an invertible function.

To find \( f^{-1}(x) \):
1. Let \( y = f(x) \).
\( y = 4x + 3 \)
2. Solve for \( x \) in terms of \( y \).
\( y - 3 = 4x \)
\( \implies x = \frac{y-3}{4} \)
3. Replace \( x \) with \( f^{-1}(y) \).
\( f^{-1}(y) = \frac{y-3}{4} \)
4. Replace \( y \) with \( x \) to get \( f^{-1}(x) \).
\( f^{-1}(x) = \frac{x-3}{4} \)
This linear function is a good example of a bijective mapping from real numbers to real numbers. Its inverse also turns out to be a linear function. The inverse function effectively reverses the operations of multiplication by 4 and addition of 3.
In simple words: We showed that \( f(x) = 4x + 3 \) is invertible because every output comes from a unique input, and every possible output is covered. To find its inverse, we swap \( x \) and \( y \) and solve, which gives us \( f^{-1}(x) = \frac{x-3}{4} \).

🎯 Exam Tip: When proving invertibility, clearly state the definitions of one-one and onto functions and show how \( f(x) = f(y) \implies x=y \) and how for every \( y \), an \( x \) exists.

 

Question 9. Show that \( f: [-1, 1] \rightarrow R \), given by \( f(x) = \frac{x}{x+2} \) is one-one. Find the inverse of the function \( f: [-1, 1] \rightarrow \text{Range } f \).
Answer:
Given \( f: [-1, 1] \rightarrow R \), defined by \( f(x) = \frac{x}{x+2} \), where \( x \neq -2 \).

For Injectivity (one-one function):
Assume \( f(x) = f(y) \) for any \( x, y \in [-1, 1] \).
\( \implies \frac{x}{x+2} = \frac{y}{y+2} \)
Cross-multiply:
\( \implies x(y+2) = y(x+2) \)
\( \implies xy + 2x = yx + 2y \)
\( \implies 2x = 2y \)
\( \implies x = y \)
Thus, \( f \) is a one-one function.

Now, we need to find the inverse of \( f: [-1, 1] \rightarrow \text{Range } f \).
Since the codomain is given as the Range of \( f \), the function is by definition onto.
Therefore, \( f \) is both one-one and onto, so its inverse \( f^{-1} \) exists.

To find \( f^{-1}(x) \):
1. Let \( y = f(x) \).
\( y = \frac{x}{x+2} \)
2. Solve for \( x \) in terms of \( y \).
\( y(x+2) = x \)
\( \implies xy + 2y = x \)
\( \implies 2y = x - xy \)
\( \implies 2y = x(1 - y) \)
\( \implies x = \frac{2y}{1-y} \)
Note that for this expression, \( y \neq 1 \).
3. Replace \( x \) with \( f^{-1}(y) \).
\( f^{-1}(y) = \frac{2y}{1-y} \)
4. Replace \( y \) with \( x \) to get \( f^{-1}(x) \).
\( f^{-1}(x) = \frac{2x}{1-x} \)
This inverse function will map values from the range of \( f \) back to the domain \( [-1, 1] \). The restriction \( y \neq 1 \) means \( x \neq 1 \) for \( f^{-1}(x) \).
In simple words: We showed that this function is "one-to-one" because different starting numbers always give different results. Since the question asks for the inverse from its domain to its range, it's also "onto". To find the inverse, we switch \( x \) and \( y \) in the formula and then solve for \( x \). This gives us the rule to go backward from the output to the original input.

🎯 Exam Tip: When the codomain is specified as the "Range \( f \)", you don't need to formally prove surjectivity; it's given by the definition of the codomain. Focus on proving injectivity and finding the inverse.

 

Question 10. Let \( A = R - \{2\} \) and \( B = R - \{1\} \). If \( f : A \rightarrow B \) is a function defined by \( f(x) = \frac{x-1}{x-2} \), then show that \( f \) is one-one and onto. Hence, find \( f^{-1} \).
Answer:
Given \( A = R - \{2\} \) (domain) and \( B = R - \{1\} \) (codomain).
The function is \( f : A \rightarrow B \) defined by \( f(x) = \frac{x-1}{x-2} \).

For Injectivity (one-one function):
Assume \( f(x) = f(y) \) for any \( x, y \in A \).
\( \implies \frac{x-1}{x-2} = \frac{y-1}{y-2} \)
Cross-multiply:
\( \implies (x-1)(y-2) = (y-1)(x-2) \)
\( \implies xy - 2x - y + 2 = xy - 2y - x + 2 \)
Subtract \( xy \) from both sides and \( 2 \) from both sides:
\( \implies -2x - y = -2y - x \)
\( \implies -2x + x = -2y + y \)
\( \implies -x = -y \)
\( \implies x = y \)
So, \( f \) is a one-one function.

For Surjectivity (onto function):
Let \( y \in B \) be any arbitrary element in the codomain \( B = R - \{1\} \). We need to find \( x \in A \) such that \( f(x) = y \).
\( y = \frac{x-1}{x-2} \)
Solve for \( x \) in terms of \( y \):
\( y(x-2) = x-1 \)
\( \implies xy - 2y = x - 1 \)
\( \implies 1 - 2y = x - xy \)
\( \implies 1 - 2y = x(1 - y) \)
\( \implies x = \frac{1-2y}{1-y} \)
Since \( y \in B \), we know \( y \neq 1 \), so the denominator \( (1-y) \) is never zero. Thus \( x \) is always a real number. We also need to ensure \( x \neq 2 \) (the excluded value from domain \( A \)).
If \( x = 2 \), then \( 2 = \frac{1-2y}{1-y} \)
\( \implies 2(1-y) = 1-2y \)
\( \implies 2 - 2y = 1 - 2y \)
\( \implies 2 = 1 \), which is false.
This means \( x \) can never be 2 for any \( y \in B \). Therefore, for every \( y \in B \), there exists an \( x \in A \) such that \( f(x) = y \).
Thus, \( f \) is an onto function.

Since \( f \) is both one-one and onto, its inverse \( f^{-1} \) exists.

To find \( f^{-1}(x) \):
Using the expression for \( x \) in terms of \( y \) we found for surjectivity:
\( x = \frac{1-2y}{1-y} \)
Replace \( x \) with \( f^{-1}(y) \):
\( f^{-1}(y) = \frac{1-2y}{1-y} \)
Replace \( y \) with \( x \) to get \( f^{-1}(x) \):
\( f^{-1}(x) = \frac{1-2x}{1-x} \)
The inverse function \( f^{-1} : B \rightarrow A \) is defined by \( f^{-1}(x) = \frac{1-2x}{1-x} \). This function effectively reverses the mapping of \( f \).
In simple words: This question asks us to prove that a function is both one-to-one (each input has a unique output) and onto (every output value in the codomain is reached). We showed this by algebra. Once proven, we found the inverse function by swapping \( x \) and \( y \) in the original function's rule and then solving for \( x \).

🎯 Exam Tip: When proving surjectivity for functions with restricted domains/codomains, ensure the \( x \) you find is not only real but also within the specified domain (e.g., \( x \neq 2 \)).

 

Question 11. Show that the function \( f \) in \( A = R - \left\{\frac{2}{3}\right\} \) defined as \( f(x) = \frac{4x+3}{6x-4} \) is one-one and onto. Hence, find \( f^{-1} \).
Answer:
Given \( A = R - \left\{\frac{2}{3}\right\} \). The function is \( f : A \rightarrow R \) defined by \( f(x) = \frac{4x+3}{6x-4} \).

For Injectivity (one-one function):
Assume \( f(x) = f(y) \) for any \( x, y \in A \).
\( \implies \frac{4x+3}{6x-4} = \frac{4y+3}{6y-4} \)
Cross-multiply:
\( \implies (4x+3)(6y-4) = (4y+3)(6x-4) \)
\( \implies 24xy - 16x + 18y - 12 = 24xy - 16y + 18x - 12 \)
Subtract \( 24xy \) and \( -12 \) from both sides:
\( \implies -16x + 18y = -16y + 18x \)
\( \implies 18y + 16y = 18x + 16x \)
\( \implies 34y = 34x \)
\( \implies y = x \)
So, \( f \) is a one-one function.

For Surjectivity (onto function):
Let \( y \in R \) be any arbitrary element in the codomain. We need to find \( x \in A \) such that \( f(x) = y \).
\( y = \frac{4x+3}{6x-4} \)
Solve for \( x \) in terms of \( y \):
\( y(6x-4) = 4x+3 \)
\( \implies 6xy - 4y = 4x + 3 \)
\( \implies 6xy - 4x = 4y + 3 \)
\( \implies x(6y - 4) = 4y + 3 \)
\( \implies x = \frac{4y+3}{6y-4} \)
For \( x \) to be defined, the denominator \( 6y-4 \) cannot be zero. So, \( 6y-4 \neq 0 \implies 6y \neq 4 \implies y \neq \frac{4}{6} \implies y \neq \frac{2}{3} \).
This implies that the range of \( f \) is \( R - \left\{\frac{2}{3}\right\} \). Since the codomain was given as \( R \), and the range is \( R - \left\{\frac{2}{3}\right\} \), this means the function is *not* onto from \( A \) to \( R \). Instead, it is onto from \( A \) to its own range \( R - \left\{\frac{2}{3}\right\} \). The question implies it is one-one and onto, meaning the codomain should effectively be \( R - \left\{\frac{2}{3}\right\} \). Let's assume the question implicitly asks to show it is onto its range.
Also, we need to ensure that \( x \neq \frac{2}{3} \).
If \( x = \frac{2}{3} \), then \( \frac{2}{3} = \frac{4y+3}{6y-4} \)
\( \implies 2(6y-4) = 3(4y+3) \)
\( \implies 12y - 8 = 12y + 9 \)
\( \implies -8 = 9 \), which is false.
This means \( x \) can never be \( \frac{2}{3} \). So, for every \( y \in R - \left\{\frac{2}{3}\right\} \), there exists an \( x \in A \) such that \( f(x) = y \).
Thus, \( f \) is onto to its range \( R - \left\{\frac{2}{3}\right\} \).

Since \( f \) is both one-one and onto (to its range), its inverse \( f^{-1} \) exists.

To find \( f^{-1}(x) \):
Using the expression for \( x \) in terms of \( y \) we found for surjectivity:
\( x = \frac{4y+3}{6y-4} \)
Replace \( x \) with \( f^{-1}(y) \):
\( f^{-1}(y) = \frac{4y+3}{6y-4} \)
Replace \( y \) with \( x \) to get \( f^{-1}(x) \):
\( f^{-1}(x) = \frac{4x+3}{6x-4} \)
Interestingly, the inverse function is the same as the original function! This is a special property of certain functions. For example, if you apply \( f \) twice, you get back to the original value: \( f(f(x)) = x \).
In simple words: We showed that this function is one-to-one (each input has a unique output) and onto (covers all possible output values in its specific range). To find its inverse, we switched \( x \) and \( y \) and solved for \( x \). It turns out the inverse function is the exact same as the original function, meaning applying it twice gets you back where you started.

🎯 Exam Tip: Sometimes a function can be its own inverse, i.e., \( f^{-1}(x) = f(x) \). This happens when \( f(f(x)) = x \). This result is valid and indicates a unique property of the function.

 

Question 12. Consider \( f : R^+ \rightarrow [4, \infty) \) given by \( f(x) = x^2 + 4 \). Show that \( f \) is invertible with the inverse \( f^{-1}(y) = \sqrt{y-4} \), where \( R^+ \) is the set of all non-negative numbers.
Answer:
Given \( f : R^+ \rightarrow [4, \infty) \) defined by \( f(x) = x^2 + 4 \). Here, \( R^+ \) denotes the set of non-negative real numbers, \( [0, \infty) \).

For Injectivity (one-one function):
Assume \( f(x) = f(y) \) for any \( x, y \in R^+ \).
\( \implies x^2 + 4 = y^2 + 4 \)
\( \implies x^2 = y^2 \)
Taking the square root of both sides:
\( \implies \sqrt{x^2} = \sqrt{y^2} \)
\( \implies |x| = |y| \)
Since \( x, y \in R^+ \) (non-negative), \( |x|=x \) and \( |y|=y \).
\( \implies x = y \)
So, \( f \) is a one-one function.

For Surjectivity (onto function):
Let \( y \in [4, \infty) \) be any arbitrary element in the codomain. We need to find \( x \in R^+ \) such that \( f(x) = y \).
\( x^2 + 4 = y \)
\( \implies x^2 = y - 4 \)
For \( x \) to be a real number, \( y - 4 \) must be non-negative. Since \( y \in [4, \infty) \), \( y \geq 4 \), so \( y - 4 \geq 0 \). This condition is satisfied.
\( \implies x = \sqrt{y-4} \)
Since \( \sqrt{y-4} \geq 0 \), this value of \( x \) is in \( R^+ \). So, for every \( y \in [4, \infty) \), there exists an \( x \in R^+ \) such that \( f(x) = y \).
Thus, \( f \) is an onto function.

Since \( f \) is both one-one and onto, \( f \) is invertible.

To find \( f^{-1}(x) \):
Using the expression for \( x \) in terms of \( y \) we found for surjectivity:
\( x = \sqrt{y-4} \)
Replace \( x \) with \( f^{-1}(y) \):
\( f^{-1}(y) = \sqrt{y-4} \)
This matches the given inverse function in terms of \( y \). The function effectively takes a positive number, squares it, and adds 4. The inverse function reverses this by subtracting 4 and then taking the positive square root.
In simple words: We proved this function is one-to-one because different positive inputs give different outputs, and it's onto because all numbers in its target range can be reached. The inverse function reverses the steps: it subtracts 4 from the input and then takes the square root to get back the original positive number.

🎯 Exam Tip: When the domain is \( R^+ \) (non-negative numbers), remember that \( \sqrt{x^2} = |x| = x \). This is crucial for proving injectivity. Also, ensure the value under the square root in the inverse is non-negative.

 

Question 13. Let \( f : R \rightarrow R : f(x) = 10x + 7 \). Find the function \( g : R \rightarrow R \) such that \( gof = fog = I_R \).
Answer:
Given \( f : R \rightarrow R \) defined by \( f(x) = 10x + 7 \).
The condition \( gof = fog = I_R \) means that \( g \) is the inverse function of \( f \), i.e., \( g(x) = f^{-1}(x) \).

First, we need to show that \( f \) is invertible.
For Injectivity (one-one function):
Assume \( f(x) = f(y) \) for any \( x, y \in R \).
\( \implies 10x + 7 = 10y + 7 \)
\( \implies 10x = 10y \)
\( \implies x = y \)
So, \( f \) is a one-one function.

For Surjectivity (onto function):
Let \( y \in R \) be any arbitrary element in the codomain. We need to find \( x \in R \) such that \( f(x) = y \).
\( 10x + 7 = y \)
\( \implies 10x = y - 7 \)
\( \implies x = \frac{y-7}{10} \)
Since \( y \in R \), \( \frac{y-7}{10} \) will also be in \( R \). So, for every \( y \) in the codomain, there is an \( x \) in the domain.
Thus, \( f \) is an onto function.

Since \( f \) is both one-one and onto, its inverse \( f^{-1} \) exists.

To find \( f^{-1}(x) \):
1. Let \( y = f(x) \).
\( y = 10x + 7 \)
2. Solve for \( x \) in terms of \( y \).
\( x = \frac{y-7}{10} \)
3. Replace \( x \) with \( f^{-1}(y) \).
\( f^{-1}(y) = \frac{y-7}{10} \)
4. Replace \( y \) with \( x \) to get \( f^{-1}(x) \).
\( f^{-1}(x) = \frac{x-7}{10} \)

Therefore, the function \( g(x) \) is \( g(x) = \frac{x-7}{10} \). This function \( g \) maps real numbers to real numbers and serves to reverse the action of \( f \).
In simple words: We are looking for a function \( g \) that "undoes" what \( f \) does. This means \( g \) must be the inverse of \( f \). We found \( g \) by setting \( f(x) = y \), solving for \( x \), and then replacing \( y \) with \( x \) in the final expression.

🎯 Exam Tip: The identity function \( I_R \) maps every element to itself, i.e., \( I_R(x) = x \). The condition \( gof = fog = I_R \) is the definition of \( g \) being the inverse of \( f \).

 

Question 14. Show that the inverse function of the function \( f \) whose rule is \( f(x) = \frac{2x+1}{3x-2} \), \( x \neq \frac{2}{3} \) is \( f \) itself.
Answer:
Given the function \( f(x) = \frac{2x+1}{3x-2} \), with \( x \neq \frac{2}{3} \).
To show that the inverse function is \( f \) itself, we need to show that \( f^{-1}(x) = f(x) \).

To find \( f^{-1}(x) \):
1. Let \( y = f(x) \).
\( y = \frac{2x+1}{3x-2} \)
2. Solve for \( x \) in terms of \( y \).
\( y(3x-2) = 2x+1 \)
\( \implies 3xy - 2y = 2x + 1 \)
\( \implies 3xy - 2x = 2y + 1 \)
\( \implies x(3y - 2) = 2y + 1 \)
\( \implies x = \frac{2y+1}{3y-2} \)
3. Replace \( x \) with \( f^{-1}(y) \).
\( f^{-1}(y) = \frac{2y+1}{3y-2} \)
4. Replace \( y \) with \( x \) to get \( f^{-1}(x) \).
\( f^{-1}(x) = \frac{2x+1}{3x-2} \)

Since \( f^{-1}(x) = \frac{2x+1}{3x-2} \), which is the same as the original function \( f(x) \), we have shown that the inverse function of \( f \) is \( f \) itself.
Alternatively, we can show that \( f(f(x)) = x \).
\( f(f(x)) = f\left(\frac{2x+1}{3x-2}\right) \)
\( = \frac{2\left(\frac{2x+1}{3x-2}\right)+1}{3\left(\frac{2x+1}{3x-2}\right)-2} \)
\( = \frac{\frac{2(2x+1) + 1(3x-2)}{3x-2}}{\frac{3(2x+1) - 2(3x-2)}{3x-2}} \)
\( = \frac{4x+2+3x-2}{6x+3-6x+4} \)
\( = \frac{7x}{7} \)
\( = x \)
Since \( f(f(x)) = x \), this confirms that \( f \) is its own inverse. This type of function is sometimes called an involution.
In simple words: We wanted to check if this function is its own inverse. We found the inverse by swapping \( x \) and \( y \) and solving, and it turned out to be the exact same formula as the original function. This means if you apply the function once, and then apply it again, you get back to your starting number.

🎯 Exam Tip: To prove a function is its own inverse, you can either find \( f^{-1}(x) \) and show it equals \( f(x) \), or directly show that the composition \( f(f(x)) = x \).

 

Question 15. Use composition to show that \( f \) and \( g \) are inverse of each other.
(i) \( f(x) = 2x - 6, g(x) = \frac{x}{2}+3 \)
(ii) \( f(x) = \frac{1}{x+1}, g(x) = \frac{1-x}{x} \)
(iii) \( f(x) = \frac{-3}{2x+5}, g(x) = \frac{-3-5x}{2x} \)
(iv) \( f(x) = x^5, g(x) = \sqrt[5]{x} \)
Answer:
To show that \( f \) and \( g \) are inverses of each other, we need to prove that their compositions \( fog(x) \) and \( gof(x) \) both result in \( x \) (the identity function).

(i) \( f(x) = 2x - 6, g(x) = \frac{x}{2}+3 \)
\( fog(x) = f(g(x)) = f\left(\frac{x}{2}+3\right) \)
\( = 2\left(\frac{x}{2}+3\right) - 6 \)
\( = x + 6 - 6 \)
\( = x \)

\( gof(x) = g(f(x)) = g(2x-6) \)
\( = \frac{2x-6}{2} + 3 \)
\( = (x-3) + 3 \)
\( = x \)
Since \( fog(x) = x \) and \( gof(x) = x \), \( f \) and \( g \) are inverses of each other.

(ii) \( f(x) = \frac{1}{x+1}, g(x) = \frac{1-x}{x} \)
\( fog(x) = f(g(x)) = f\left(\frac{1-x}{x}\right) \)
\( = \frac{1}{\left(\frac{1-x}{x}\right)+1} \)
\( = \frac{1}{\frac{1-x+x}{x}} \)
\( = \frac{1}{\frac{1}{x}} \)
\( = x \)

\( gof(x) = g(f(x)) = g\left(\frac{1}{x+1}\right) \)
\( = \frac{1-\left(\frac{1}{x+1}\right)}{\left(\frac{1}{x+1}\right)} \)
\( = \frac{\frac{x+1-1}{x+1}}{\frac{1}{x+1}} \)
\( = \frac{\frac{x}{x+1}}{\frac{1}{x+1}} \)
\( = x \)
Since \( fog(x) = x \) and \( gof(x) = x \), \( f \) and \( g \) are inverses of each other.

(iii) \( f(x) = \frac{-3}{2x+5}, g(x) = \frac{-3-5x}{2x} \)
\( fog(x) = f(g(x)) = f\left(\frac{-3-5x}{2x}\right) \)
\( = \frac{-3}{2\left(\frac{-3-5x}{2x}\right)+5} \)
\( = \frac{-3}{\frac{-3-5x}{x}+5} \)
\( = \frac{-3}{\frac{-3-5x+5x}{x}} \)
\( = \frac{-3}{\frac{-3}{x}} \)
\( = -3 \times \frac{x}{-3} \)
\( = x \)

\( gof(x) = g(f(x)) = g\left(\frac{-3}{2x+5}\right) \)
\( = \frac{-3-5\left(\frac{-3}{2x+5}\right)}{2\left(\frac{-3}{2x+5}\right)} \)
\( = \frac{\frac{-3(2x+5)-5(-3)}{2x+5}}{\frac{-6}{2x+5}} \)
\( = \frac{-6x-15+15}{-6} \)
\( = \frac{-6x}{-6} \)
\( = x \)
Since \( fog(x) = x \) and \( gof(x) = x \), \( f \) and \( g \) are inverses of each other.

(iv) \( f(x) = x^5, g(x) = \sqrt[5]{x} \)
\( fog(x) = f(g(x)) = f(\sqrt[5]{x}) \)
\( = (\sqrt[5]{x})^5 \)
\( = x \)

\( gof(x) = g(f(x)) = g(x^5) \)
\( = \sqrt[5]{x^5} \)
\( = x \)
Since \( fog(x) = x \) and \( gof(x) = x \), \( f \) and \( g \) are inverses of each other.
The method of composition provides a straightforward way to verify if two functions are indeed inverses, by checking if they "cancel each other out".
In simple words: To show that two functions are inverses, we put one inside the other in two ways: \( f \) inside \( g \) (called \( gof \)) and \( g \) inside \( f \) (called \( fog \)). If both of these compositions result in just \( x \), it means the functions undo each other perfectly and are therefore inverses.

🎯 Exam Tip: Remember to calculate both \( fog(x) \) and \( gof(x) \). Both compositions must yield \( x \) for \( f \) and \( g \) to be considered inverses of each other.

The provided content from pages 15 and 16 consists entirely of website navigation links, page metadata, a comment section, and footer information, not actual educational questions or answers. As per the content processing rules (specifically "IGNORE AND SKIP — PAGE HEADER / SEO TITLES" and "IGNORE AND SKIP — FOOTER / NAVIGATION"), this type of content must be completely ignored. Therefore, no HTML output is generated for this specific range.