RBSE Solutions Class 12 Maths Chapter 1 Composite Functions Exercise 1.2

Question 1. If A = {1, 2, 3,4}, B = {a,b,c,d}, then define four bijection from A to B and also find their inverse functions.
Answer:
Given:
\( A = \{1, 2, 3, 4\} \)
\( B = \{a, b, c, d\} \)
(a) A bijection from A to B is \( f_1 = \{(1, a),(2, b), (3, c), (4, d)\} \).
Its inverse function is \( f_1^{-1} = \{(a, 1), (b, 2), (c, 3), (d, 4)\} \).
(b) A second bijection from A to B is \( f_2 = \{(1, a), (2, c), (3, b), (4, d)\} \).
Its inverse function is \( f_2^{-1} = \{(a, 1), (c, 2), (b, 3), (d, 4)\} \).
(c) A third mapping from A to B is \( f_3 = \{(1, b), (2, a), (3, d), (4, b)\} \).
Its inverse function is \( f_3^{-1} = \{(b, 1), (a, 2), (d, 3), (6,4)\} \). This mapping has an element in the codomain 'b' mapped twice, which means it is not a bijection.
(d) The inverse for a fourth mapping is given as \( f_4^{-1} = \{(c, 1), (a, z), (a, 3), (b, 4)\} \). This part of the problem statement is corrupted in the source text, so the exact mapping for \( f_4 \) is unclear.
In simple words: A bijection links each item in one set to exactly one item in another set. For each bijection, we also find its inverse, which simply swaps the pairs. For instance, if '1' maps to 'a', the inverse maps 'a' back to '1'.

🎯 Exam Tip: To define a bijection, ensure every element in the first set (domain) maps to a unique element in the second set (codomain), and every element in the codomain is mapped to by exactly one element from the domain. Always list the pairs correctly for both the function and its inverse.

Question 2. (i) f(x) = 2x - 3 (ii) f (x) = x³ + 5 Then, prove that f is bijection in both conditions, Also find f-1.
Answer:
(i) Given function:
\( f : R \rightarrow R, f(x) = 2x - 3 \)
**For One-one function:**
Let \( a, b \in R \).
Assume \( f(a) = f(b) \).
\( \implies 2a - 3 = 2b - 3 \)
\( \implies 2a = 2b \)
\( \implies a = b \)
So, \( f(a) = f(b) \implies a = b \), for all \( a, b \in R \).
Therefore, \( f \) is a one-one function.
**For Onto function:**
Let \( y \in R \) (co-domain).
Assume \( f(x) = y \).
\( \implies 2x - 3 = y \)
\( \implies 2x = y + 3 \)
\( \implies x = \frac{y+3}{2} \)
Since \( x = \frac{y+3}{2} \in R \) for every \( y \in R \), a pre-image exists for each value of \( y \) in the codomain. This shows that the range of \( f \) is equal to its co-domain.
Therefore, \( f \) is an onto function.
Since \( f \) is both one-one and onto, it is a bijection. A function that is both injective (one-one) and surjective (onto) is called a bijective function.
**To find \( f^{-1} \):**
Let \( f(x) = y \). Then \( f^{-1}(y) = x \).
We have \( y = 2x - 3 \).
Solving for \( x \): \( 2x = y + 3 \implies x = \frac{y+3}{2} \).
So, \( f^{-1}(y) = \frac{y+3}{2} \).
Replacing \( y \) with \( x \), we get \( f^{-1}(x) = \frac{x+3}{2} \).

(ii) Given function:
\( f : R \rightarrow R, f(x) = x^3 + 5 \)
**For One-one function:**
Let \( a, b \in R \).
Assume \( f(a) = f(b) \).
\( \implies a^3 + 5 = b^3 + 5 \)
\( \implies a^3 = b^3 \)
\( \implies a = b \)
So, \( f(a) = f(b) \implies a = b \), for all \( a, b \in R \).
Therefore, \( f \) is a one-one function.
**For Onto function:**
Let \( y \in R \) (co-domain).
Assume \( f(x) = y \).
\( \implies x^3 + 5 = y \)
\( \implies x^3 = y - 5 \)
\( \implies x = (y - 5)^{1/3} \)
Since \( x = (y - 5)^{1/3} \in R \) for every \( y \in R \), a pre-image exists for each value of \( y \) in the codomain. This shows that the range of \( f \) is equal to its co-domain.
Therefore, \( f \) is an onto function.
Since \( f \) is both one-one and onto, it is a bijection. A function is bijective when every element of the codomain is mapped to by exactly one element of the domain.
**To find \( f^{-1} \):**
Let \( f(x) = y \). Then \( f^{-1}(y) = x \).
We have \( y = x^3 + 5 \).
Solving for \( x \): \( x^3 = y - 5 \implies x = (y - 5)^{1/3} \).
So, \( f^{-1}(y) = (y - 5)^{1/3} \).
Replacing \( y \) with \( x \), we get \( f^{-1}(x) = (x - 5)^{1/3} \).
In simple words: To prove a function is a bijection, we first show it's "one-one" (each input gives a unique output) and then "onto" (every possible output can be reached). To find the inverse, we swap x and y in the function's equation and solve for y.

🎯 Exam Tip: Remember to clearly state the domain and codomain of the function. For polynomials, \( f(x) = x^n + c \) is one-one if \( n \) is odd, but not if \( n \) is even (e.g., \( x^2 \) maps both positive and negative x to the same value). When finding the inverse, make sure to express \( x \) in terms of \( y \) and then replace \( y \) with \( x \) in the final inverse function.

Question 5. If f : R → R, such that f(x) = ax + b, a ≠ 0, then prove that f is a bijection function. Also, find f-¹.
Answer:
Given function: \( f : R \rightarrow R \), \( f(x) = ax + b \), where \( a \neq 0 \).
We need to prove that \( f \) is a bijection function and then find its inverse \( f^{-1} \).
**For One-one function:**
Let \( p, q \in R \).
Assume \( f(p) = f(q) \).
\( \implies ap + b = aq + b \)
\( \implies ap = aq \)
Since \( a \neq 0 \), we can divide both sides by \( a \).
\( \implies p = q \)
So, \( f(p) = f(q) \implies p = q \), for all \( p, q \in R \).
Therefore, \( f \) is a one-one function. This means different inputs always give different outputs.
**For Onto function:**
Let \( y \in R \) (co-domain).
Assume \( f(x) = y \).
\( \implies ax + b = y \)
\( \implies ax = y - b \)
\( \implies x = \frac{y-b}{a} \)
Since \( a \neq 0 \), for every real value of \( y \), we can find a unique real value \( x = \frac{y-b}{a} \) in the domain. This means that every element in the codomain has a pre-image in the domain.
Therefore, \( f \) is an onto function. This ensures that the function covers all values in its codomain.
Since \( f \) is both one-one and onto, it is a bijection function.
**To find \( f^{-1} \):**
Let \( f(x) = y \). Then \( f^{-1}(y) = x \).
We have \( y = ax + b \).
Solving for \( x \):
\( ax = y - b \)
\( x = \frac{y-b}{a} \)
So, \( f^{-1}(y) = \frac{y-b}{a} \).
Replacing \( y \) with \( x \), the inverse function is \( f^{-1}(x) = \frac{x-b}{a} \).
In simple words: For a linear function like \( f(x) = ax + b \) (where 'a' is not zero), it is always a bijection because each input gives a unique output, and every output is possible. To find its inverse, we just swap x and y and solve for x.

🎯 Exam Tip: When proving a function is one-one, always start by assuming \( f(p) = f(q) \) and work to show \( p = q \). For onto, assume \( f(x) = y \) and solve for \( x \) in terms of \( y \), then show \( x \) is always in the domain for all \( y \) in the codomain. Remember that for linear functions, the inverse is also linear.

Question 6. If f : R → R,f(x) = cos (x + 2), is f¯¹ exists.
Answer:
Given function: \( f : R \rightarrow R, f(x) = \cos(x + 2) \).
For the inverse function \( f^{-1} \) to exist, the function \( f \) must be a bijection (both one-one and onto).
Let's check if \( f \) is a one-one function.
Consider two different input values for \( x \): \( x_1 = 0 \) and \( x_2 = 2\pi \).
For \( x_1 = 0 \):
\( f(0) = \cos(0 + 2) = \cos(2) \)
For \( x_2 = 2\pi \):
\( f(2\pi) = \cos(2\pi + 2) \)
Since the cosine function has a period of \( 2\pi \), \( \cos(2\pi + 2) = \cos(2) \).
So, \( f(0) = \cos(2) \) and \( f(2\pi) = \cos(2) \).
Here, we have different input values \( 0 \) and \( 2\pi \) that produce the same output value \( \cos(2) \).
This means that \( f \) is not a one-one function.
Since \( f \) is not a one-one function, it cannot be a bijection.
Therefore, the inverse function \( f^{-1} \) does not exist. A function must be both one-to-one and onto for its inverse to exist.
In simple words: For a function to have an inverse, it must give a different output for every different input. The cosine function, like \( \cos(x+2) \), repeats its values (for example, \( \cos(0+2) \) is the same as \( \cos(2\pi+2) \)), so it's not "one-one". Because it's not one-one, it cannot have an inverse.

🎯 Exam Tip: To prove a function is NOT one-one, you only need to find one counterexample: two different inputs that produce the same output. Trigonometric functions like sine and cosine are generally not one-one over their entire domain \( R \) due to their periodic nature, which prevents them from having an inverse unless their domain is restricted.

Question 7. Find f¯¹ (if exists), where f: A → B, such that
(i) A = {0,-1,- 3, 2}, B = {-9, – 3,0, 6}, f(x) = 3x
(ii) A = {1, 3, 5, 7,9}, B = {0, 1, 9, 25, 49, 81), f(x) = x²
(iii) A = B = R, f(x) = x3
Answer:
(i) Given function: \( f: A \rightarrow B \), where \( A = \{0, -1, -3, 2\} \), \( B = \{-9, -3, 0, 6\} \), and \( f(x) = 3x \).
Let's find the images of the elements in set A:
\( f(0) = 3 \times 0 = 0 \)
\( f(-1) = 3 \times (-1) = -3 \)
\( f(-3) = 3 \times (-3) = -9 \)
\( f(2) = 3 \times 2 = 6 \)
So, \( f = \{(0, 0), (-1, -3), (-3, -9), (2, 6)\} \).
**One-one function:** Each element in A maps to a unique element in B.
For example, \( 0 \rightarrow 0 \), \( -1 \rightarrow -3 \), \( -3 \rightarrow -9 \), \( 2 \rightarrow 6 \). All outputs are distinct.
Thus, \( f \) is a one-one function.
**Onto function:** The range of \( f \) is \( \{0, -3, -9, 6\} \), which is exactly equal to set B. So, every element in B has a pre-image in A.
Thus, \( f \) is an onto function.
Since \( f \) is both one-one and onto, \( f \) is a bijection, and its inverse \( f^{-1} \) exists.
To find \( f^{-1} \), we swap the elements in each pair of \( f \):
\( f^{-1} = \{(0, 0), (-3, -1), (-9, -3), (6, 2)\} \).

(ii) Given function: \( f: A \rightarrow B \), where \( A = \{1, 3, 5, 7, 9\} \), \( B = \{0, 1, 9, 25, 49, 81\} \), and \( f(x) = x^2 \).
Let's find the images of the elements in set A:
\( f(1) = 1^2 = 1 \)
\( f(3) = 3^2 = 9 \)
\( f(5) = 5^2 = 25 \)
\( f(7) = 7^2 = 49 \)
\( f(9) = 9^2 = 81 \)
So, \( f = \{(1, 1), (3, 9), (5, 25), (7, 49), (9, 81)\} \).
**One-one function:** Each element in A maps to a unique element in B. For example, \( 1 \rightarrow 1 \), \( 3 \rightarrow 9 \), \( 5 \rightarrow 25 \), \( 7 \rightarrow 49 \), \( 9 \rightarrow 81 \). All outputs are distinct.
Thus, \( f \) is a one-one function.
**Onto function:** The range of \( f \) is \( \{1, 9, 25, 49, 81\} \).
The set B is \( \{0, 1, 9, 25, 49, 81\} \).
Since \( \{1, 9, 25, 49, 81\} \neq \{0, 1, 9, 25, 49, 81\} \), the range of \( f \) is not equal to the codomain B. Specifically, the element \( 0 \) in B does not have a pre-image in A.
Thus, \( f \) is not an onto function.
Since \( f \) is not an onto function, it is not a bijection, and its inverse \( f^{-1} \) does not exist.

(iii) Given function: \( f: A \rightarrow B \), where \( A = B = R \) (all real numbers), and \( f(x) = x^3 \).
**One-one function:**
Let \( a, b \in R \).
Assume \( f(a) = f(b) \).
\( \implies a^3 = b^3 \)
\( \implies a = b \) (because for real numbers, if cubes are equal, the numbers themselves must be equal).
Thus, \( f \) is a one-one function.
**Onto function:**
Let \( y \in R \) (co-domain).
Assume \( f(x) = y \).
\( \implies x^3 = y \)
\( \implies x = y^{1/3} \)
For every real value of \( y \), there exists a unique real value \( x = y^{1/3} \) in the domain. This means every element in the codomain has a pre-image.
Thus, \( f \) is an onto function.
Since \( f \) is both one-one and onto, it is a bijection, and its inverse \( f^{-1} \) exists.
To find \( f^{-1} \):
Let \( f(x) = y \). Then \( f^{-1}(y) = x \).
We have \( y = x^3 \).
Solving for \( x \): \( x = y^{1/3} \).
So, \( f^{-1}(y) = y^{1/3} \).
Replacing \( y \) with \( x \), the inverse function is \( f^{-1}(x) = x^{1/3} \).
In simple words: For each part, we check if the function is "one-one" (unique outputs for unique inputs) and "onto" (all possible outputs are covered). If both are true, an inverse exists, and we find it by reversing the input-output pairs. If not, the inverse does not exist.

🎯 Exam Tip: Always verify both one-one and onto conditions to determine if an inverse exists. A common mistake is to only check one condition. Remember that for finite sets, if a function is one-one, its range equals the codomain, implying it's also onto. For infinite sets, like \( R \), both conditions must be explicitly checked.