CBSE Class 12 Mathematics Relations And Functions Class Test Set 09

Question. Define 1-1 and onto function. also represent it diametrically.
Answer:
1-1 (Injective) Function:
A function \( f: A \to B \) is called a one-to-one (or injective) function if distinct elements of \( A \) have distinct images in \( B \).
Mathematically, for all \( x_1, x_2 \in A \), if \( f(x_1) = f(x_2) \), then \( x_1 = x_2 \).

Onto (Surjective) Function:
A function \( f: A \to B \) is called an onto (or surjective) function if every element of \( B \) is the image of at least one element of \( A \) under \( f \).
Mathematically, for every \( y \in B \), there exists \( x \in A \) such that \( f(x) = y \). This implies that the range of \( f \) is equal to the codomain \( B \).

Diagrammatic Representation:
- 1-1 Function: Let \( A = \{1, 2, 3\} \) and \( B = \{x, y, z, w\} \). Let \( f = \{(1, x), (2, y), (3, z)\} \). Since every element in the domain maps to a unique, distinct element in the codomain, it is a 1-1 function.
- Onto Function: Let \( A = \{1, 2, 3\} \) and \( B = \{x, y\} \). Let \( f = \{(1, x), (2, y), (3, y)\} \). Since every element in the codomain has a pre-image in the domain (Range \( = B \)), it is an onto function.

Question. Prove that the function \( f: \mathbb{R} \to \mathbb{R} : f(x) = 2x \) is one-one and onto.
Answer:
Given function: \( f: \mathbb{R} \to \mathbb{R} \) defined by \( f(x) = 2x \).

1. One-one (Injectivity):
Let \( x_1, x_2 \in \mathbb{R} \) (domain) such that:
\( f(x_1) = f(x_2) \)
\( \implies 2x_1 = 2x_2 \)
\( \implies x_1 = x_2 \)
Since \( f(x_1) = f(x_2) \implies x_1 = x_2 \), the function \( f \) is one-one.

2. Onto (Surjectivity):
Let \( y \in \mathbb{R} \) (codomain).
We need to find \( x \in \mathbb{R} \) (domain) such that \( f(x) = y \).
\( 2x = y \implies x = \frac{y}{2} \)
Since \( y \in \mathbb{R} \), it follows that \( \frac{y}{2} \in \mathbb{R} \), which belongs to the domain.
Moreover, \( f\left(\frac{y}{2}\right) = 2\left(\frac{y}{2}\right) = y \).
Thus, for every \( y \in \mathbb{R} \) (codomain), there exists a pre-image \( x = \frac{y}{2} \in \mathbb{R} \) (domain) such that \( f(x) = y \).
Therefore, the function \( f \) is onto.

Since the function \( f \) is both one-one and onto, the proof is complete.

Question. Let \( A = \{1, 2, 3\} \), \( B = \{4, 5, 6, 7\} \) and let \( f = \{(1, 4), (2, 5), (3, 6)\} \) be a function from \( A \) to \( B \). Show that \( f \) is 1-1 function.
Answer:
Given:
\( A = \{1, 2, 3\} \)
\( B = \{4, 5, 6, 7\} \)
\( f = \{(1, 4), (2, 5), (3, 6)\} \)

From the definition of \( f \), we have:
- The image of \( 1 \) is \( f(1) = 4 \)
- The image of \( 2 \) is \( f(2) = 5 \)
- The image of \( 3 \) is \( f(3) = 6 \)

We observe that the images of distinct elements in the domain \( A \) are distinct under \( f \). Specifically:
\( 1 \neq 2 \implies f(1) \neq f(2) \) (since \( 4 \neq 5 \delta \))
\( 2 \neq 3 \implies f(2) \neq f(3) \) (since \( 5 \neq 6 \))
\( 1 \neq 3 \implies f(1) \neq f(3) \) (since \( 4 \neq 6 \))

Since no two distinct elements of \( A \) have the same image in \( B \), the function \( f \) is a 1-1 function.

Question. Let \( A = \mathbb{R} - \{3\} \) and \( B = \mathbb{R} - \{1\} \). \( f: A \to B \) defined by \( f(x) = \frac{x-2}{x-3} \). Is \( f \) One-One and onto ? Justify your answer.
Answer:
Given:
\( A = \mathbb{R} - \{3\} \)
\( B = \mathbb{R} - \{1\} \)
\( f: A \to B \) is defined by \( f(x) = \frac{x-2}{x-3} \).

1. Checking for One-One (Injectivity):
Let \( x_1, x_2 \in A \) such that \( f(x_1) = f(x_2) \).
\( \implies \frac{x_1 - 2}{x_1 - 3} = \frac{x_2 - 2}{x_2 - 3} \)
Cross-multiplying, we get:
\( (x_1 - 2)(x_2 - 3) = (x_2 - 2)(x_1 - 3) \)
\( \implies x_1 x_2 - 3x_1 - 2x_2 + 6 = x_1 x_2 - 3x_2 - 2x_1 + 6 \)
Subtracting \( x_1 x_2 + 6 \) from both sides:
\( \implies -3x_1 - 2x_2 = -3x_2 - 2x_1 \)
\( \implies -3x_1 + 2x_1 = -3x_2 + 2x_2 \)
\( \implies -x_1 = -x_2 \)
\( \implies x_1 = x_2 \)
Since \( f(x_1) = f(x_2) \implies x_1 = x_2 \), \( f \) is a One-One function.

2. Checking for Onto (Surjectivity):
Let \( y \in B \), which means \( y \in \mathbb{R} \) and \( y \neq 1 \).
We want to find \( x \in A \) such that \( f(x) = y \).
\( \implies \frac{x-2}{x-3} = y \)
\( \implies x - 2 = y(x - 3) \)
\( \implies x - 2 = xy - 3y \)
\( \implies x - xy = 2 - 3y \)
\( \implies x(1 - y) = 2 - 3y \)
\( \implies x = \frac{2 - 3y}{1 - y} = \frac{3y - 2}{y - 1} \)
Since \( y \neq 1 \), the denominator \( y - 1 \neq 0 \), so \( x \) is a well-defined real number.

Next, we must check if \( x \in A \) (i.e., \( x \neq 3 \)).
Suppose \( x = 3 \):
\( \implies \frac{3y - 2}{y - 1} = 3 \)
\( \implies 3y - 2 = 3(y - 1) \)
\( \implies 3y - 2 = 3y - 3 \)
\( \implies -2 = -3 \), which is a contradiction.
Hence, \( x \neq 3 \), meaning \( x \in A \).

Let us verify \( f(x) = y \):
\( f\left(\frac{3y-2}{y-1}\right) = \frac{\frac{3y-2}{y-1} - 2}{\frac{3y-2}{y-1} - 3} = \frac{\frac{3y - 2 - 2(y - 1)}{y - 1}}{\frac{3y - 2 - 3(y - 1)}{y - 1}} = \frac{3y - 2 - 2y + 2}{3y - 2 - 3y + 3} = \frac{y}{1} = y \).
Thus, for every \( y \in B \), there exists \( x = \frac{3y - 2}{y - 1} \in A \) such that \( f(x) = y \).
Hence, \( f \) is an onto function.

Conclusion:
Yes, \( f \) is both One-One and onto (i.e., bijective).