CBSE Class 12 Mathematics Relations And Functions Class Test Set 07

XII - Relations and Functions - Test No. 3

Q1. Define 1-1 and onto function. also represent it diametrically.
Answer:
1-1 (One-One or Injective) Function: A function \(f: A \to B\) is defined to be one-one if distinct elements of \(A\) have distinct images in \(B\). That is, for all \(x_1, x_2 \in A\),
\(f(x_1) = f(x_2) \implies x_1 = x_2\)
Alternatively, \(x_1 \neq x_2 \implies f(x_1) \neq f(x_2)\).

Onto (Surjective) Function: A function \(f: A \to B\) is defined to be onto if every element of \(B\) is the image of some element of \(A\). That is, for every \(y \in B\), there exists an element \(x \in A\) such that \(f(x) = y\).

Diagrammatic Representation:
For a one-one function, arrows from distinct domain elements point to distinct codomain elements. For an onto function, every element in the codomain has at least one incoming arrow from the domain.
In simple words: A one-one function never maps two different inputs to the same output, and an onto function covers every single possible output value in the target set.

Exam Tip: To prove a function is one-one, always start by assuming \(f(x_1) = f(x_2)\) and show that \(x_1 = x_2\). To prove onto, express \(x\) in terms of \(y\) and check if every \(y\) in the codomain gives a valid \(x\) in the domain.

Q2. Prove that the function \(f: R \to R : f(x) = 2x\) is one-one and onto.
Answer:
Injectivity (One-One):
Let \(x_1, x_2 \in R\) such that:
\(f(x_1) = f(x_2)\)
Substitute the function definition:
\(2x_1 = 2x_2\)
Divide both sides by 2:
\(x_1 = x_2\)
Since \(f(x_1) = f(x_2) \implies x_1 = x_2\), the function \(f\) is one-one.

Surjectivity (Onto):
Let \(y\) be an arbitrary element in the codomain \(R\).
We need to find an element \(x \in R\) such that \(f(x) = y\):
\(2x = y\)
\(x = \frac{y}{2}\)
Since \(y \in R\), its half \(\frac{y}{2}\) is also a real number in the domain \(R\).
Thus, for every \(y \in R\), there exists \(x = \frac{y}{2} \in R\) such that:
\(f(x) = f\left(\frac{y}{2}\right) = 2\left(\frac{y}{2}\right) = y\)
Therefore, \(f\) is onto.

In simple words: Since doubling any two different numbers gives two different results, it is one-one. Since every real number can be halved to get back the original number, it is onto.

Exam Tip: Linear functions of the form \(f(x) = ax + b\) where \(a \neq 0\) over real numbers are always bijective (both one-one and onto).

Q3. 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:
The given domain is \(A = \{1, 2, 3\}\) and codomain is \(B = \{4, 5, 6, 7\}\).
The function values are given by the set of ordered pairs:
\(f(1) = 4\)
\(f(2) = 5\)
\(f(3) = 6\)
We observe that the images of distinct elements of \(A\) are:
Distinct outputs: \(4 \neq 5 \neq 6\)
Since no two distinct elements in domain \(A\) have the same image in codomain \(B\), the function \(f\) is one-one.
In simple words: Every element in set A is mapped to a completely different element in set B, so it is a one-one function.

Exam Tip: For finite sets, checking if all second coordinates in the ordered pairs are distinct is the quickest way to prove a function is one-one.

Q4. Let \(A = R - \{3\}\) and \(B = R - \{1\}\). \(f: A \to B\) defined by \(f(x) = \frac{x-2}{x-3}\). Is \(f\) one-one and onto? Justify your answer.
Answer:
Check for One-One:
Let \(x_1, x_2 \in A\) such that:
\(f(x_1) = f(x_2)\)
\(\frac{x_1 - 2}{x_1 - 3} = \frac{x_2 - 2}{x_2 - 3}\)
Cross-multiply to solve:
\((x_1 - 2)(x_2 - 3) = (x_2 - 2)(x_1 - 3)\)
\(x_1x_2 - 3x_1 - 2x_2 + 6 = x_1x_2 - 3x_2 - 2x_1 + 6\)
Cancel common terms from both sides:
\(-3x_1 - 2x_2 = -3x_2 - 2x_1\)
\(-3x_1 + 2x_1 = -3x_2 + 2x_2\)
\(-x_1 = -x_2\)
\(x_1 = x_2\)
Since \(f(x_1) = f(x_2) \implies x_1 = x_2\), the function \(f\) is one-one.

Check for Onto:
Let \(y \in B\) (where \(y \neq 1\)). We need to find \(x \in A\) such that \(f(x) = y\):
\(\frac{x - 2}{x - 3} = y\)
\(x - 2 = y(x - 3)\)
\(x - 2 = xy - 3y\)
\(x - xy = 2 - 3y\)
\(x(1 - y) = 2 - 3y\)
\(x = \frac{2 - 3y}{1 - y}\)
Check if \(x \in A\) (i.e., \(x \neq 3\)):
If \(\frac{2 - 3y}{1 - y} = 3\), then:
\(2 - 3y = 3(1 - y)\)
\(2 - 3y = 3 - 3y\)
\(2 = 3\) (which is a contradiction)
Thus, for every \(y \in R - \{1\}\), there exists a unique real number \(x = \frac{2 - 3y}{1 - y} \neq 3\) such that \(f(x) = y\).
Therefore, \(f\) is onto.

In simple words: By setting the formula equal to each other and solving, we showed unique inputs give unique outputs (one-one), and every target value has a valid pre-image (onto).

Exam Tip: When proving onto for rational functions, always ensure the derived expression for \(x\) does not equal the excluded value in the domain.