CBSE Class 12 Mathematics Relations And Functions Class Test Set 08

XII - Relations and Functions - Test No. 3

Q1. Define 1-1 and onto function, also represent it diametrically.
Answer: 1. One-One Function (Injective): 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 every \(x_1, x_2 \in A\),
\(f(x_1) = f(x_2) \implies x_1 = x_2\)
Alternatively, \(x_1 \neq x_2 \implies f(x_1) \neq f(x_2)\).
2. Onto Function (Surjective): A function \(f: A \to B\) is said 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 \(f(x) = y\).
3. Diagrammatic Representation:
For a one-one function, arrows from distinct elements in set A point to different elements in set B without converging on the same target element.
For an onto function, every element in set B has at least one incoming arrow from set A.
In simple words: A one-one function maps different inputs to different outputs, and an onto function covers every possible target value without leaving any element unmapped in the codomain.

Exam Tip: Always use the algebraic method \(f(x_1) = f(x_2) \to x_1 = x_2\) to prove one-one, and solve \(f(x) = y\) in terms of \(x\) to prove onto.

Q2. Prove that the function \(f: R \to R: f(x) = 2x\) is one-one and onto.
Answer: Given function: \(f: R \to R\) defined by \(f(x) = 2x\).
1. Injectivity (One-One):
Let \(x_1, x_2 \in R\) be 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.
2. Surjectivity (Onto):
Let \(y \in R\) be any arbitrary element in the codomain \(R\).
We need to find an element \(x \in R\) in the domain such that:
\(f(x) = y\)
\(2x = y\)
\(x = \frac{y}{2}\)
Since \(y \in R\) (real number), \(\frac{y}{2}\) is also a real number. 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.
Since \(f\) is both one-one and onto, it is a bijective function.
In simple words: Multiplying real numbers by 2 always gives unique answers for unique inputs, and every real number can be halved to find its pre-image.

Exam Tip: For linear functions over real numbers like \(ax + b\) (where \(a \neq 0\)), the function is 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: Given sets: \(A = \{1, 2, 3\}\) and \(B = \{4, 5, 6, 7\}\).
Given function represented as set of ordered pairs:
\(f = \{(1, 4), (2, 5), (3, 6)\}\)
The images of the elements of domain \(A\) under \(f\) are:
\(f(1) = 4\)
\(f(2) = 5\)
\(f(3) = 6\)
Observe that all the elements in domain \(A\) have distinct images in codomain \(B\):
\(4 \neq 5 \neq 6\)
Since distinct elements of \(A\) have distinct images in \(B\), the function \(f\) is a one-one function.
In simple words: Since no two numbers in set A point to the same number in set B, the function is one-one.

Exam Tip: When a function is given in roster form as ordered pairs, check if all second coordinates (outputs) are distinct to prove it 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: Given function: \(f: A \to B\) where \(A = R - \{3\}\) and \(B = R - \{1\}\), defined by:
\(f(x) = \frac{x-2}{x-3}\)
1. Injectivity (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:
\((x_1 - 2)(x_2 - 3) = (x_2 - 2)(x_1 - 3)\)
\(x_1x_2 - 3x_1 - 2x_2 + 6 = x_1x_2 - 3x_2 - 2x_1 + 6\)
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.
2. Surjectivity (Onto):
Let \(y \in B\) such that \(y \in R - \{1\}\).
We write \(f(x) = y\):
\(\frac{x-2}{x-3} = y\)
\(x - 2 = y(x - 3)\)
\(x - 2 = xy - 3y\)
Group terms with \(x\) on one side:
\(x - xy = 2 - 3y\)
\(x(1 - y) = 2 - 3y\)
\(x = \frac{2 - 3y}{1 - y}\)
Since \(y \neq 1\) (as \(y \in B = R - \{1\}\)), the denominator \(1 - y \neq 0\).
Also, let us check if \(x = 3\):
If \(\frac{2 - 3y}{1 - y} = 3\), then \(2 - 3y = 3 - 3y \implies 2 = 3\), which is a contradiction.
Thus, for every \(y \in B\), there exists a unique real number \(x = \frac{2 - 3y}{1 - y} \in A\) such that \(f(x) = y\).
Therefore, \(f\) is onto.
Conclusion: The function \(f\) is both one-one and onto.
In simple words: By solving the equation algebraically, we showed that every distinct input gives a unique output, and every output in the target set has a valid matching input.

Exam Tip: Always state domain restrictions clearly when testing for onto; here, ensure the denominator for the pre-field formula does not become zero or equal to excluded domain points.