CBSE Class 12 Mathematics Relations And Functions Assignment Set 12

Question 1. Define one-one and onto function. Give example and also draw diagram.
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 all \(x_1, x_2 \in A\), \(f(x_1) = f(x_2) \implies x_1 = x_2\) (or equivalently, \(x_1 \neq x_2 \implies f(x_1) \neq f(x_2)\)).
Example: Let \(f: \mathbb{R} \to \mathbb{R}\) be defined by \(f(x) = 2x\). If \(f(x_1) = f(x_2)\), then \(2x_1 = 2x_2 \implies x_1 = x_2\). Thus, \(f\) is one-one.
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 under \(f\). That is, for every \(y \in B\), there exists an element \(x \in A\) such that \(f(x) = y\).
Example: Let \(f: \mathbb{R} \to \mathbb{R}\) be defined by \(f(x) = x + 1\). For any \(y \in \mathbb{R}\), we can choose \(x = y - 1 \in \mathbb{R}\) such that \(f(x) = (y - 1) + 1 = y\). Thus, \(f\) is onto.
Diagram Representation for One-One:
In simple words: A one-one function maps different inputs to different outputs, while an onto function covers every possible output in the target set.

Exam Tip: To prove 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 \(y\) from \(y = f(x)\).

Question 2. What is the range of the function \(f(x) = \frac{|x-1|}{x-1}\)?
Answer: The given function is \(f(x) = \frac{|x-1|}{x-1}\), defined for all \(x \neq 1\).
Case 1: When \(x > 1\), \(|x-1| = x-1\)
\(f(x) = \frac{x-1}{x-1} = 1\)
Case 2: When \(x < 1\), \(|x-1| = -(x-1)\)
\(f(x) = \frac{-(x-1)}{x-1} = -1\)
Thus, the range of the function is the set \(\{-1, 1\}\).
In simple words: The function only outputs 1 when values are greater than 1, and -1 when values are less than 1.

Exam Tip: Remember to split absolute value functions into positive and negative cases based on the domain restrictions.

Question 3. 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. State whether f is one-one or not.
Answer: Given set \(A = \{1, 2, 3\}\) and set \(B = \{4, 5, 6, 7\}\).
The function is given by \(f = \{(1, 4), (2, 5), (3, 6)\}\).
Images of the elements of A are:
\(f(1) = 4\)
\(f(2) = 5\)
\(f(3) = 6\)
Since distinct elements of A have distinct images in B, the function \(f\) is one-one.
In simple words: Every element in set A points to a unique element in set B, so it is one-one.

Exam Tip: For a finite set given as ordered pairs, check if all second elements (outputs) are distinct to determine injectivity.

Question 4. Let \(f: \mathbb{N} \to \mathbb{N}\) be defined as \(f(n) = \begin{cases} \frac{n+1}{2}, & \text{if } n \text{ is odd} \\ \frac{n}{2}, & \text{if } n \text{ is even} \end{cases}\) for all \(n \in \mathbb{N}\). Find whether the function is bijective.
Answer: To check if \(f\) is one-one:
Let us test with two distinct elements, say \(n_1 = 1\) (odd) and \(n_2 = 2\) (even).
\(f(1) = \frac{1+1}{2} = 1\)
\(f(2) = \frac{2}{2} = 1\)
Here, \(f(1) = f(2) = 1\) even though \(1 \neq 2\).
Therefore, \(f\) is not one-one (not injective).
Since a bijective function must be both one-one and onto, and \(f\) is not one-one, the function \(f\) is not bijective.
In simple words: Two different numbers (1 and 2) give the exact same output (1), meaning the function fails the one-one test and cannot be bijective.

Exam Tip: To show a function is not one-one, finding a single counterexample where two different inputs yield the same output is sufficient.

Question 5. Let \(A = \mathbb{R} - \{3\}\) and \(B = \mathbb{R} - \{1\}\). The function \(f: A \to B\) defined by \(f(x) = \frac{x-2}{x-3}\). Is f one-one and onto? Justify your answer.
Answer: 1. Check for one-one:
Let \(x_1, x_2 \in A\) such that \(f(x_1) = f(x_2)\)
\(\frac{x_1 - 2}{x_1 - 3} = \frac{x_2 - 2}{x_2 - 3}\)
\((x_1 - 2)(x_2 - 3) = (x_2 - 2)(x_1 - 3)\)
\(x_1 x_2 - 3x_1 - 2x_2 + 6 = x_1 x_2 - 3x_2 - 2x_1 + 6\)
\(-3x_1 - 2x_2 = -3x_2 - 2x_1\)
\(x_1 = x_2\)
Thus, \(f\) is one-one.
2. Check for onto:
Let \(y \in B\) (where \(y \neq 1\)). Let \(f(x) = y\)
\(\frac{x-2}{x-3} = y\)
\(x - 2 = y(x - 3)\)
\(x - 2 = xy - 3y\)
\(x - xy = 2 - 3y\)
\(x(1 - y) = 2 - 3y\)
\(x = \frac{2 - 3y}{1 - y}\)
Since \(y \neq 1\), \(x\) is well-defined and \(x \in \mathbb{R}\). Also, \(x \neq 3\) because if \(\frac{2 - 3y}{1 - y} = 3\), then \(2 - 3y = 3 - 3y \implies 2 = 3\) (which is impossible).
Thus, for every \(y \in B\), there exists \(x = \frac{2 - 3y}{1 - y} \in A\) such that \(f(x) = y\).
Therefore, \(f\) is onto.
In simple words: The function successfully passes both tests: different inputs give different outputs, and every target value has a corresponding input.

Exam Tip: Always verify that the derived expression for \(x\) in terms of \(y\) does not violate the domain restrictions of set A.

Question 6. \(f: \mathbb{R} \to \mathbb{R}\) defined by \(f(x) = 3 - 4x\). Prove f is a bijective function i.e., one-one and onto.
Answer: 1. Injectivity (One-one):
Let \(x_1, x_2 \in \mathbb{R}\) such that \(f(x_1) = f(x_2)\)
\(3 - 4x_1 = 3 - 4x_2\)
\(-4x_1 = -4x_2\)
\(x_1 = x_2\)
Therefore, \(f\) is one-one.
2. Surjectivity (Onto):
Let \(y \in \mathbb{R}\) (codomain). Let \(f(x) = y\)
\(3 - 4x = y\)
\(4x = 3 - y\)
\(x = \frac{3 - y}{4}\)
Since \(y \in \mathbb{R}\), clearly \(x = \frac{3 - y}{4} \in \mathbb{R}\).
For this \(x\), \(f(x) = 3 - 4\left(\frac{3 - y}{4}\right) = 3 - (3 - y) = y\).
Therefore, \(f\) is onto.
Since \(f\) is both one-one and onto, it is a bijective function.
In simple words: Linear functions of the form \(ax + b\) (where \(a \neq 0\)) over real numbers are always bijective.

Exam Tip: When proving bijectivity, clearly label the two parts as One-one and Onto to secure full marks.

Question 7. Is the following function \(f(x) = 1 + x^2\), \(f: \mathbb{R} \to \mathbb{R}\) is one-one, onto. [No]
Answer: 1. Check for one-one:
Let \(x_1 = 1\) and \(x_2 = -1\).
\(f(1) = 1 + (1)^2 = 2\)
\(f(-1) = 1 + (-1)^2 = 2\)
Since \(f(1) = f(-1) = 2\) but \(1 \neq -1\), the function \(f\) is not one-one.
2. Check for onto:
For any \(x \in \mathbb{R}\), \(x^2 \ge 0 \implies 1 + x^2 \ge 1\).
Therefore, the range of \(f\) is \([1, \infty)\), which is a subset of the codomain \(\mathbb{R}\).
Negative real numbers in the codomain have no pre-image in the domain (e.g., \(y = -1\) gives \(1 + x^2 = -1 \implies x^2 = -2\), which has no real solution).
Thus, \(f\) is not onto.
Since \(f\) is neither one-one nor onto, the answer is No.
In simple words: Squaring numbers means negative outputs are impossible, and both positive and negative inputs give the same result, so it is neither one-one nor onto.

Even powers like \(x^2\) on real domains generally fail both injectivity and surjectivity tests.

Question 8. Test the injectivity and surjectivity of \(f: \mathbb{N} \to \mathbb{N}\) : \(f(x) = x^2\). [ANS: One-one, not onto]
Answer: 1. Injectivity (One-one):
Let \(x_1, x_2 \in \mathbb{N}\) such that \(f(x_1) = f(x_2)\)
\(x_1^2 = x_2^2\)
Since \(x_1, x_2 \in \mathbb{N}\) (positive integers), this gives \(x_1 = x_2\).
Therefore, \(f\) is one-one.
2. Surjectivity (Onto):
Consider an element in the codomain, say \(y = 2 \in \mathbb{N}\).
If \(f(x) = 2\), then \(x^2 = 2 \implies x = \sqrt{2}\).
Since \(\sqrt{2} \notin \mathbb{N}\), there is no natural number \(x\) whose square is 2.
Therefore, \(f\) is not onto.
In simple words: Different natural numbers give different squares (one-one), but not all natural numbers are perfect squares like 2, 3, 5, etc. (not onto).

Exam Tip: Always pay close attention to the domain and codomain sets (\(\mathbb{N}\), \(\mathbb{Z}\), or \(\mathbb{R}\)) as they entirely change surjectivity results.

Question 9. Show that the function \(f: \mathbb{R} \to \mathbb{R}\) given by \(f(x) = x^3\) is injective.
Answer: Let \(x_1, x_2 \in \mathbb{R}\) such that \(f(x_1) = f(x_2)\)
\(x_1^3 = x_3^2 \dots\) -> \(x_1^3 = x_2^3\)
Taking the cube root on both sides:
\(x_1 = x_2\)
Since \(f(x_1) = f(x_2) \implies x_1 = x_2\) for all \(x_1, x_2 \in \mathbb{R}\), the function \(f\) is injective (one-one).
In simple words: Cube roots are unique for every real number, so distinct inputs always produce distinct cubes.

Exam Tip: Odd powers like \(x^3, x^5\) are always injective over the set of real numbers, unlike even powers.

Question 10. If \(f: \mathbb{R} \to \mathbb{R}\) such that \(f(x) = x^2 - 3x + 2\). Find \(f(f(x))\) [ANS: \(x^4 - 6x^3 + 10x^2 - 3x\)]
Answer: Given function: \(f(x) = x^2 - 3x + 2\)
We need to find \(f(f(x))\), which means substituting \(f(x)\) into itself:
\(f(f(x)) = (f(x))^2 - 3(f(x)) + 2\)
Substitute \(f(x) = x^2 - 3x + 2\):
\(f(f(x)) = (x^2 - 3x + 2)^2 - 3(x^2 - 3x + 2) + 2\)
Expand the terms:
1. \((x^2 - 3x + 2)^2 = (x^2)^2 + (-3x)^2 + (2)^2 + 2(x^2)(-3x) + 2(-3x)(2) + 2(x^2)(2)\)
\(= x^4 + 9x^2 + 4 - 6x^3 - 12x + 4x^2\)
\(= x^4 - 6x^3 + 13x^2 - 12x + 4\)
2. \(-3(x^2 - 3x + 2) = -3x^2 + 9x - 6\)
3. Constant term = \(+2\)
Now, combine all expanded parts:
\(f(f(x)) = (x^4 - 6x^3 + 13x^2 - 12x + 4) + (-3x^2 + 9x - 6) + 2\)
Group like powers of \(x\):
\(= x^4 - 6x^3 + (13x^2 - 3x^2) + (-12x + 9x) + (4 - 6 + 2)\)
\(= x^4 - 6x^3 + 10x^2 - 3x + 0\)
\(= x^4 - 6x^3 + 10x^2 - 3x\)
In simple words: Replace every \(x\) in the original function expression with the function itself and carefully expand and simplify the algebra.

Exam Tip: Use the trinomial square formula \((a+b+c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca\) to quickly expand expressions like \((x^2 - 3x + 2)^2\).