CBSE Class 12 Mathematics Relations And Functions Assignment Set 14

Question 1. Define one-one and onto function. Give example and also draw diagram.
Answer:
1. One-One Function (Injective Function):
A function \(f: A \to B\) is defined to be one-one (or injective) 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, \(x_1 \neq x_2 \implies f(x_1) \neq f(x_2)\).

Example: Let \(f: \mathbb{R} \to \mathbb{R}\) be defined by \(f(x) = 2x\).
If \(f(x_1) = f(x_2)\), then \(2x_1 = 2x_2 \implies x_1 = x_2\). Hence, \(f\) is one-one.

2. Onto Function (Surjective Function):
A function \(f: A \to B\) is said to be onto (or surjective) 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 \(f(x) = y\).

Example: Let \(f: \mathbb{R} \to \mathbb{R}\) be defined by \(f(x) = x + 1\).
For any \(y \in \mathbb{R}\) (co-domain), we can find \(x = y - 1 \in \mathbb{R}\) such that \(f(x) = (y - 1) + 1 = y\). Hence, \(f\) is onto.

In simple words: A one-one function maps different inputs to different outputs, while an onto function ensures no element in the target set is left unmapped.

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 it is onto, solve the equation \(f(x) = y\) for \(x\) in terms of \(y\) and check if \(x\) belongs to the domain for all \(y\) in the co-domain.

Question 2. What is the range of the function \(f(x) = \frac{|x-1|}{x-1}\)?
Answer:
Given function is \(f(x) = \frac{|x-1|}{x-1}\), defined for \(x \neq 1\).
We know the definition of the absolute value function:
\(|x-1| = x - 1\) if \(x > 1\)
\(|x-1| = -(x - 1)\) if \(x < 1\)
Case 1: When \(x > 1\), \(f(x) = \frac{x-1}{x-1} = 1\)
Case 2: When \(x < 1\), \(f(x) = \frac{-(x-1)}{x-1} = -1\)
Therefore, the range of the function is \(\{-1, 1\}\).
In simple words: The function outputs either 1 or -1 depending on whether the input is greater than or less than 1.

Exam Tip: Always break modulus functions into cases based on the critical points where the expression inside the modulus becomes zero.

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 sets: \(A = \{1, 2, 3\}\) and \(B = \{4, 5, 6, 7\}\)
Function \(f = \{(1, 4), (2, 5), (3, 6)\}\)
Here, images of elements of \(A\) under \(f\) are:
\(f(1) = 4\)
\(f(2) = 5\)
\(f(3) = 6\)
Since distinct elements of \(A\) have distinct images in \(B\) (no two elements map to the same output), 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 the function is one-one.

Exam Tip: For finite sets given in roster form, check if all second coordinates in the ordered pairs are distinct to confirm 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 \(n \in \mathbb{N}\)
Find whether the function is bijective.
Answer:
To check for injectivity (one-one):
Let us test with two distinct natural numbers, say \(n_1 = 1\) (odd) and \(n_2 = 2\) (even).
\(f(1) = \frac{1+1}{2} = 1\)
\(f(2) = \frac{2}{2} = 1\)
Here, \(f(1) = f(2) = 1\), but \(1 \neq 2\).
Since distinct elements 1 and 2 have the same image 1, the function \(f\) is not one-one.
To check for surjectivity (onto):
For any \(y \in \mathbb{N}\) (co-domain), let us choose an even number \(n = 2y \in \mathbb{N}\).
Then \(f(2y) = \frac{2y}{2} = y\).
Thus, every element \(y \ln \mathbb{N}\) has a pre-image \(2y \in \mathbb{N}\). Hence, \(f\) is onto.
Since the function is not one-one, it is not bijective.
In simple words: The function is onto because every counting number has a pre-image, but it is not one-one because both 1 and 2 give an output of 1. Therefore, it is not 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 \(f(x_1) = f(x_2)\).
\(\frac{x_1 - 2}{x_1 - 3} = \frac{x_2 - 2}{x_2 - 3}\)
\((x_1 - 2)(x_2 - 3) = (x_2 - 2)(x_1 - 3)\)
\(x_1 x_2 - 3x_1 - 2x_2 + 6 = x_1 x_2 - 3x_2 - 2x_1 + 6\)
\(-3x_1 - 2x_2 = -3x_2 - 2x_1\)
\(-x_1 = -x_2 \implies x_1 = x_2\)
Therefore, \(f\) is one-one.

2. Check for onto:
Let \(y \in B\) (co-domain), where \(y \neq 1\).
Let \(f(x) = y \implies \frac{x-2}{x-3} = y\)
\(x - 2 = y(x - 3)\)
\(x - 2 = xy - 3y\)
\(x - xy = 2 - 3y\)
\(x(1 - y) = 2 - 3y\)
\(x = \frac{2 - 3y}{1 - y}\)
Since \(y \neq 1\), the denominator \(1 - y \neq 0\), and since \(y \in \mathbb{R} - \{1\}\), \(x = \frac{3y - 2}{y - 1} \neq 3\), so \(x \in A\).
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 is both one-one and onto because every distinct input gives a unique output, and every number in the target set has a valid matching input.

Exam Tip: When proving onto for rational functions, express \(x\) in terms of \(y\) and verify that the resulting expression is well-defined for all values in the co-domain.

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 \(f(x_1) = f(x_2)\).
\(3 - 4x_1 = 3 - 4x_2\)
Subtract 3 from both sides: \(-4x_1 = -4x_2\)
Divide by -4: \(x_1 = x_2\)
Hence, \(f\) is one-one.

2. Surjectivity (Onto):
Let \(y \in \mathbb{R}\) be any element in the co-domain.
Let \(f(x) = y \implies 3 - 4x = y\)
\(4x = 3 - y\)
\(x = \frac{3 - y}{4}\)
Since \(y \in \mathbb{R}\), \(x = \frac{3 - y}{4}\) is also a real number (\(x \in \mathbb{R}\)).
Thus, for every \(y \in \mathbb{R}\), there exists \(x = \frac{3 - y}{4} \in \mathbb{R}\) such that:
\(f\left(\frac{3 - y}{4}\right) = 3 - 4\left(\frac{3 - y}{4}\right) = 3 - (3 - y) = y\)
Hence, \(f\) 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: Linear functions from \(\mathbb{R} to \mathbb{R}\) with non-zero slope are always bijections because lines are strictly monotonic and cover all real heights.

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 \(x_2 = -1\). Both belong to \(\mathbb{R}\).
\(f(1) = 1 + (1)^2 = 2\)
\(f(-1) = 1 + (-1)^2 = 2\)
Here, \(f(1) = f(-1) = 2\), but \(1 \neq -1\).
Therefore, \(f\) is not one-one.

2. Check for onto:
For any \(x \in \mathbb{R}\), we know that \(x^2 \ge 0\).
Adding 1 to both sides: \(1 + x^2 \ge 1\)
Thus, the range of \(f\) is \([1, \infty)\), which is a proper subset of the co-domain \(\mathbb{R}\).
Negative real numbers like -1 have no pre-image in \(\mathbb{R}\). For instance, \(1 + x^2 = -1 \implies x^2 = -2\) (not possible in real numbers).
Therefore, \(f\) is not onto.
Conclusion: The function is neither one-one nor onto.
In simple words: Negative outputs are impossible because of \(x^2\), and both positive and negative inputs give the same output (like 1 and -1 both giving 2).

Exam Tip: Even powers like \(x^2\) on real domains immediately fail injectivity due to symmetry around zero, and fail surjectivity due to bounded range.

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. Test for 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 domain is \(\mathbb{N}\) (positive integers), \(x_1 = x_2\).
Therefore, \(f\) is one-one.

2. Test for surjectivity (Onto):
Consider the co-domain element \(2 \in \mathbb{N}\).
Let \(f(x) = 2 \implies x^2 = 2\)
\(x = \sqrt{2}\)
Since \(\sqrt{2} \notin \mathbb{N}\), there is no natural number whose square is 2.
Thus, element 2 in the co-domain has no pre-image in \(\mathbb{N}\).
Therefore, \(f\) is not onto.
In simple words: Different natural numbers give different squares (making it one-one), but not all natural numbers are perfect squares like 2 or 3 (making it not onto).

Exam Tip: Always pay close attention to the domain and co-domain sets (\(\mathbb{N}, \mathbb{Z}, \mathbb{R}\)); they completely change whether a function is onto.

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 \(f(x_1) = f(x_2)\).
\(x_1^3 = x_2^3\)
\(x_1^3 - x_2^3 = 0\)
\((x_1 - x_2)(x_1^2 + x_1 x_2 + x_2^2) = 0\)
For real numbers, the quadratic factor \(x_1^2 + x_1 x_2 + x_2^2 = \left(x_1 + \frac{x_2}{2}\right)^2 + \frac{3}{4}x_2^2\) is always greater than or equal to 0, and is non-zero for all non-zero \(x_1, x_2\).
Thus, \(x_1 - x_2 = 0 \implies x_1 = x_2\).
Therefore, \(f\) is injective (one-one).
In simple words: Cubing a real number preserves its sign and uniqueness, meaning different numbers always result in different cubes.

Exam Tip: Odd-degree polynomial functions like \(x^3\) over real numbers are always strictly increasing and hence always injective.

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 \(f(x)\) in place of \(x\) in the function definition:
\(f(f(x)) = (f(x))^2 - 3(f(x)) + 2\)
Substitute \(f(x) = x^2 - 3x + 2\):
\(f(f(x)) = (x^2 - 3x + 2)^2 - 3(x^2 - 3x + 2) + 2\)
Now, expand the square term using \((a+b+c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca\):
\((x^2 - 3x + 2)^2 = (x^2)^2 + (-3x)^2 + (2)^2 + 2(x^2)(-3x) + 2(-3x)(2) + 2(2)(x^2)\)
\(= x^4 + 9x^2 + 4 - 6x^3 - 12x + 4x^2\)
\(= x^4 - 6x^3 + 13x^2 - 12x + 4\)
Next, expand the second term:ப்பிட
\(-3(x^2 - 3x + 2) = -3x^2 + 9x - 6\)
Combine all the terms together:
\(f(f(x)) = (x^4 - 6x^3 + 13x^2 - 12x + 4) + (-3x^2 + 9x - 6) + 2\)
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: To find composite function \(f(f(x))\), replace every \(x\) in the original equation with the entire expression of \(f(x)\) and simplify algebraically.

Exam Tip: Be extremely careful with signs when expanding trinomial squares and distributing negative multipliers.