RBSE Solutions Class 11 Maths Chapter 2 Relations and Functions Exercise 2.3

Question 1. Examine which of the following is/are functions:
(i) {(1, 2), (2, 3), (3, 4), (2, 1)}
(ii) {(a, 0), (b, 0), (c, 1), (d, 1)}
(iii) {(1, a), (2, b), (1, b), (2, a)}
(iv) {(a, a), (b, b), (c, c)}
(v) {(a, b)}
(vi) {(4, 1), (4, 2), (4, 3), (4, 4)}
(vii) {(1, 4), (2, 4), (3, 4), (4, 4)}
(viii) {(x, y) | x, y ∈ R \( \wedge y^2 = x \)}
(ix) {(x, y) | x, y ∈ R \( \wedge x^2 = y \)}
(x) {(x, y) | x, y ∈ R \( \wedge x = y^3 \)}
(xi) {(x, y) | x, y ∈ R \( \wedge y = x^3 \)}
Answer:
(i) {(1, 2), (2, 3), (3, 4), (2, 1)}
This is not a function. An input (like 2) maps to two different outputs (3 and 1), which is against the rule for functions. Functions are important in math because they represent predictable relationships where each input has a unique result.
(ii) {(a, 0), (b, 0), (c, 1), (d, 1)}
This is a function. Each unique first element (a, b, c, d) has only one corresponding second element (0 or 1). This type of mapping, where multiple inputs can share the same output, is common in many real-world scenarios.
(iii) {(1, a), (2, b), (1, b), (2, a)}
This is not a function. The input '1' maps to both 'a' and 'b'. Also, input '2' maps to both 'b' and 'a'. Functions require consistency; if an input gives different outputs, it's not a function.
(iv) {(a, a), (b, b), (c, c)}
This is a function. Each first element (a, b, c) is unique and maps to exactly one second element (a, b, c respectively). This is an example of an identity function, where the output is always the same as the input.
(v) {(a, b)}
This is a function. This set has only one ordered pair. The input 'a' maps to exactly one output 'b'. Even a single pair can form a valid function, representing a simple, direct relationship.
(vi) {(4, 1), (4, 2), (4, 3), (4, 4)}
This is not a function. The first element '4' maps to multiple different second elements (1, 2, 3, and 4). A common way to visualize this is the vertical line test; if a vertical line crosses the graph more than once, it's not a function.
(vii) {(1, 4), (2, 4), (3, 4), (4, 4)}
This is a function. Each first element (1, 2, 3, 4) is unique, and each maps to a single output (which is 4 for all of them). This is an example of a constant function, where all inputs produce the same output.
(viii) {(x, y) | x, y ∈ R \( \wedge y^2 = x \)}
This is not a function. The relation \( y^2 = x \) means that for a single positive value of \( x \), \( y \) can have two different real values (e.g., if \( x=4 \), then \( y \) can be \( 2 \) or \( -2 \)). For a function, every element in the domain must map to exactly one element in the codomain.
(ix) {(x, y) | x, y ∈ R \( \wedge x^2 = y \)}
This is a function. For any real number \( x \), the value of \( y = x^2 \) will always be a single, unique real number. This represents a parabola opening upwards, and it passes the vertical line test.
(x) {(x, y) | x, y ∈ R \( \wedge x = y^3 \)}
This is a function. The relation \( x = y^3 \) can be rewritten as \( y = \sqrt[3]{x} \). For every real number \( x \), there is only one unique real cube root \( y \). The cube root function is defined for all real numbers, unlike the square root function.
(xi) {(x, y) | x, y ∈ R \( \wedge y = x^3 \)}
This is a function. For every real number \( x \), the value of \( y = x^3 \) will be a single, unique real number. This is a cubic function, which also passes the vertical line test.
In simple words: A relation is a function if each input has exactly one output. If an input leads to more than one output, it is not a function.

🎯 Exam Tip: To quickly check if a relation is a function, look for any input value (the first element in an ordered pair or 'x' in an equation) that corresponds to more than one output value (the second element or 'y').

Question 2. If f : R \( \to \) R, f(x) = x², then find
(i) Range of f,
(ii) {x | f(x) = 4},
(iii) {y | f(y) = -1}
Answer:
(i) Range of f:
Since \( f(x) = x^2 \), and any real number squared is always non-negative (zero or positive), the range of \( f \) includes all real numbers greater than or equal to zero. So, the range is \( [0, \infty) \). The squaring operation always turns negative numbers positive and keeps positive numbers positive, while zero remains zero.
(ii) {x | f(x) = 4}:
We need to find the values of \( x \) for which \( f(x) = 4 \). Since \( f(x) = x^2 \), we set \( x^2 = 4 \). Taking the square root of both sides gives \( x = \pm 2 \). So, the set of such \( x \) values is \( \{-2, 2\} \). Both positive and negative numbers, when squared, result in a positive number.
(iii) {y | f(y) = -1}:
We are looking for values of \( y \) where \( f(y) = -1 \). Substituting into the function, we get \( y^2 = -1 \). In real numbers, there is no number whose square is negative. Therefore, the set of such \( y \) values is an empty set or null set, denoted by \( \Phi \). In the real number system, the square of any real number can never be negative.
In simple words: The range of \( f(x) = x^2 \) is all numbers from zero to infinity because squaring any real number always gives a positive result or zero. For \( f(x) = 4 \), \( x \) can be 2 or -2. For \( f(y) = -1 \), there is no real number solution.

🎯 Exam Tip: Remember that the square of any real number is always non-negative. This is a key property when determining ranges or solving equations involving squares in real number systems.

Question 3. Let A = {-2, -1, 0, 1, 2} and function f is defined in A to R by f(x) = x² + 1. Find the range of f.
Answer: To find the range, we apply the function \( f(x) = x^2 + 1 \) to each element in the set \( A \):
For \( x = -2 \), \( f(-2) = (-2)^2 + 1 = 4 + 1 = 5 \).
For \( x = -1 \), \( f(-1) = (-1)^2 + 1 = 1 + 1 = 2 \).
For \( x = 0 \), \( f(0) = (0)^2 + 1 = 0 + 1 = 1 \).
For \( x = 1 \), \( f(1) = (1)^2 + 1 = 1 + 1 = 2 \).
For \( x = 2 \), \( f(2) = (2)^2 + 1 = 4 + 1 = 5 \).
The unique output values are 1, 2, and 5. So, the range of \( f \) is \( \{1, 2, 5\} \). The range of a function is the set of all possible output values from its domain.
In simple words: We put each number from set A into the rule \( x^2 + 1 \) and write down all the different answers we get. These answers make up the range.

🎯 Exam Tip: When finding the range for a specific domain, always list the unique output values. Duplicates should only be counted once in the final set.

Question 4. Let A = {-2, -1, 0, 1, 2} and f : A \( \to \) Z where f(x) = x² + 2x – 3, then find
(i) Range of f
(ii) Pre image of 6, -3 and 5
Answer:
(i) Range of f:
To find the range, we calculate \( f(x) \) for each element in set \( A \):
For \( x = -2 \), \( f(-2) = (-2)^2 + 2(-2) - 3 = 4 - 4 - 3 = -3 \).
For \( x = -1 \), \( f(-1) = (-1)^2 + 2(-1) - 3 = 1 - 2 - 3 = -4 \).
For \( x = 0 \), \( f(0) = (0)^2 + 2(0) - 3 = 0 + 0 - 3 = -3 \).
For \( x = 1 \), \( f(1) = (1)^2 + 2(1) - 3 = 1 + 2 - 3 = 0 \).
For \( x = 2 \), \( f(2) = (2)^2 + 2(2) - 3 = 4 + 4 - 3 = 5 \).
The distinct values in the range are \( \{-4, -3, 0, 5\} \). Quadratic functions like this produce a parabolic shape, and their range is determined by the vertex and the direction it opens.
(ii) Pre image of 6, -3 and 5:
To find the pre-image of 6, we set \( f(x) = 6 \):
\( x^2 + 2x - 3 = 6 \)
\( \implies x^2 + 2x - 9 = 0 \)
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\( x = \frac{-2 \pm \sqrt{2^2 - 4(1)(-9)}}{2(1)} \)
\( \implies x = \frac{-2 \pm \sqrt{4 + 36}}{2} \)
\( \implies x = \frac{-2 \pm \sqrt{40}}{2} \)
\( \implies x = \frac{-2 \pm 2\sqrt{10}}{2} \)
\( \implies x = -1 \pm \sqrt{10} \)
Since neither \( -1 + \sqrt{10} \) nor \( -1 - \sqrt{10} \) are in the domain \( A = \{-2, -1, 0, 1, 2\} \), the pre-image of 6 is the empty set, \( \Phi \). A pre-image is the input value that produces a specific output value in a function.
To find the pre-image of -3, we set \( f(x) = -3 \):
\( x^2 + 2x - 3 = -3 \)
\( \implies x^2 + 2x = 0 \)
\( \implies x(x + 2) = 0 \)
This gives two solutions: \( x = 0 \) or \( x + 2 = 0 \implies x = -2 \). Both \( 0 \) and \( -2 \) are elements of the domain \( A \). So, the pre-image of -3 is \( \{-2, 0\} \). Sometimes, different inputs can lead to the same output value in a function.
To find the pre-image of 5, we set \( f(x) = 5 \):
\( x^2 + 2x - 3 = 5 \)
\( \implies x^2 + 2x - 8 = 0 \)
Factoring the quadratic equation:
\( (x+4)(x-2) = 0 \)
This gives two solutions: \( x + 4 = 0 \implies x = -4 \) or \( x - 2 = 0 \implies x = 2 \). From the domain \( A = \{-2, -1, 0, 1, 2\} \), only \( x = 2 \) is an element. So, the pre-image of 5 is \( \{2\} \). Always check if the calculated pre-image values actually belong to the given domain set.
In simple words: First, we find the range by putting each number from set A into the function rule and collecting all the output numbers. Then, to find the pre-image, we set the function equal to the target number (6, -3, or 5) and solve for \( x \). We then check if these \( x \) values are part of the original set A.

🎯 Exam Tip: When finding pre-images, always make sure your calculated \( x \) values actually belong to the given domain of the function. If they don't, they are not part of the pre-image set.