CBSE Class 11 Mathematics Relations and Functions Assignment Set 02

Relations and Functions MCQ Questions with Answers Class 11 Mathematics

RELATIONS & FUNCTIONS

Question 1. If \(A = \{2,3,5\}\), \(B = \{2,4,6\}\) & \(R\) is a relation from \(A\) to \(B\) defined by \(R = \{(x, y) : x \in A, y \in B \text{ and } x > y\}\), then write \(R\) in the roster form.
Answer: Given sets are \(A = \{2, 3, 5\}\) and \(B = \{2, 4, 6\}\).
The relation \(R\) consists of pairs \((x, y)\) such that \(x \in A\), \(y \in B\), and \(x > y\).
Checking all possible pairs \((x, y)\):
For \(x = 2\): No element in \(B\) is strictly less than 2.
For \(x = 3\): \(3 > 2\), so \((3, 2) \in R$.
For \(x = 5\): \(5 > 2\) and \(5 > 4\), so \((5, 2) \in R\) and \((5, 4) \in R$.
Therefore, in roster form: \(R = \{(3, 2), (5, 2), (5, 4)\}\).
In simple words: We find pairs from set \(A\) and set \(B\) where the first number is greater than the second number.
Exam Tip: Always test every element of set \(A\) against every element of set \(B\) systematically to avoid missing any ordered pair.

Question 2. If \((a, -2)\) & \((4, b^2)\) belongs to the relation \(R\) on \(Z\) where \(R = \{(x, y); x, y \in Z \text{ and } y = 2x - 4\}\) then find the values of \(a\) & \(b\).
Answer: Given relation is \(y = 2x - 4\).
Since \((a, -2) \in R\), substitute \(x = a\) and \(y = -2\):
\(-2 = 2a - 4\)
\(2a = 2 \implies a = 1\)
Since \((4, b^2) \in R\), substitute \(x = 4\) and \(y = b^2\):
\(b^2 = 2(4) - 4\)
\(b^2 = 8 - 4 = 4\)
\(b = \pm 2\)
Therefore, \(a = 1\) and \(b = \pm 2\).
In simple words: Substitute the given coordinates into the relation equation to find the unknown variables.

Exam Tip: Remember that \(b^2 = 4\) gives two values for \(b\): both \(2\) and \(-2\).

Question 3. Express the following relation in roster form: \(R = \{(x, y) : y + 2x = 5, x, y \in W\}\)
Answer: Given equation is \(y + 2x = 5\), which can be written as \(y = 5 - 2x\).
Given \(x, y \in W\) (Whole numbers: \(0, 1, 2, 3, \dots\)).
Substitute values of \(x\) from whole numbers:
If \(x = 0\): \(y = 5 - 2(0) = 5\) (Since \(5 \in W\), \((0, 5) \in R\))
If \(x = 1\): \(y = 5 - 2(1) = 3\) (Since \(3 \in W\), \((1, 3) \in R\))
If \(x = 2\): \(y = 5 - 2(2) = 1\) (Since \(1 \in W\), \((2, 1) \in R\))
If \(x = 3\): \(y = 5 - 2(3) = -1\) (Since \(-1 \notin W\), stop here)
Therefore, roster form: \(R = \{(0, 5), (1, 3), (2, 1)\}\).
In simple words: Find whole number values of \(x\) that give whole number values for \(y\) using the equation \(y = 5 - 2x\).

Exam Tip: Pay close attention to the set domain (\(W\) in this case), as it restricts which calculated values are valid.

Question 4. If \(R\) is the set of all real numbers, what do the Cartesian product \(R \times R\) & \(R \times R \times R\) represent?
Answer: \(R \times R\) represents the Cartesian plane (or two-dimensional space, \(\mathbb{R}^2\)), consisting of all ordered pairs \((x, y)\) where \(x, y \in R\).
\(R \times R \times R\) represents three-dimensional space (or \(\mathbb{R}^3\)), consisting of all ordered triplets \((x, y, z)\) where \(x, y, z \in R\).
In simple words: \(R \times R\) is the 2D coordinate plane and \(R \times R \times R\) is 3D space.

Exam Tip: Associate the number of set repetitions in the Cartesian product with the geometric dimension (\(2\) sets = 2D plane, \(3\) sets = 3D space).

Question 5. If \(A\) and \(B\) are finite sets such that \(n(A) = m\) and \(n(B) = k\), find the number of relations from \(A\) to \(B\).
Answer: Total number of elements in the Cartesian product \(A \times B\) is \(n(A \times B) = n(A) \times n(B) = m \times k = mk\).
A relation from \(A\) to \(B\) is a subset of \(A \times B\).
The total number of subsets of a set with \(mk\) elements is \(2^{mk}\).
Therefore, the total number of relations from \(A\) to \(B\) is \(2^{mk}\).
In simple words: The number of relations is equal to the total number of subsets of \(A \times B\), which is \(2^{mk}\).

Exam Tip: Remember the distinction: total relations = \(2^{mn}\), total functions = \(n^m\).

Question 7. A relation \(f\) defined by \(f : x \to x^2 - 2\), where \(x \in \{-1, -2, 0, 2\}\)
(a) List all the elements of \(f\).
(b) Is \(f\) a function?
Answer: Given \(f(x) = x^2 - 2\) for \(x \in \{-1, -2, 0, 2\}\).
Calculating values for each \(x\):
For \(x = -1\): \(f(-1) = (-1)^2 - 2 = 1 - 2 = -1\)
For \(x = -2\): \(f(-2) = (-2)^2 - 2 = 4 - 2 = 2\)
For \(x = 0\): \(f(0) = (0)^2 - 2 = 0 - 2 = -2\)
For \(x = 2\): \(f(2) = (2)^2 - 2 = 4 - 2 = 2\)
(a) Elements of \(f\) in roster form: \(\{(-1, -1), (-2, 2), (0, -2), (2, 2)\}\).
(b) Yes, \(f\) is a function because every element in the domain has a unique image in the codomain.
In simple words: Plug each given number into the formula to find its output, then check if every input gives exactly one output.

Exam Tip: A relation is a function if no two distinct ordered pairs have the same first element.

Question 8. Let \(f(x) = \sqrt{x}\), \(g(x) = x\) be two functions. Find the following functions:
(a) \(f + g\)
(b) \(f - g\)
(c) \(f \cdot g\)
(d) \(f / g\)
Answer: Given \(f(x) = \sqrt{x}\) (domain \([0, \infty)\)) and \(g(x) = x\) (domain \((-\infty, \infty)\)). The common domain is \([0, \infty)\).
(a) \((f + g)(x) = f(x) + g(x) = \sqrt{x} + x\) for \(x \ge 0\)
(b) \((f - g)(x) = f(x) - g(x) = \sqrt{x} - x\) for \(x \ge 0\)
(c) \((f \cdot g)(x) = f(x) \cdot g(x) = \sqrt{x} \cdot x = x^{3/2}\) for \(x \ge 0\)
(d) \((f / g)(x) = \frac{f(x)}{g(x)} = \frac{\sqrt{x}}{x} = \frac{1}{\sqrt{x}}\) for \(x > 0\)
In simple words: Combine the two function formulas using addition, subtraction, multiplication, or division.

Exam Tip: For division \((f/g)(x)\), remember to exclude points where the denominator \(g(x) = 0\).

Question 9. Let \(A = \{3, 5\}\) and \(B = \{7, 11\}\). Let \(R = \{(a, b) : a \in A, b \in B, a - b \text{ is odd}\}\). Show that \(R\) is an empty relation from \(A\) to \(B\).
Answer: Let us test all possible pairs \((a, b)\) where \(a \in \{3, 5\}\) and \(b \in \{7, 11\pi\)$:

- For \((3, 7)\): \(3 - 7 = -4\) (Even)
- For \((3, 11)\): \(3 - 11 = -8\) (Even)
- For \((5, 7)\): \(5 - 7 = -2\) (Even)
- For \((5, 11)\): \(5 - 11 = -6\) (Even)
Since no pair satisfies the condition that \(a - b\) is odd, \(R = \emptyset\).
Therefore, \(R\) is an empty relation from \(A\) to \(B\).
In simple words: Subtract every element of \(B\) from every element of \(A\); since all differences are even, no pairs satisfy the condition.
Exam Tip: An empty relation contains no elements and is denoted by the empty set symbol \(\emptyset\).

Question 10. Find the range of the following function:
(a) \(F(x) = x^2 + 3$
(b) \(g(x) = \frac{x^2 + 1}{3x - 2}\) [Note: OCR corrected expression]
Answer:
(a) Given \(F(x) = x^2 + 3\).
Since \(x^2 \ge 0\) for all real \(x\), adding \(3\) gives:
\(x^2 + 3 \ge 3$
Range = \([3, \infty)\).
(b) Given \(g(x) = \frac{x^2 + 1}{3x - 2}\).
Let \(y = \frac{x^2 + 1}{3x - 2}\)
\(y(3x - 2) = x^2 + 1$
\(3xy - 2y = x^2 + 1$
\(x^2 - 3yx + (2y + 1) = 0$
Since \(x\) is real, the discriminant (\(\Delta\)) of this quadratic equation must be greater than or equal to \(0\):
\(\Delta = (-3y)^2 - 4(1)(2y + 1) \ge 0$
\(9y^2 - 8y - 4 \ge 0$
Solving the quadratic equation \(9y^2 - 8y - 4 = 0\) using the quadratic formula:
\(y = \frac{8 \pm \sqrt{(-8)^2 - 4(9)(-4)}}{2(9)} = \frac{8 \pm \sqrt{64 + 144}}{18} = \frac{8 \pm \sqrt{208}}{18} = \frac{8 \pm 4\sqrt{13}}{18} = \frac{4 \pm 2\sqrt{13}}{9}\)
Thus, the range is \(\left(-\infty, \frac{4 - 2\sqrt{13}}{9}\right] \cup \left[\frac{4 + 2\sqrt{13}}{9}, \infty\right)\).
In simple words: For part (a), use minimum values of squared terms. For part (b), set the function equal to \(y\) and use the discriminant condition for real roots.

Exam Tip: When finding the range of rational functions of the form \(\frac{quadratic}{linear}\) or \(\frac{quadratic}{quadratic}\), express \(x\) in terms of \(y\) and apply \(\Delta \ge 0\).

Question 11. Draw the graph of the following functions & determine their range.
a. Greatest integer function \(f(x) = [x]\)
b. \(f(x) = \begin{cases} 1 - x, & x < 0 \\ 1, & x = 0 \\ x + 1, & x > 0 \end{cases}\)
Answer:
a. Greatest integer function \(f(x) = [x]\):
- Graph: Consists of a series of horizontal line segments (steps) of length \(1\) at integer heights, with a closed circle at the left end and an open circle at the right end.
- Range: The set of all integers \(\mathbb{Z}\).
b. Piecewise function \(f(x)\):
- For \(x < 0\), \(f(x) = 1 - x\) (line sloping downwards from \((0,1)\)).
- At \(x = 0\), \(f(0) = 1\).
- For \(x > 0\), \(f(x) = x + 1\) (line sloping upwards from \((0,1)\)).
- Range: \((0, \infty)\).
In simple words: Plot each piece of the function separately according to the given conditions.

Exam Tip: Use solid dots for points included in the domain (\(\le\) or \(\ge\)) and hollow circles for excluded points (\(<\) or \(>\)).

Question 12. Find the domain of the following function:
a. \(f(x) = \sqrt{x^2 - 4}\)
b. \(g(x) = 1 - |x - 3|\)
c. \(h(x) = \frac{x + 7}{x^2 - 8x + 4}\) [Note: denominator normalized from OCR]
Answer:
a. For \(f(x) = \sqrt{x^2 - 4}\) to be defined, the expression under the square root must be non-negative:
\(x^2 - 4 \ge 0 \implies (x - 2)(x + 2) \ge 0$
Domain = \((-\infty, -2] \cup [2, \infty)\).
b. For \(g(x) = 1 - |x - 3|\):
Since the absolute value function is defined for all real numbers, Domain = \((-\infty, \infty)\) (or \(R\)).
c. For \(h(x) = \frac{x + 7}{x^2 - 8x + 4}\):
The denominator cannot be zero:
\(x^2 - 8x + 4 \neq 0$
Using the quadratic formula to find roots: \(x = \frac{8 \pm \sqrt{(-8)^2 - 4(1)(4)}}{2} = \frac{8 \pm \sqrt{48}}{2} = 4 \pm 2\sqrt{3}\).
Domain = \(R \setminus \{4 - 2\sqrt{3}, 4 + 2\sqrt{3}\}\).
In simple words: Exclude values where denominators are zero or values that make square roots negative.

Exam Tip: Always check two main restrictions for domains: denominators cannot be \(0$, and expressions inside even roots must be \(\ge 0\).

Question 13. If \(A = \{1, 2, 3\}\), \(B = \{4, 5\}\) and \(C = \{5, 6\}\), then verify that : \(A \times (B - C) = (A \times B) - (A \times C)\)
Answer: Given \(A = \{1, 2, 3\}\), \(B = \{4, 5\}\), \(C = \{5, 6\}\).
Step 1: Find \(B - C\)
\(B - C = \{4, 5\} \setminus \{5, 6\} = \{4\}\)
Step 2: Find LHS = \(A \times (B - C)\)
\(A \times (B - C) = \{1, 2, 3\} \times \{4\} = \{(1, 4), (2, 4), (3, 4)\}\)
Step 3: Find \(A \times B\) and \(A \times C$
\(A \times B = \{1, 2, 3\} \times \{4, 5\} = \{(1, 4), (1, 5), (2, 4), (2, 5), (3, 4), (3, 5)\}\)
\(A \times C = \{1, 2, 3\} \times \{5, 6\} = \{(1, 5), (1, 6), (2, 5), (2, 6), (3, 5), (3, 6)\}\)
Step 4: Find RHS = \((A \times B) - (A \times C)\)
Removing common pairs present in \(A \times C\) from \(A \times B\):
RHS = \(\{(1, 4), (2, 4), (3, 4)\}\)
Since LHS = RHS, the property is verified.
In simple words: Calculate both sides of the equation step-by-step using set operations and Cartesian products to show they result in identical sets.

Exam Tip: Cartesian product distributes over set difference: \(A \times (B - C) = (A \times B) - (A \times C)\).

Question 14. If \(A = \{1, 2\}\), \(B = \{2, 3\}\) and \(C = \{0\}\), Form the set \(A \times B \times C\).
Answer: First, find \(A \times B\):
\(A \times B = \{1, 2\} \times \{2, 3\} = \{(1, 2), (1, 3), (2, 2), (2, 3)\}\)
Now, find \((A \times B) \times C\):
\(A \times B \times C = \{(1, 2, 0), (1, 3, 0), (2, 2, 0), (2, 3, 0)\}\).
In simple words: Combine every pair from \(A \times B\) with each element of set \(C\) to form ordered triplets.

Exam Tip: The Cartesian product of three sets results in ordered triplets \((x, y, z)\).

Question 15. In a class of 25 students, 12 had taken maths., 8 had taken Maths. but not Biology. If each student has taken at least one subject, find the number of students who had taken:
a. Biology but not math.
b. Both Maths. & Biology.
Answer: Let \(M\) be the set of students who took Mathematics and \(B\) be the set of students who took Biology.
Given:
Total students \(n(M \cup B) = 25$
Students who took Maths \(n(M) = 12$
Students who took Maths but not Biology \(n(M \setminus B) = 8$
We know that \(n(M) = n(M \cap B) + n(M \setminus B)\):
\(12 = n(M \cap B) + 8 \implies n(M \cap B) = 12 - 8 = 4$
(b) Number of students who took both Maths and Biology = \(4$.
Total students formula: \(n(M \cup B) = n(M) + n(B) - n(M \cap B)\)
\(25 = 12 + n(B) - 4$
\(25 = 8 + n(B) \implies n(B) = 17$
(a) Number of students who took Biology but not math \(n(B \setminus M) = n(B) - n(M \cap B) = 17 - 4 = 13\).
In simple words: Use set difference formulas and Venn diagram principles to find intersection and disjoint portions.

Exam Tip: Remember the fundamental relation: \(n(A \setminus B) = n(A) - n(A \cap B)\).

Question 16. A and B are two sets such that \(n(A - B) = 14 + x\), \(n(B - A) = 3x\) and \(n(A \cap B) = x\). If \(n(A) = n(B)\), find the value of \(x\) and \(n(A \cup B)\).
Answer: Given:
\(n(A \setminus B) = 14 + x$
\(n(B \setminus A) = 3x$
\(n(A \cap B) = x$
Since \(n(A) = n(A \setminus B) + n(A \cap B)\):
\(n(A) = (14 + x) + x = 14 + 2x$
Since \(n(B) = n(B \setminus A) + n(A \cap B)\):
\(n(B) = 3x + x = 4x$
Given \(n(A) = n(B)\):
\(14 + 2x = 4x \implies 2x = 14 \implies x = 7$
Now find \(n(A)\) and \(n(B)\):
\(n(A) = 14 + 2(7) = 28\), \(n(B) = 4(7) = 28$
\(n(A \cup B) = n(A \setminus B) + n(B \setminus A) + n(A \cap B) = (14 + x) + 3x + x = 14 + 5x = 14 + 5(7) = 14 + 35 = 49\).
In simple words: Set the expressions for \(n(A)\) and \(n(B)\) equal to each other to solve for \(x\).

Exam Tip: Total union \(n(A \cup B)\) is the sum of the three disjoint regions: only \(A\), only \(B\), and both \(A\) and \(B\).

Question 17. If \(A \times B = \{(a, 1), (b, 3), (a, 3), (b, 1), (a, 2), (b, 2)\}\). Find \(B \times A\).
Answer: From the given Cartesian product \(A \times B\):
Set \(A\) contains all first elements: \(A = \{a, b\}\)
Set \(B\) contains all second elements: \(B = \{1, 3, 2\} = \{1, 2, 3\}\)
Now, find \(B \times A\) by reversing each ordered pair in \(A \times B\):
\(B \times A = \{(1, a), (3, b), (3, a), (1, b), (2, a), (2, b)\}\).
In simple words: Swap the coordinates in every pair of \(A \times B\) to get \(B \times A\).

Exam Tip: The first elements of \(A \times B\) form set \(A\), and the second elements form set \(B\).

Question 18. Find the domain and range of the following function:
a. \(\sqrt{16 - x^2}\)
b. \(\frac{x^2 - 4}{x - 2}\)
Answer:
a. Let \(f(x) = \sqrt{16 - x^2}\).
- Domain: For \(f(x)\) to be real, \(16 - x^2 \ge 0 \implies x^2 \le 16 \implies -4 \le x \le 4\). Domain = \([-4, 4]\).
- Range: As \(x\) varies from \(-4\) to \(4\), \(x^2\) ranges from \(0\) to \(16\). Thus \(16 - x^2\) ranges from \(0\) to \(16\), and its square root ranges from \(0\) to \(4\). Range = \([0, 4]\).
b. Let \(g(x) = \frac{x^2 - 4}{x - 2}\).
- Domain: Denominator cannot be zero, so \(x - 2 \neq 0 \implies x \neq 2\). Domain = \(R \setminus \{2\}\).
- Range: Simplify the function for \(x \neq 2\):
\(g(x) = \frac{(x - 2)(x + 2)}{x - 2} = x + 2$
Since \(x \neq 2\), \(g(x) \neq 2 + 2 = 4\). Range = \(R \setminus \{4\}\).
In simple words: Find valid inputs for the domain, then simplify or analyze output limits for the range.

Exam Tip: Always simplify rational functions before determining their range, and remember to exclude the hole created by canceling common factors from the range.