CBSE Class 11 Mathematics Relations and Functions Assignment Set 04

VERY SHORT ANSWER TYPE QUESTIONS (1 MARK)

Question 1. Find a and b if \((a - 1, b + 5) = (2, 3)\)
Answer: Equating the corresponding components of the two equal ordered pairs:
\(a - 1 = 2 \implies a = 3\)
\(b + 5 = 3 \implies b = -2\)
Thus, \(a = 3\) and \(b = -2\).
In simple words: When two ordered pairs are equal, their corresponding coordinates are equal.

Exam Tip: Set each coordinate of the first pair equal to the corresponding coordinate of the second pair to form linear equations in one variable.

Question 2. If \(A = \{1, 3, 5\}\), \(B = \{2, 3\}\) find \(A \times B\)
Answer: \(A \times B\) is the set of all ordered pairs whose first element belongs to \(A\) and second element belongs to \(B\).
\(A \times B = \{(1, 2), (1, 3), (3, 2), (3, 3), (5, 2), (5, 3)\}\)
In simple words: Pair every element of set A with every element of set B.

Exam Tip: The total number of ordered pairs in \(A \times B\) is \(n(A) \times n(B) = 3 \times 2 = 6\).

Question 3. If \(A = \{1, 3, 5\}\), \(B = \{2, 3\}\) find \(B \times A\)
Answer: \(B \times A\) is the set of all ordered pairs whose first element belongs to \(B\) and second element belongs to \(A\).
\(B \times A = \{(2, 1), (2, 3), (2, 5), (3, 1), (3, 3), (3, 5)\}\)
In simple words: Pair every element of set B with every element of set A.

Exam Tip: Cartesian product is not commutative, meaning \(A \times B \neq B \times A\).

Question 4. Let \(A = \{1, 2\}\), \(B = \{2, 3, 4\}\), \(C = \{4, 5\}\), find \(A \times (B \cap C)\)
Answer: First, find the intersection of sets \(B\) and \(C\):
\(B \cap C = \{2, 3, 4\} \cap \{4, 5\} = \{4\}\)
Now, find the Cartesian product of \(A\) and \((B \cap C)\):
\(A \times (B \cap C) = \{1, 2\} \times \{4\} = \{(1, 4), (2, 4)\}\)
In simple words: Find common elements of B and C first, then form ordered pairs with elements of A.

Exam Tip: Always evaluate set operations inside parentheses before applying the Cartesian product.

Question 5. Let \(A = \{1, 2\}\), \(B = \{2, 3, 4\}\), \(C = \{4, 5\}\), find \(A \times (B \cup C)\)
Answer: First, find the union of sets \(B\) and \(C\):
\(B \cup C = \{2, 3, 4\} \cup \{4, 5\} = \{2, 3, 4, 5\}\)
Now, find the Cartesian product of \(A\) and \((B \cup C)\):
\(A \times (B \cup C) = \{1, 2\} \times \{2, 3, 4, 5\}\)
\(A \times (B \cup C) = \{(1, 2), (1, 3), (1, 4), (1, 5), (2, 2), (2, 3), (2, 4), (2, 5)\}\)
In simple words: Combine all elements of B and C without repetition, then pair them with elements of A.

Exam Tip: Verify that the number of elements equals \(n(A) \times n(B \cup C) = 2 \times 4 = 8\).

Question 6. If \(P = \{1, 3\}\), \(Q = \{2, 3, 5\}\), find the number of relations from P to Q
Answer: Given \(n(P) = 2\) and \(n(Q) = 3\).
The number of elements in the Cartesian product \(P \times Q\) is \(n(P \times Q) = n(P) \times n(Q) = 2 \times 3 = 6\).
The total number of relations from set P to set Q is equal to the number of subsets of \(P \times Q\), which is given by \(2^{n(P \times Q)}\).
Number of relations = \(2^6 = 64\).
In simple words: The total number of relations is 2 raised to the power of the product of the cardinalities of the two sets.

Exam Tip: Remember the formula: if \(n(A) = p\) and \(n(B) = q\), then number of relations = \(2^{pq}\).

Question 7. If \(A = \{1, 2, 3, 5\}\) and \(B = \{4, 6, 9\}\), \(R = \{(x, y) : |x - y| \text{ is odd, } x \in A, y \in B\}\). Write R in roster form
Answer: We test all pairs \((x, y)\) from \(A \times B\) where the absolute difference \(|x - y|\) is an odd number:
For \(x = 1\): \(|1 - 4| = 3\) (odd), \(|1 - 6| = 5\) (odd), \(|1 - 9| = 8\) (even)
For \(x = 2\): \(|2 - 4| = 2\) (even), \(|2 - 6| = 4\) (even), \(|2 - 9| = 7\) (odd)
For \(x = 3\): \(|3 - 4| = 1\) (odd), \(|3 - 6| = 3\) (odd), \(|3 - 9| = 6\) (even)
For \(x = 5\): \(|5 - 4| = 1\) (odd), \(|5 - 6| = 1\) (odd), \(|5 - 9| = 4\) (even)
Therefore, \(R = \{(1, 4), (1, 6), (2, 9), (3, 4), (3, 6), (5, 4), (5, 6)\}\).
In simple words: Select only those pairs whose subtraction results in an odd number.

Exam Tip: The difference between an odd and an even number is always odd. Use this property to quickly filter pairs.

Question 8. Which of the following relations are functions. Give reason. \(R = \{(1, 1), (2, 2), (3, 3), (4, 4), (4, 5)\}\)
Answer: Not a function.
Reason: The element \(4\) in the domain has two different images, \(4\) and \(5\), which violates the definition of a function where each element must have a unique image.
In simple words: A relation is not a function if one input gives multiple outputs.

Exam Tip: Check if any first element in the ordered pairs repeats with a different second element.

Question 9. Which of the following relations are functions. Give reason. \(R = \{(2, 1), (2, 2), (2, 3), (2, 4)\}\)
Answer: Not a function.
Reason: The element \(2\) in the domain has four different images (\(1, 2, 3, 4\)), whereas a function must assign a unique image to every domain element.
In simple words: A single input cannot have multiple outputs in a function.

Exam Tip: Look for repeated x-coordinates with different y-coordinates.

Question 10. Which of the following relations are functions. Give reason. \(R = \{(1, 2), (2, 5), (3, 8), (4, 10), (5, 12), (6, 12)\}\)
Answer: Function.
Reason: Every element in the domain \(\{1, 2, 3, 4, 5, 6\}\) has one and only one unique image in the codomain. (Note: different domain elements having the same image \(12\) is permitted).
In simple words: It is a function because each input has exactly one output, even if two inputs share the same output.

Exam Tip: Distinct inputs mapping to the same output still form a valid function (many-to-one mapping).

Question 11. Which of the following arrow diagrams represent a function? Why? (Diagram with domain X = {a, b, c, d} and codomain Y = {0, 1, 2, 3, 4})
Answer: Function.
Reason: Every element of set X has a unique image in set Y, and no element of X is left unmapped or has multiple arrows coming out of it.
In simple words: Every element in the first set shoots out exactly one arrow.

Exam Tip: In an arrow diagram, a relation is a function if every node in the source set has precisely one outgoing arrow.

Question 12. Which of the following arrow diagrams represent a function? Why? (Diagram with domain X = {2, 3, 5, 7} and codomain Y = {-1, 0, 2, 4, 3})
Answer: Not a function.
Reason: Element \(2\) in domain X has two outgoing arrows leading to two different images in Y.
In simple words: It is not a function because one input element has multiple output connections.

Exam Tip: If any element in the domain has zero or more than one outgoing arrow, the relation fails to be a function.

Question 13. Let f and g be two real valued functions, defined by, \(f(x) = x^2\), \(g(x) = 3x + 2\), find : \((f + g)(-2)\)
Answer: By definition of algebra of functions:
\((f + g)(-2) = f(-2) + g(-2)\)
Calculate \(f(-2) = (-2)^2 = 4\)
Calculate \(g(-2) = 3(-2) + 2 = -6 + 2 = -4\)
\((f + g)(-2) = 4 + (-4) = 0\)
In simple words: Evaluate each function at \(x = -2\) separately and add the results.

Exam Tip: Use the property \((f \pm g)(x) = f(x) \pm g(x)\).

Question 14. Let f and g be two real valued functions, defined by, \(f(x) = x^2\), \(g(x) = 3x + 2\), find : \((f - g)(1)\)
Answer: By definition:
\((f - g)(1) = f(1) - g(1)\)
Calculate \(f(1) = (1)^2 = 1\)
Calculate \(g(1) = 3(1) + 2 = 5\)
\((f - g)(1) = 1 - 5 = -4\)
In simple words: Find the value of f(1) and g(1) and subtract them.

Exam Tip: Be careful with subtraction signs when evaluating \(g(1)\).

Question 15. Let f and g be two real valued functions, defined by, \(f(x) = x^2\), \(g(x) = 3x + 2\), find : \((fg)(-1)\)
Answer: By definition:
\((fg)(-1) = f(-1) \cdot g(-1)\)
Calculate \(f(-1) = (-1)^2 = 1\)
Calculate \(g(-1) = 3(-1) + 2 = -3 + 2 = -1\)
\((fg)(-1) = 1 \cdot (-1) = -1\)
In simple words: Evaluate both functions at \(x = -1\) and multiply their values.

Exam Tip: Product of functions means multiplying their functional outputs, not multiplying the functions algebraically first (though both give the same result).

Question 16. Let f and g be two real valued functions, defined by, \(f(x) = x^2\), \(g(x) = 3x + 2\), find : \(\left(\frac{f}{g}\right)(0)\)
Answer: By definition:
\(\left(\frac{f}{g}\right)(0) = \frac{f(0)}{g(0)}\), provided \(g(0) \neq 0\)
Calculate \(f(0) = (0)^2 = 0\)
Calculate \(g(0) = 3(0) + 2 = 2\)
\(\left(\frac{f}{g}\right)(0) = \frac{0}{2} = 0\)
In simple words: Divide the value of f(0) by g(0).

Exam Tip: Always verify that the denominator function evaluated at the given point is non-zero.

Question 17. If \(f(x) = x^3\), find the value of \(\frac{f(5) - f(1)}{5 - 1}\)
Answer: Calculate \(f(5) = (5)^3 = 125\)
Calculate \(f(1) = (1)^3 = 1\)
Substitute the values into the expression:
\(\frac{f(5) - f(1)}{5 - 1} = \frac{125 - 1}{4} = \frac{124}{4} = 31\)
In simple words: Cube 5 and 1, subtract them, and then divide by 4.

Exam Tip: This expression represents the average rate of change of the function over the interval [1, 5].

Question 18. Find the domain of the real function, \(f(x) = \sqrt{x^2 - 4}\)
Answer: For the real-valued function \(f(x)\) to be defined, the expression inside the square root must be greater than or equal to zero:
\(x^2 - 4 \geq 0\)
\((x - 2)(x + 2) \geq 10 \text{ (or } 0\text{)}\)
Using the sign scheme method, the inequality holds when \(x \leq -2\) or \(x \geq 2\).
Therefore, the domain is \((-\infty, -2] \cup [2, \infty)\).
In simple words: The value under the square root cannot be negative, so \(x^2\) must be at least 4.

Exam Tip: For even roots like square roots, set the radicand \(\geq 0\) and solve the resulting inequality.

Question 19. Find the domain of the function, \(f(x) = \frac{x^2 + 2x + 3}{x^2 - 5x + 6}\)
Answer: A rational function is defined for all real numbers except where the denominator is zero.
Set the denominator equal to zero:
\(x^2 - 5x + 6 = 0\)
\((x - 2)(x - 3) = 0\)
\(x = 2 \text{ or } x = 3\)
Thus, the function is defined for all real numbers except \(2\) and \(3\).
Domain = \(R - \{2, 3\}\)
In simple words: Find where the denominator becomes zero and exclude those points from the real number line.

Exam Tip: Factorize quadratic denominators completely to find all restricted values.

Question 20. Find the range of the following functions: \(f(x) = \frac{1}{1 - x^2}\)
Answer: Let \(y = \frac{1}{1 - x^2}\). We solve for \(x^2\) in terms of \(y\):
\(y(1 - x^2) = 1\)
\(y - yx^2 = 1\)
\(yx^2 = y - 1\)
\(x^2 = \frac{y - 1}{y}\)
Since \(x^2 \geq 0\) for all real \(x\):
\(\frac{y - 1}{y} \geq 0\)
Using the wavy curve method for inequalities, this holds when \(y < 0\) or \(y \geq 1\).
Also, when \(y = 0\), \(\frac{1}{1 - x^2} = 0\) has no real solution for \(x\), so \(y \neq 0\).
Therefore, the range is \((-\infty, 0) \cup [1, \infty)\).
In simple words: Express x in terms of y and find the valid output values for y.

Exam Tip: For rational functions, expressing x as a function of y is a standard technique to determine the range.

Question 21. Find the range of the following functions: \(f(x) = x^2 + 2\)
Answer: We know that for any real number \(x\):
\(x^2 \geq 0\)
Adding 2 to both sides:
\(x^2 + 2 \geq 2\)
Therefore, \(f(x) \geq 2\).
Range = \([2, \infty)\)
In simple words: Since the square of any number is at least 0, adding 2 makes the minimum value 2.

Exam Tip: Use standard inequalities like \(x^2 \geq 0\) to build up the expression for \(f(x)\).

Question 22. Find the domain of the relation, \(R = \{(x, y) : x, y \in Z, xy = 4\}\)
Answer: Given \(xy = 4\) where \(x, y \in Z\) (integers).
Solving for \(y\), we get \(y = \frac{4}{x}\).
For \(y\) to be an integer, \(x\) must be a divisor of \(4\).
The divisors of \(4\) in integers are \(\pm 1, \pm 2, \pm 4\).
Thus, the domain is the set of all possible first elements \(x\):
Domain = \(\{-4, -2, -1, 1, 2, 4\}\)
In simple words: Domain is the set of integers that can divide 4 evenly.

Exam Tip: Domain of a relation defined by an equation consists of all inputs for which the corresponding output exists in the specified set.

Question 23. Find the range of the following relations: \(R = \{(a, b) : a, b \in N \text{ and } 2a + b = 10\}\)
Answer: Given \(2a + b = 10\), where \(a, b \in N\) (Natural numbers).
Express \(b\) in terms of \(a\):
\(b = 10 - 2a\)
Since \(a \in N\), we test positive integer values for \(a\) such that \(b \in N\) (\(b > 0\)):
If \(a = 1\): \(b = 10 - 2(1) = 8\)
If \(a = 2\): \(b = 10 - 2(2) = 6\)
If \(a = 3\): \(b = 10 - 2(3) = 4\)
If \(a = 4\): \(b = 10 - 2(4) = 2\)
If \(a \geq 5\), \(b \leq 0\), which is not in \(N\).
Thus, the relation in roster form is \(R = \{(1, 8), (2, 6), (3, 4), (4, 2)\}\).
The range is the set of all second elements \(b\):
Range = \(\{2, 4, 6, 8\}\)
In simple words: Substitute natural numbers for a to find valid natural number outputs for b.

Exam Tip: Always check the given set constraints (like \(N\) or \(Z\)) when listing relation elements.

Question 24. Find the range of the following relations: \(R = \left\{\left(x, \frac{1}{x}\right) : x \in Z, 0 < x < 6\right\}\)
Answer: Given integers \(x\) such that \(0 < x < 6\).
The possible integer values for \(x\) are \(1, 2, 3, 4, 5\).
Corresponding values of \(\frac{1}{x}\) for each \(x\) are:
For \(x = 1\): \(\frac{1}{1} = 1\)
For \(x = 2\): \(\frac{1}{2}\)
For \(x = 3\): \(\frac{1}{3}\)
For \(x = 4\): \(\frac{1}{4}\)
For \(x = 5\): \(\frac{1}{5}\)
Thus, the range is \(\left\{1, \frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \frac{1}{5}\right\}\).
In simple words: Substitute each integer between 0 and 6 into the reciprocal expression to get the outputs.

Exam Tip: Range consists exclusively of the second coordinates of the ordered pairs in the relation.

SHORT ANSWER TYPE QUESTIONS (4 MARKS)

Question 25. Let \(A = \{1, 2, 3, 4\}\), \(B = \{1, 4, 9, 16, 25\}\) and R be a relation defined from A to B as, \(R = \{(x, y) : x \in A, y \in B \text{ and } y = x^2\}\)
(a) Depict this relation using arrow diagram.
(b) Find domain of R.
(c) Find range of R.
(d) Write co-domain of R.
Answer:
First, let us write the relation R in roster form by checking \(y = x^2\) for each \(x \in A\):
\(R = \{(1, 1), (2, 4), (3, 9), (4, 16)\}\)
(a) Arrow diagram description:
Draw two ovals representing sets A and B. Draw arrows from \(1 \to 1\), \(2 \to 4\), \(3 \to 9\), and \(4 \to 16\). Element \(25\) in set B will have no incoming arrow.
(b) Domain of R: Set of all first elements = \(\{1, 2, 3, 4\}\)
(c) Range of R: Set of all second elements = \(\{1, 4, 9, 16\}\)
(d) Co-domain of R: The entire set B = \(\{1, 4, 9, 16, 25\}\)
In simple words: Square each element of set A to find its mapped partner in set B.

Exam Tip: Co-domain is always the entire second set given in the definition, whereas range is only the subset of elements that actually get mapped to.

Question 26. Let \(R = \{(x, y) : x, y \in N \text{ and } y = 2x\}\) be a relation on N. Find :
(i) Domain
(ii) Codomain
(iii) Range
Is this relation a function from N to N?
Answer:
The relation in roster form is \(R = \{(1, 2), (2, 4), (3, 6), (4, 8), \dots\}\).
(i) Domain: The set of all first coordinates \(x \in N\), which is \(N\).
(ii) Codomain: The second set in the relation definition, which is \(N\).
(iii) Range: The set of all second coordinates \(y = 2x \text{ for } x \in N\), which is the set of all even natural numbers.
Is it a function?
Yes, every natural number \(x\) has one and only one unique image \(2x\) in \(N\).
In simple words: Every natural number has exactly one double, making it a valid function.

Exam Tip: To prove a relation is a function, verify that every element in the domain has a unique image in the codomain.

Question 27. Let \(f(x) = \begin{cases} x^2, & \text{when } 0 \leq x \leq 2 \\ 2x, & \text{when } 2 \leq x \leq 5 \end{cases}\) and \(g(x) = \begin{cases} x^2, & \text{when } 0 \leq x \leq 3 \\ 2x, & \text{when } 3 \leq x \leq 5 \end{cases}\). Show that f is a function while g is not a function.
Answer:
For \(f(x)\):
At the overlapping point \(x = 2\):
Using the first branch: \(f(2) = 2^2 = 4\)
Using the second branch: \(f(2) = 2(2) = 4\)
Since both branches yield the exact same value at \(x = 2\), the image is unique. Thus, \(f\) is a function.
For \(g(x)\):
At the overlapping point \(x = 3\):
Using the first branch: \(g(3) = 3^2 = 9\)
Using the second branch: \(g(3) = 2(3) = 6\)
Since \(g(3)\) gives two different values (\(9\) and \(6\)) at the same point \(x = 3\), an element has two images. Thus, \(g\) is not a function.
In simple words: A piecewise-defined rule fails to be a function if the branches give conflicting answers at the boundary point.

Exam Tip: Always check the boundary/overlapping points of piecewise functions to ensure consistency.

Question 28. Find the domain and range of \(f(x) = |2x - 3| - 3\)
Answer:
Domain:
Since the modulus function is defined for all real numbers, the domain of \(f(x)\) is \(R\).
Range:
We know that for any real number \(x\):
\(|2x - 3| \geq 0\)
Subtracting 3 from both sides:
\(|2x - 3| - 3 \geq -3\)
Therefore, \(f(x) \geq -3\).
Range = \([-3, \infty)\)
In simple words: Absolute value is never negative, so the smallest possible value for the expression is -3.

Exam Tip: Use the property that modulus expressions have a minimum value of zero.

Question 29. Draw the graph of the Greatest Integer function
Answer:
The greatest integer function is defined as \(f(x) = [x]\), which represents the greatest integer less than or equal to \(x\).
- For \(x \in [0, 1)\), \(f(x) = 0\)
- For \(x \in [1, 2)\), \(f(x) = 1\)
- For \(x \in [2, 3)\), \(f(x) = 2\)
- For \(x \in [-1, 0)\), \(f(x) = -1\)
The graph consists of a series of step-like horizontal line segments (staircase pattern) with closed circles at the left endpoints and open circles at the right endpoints.
In simple words: The graph looks like a staircase where each step is closed on the left and open on the right.

Exam Tip: Clearly mark open circles at integer boundaries on the right to indicate that the endpoint is not included in that step.

Question 30. Draw the graph of the Constant function, \(f : R \to R; f(x) = 2 \forall x \in R\). Also find its domain and range.
Answer:
Graph description:
The graph of \(f(x) = 2\) is a horizontal straight line parallel to the X-axis passing through the point \((0, 2)\) at a constant height of 2 units above the X-axis.
Domain: Since the function is defined for all real numbers, Domain = \(R\).
Range: Since the output is permanently 2, Range = \(\{2\}\).
In simple words: A constant function produces a flat horizontal line at y = 2.

Exam Tip: The range of any non-zero constant function \(f(x) = c\) is simply the singleton set \(\{c\}\).