CBSE Class 11 Mathematics Relations Functions Worksheet Set 04

CBSE Class 11 Mathematics Worksheet - Relations Functions (2). Students can download these worksheets and practice them. This will help them to get better marks in examinations. Also refer to other worksheets for the same chapter and other subjects too. Use them for better understanding of the subjects.

Question 11. Find the domain of \( f(x) = \frac{1}{\sqrt{1-\cos x}} \)
Answer: For the function \( f(x) \) to be real and defined, the term inside the square root in the denominator must be strictly greater than zero: \[ 1 - \cos x > 0 \] Since the range of the cosine function is \( -1 \le \cos x \le 1 \) for all real numbers \( x \), the inequality \( 1 - \cos x \ge 0 \) is always satisfied. However, since this expression is in the denominator, it cannot equal zero:
\( \implies \cos x \neq 1 \)
We know that \( \cos x = 1 \) at even multiples of \( \pi \): \[ x = 2n\pi \quad \text{for } n \in \mathbb{Z} \] Thus, the domain of the function includes all real numbers except these points.
\( \text{Domain} = \mathbb{R} - \{2n\pi : n \in \mathbb{Z}\} \)
In simple words: To avoid dividing by zero or taking the square root of a negative value, the term \( 1 - \cos x \) must be positive. This is true for every real number except when \( \cos x = 1 \), which occurs at \( 2n\pi \).

Exam Tip: When a square root is in the denominator, remember that the expression inside must be strictly positive (\( > 0 \)), whereas a square root in the numerator only requires the expression to be non-negative (\( \ge 0 \)).

Question 12. Let \( f = \{(2,4), (5,6), (8, -1), (10, -3)\} \), \( g = \{(2,5), (7,1), (8,4), (10,13), (11, -5)\} \). Find the domain of \( f + g, f - g \) & \( fg \).
Answer: The domain of a function defined as a set of ordered pairs is the set of all first elements. The domain of function \( f \), denoted as \( D_f \), is: \[ D_f = \{2, 5, 8, 10\} \] The domain of function \( g \), denoted as \( D_g \), is: \[ D_g = \{2, 7, 8, 10, 11\} \] The operations of addition \( (f + g) \), subtraction \( (f - g) \), and multiplication \( (fg) \) are defined only on those values that are common to both domains. Hence, the domain for these combined functions is the intersection of \( D_f \) and \( D_g \): \[ D_{f+g} = D_{f-g} = D_{fg} = D_f \cap D_g \] \[ D_f \cap D_g = \{2, 5, 8, 10\} \cap \{2, 7, 8, 10, 11\} \]
\( \implies D_f \cap D_g = \{2, 8, 10\} \)
In simple words: To perform operations like adding, subtracting, or multiplying two functions, we can only use inputs that are valid for both functions at the same time. This is why we find the common numbers in both domains.

Exam Tip: For any algebraic combination of functions like \( f+g \), \( f-g \), or \( fg \), the domain is always \( D_f \cap D_g \). However, for division \( \frac{f}{g} \), you must also exclude points where \( g(x) = 0 \).

Question 13. Draw the graph of \( f(x) = 1 + |x - 2| \)
Answer: The given function is: \[ f(x) = 1 + |x - 2| \] Using the definition of the absolute value function: \[ |x - 2| = \begin{cases} x - 2 & \text{if } x - 2 \ge 0 \implies x \ge 2 \\ -(x - 2) & \text{if } x - 2 < 0 \implies x < 2 \end{cases} \] Substituting this into our function, we get: \[ f(x) = \begin{cases} 1 + (x - 2) & \text{if } x \ge 2 \\ 1 - (x - 2) & \text{if } x < 2 \end{cases} \]
\( \implies f(x) = \begin{cases} x - 1 & \text{if } x \ge 2 \\ -x + 3 & \text{if } x < 2 \end{cases} \)
To plot the graph, we can find several key coordinates:

• For \( x = 2 \): \( f(2) = 1 + |2 - 2| = 1 \). Point: \( (2, 1) \)
• For \( x = 3 \): \( f(3) = 1 + |3 - 2| = 2 \). Point: \( (3, 2) \)
• For \( x = 4 \): \( f(4) = 1 + |4 - 2| = 3 \). Point: \( (4, 3) \)
• For \( x = 1 \): \( f(1) = 1 + |1 - 2| = 2 \). Point: \( (1, 2) \)
• For \( x = 0 \): \( f(0) = 1 + |0 - 2| = 3 \). Point: \( (0, 3) \)
• For \( x = -1 \): \( f(-1) = 1 + |-1 - 2| = 4 \). Point: \( (-1, 4) \)

Plotting these points yields a continuous, V-shaped graph with its vertex at \( (2, 1) \).
Domain: \( \mathbb{R} \) (all real numbers)
Range: \( [1, \infty) \)
In simple words: The graph has a sharp turning point at \( (2, 1) \) because of the absolute value. The y-value is at its lowest here and grows larger on both sides, creating a perfect V-shape.

Exam Tip: Find the vertex of a modulus function \( a|x - h| + k \) by setting the term inside the absolute value to zero, which gives the coordinate \( (h, k) \). Plotting a couple of points on each side then guarantees a precise sketch.

Question 14. Let R be the relation on Set N (natural no.s) defined by \( R = \{(a, b): a + 3b = 12; a \in N \text{ and } b \in N\} \). Find R (roster form), domain, Range, Co-domain and arrow diagram.
Answer: We are given the relation defined on the set of natural numbers \( \mathbb{N} \): \[ R = \{(a, b) : a + 3b = 12 \text{ where } a, b \in \mathbb{N}\} \] To write the relation in roster form, we rewrite the equation as: \[ a = 12 - 3b \] Since both \( a \) and \( b \) must be natural numbers (positive integers starting from 1):

• If \( b = 1 \): \( a = 12 - 3(1) = 9 \in \mathbb{N} \). Pair: \( (9, 1) \)
• If \( b = 2 \): \( a = 12 - 3(2) = 6 \in \mathbb{N} \). Pair: \( (6, 2) \)
• If \( b = 3 \): \( a = 12 - 3(3) = 3 \in \mathbb{N} \). Pair: \( (3, 3) \)
• If \( b \ge 4 \): \( a \le 0 \), which is not a natural number.

(i) **Roster Form:** \[ R = \{(9, 1), (6, 2), (3, 3)\} \] (ii) **Domain:** The set of all first components of the ordered pairs in \( R \): \[ \text{Domain} = \{3, 6, 9\} \] (iii) **Range:** The set of all second components of the ordered pairs in \( R \): \[ \text{Range} = \{1, 2, 3\} \] (iv) **Co-domain:** Since the relation is defined from \( \mathbb{N} \) to \( \mathbb{N} \), the co-domain is: \[ \text{Co-domain} = \mathbb{N} \] (v) **Arrow Diagram:**
In simple words: We find pairs of natural numbers that satisfy the equation. Testing positive whole numbers starting from 1 for b gives three valid pairs, shown with arrows connecting them.

Exam Tip: Be careful with the set definition. Natural numbers are only positive integers (\( 1, 2, 3, \dots \)), which means zero and negative values cannot be part of the domain or range sets.

Question 15. Let \( A = \{1,2,3,4,6\} \), Let R is a relation on A defined by \( R = \{(a, b): b \text{ is exactly divisible by } a\} \). Find R, domain, Range, Co-domain and Arrow diagram.
Answer: We are given the relation: \[ R = \{(a, b) : b \text{ is exactly divisible by } a \text{ where } a, b \in A\} \] To determine the pairs, we check divisibility for each element in \( A = \{1, 2, 3, 4, 6\} \):

• 1 divides all elements: \( (1,1), (1,2), (1,3), (1,4), (1,6) \)
• 2 divides 2, 4, and 6: \( (2,2), (2,4), (2,6) \)
• 3 divides 3 and 6: \( (3,3), (3,6) \)
• 4 divides 4: \( (4,4) \)
• 6 divides 6: \( (6,6) \)

(i) **Roster Form:** \[ R = \{(1,1), (1,2), (1,3), (1,4), (1,6), (2,2), (2,4), (2,6), (3,3), (3,6), (4,4), (6,6)\} \] (ii) **Domain:** The set of first components in the pairs: \[ \text{Domain} = \{1, 2, 3, 4, 6\} \] (iii) **Range:** The set of second components in the pairs: \[ \text{Range} = \{1, 2, 3, 4, 6\} \] (iv) **Co-domain:** Since the relation is from \( A \) to \( A \), the co-domain is: \[ \text{Co-domain} = A = \{1, 2, 3, 4, 6\} \] (v) **Arrow Diagram:**
In simple words: This relation connects any number on the left oval to a number on the right oval that can be divided by it without leaving a remainder. For instance, 2 is linked to 2, 4, and 6.

Exam Tip: Every non-zero integer is exactly divisible by itself. Thus, reflexive pairs like \( (a, a) \) for all \( a \in A \) must always be included in the roster form.

Question 16. Given Arrow diagram. Find relation in set builder & Roster form also find Domain, Range, co-domain.
Answer: From the given arrow diagram: (i) **Roster Form:** We directly write down the pairs represented by the arrows: \[ R = \{(25, 5), (25, -5), (9, 3), (9, -3), (4, 2), (4, -2)\} \] (ii) **Set Builder Form:** We observe that the elements in the domain set \( P \) are the squares of the corresponding elements in the codomain set \( Q \): \[ 25 = 5^2 \quad \text{and} \quad 25 = (-5)^2 \] \[ 9 = 3^2 \quad \text{and} \quad 9 = (-3)^2 \] \[ 4 = 2^2 \quad \text{and} \quad 4 = (-2)^2 \] Therefore, the relation can be expressed as: \[ R = \{(x, y) : x = y^2 \text{ where } x \in P \text{ and } y \in Q\} \] (iii) **Domain:** The set of elements in \( P \) that are mapped to \( Q \): \[ \text{Domain} = \{4, 9, 25\} \] (iv) **Range:** The set of elements in \( Q \) that receive arrows from \( P \): \[ \text{Range} = \{-5, -3, -2, 2, 3, 5\} \] (v) **Co-domain:** The entire target set \( Q \): \[ \text{Co-domain} = \{-5, -3, -2, 1, 2, 3, 5\} \]
In simple words: Each number on the left is matched with its positive and negative square roots on the right. Note that 1 is in the co-domain because it is in set Q, but not in the range because no arrow points to it.

Exam Tip: Remember that the Range is a subset of the Co-domain. Only include elements in the range that are actually mapped to by some element of the domain.

Question 17. Find the domain of the function, \( f(x) = \sqrt{x - 3 - 2\sqrt{x - 4}} - \sqrt{x - 3 + 2\sqrt{x - 4}} \)
Answer: For the function \( f(x) \) to yield real values, the terms inside all square roots must be non-negative: 1. The innermost square root term requires: \[ x - 4 \ge 0 \implies x \ge 4 \] 2. The outer square roots require: \[ x - 3 - 2\sqrt{x - 4} \ge 0 \quad \text{and} \quad x - 3 + 2\sqrt{x - 4} \ge 0 \] Let us simplify these expressions by expressing them as perfect squares: \[ x - 3 - 2\sqrt{x - 4} = (x - 4) - 2\sqrt{x - 4} + 1 = \left(\sqrt{x - 4} - 1\right)^2 \] \[ x - 3 + 2\sqrt{x - 4} = (x - 4) + 2\sqrt{x - 4} + 1 = \left(\sqrt{x - 4} + 1\right)^2 \] Since the square of any real number is always non-negative: \[ \left(\sqrt{x - 4} - 1\right)^2 \ge 0 \quad \text{and} \quad \left(\sqrt{x - 4} + 1\right)^2 \ge 0 \] These inequalities are always satisfied for any real value of \( \sqrt{x-4} \). Therefore, the only constraint on \( x \) is the condition for the innermost square root to be defined, which is: \[ x \ge 4 \] Hence, the domain of the function is: \[ x \in [4, \infty) \]
In simple words: The expressions under the outer square roots are perfect square trinomials, which are always non-negative. This means our only constraint is making sure the term inside the inner square root, \( \sqrt{x - 4} \), is defined. Thus, \( x \) must be 4 or greater.

Exam Tip: Recognizing algebraic identities like \( a^2 - 2ab + b^2 = (a-b)^2 \) under radical signs is a very powerful technique for simplifying complex functions and finding their domains easily.

Question 18. Find the domain of the function, \( f(x) = \sqrt{\frac{x+3}{(2-x)(x-5)}} \)
Answer: For \( f(x) \) to be real and defined, the term inside the square root must be non-negative and the denominator cannot be zero: \[ \frac{x+3}{(2-x)(x-5)} \ge 0 \quad \text{and} \quad (2-x)(x-5) \neq 0 \] We can write the denominator in standard form by factoring out a negative sign: \[ \frac{x+3}{-(x-2)(x-5)} \ge 0 \] Multiplying both sides of the inequality by \( -1 \) reverses the inequality direction: \[ \frac{x+3}{(x-2)(x-5)} \le 0 \quad \text{where } x \neq 2 \text{ and } x \neq 5 \] Using the sign scheme (wavy curve) method, the critical points are \( x = -3 \), \( x = 2 \), and \( x = 5 \):

• For \( x > 5 \): the expression is positive \( (+) \)
• For \( 2 < x < 5 \): the expression is negative \( (-) \)
• For \( -3 < x < 2 \): the expression is positive \( (+) \)
• For \( x < -3 \): the expression is negative \( (-) \)

Since we require the expression to be less than or equal to zero \( (\le 0) \), the solution intervals are: \[ x \in (-\infty, -3] \cup (2, 5) \] Note that open intervals are used for \( 2 \) and \( 5 \) to prevent the denominator from becoming zero.
\( \text{Domain} = (-\infty, -3] \cup (2, 5) \)
In simple words: To keep the square root valid and avoid division by zero, the fraction must be positive or zero. We find the intervals where the numerator and denominator have different signs, keeping the denominator's critical values open.

Exam Tip: Never include boundary points that make the denominator zero in the final domain. Always use open brackets (parentheses) for denominator roots.

Question 19. \( f(x) = \begin{cases} 1 - x & : x < 0 \\ 1 & : x = 0 \\ x + 1 & : x > 0 \end{cases} \) draw graph of \( f(x) \).
Answer: We analyze the piecewise function in three sections:
1. **For \( x < 0 \):** The function is a straight line defined by \( f(x) = 1 - x \).

• For \( x = -1 \): \( f(-1) = 1 - (-1) = 2 \). Point: \( (-1, 2) \)
• For \( x = -2 \): \( f(-2) = 1 - (-2) = 3 \). Point: \( (-2, 3) \)
• For \( x = -3 \): \( f(-3) = 1 - (-3) = 4 \). Point: \( (-3, 4) \)

2. **For \( x = 0 \):** The value of the function is defined as \( f(0) = 1 \). Point: \( (0, 1) \)
3. **For \( x > 0 \):** The function is a straight line defined by \( f(x) = x + 1 \).

• For \( x = 1 \): \( f(1) = 1 + 1 = 2 \). Point: \( (1, 2) \)
• For \( x = 2 \): \( f(2) = 2 + 1 = 3 \). Point: \( (2, 3) \)
• For \( x = 3 \): \( f(3) = 3 + 1 = 4 \). Point: \( (3, 4) \)

Since \( \lim_{x \to 0^{-}} (1-x) = 1 \) and \( \lim_{x \to 0^{+}} (x+1) = 1 \), the two lines meet continuously at \( (0, 1) \).
In simple words: This piecewise function can be graphed by plotting each segment. They meet seamlessly at \( (0, 1) \) to form a V-shape, which is equivalent to the graph of \( f(x) = |x| + 1 \).

Exam Tip: For piecewise graphs, always check the boundary points. Calculate the limit from the left and the right of the boundary value to verify if the segments meet at the same point.

Question 20. Find the domain of the function, \( f(x) = \frac{1}{\sqrt{[x]^2-[x]-6}} \).
Answer: For \( f(x) \) to be defined, the expression inside the square root in the denominator must be strictly positive: \[ [x]^2 - [x] - 6 > 0 \] Let us factor this quadratic expression in terms of the greatest integer function \( [x] \) by splitting the middle term: \[ [x]^2 - 3[x] + 2[x] - 6 > 0 \]
\( \implies [x]([x] - 3) + 2([x] - 3) > 0 \)
\( \implies ([x] - 3)([x] + 2) > 0 \)
Solving this quadratic inequality gives: \[ [x] < -2 \quad \text{or} \quad [x] > 3 \] Now, we analyze both intervals using the properties of the greatest integer function (where \( [x] \) must be an integer): 1. **For \( [x] < -2 \):** This means: \[ [x] \le -3 \implies x < -2 \] 2. **For \( [x] > 3 \):** This means: \[ [x] \ge 4 \implies x \ge 4 \] Combining both solutions, we obtain: \[ x \in (-\infty, -2) \cup [4, \infty) \] Therefore, the domain of the function is: \[ x \in (-\infty, -2) \cup [4, \infty) \]
In simple words: First we solve the quadratic inequality for the step function \( [x] \), which gives \( [x] < -2 \) or \( [x] > 3 \). Translating these conditions to real values of \( x \) gives all real numbers below -2 and all real numbers 4 or above.

Exam Tip: Pay careful attention to greatest integer function inequalities. For example, the inequality \( [x] > k \) translates to \( [x] \ge k+1 \implies x \ge k+1 \), which yields a closed interval boundary, while \( [x] < k \implies [x] \le k-1 \implies x < k \), which yields an open interval boundary.