Samacheer Kalvi Class 10 Maths Solutions Chapter 1 Relations and Functions More Ques

I. Multiple Choice Questions.

Question 1. If n(A × B) = 15 and B = {1, 3, 7} then n(A) is
(a) 3
(b) 5
(c) 1
(d) 7
Answer: (b) 5
In simple words: To find the number of elements in set A, we divide the total number of elements in the Cartesian product A × B by the number of elements in set B. This is because the product of the number of elements in two sets gives the number of elements in their Cartesian product.

🎯 Exam Tip: Remember the formula \( n(A \times B) = n(A) \times n(B) \). This helps to quickly find the number of elements in one set if the others are known.

Question 2. If A = {a, b, c}, B = {b, d, e}, C = {a, e, i, o, u} then n[(A \( \cap \) C) × B] is
(a) 18
(b) 36
(c) 9
(d) 3
Answer: (d) 3
In simple words: First, find the common elements between set A and set C. Then, count how many elements are in this common set. Finally, multiply this count by the number of elements in set B to get the total number of pairs.

🎯 Exam Tip: Always calculate the intersection \( (A \cap C) \) first, then find its cardinality \( n(A \cap C) \), and finally multiply it by \( n(B) \).

Question 3. If there are 28 relations from a set A = {2,4, 6, 8} to a set B, then the number of elements in B is
(a) 3
(b) 14
(c) 5
(d) 4
Answer: (The calculated answer is 7, which is not among the given options)
Solution:
Given \( n(A) = 4 \) (since A = {2, 4, 6, 8}).
The number of relations from set A to set B is given by \( 2^{n(A) \times n(B)} \).
We are given that the number of relations is 28. This implies an error in the question or options, as 28 is not a power of 2. For the number of relations to be 28, the value \( n(A) \times n(B) \) would not be an integer if 28 were \( 2^{n(A) \times n(B)} \).
However, if we interpret "28 relation" as \( n(A \times B) = 28 \), which is common in some problem styles, then:
\( n(A \times B) = n(A) \times n(B) \)
\( 28 = 4 \times n(B) \)
\( n(B) = \frac{28}{4} \)
\( n(B) = 7 \)
Thus, the number of elements in B is 7.
In simple words: The number of possible relations between two sets depends on how many pairs can be made. If we assume the total number of pairs is 28, and set A has 4 elements, then set B must have 7 elements to make 28 pairs (4 multiplied by 7).

🎯 Exam Tip: Be careful with problems about relations. The total number of relations from set A to set B is \( 2^{n(A) \times n(B)} \). If the question is about the number of elements in the Cartesian product \( n(A \times B) \), then it is simply \( n(A) \times n(B) \).

Question 4. The ordered pairs (a + 1, 4) (3, 4a + b) are equal then (a, b) is
(a) (4, 20)
(b) (20, 4)
(c) (-4, 20)
(d) (20, -4)
Answer: (c) (-4, 20)
In simple words: When two ordered pairs are equal, it means their first parts are equal and their second parts are equal. So, we set \( a+1 \) equal to 3, and 4 equal to \( 4a+b \). We solve these two small equations to find the values of 'a' and 'b'.

🎯 Exam Tip: Always equate the corresponding components of the ordered pairs. First component equals first component, and second component equals second component. Solve the simpler equation first.

Question 5. The range of the relation R = {(x, x³) / x is a prime number less than 13} is
(a) (2,3,5,7,11)
(b) (4,9,25,49,121)
(c) (8,27, 125,343, 1331)
(d) (1,8,27, 125,343, 1331)
Answer: (c) (8, 27, 125,343, 1331)
In simple words: First, list all the prime numbers that are smaller than 13. Then, for each of these prime numbers, calculate its cube (multiply it by itself three times). The list of these cubed numbers will be the range of the relation.

🎯 Exam Tip: Clearly identify the domain (prime numbers less than 13) first. Then apply the rule \( x \to x^3 \) to each element in the domain to find the range elements.

Question 6. If {(x, 2), (4, y) } represents an identity function, then (x, y) is
(a) (2, 4)
(b) (4, 2)
(c) (2, 2)
(d) (4, 4)
Answer: (a) (2, 4)
In simple words: For an identity function, every input must match its output exactly. So, if the output is 2, the input 'x' must also be 2. If the input is 4, the output 'y' must also be 4. This ensures that the function maps each element to itself.

🎯 Exam Tip: Remember the definition of an identity function: \( f(x) = x \). This means that for any ordered pair \( (a, b) \) in an identity function, \( a \) must always be equal to \( b \).

Question 7. If {(7, 11), (5, a)} represents a constant function, then the value of 'a' is
(a) 7
(b) 11
(d) 9
Answer: (b) 11
In simple words: In a constant function, all different input values always give the same output value. Since the first pair shows an output of 11, the output for any other input, like 5, must also be 11.

🎯 Exam Tip: For a constant function, the range contains only one element. All elements from the domain map to this single element in the range.

Question 8. Given f(x) = (- 1)^x is a function from N to Z. Then the range of f is
(a) {1}
(b) N
(c) { 1,- 1 }
(d) Z
Answer: (c) {1, - 1}
In simple words: When you raise -1 to a power, the answer will always be either 1 (if the power is an even number) or -1 (if the power is an odd number). Since natural numbers can be both odd and even, the function's output can only be 1 or -1.

🎯 Exam Tip: Test the function with a few natural numbers (e.g., \( x=1, 2, 3 \)) to see the pattern of the outputs and determine the range correctly.

Question 9. If f = { (6, 3), (8, 9), (5, 3), (-1, 6) }, then the pre-images of 3 are
(a) 5 and-1
(b) 6 and 8
(c) 8 and-1
(d) 6 and 5
Answer: (d) 6 and 5
In simple words: The pre-image of a number is the input that gives that number as an output. Look through the given pairs and find all the inputs that have 3 as their output. Both 6 and 5 lead to an output of 3.

🎯 Exam Tip: The pre-image refers to the input \( x \) values for a given output \( y \). Look for pairs \( (x, y) \) where \( y \) is the target value (in this case, 3).

Question 10. Let A= { 1, 3, 4, 7, 11 }, B = {-1, 1, 2, 5, 7, 9 } and f : A\( \to \)B be given by f = {(1, -1), (3,2), (4, 1), (7, 5), (11, 9)}.
(a) one-one
(b) onto
(c) bijective
(d) not a function
Answer: (a) one - one
In simple words: A one-one function means each unique input gives a unique output. In this case, every number from set A maps to a different number in set B. It is not "onto" because some numbers in B (like 7) are not used as outputs.

🎯 Exam Tip: For a function to be one-one, every element in the domain must map to a distinct element in the codomain. If any two elements in the domain map to the same element in the codomain, it is not one-one.

Question 11. The given diagram represents

(a) an onto function
(b) a constant function
(c) an one-one function
(d) not a function
Answer: (d) not a function
In simple words: A diagram shows a function if each item in the first set (domain) goes to only one item in the second set (codomain). Here, the number 2 in set A points to two different numbers (1 and 2) in set B. This means it is not a function.

🎯 Exam Tip: The vertical line test or checking if each domain element has exactly one image is crucial for identifying functions. If any element in the domain has multiple images, it is not a function.

Question 12. Let A = { 1, 2, 3, 4, 5 } and f: A \( \to \) B is defined by f(x) = x – 2, then the range of f is
(a) {1,4, 5}
(b) {1,2, 3, 4, 5}
(c) { 2, 3, 4 }
(d) { 3, 4, 5}
Answer: (d) {3, 4, 5}
In simple words: To find the range, you take each number from set A and apply the rule \( f(x) = x-2 \) to it. For example, if you take 5 from A, the result is \( 5-2 = 3 \). Do this for all elements of A, and list all the unique results you get.

🎯 Exam Tip: Always substitute each element from the domain into the function rule to find all possible image values. The set of these image values is the range.

Question 13. If f(x) = x² + 5, then f(-4) =
(a) 26
(b) 21
(c) 20
(d) – 20
Answer: (b) 21
In simple words: To find \( f(-4) \), simply replace every 'x' in the function's rule with -4. Remember that when you square a negative number, the result is positive. So, \( (-4)^2 \) becomes 16, and then you add 5.

🎯 Exam Tip: Pay close attention to signs, especially when squaring negative numbers. \( (-a)^2 = a^2 \), not \( -a^2 \).

Question 14. If the range of a function is a singleton set, then it is
(a) a constant function
(b) an identity function
(c) a bijective function
Answer: (a) a constant function
In simple words: A singleton set means there is only one element. If the range of a function has only one element, it means every input to the function produces that exact same single output. This is the definition of a constant function.

🎯 Exam Tip: A constant function maps all elements of its domain to a single element in its codomain, resulting in a range containing only that one element.

Question 15. If f : A\( \to \)B is a bijective function and if n(A) = 5, then n(B) is equal to
(a) 10
(b) 4
(c) 5
(d) 25
Answer: (c) 5
In simple words: A bijective function is both one-to-one and onto. This means every element in set A has a unique partner in set B, and every element in set B has a unique partner in set A. Because of this perfect matching, both sets must have the same number of elements.

🎯 Exam Tip: For a function to be bijective, the number of elements in the domain \( n(A) \) must be equal to the number of elements in the codomain \( n(B) \).

Question 16. If f: R \( \to \) R defined by f(x) = 3x – 6 and g : R \( \to \) R defined by g(x) = 3x + k if fog = gof then the value of k is
(a) – 5
(b) 5
(c) 6
(d) -6
Answer: (d) -6
In simple words: We need to find the value of 'k' that makes the order of applying the functions not matter. This means if you apply 'g' then 'f' (which is fog), it should be the same as applying 'f' then 'g' (which is gof). We calculate both combinations, set them equal to each other, and then solve for 'k'.

🎯 Exam Tip: When evaluating composite functions like \( f \circ g(x) \), remember it means \( f(g(x)) \). Substitute \( g(x) \) into \( f(x) \). For \( g \circ f(x) \), it means \( g(f(x)) \), so substitute \( f(x) \) into \( g(x) \). Always perform these substitutions carefully.

Question 17. If f(x) = x² – x then f (x – 1) – f(x + 1) is
(a) 4x
(b) 4x + 2
(c) 2 - 4x
(d) 4x - 2
Answer: (c) 2 - 4x
In simple words: We need to calculate the function's value at \( (x-1) \) and then at \( (x+1) \), and finally subtract the second result from the first. This involves careful substitution and algebraic simplification, especially with squaring terms like \( (x-1)^2 \) and \( (x+1)^2 \).

🎯 Exam Tip: Be careful with algebraic expansions, especially with terms like \( (x-1)^2 = x^2 - 2x + 1 \) and \( (x+1)^2 = x^2 + 2x + 1 \). Double-check your signs when subtracting expressions.

Question 19. Composition of function is
(a) commutative
(b) associative
(c) commutative and associative
(d) not associative
Answer: (b) associative
In simple words: Composition of functions means doing one function after another. Associative means that if you have three functions, the way you group them (which two you compose first) does not change the final result. For example, \( f \circ (g \circ h) \) is the same as \( (f \circ g) \circ h \).

🎯 Exam Tip: While composition of functions is associative, it is generally not commutative. This means \( f \circ g \) is usually not equal to \( g \circ f \).

Question 20. A comet is heading for Jupiter with acceleration a = 50 kms⁻². The velocity of the comet at time "t" is given by f(t) = at² – at + 1. Then the velocity at time t = 5 seconds is
(a) 900kms⁻¹
(c) 2001 kms⁻¹
(d) 50 kms⁻¹
Answer: (b) 1001 kms⁻¹
In simple words: To find the velocity at 5 seconds, you simply put \( t=5 \) into the given velocity formula. Remember to also substitute the value of 'a' as 50 km/s\( ^2 \). Perform the calculations step-by-step to get the final velocity.

🎯 Exam Tip: Always substitute the given values (like 'a' and 't') into the formula carefully. Pay attention to the order of operations (exponents before multiplication/subtraction).

II. Answer the following questions.

Question 1. If f(x) = (1 + x) and g(x) = (2x – 1), show that fo(g(x)) = gof(x)
Answer:
Given functions are \( f(x) = 1 + x \) and \( g(x) = 2x - 1 \).
First, calculate \( f \circ g(x) \):
\( f \circ g(x) = f(g(x)) \)
\( = f(2x - 1) \)
Now, substitute \( (2x - 1) \) into \( f(x) \):
\( = 1 + (2x - 1) \)
\( = 1 + 2x - 1 \)
\( = 2x \) (Equation 1)

Next, calculate \( g \circ f(x) \):
\( g \circ f(x) = g(f(x)) \)
\( = g(1 + x) \)
Now, substitute \( (1 + x) \) into \( g(x) \):
\( = 2(1 + x) - 1 \)
\( = 2 + 2x - 1 \)
\( = 2x + 1 \) (Equation 2)

Comparing Equation 1 and Equation 2:
\( 2x \neq 2x + 1 \)
Thus, it is found that \( f \circ g(x) \neq g \circ f(x) \). This calculation verifies the relationship between the composite functions.
In simple words: We calculate \( f(g(x)) \) by putting the rule for \( g(x) \) into \( f(x) \), which gives \( 2x \). Then we calculate \( g(f(x)) \) by putting the rule for \( f(x) \) into \( g(x) \), which gives \( 2x+1 \). Since these two results are not the same, it means \( f \circ g(x) \) is not equal to \( g \circ f(x) \).

🎯 Exam Tip: Always show both composite functions separately. Evaluate \( f(g(x)) \) and \( g(f(x)) \) step-by-step. Remember that composition of functions is generally not commutative.

Question 2. If A × B = {(a, x) (a, y) (b, x) (b, y) (c, x) (c, y)} then find A and B
Answer:
Given \( A \times B = \{(a, x), (a, y), (b, x), (b, y), (c, x), (c, y)\} \).
The set A consists of all the first elements of the ordered pairs in \( A \times B \).
The first elements are a, b, and c.
So, \( A = \{a, b, c\} \).
The set B consists of all the second elements of the ordered pairs in \( A \times B \).
The second elements are x and y.
So, \( B = \{x, y\} \).
This demonstrates how the individual sets can be determined from their Cartesian product.
In simple words: Set A is made of all the first items from each pair. Set B is made of all the second items from each pair. We just collect these unique items to form sets A and B.

🎯 Exam Tip: To find set A from \( A \times B \), collect all unique first components of the ordered pairs. To find set B, collect all unique second components of the ordered pairs.

Question 3. Let A = {x \( \in \) w/3 < x < 7}, B = {x \( \in \) N/0 < x < 3}, C = {x \( \in \) w/x < 2} verify A × (B \( \cap \) C) = (A × B) \( \cap \) (A x C)
Answer:
First, let's define the sets A, B, and C based on the given conditions:
Set A: Whole numbers (w) such that 3 < x < 7.
\( A = \{4, 5, 6\} \).

Set B: Natural numbers (N) such that 0 < x < 3.
\( B = \{1, 2\} \).

Set C: Whole numbers (w) such that x < 2.
\( C = \{0, 1\} \).

Now, let's verify the given equation: \( A \times (B \cap C) = (A \times B) \cap (A \times C) \).

**Step 1: Calculate the Left Hand Side (LHS) - \( A \times (B \cap C) \)**
First, find \( B \cap C \):
\( B \cap C = \{1, 2\} \cap \{0, 1\} = \{1\} \).

Next, find \( A \times (B \cap C) \):
\( A \times (B \cap C) = \{4, 5, 6\} \times \{1\} \)
\( = \{(4, 1), (5, 1), (6, 1)\} \) (Equation 1)

**Step 2: Calculate the Right Hand Side (RHS) - \( (A \times B) \cap (A \times C) \)**
First, find \( A \times B \):
\( A \times B = \{4, 5, 6\} \times \{1, 2\} \)
\( = \{(4, 1), (4, 2), (5, 1), (5, 2), (6, 1), (6, 2)\} \).

Next, find \( A \times C \):
\( A \times C = \{4, 5, 6\} \times \{0, 1\} \)
\( = \{(4, 0), (4, 1), (5, 0), (5, 1), (6, 0), (6, 1)\} \).

Finally, find \( (A \times B) \cap (A \times C) \):
This involves finding common ordered pairs in both \( A \times B \) and \( A \times C \).
\( (A \times B) \cap (A \times C) = \{(4, 1), (5, 1), (6, 1)\} \) (Equation 2)

**Step 3: Compare LHS and RHS**
From Equation 1 and Equation 2, we can see that:
\( A \times (B \cap C) = (A \times B) \cap (A \times C) \).
The equation is therefore verified. This property is known as the distributivity of Cartesian product over intersection.
In simple words: First, we list the numbers in each set (A, B, C) based on their rules. Then, we check if multiplying set A by the common part of B and C gives the same answer as finding the common part of (A times B) and (A times C). We found both sides give the same set of pairs, so the statement is true.

🎯 Exam Tip: Clearly list the elements of each set first. Then, perform the operations on each side of the equation step-by-step. Remember that set intersection \( (\cap) \) means finding common elements, and Cartesian product \( (\times) \) means forming all possible ordered pairs.

Question 4. Let A = {10, 11, 12, 13, 14}; B = {0, 1, 2, 3, 5} and f₁: A \( \to \) B, i = 1, 2, 3. State the type of function for the following (give reason): (i) f₁ = {(10,1), (11,2), (12,3), (13,5), (14,3)}
Answer:
Given \( f_1 = \{(10,1), (11,2), (12,3), (13,5), (14,3)\} \).


**Type of Function: Not a one-one function.**
**Reason:** The elements 12 and 14 from set A both have the same image, 3, in set B. For a function to be one-one, each distinct element in the domain must map to a distinct element in the codomain.

**Type of Function: Not an onto function.**
**Reason:** The element 0 in set B has no pre-image in set A (no element from A maps to 0). For a function to be onto, every element in the codomain (set B) must have at least one pre-image in the domain (set A).

Therefore, the given function \( f_1 \) is neither one-one nor onto, so it is also not a bijective function. This function simply maps elements from one set to another based on the given pairs.
In simple words: This function is not "one-to-one" because two different numbers from set A (12 and 14) both lead to the same number (3) in set B. It is also not "onto" because the number 0 in set B is not an output for any number in set A. So, it is not a perfect match.

🎯 Exam Tip: To check if a function is one-one, ensure no two distinct domain elements map to the same codomain element. To check if it's onto, ensure every codomain element has at least one pre-image in the domain.

Question 4. (ii) f2 = {(10,1), (11,1), (12,1), (13,1), (14,1)}
Answer:
Given \( f_2 = \{(10,1), (11,1), (12,1), (13,1), (14,1)\} \).


**Type of Function: Constant function.**
**Reason:** Every element from set A (the domain) maps to the same single element, 1, in set B (the codomain). This means the output is always constant, regardless of the input from A. This function is also many-to-one because multiple inputs map to the same output.
In simple words: No matter what number you pick from set A, the function always gives the answer 1 in set B. This kind of function, where all inputs give the same output, is called a constant function.

🎯 Exam Tip: A key characteristic of a constant function is that its range contains only one element. All elements of the domain map to this single element.

Question 4. (iii) f3 = {(10,0), (11,1), (12,2), (13,3), (14,5)}
Answer:
Given \( f_3 = \{(10,0), (11,1), (12,2), (13,3), (14,5)\} \).


**Type of Function: One-one and onto function (or) Bijective function.**
**Reason:**
* **One-one:** Each distinct element in set A maps to a distinct element in set B. No two elements in A share the same image in B.
* **Onto:** Every element in set B (the codomain) has at least one pre-image in set A (the domain). The range of \( f_3 \) is {0, 1, 2, 3, 5}, which is equal to set B.
Since \( f_3 \) is both one-one and onto, it is a bijective function. This type of function establishes a perfect pairing between the elements of the two sets.
In simple words: This function is like a perfect matching. Each number in set A goes to a different number in set B (it's "one-to-one"), and every number in set B has a number from set A that leads to it (it's "onto"). When a function is both of these, it's called "bijective".

🎯 Exam Tip: A function is bijective if and only if it is both injective (one-one) and surjective (onto). When a function is bijective, the number of elements in the domain equals the number of elements in the codomain, and there's a unique mapping both ways.

Question 5. If X = {1, 2, 3, 4, 5}, Y = {1, 3, 5, 7, 9} determine which of the following relations from X to Y are functions? Give reason for your answer. If it is a function, state its type. (i) R1 = {(x,y)| y = x + 2,x \( \in \) X,y \( \in \) Y}
Answer:
Given sets: \( X = \{1, 2, 3, 4, 5\} \) and \( Y = \{1, 3, 5, 7, 9\} \).
Relation \( R_1 \) is defined as \( y = x + 2 \). Let's find the images for each element in X:
When \( x = 1 \), \( y = 1 + 2 = 3 \). (3 is in Y)
When \( x = 2 \), \( y = 2 + 2 = 4 \). (4 is NOT in Y)
When \( x = 3 \), \( y = 3 + 2 = 5 \). (5 is in Y)
When \( x = 4 \), \( y = 4 + 2 = 6 \). (6 is NOT in Y)
When \( x = 5 \), \( y = 5 + 2 = 7 \). (7 is in Y)

So, the set of ordered pairs for \( R_1 \) is \( \{(1,3), (2,4), (3,5), (4,6), (5,7)\} \).
From the calculated pairs, we see that for \( x=2 \), the image \( y=4 \) is not in set Y. Similarly, for \( x=4 \), the image \( y=6 \) is not in set Y. Therefore, not all elements of X have images within Y.


**Conclusion: R1 is NOT a function from X to Y.**
**Reason:** For a relation to be a function from set X to set Y, every element in X must have exactly one image in Y. Here, the elements 2 and 4 in X do not have images that are part of Y (since 4 and 6 are not in Y). Therefore, \( R_1 \) is not a function from X to Y. The elements 2 and 4 in X are not mapped to any element within Y.
In simple words: A relation is a function if every number in the first set (X) is linked to exactly one number in the second set (Y). In this problem, when we add 2 to some numbers in X (like 2 and 4), the answers (4 and 6) are not found in set Y. Because not every number in X has a partner in Y, this relation is not a function from X to Y.

🎯 Exam Tip: Always verify that for every element in the domain, its image exists within the specified codomain. If even one domain element maps outside the codomain, the relation is not a function *to* that codomain.