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NCERT Solutions Class 11 Maths

Chapter 2: Relations and Functions

Complete Guide with Expert Tips & Exam Tricks

Chapter 2 is the foundation for Calculus in Class 12. Mastering Cartesian Products, Domain, and Range now is crucial. This guide provides detailed, step-by-step solutions to every question in the NCERT textbook, designed to act as your personal tutor.

Exercise 2.1: Cartesian Products of Sets

Topic: Understanding ordered pairs (x, y) and how to cross-multiply sets.

Q1. If (x/3 + 1, y - 2/3) = (5/3, 1/3), find the values of x and y.

Solution:
Since ordered pairs are equal, their corresponding parts must be equal.
1. x/3 + 1 = 5/3
  x/3 = 5/3 - 1
  x/3 = 2/3
  x = 2
2. y - 2/3 = 1/3
  y = 1/3 + 2/3
  y = 3/3
  y = 1
  Answer: x = 2, y = 1

Q2. If the set A has 3 elements and the set B = {3, 4, 5}, then find the number of elements in (A × B).

Solution:
- n(A) = 3
- n(B) = 3 (since B has 3 elements)
- n(A × B) = n(A) × n(B) = 3 × 3 = 9
  Answer: 9 elements

Q3. If G = {7, 8} and H = {5, 4, 2}, find G × H and H × G.

Solution:
- G × H = {(7,5), (7,4), (7,2), (8,5), (8,4), (8,2)}
- H × G = {(5,7), (5,8), (4,7), (4,8), (2,7), (2,8)}
💡 Expert Tip: Notice that G × H is NOT equal to H × G because (7,5) is different from (5,7). Order matters!

Q4. True or False Statements:

If P = {m, n} and Q = {n, m}, then P × Q = {(m, n), (n, m)}.
- False. P × Q should have 2 × 2 = 4 pairs.
- Correct: P × Q = {(m, n), (m, m), (n, n), (n, m)}
If A and B are non-empty sets, then A × B is a non-empty set of ordered pairs (x, y) such that x ∈ A and y ∈ B.
- True. This is the exact definition of Cartesian product.
If A = {1, 2}, B = {3, 4}, then A × (B ∩ ∅) = ∅.
- True. The intersection of B with an empty set (∅) is empty. The cross product with an empty set is always empty. 1

Q5. If A = {-1, 1}, find A × A × A.

Solution:
- First, A × A = {(-1, -1), (-1, 1), (1, -1), (1, 1)}
- Now cross this with A again:
  Answer: {(-1,-1,-1), (-1,-1,1), (-1,1,-1), (-1,1,1), (1,-1,-1), (1,-1,1), (1,1,-1), (1,1,1)}

Q6. If A × B = {(a, x), (a, y), (b, x), (b, y)}, find A and B.

Solution:
- Set A contains all first elements: A = {a, b}
- Set B contains all second elements: B = {x, y}

Q8. Let A = {1, 2} and B = {3, 4}. Write A × B. How many subsets will A × B have?

Solution:
- A × B = {(1, 3), (1, 4), (2, 3), (2, 4)}
- Number of elements n = 4.
- Number of subsets = 2⁴ = 16.

Q9. If n(A) = 3 and n(B) = 2. If (x, 1), (y, 2), (z, 1) are in A × B, find A and B.

Solution:
- A must contain the first components: x, y, z. Since n(A)=3, A = {x, y, z}.
- B must contain the second components: 1, 2. Since n(B)=2, B = {1, 2}.

Exercise 2.2: Relations

Topic: Domain, Range, and Codomain.

Q1. Let A = {1, 2, ..., 14}. Define R = {(x, y) : 3x - y = 0, where x, y ∈ A}. Write Domain, Codomain, Range.

Solution:
- The relation is 3x - y = 0, which means y = 3x.
- If x=1, y=3 (Valid)
- If x=2, y=6 (Valid)
- If x=3, y=9 (Valid)
- If x=4, y=12 (Valid)
- If x=5, y=15 (Invalid, because 15 is not in set A)
- R = {(1, 3), (2, 6), (3, 9), (4, 12)}
- Domain (Inputs): {1, 2, 3, 4}
- Range (Outputs): {3, 6, 9, 12}
- Codomain (Total Set A): {1, 2, ..., 14}
- ⚠️ Common Mistake: Students often confuse Range (actual outputs) with Codomain (all possible outputs). Here, Codomain is the entire set A.

Q2. Define a relation R on N by R = {(x, y) : y = x + 5, x is a natural number < 4}.

Solution:
- Possible values for x: 1, 2, 3 (since x < 4).
- When x=1, y = 1+5 = 6
- When x=2, y = 2+5 = 7
- When x=3, y = 3+5 = 8
- Roster Form: {(1, 6), (2, 7), (3, 8)}
- Domain: {1, 2, 3}
- Range: {6, 7, 8}

Q3. A = {1, 2, 3, 5}, B = {4, 6, 9}. R = {(x, y) : difference between x and y is odd}.

Solution:
- We need pairs where one number is odd and the other is even (Odd - Even = Odd).
- R = {(1, 4), (1, 6), (2, 9), (3, 4), (3, 6), (5, 4), (5, 6)}

Q5. Let A = {1, 2, 3, 4, 6}. R = {(a, b) : b is exactly divisible by a}.

Solution:
- Pairs where 'b' divides by 'a':
- (1, 1), (1, 2), (1, 3), (1, 4), (1, 6) [1 divides everything]
- (2, 2), (2, 4), (2, 6)
- (3, 3), (3, 6)
- (4, 4)
- (6, 6)
- Domain: {1, 2, 3, 4, 6}
- Range: {1, 2, 3, 4, 6}

Q7. Write R = {(x, x³) : x is a prime number less than 10} in roster form.

Solution:
- Primes < 10 are 2, 3, 5, 7.
- Calculate cubes: 2³=8, 3³=27, 5³=125, 7³=343.
- R = {(2, 8), (3, 27), (5, 125), (7, 343)}

Exercise 2.3: Functions

Topic: Identify Functions, Domain & Range of Real Functions.

Q1. Which relations are functions?

{(2,1), (5,1), (8,1), ...}: Yes, because every input (2, 5, 8...) has a unique output (1). (Many-to-one is allowed).
{(2,1), (4,2), (6,3), ...}: Yes, unique inputs have unique outputs.
{(1,3), (1,5), (2,5)}: No, because the input '1' has two different outputs (3 and 5).

Q2. Find the Domain and Range:

(i) f(x) = -|x|

Domain: x can be any real number (R).
Range: Since |x| is always positive or zero, -|x| must be negative or zero.
Range is (-∞, 0].

(ii) f(x) = √(9 - x²)

Domain: The term inside the root must be ≥ 0.
9 - x² ≥ 0
x² ≤ 9
-3 ≤ x ≤ 3
Domain: [-3, 3]
Range:
Let y = √(9 - x²). Since square roots are non-negative, y ≥ 0.
The maximum value is when x=0, y=3.
Range: [0, 3]

Q5. Find the range:

(i) f(x) = 2 - 3x, x > 0

Since x > 0, multiply by -3 -> -3x < 0.
Add 2 -> 2 - 3x < 2.
So f(x) < 2.
Range: (-∞, 2).

(ii) f(x) = x² + 2

x² is always ≥ 0.
So x² + 2 is always ≥ 2.
Range: [2, ∞).

Miscellaneous Exercise on Chapter 2

Topic: Advanced problems on Functions.

Q1. Show that f(x) is a function and g(x) is not.

f(x): Defined as x² for 0 ≤ x ≤ 3 and 3x for 3 ≤ x ≤ 10.
- At the "joint" x=3:
  Left side: 3² = 9
  Right side: 3(3) = 9.
  The outputs match. It is a function.
g(x): Defined as x² for 0 ≤ x ≤ 2 and 3x for 2 ≤ x ≤ 10.
- At the "joint" x=2:
  Left side: 2² = 4
  Right side: 3(2) = 6.
  Input '2' produces two different outputs (4 and 6). Not a function.

Q2. If f(x) = x², find [f(1.1) - f(1)] / [1.1 - 1].

Solution:
= (1.1² - 1²) / 0.1
= (1.21 - 1) / 0.1
= 0.21 / 0.1
= 2.1

Q3. Find Domain of f(x) = (x² + 2x + 1) / (x² - 8x + 12).

Solution:
- The denominator cannot be zero.
- x² - 8x + 12 = 0
- (x - 6)(x - 2) = 0
- So, x cannot be 2 or 6.
- Domain: R - {2, 6} (All Real numbers except 2 and 6).

Q4. Find Domain and Range of f(x) = √(x - 1).

Domain: x - 1 ≥ 0 -> x ≥ 1. [1, ∞).
Range: Square roots produce non-negative results. [0, ∞).

Q5. Find Domain and Range of f(x) = |x - 1|.

Domain: Defined for all Real Numbers R.
Range: Absolute value is always non-negative. [0, ∞).

Q6. Find Range of f(x) = x² / (1 + x²).

Solution:
- Let y = x² / (1 + x²).
- Since x² ≥ 0, the numerator is positive.
- The denominator (1 + x²) is always greater than the numerator (x²).
- Therefore, the fraction is always less than 1.
- Since x² is non-negative, the result is also ≥ 0.
- Range: [0, 1)
- 💡 Expert Tip: Notice '1' is strictly excluded (use a round bracket) because the numerator can never quite reach the value of the denominator.

Q8. Let f = {(1,1), (2,3), (0,-1), (-1,-3)} be a function Z to Z defined by f(x) = ax + b. Find a, b.

Solution:
- Using point (1, 1): a(1) + b = 1 -> a + b = 1.
- Using point (0, -1): a(0) + b = -1 -> b = -1.
- Substitute b = -1 into the first equation: a - 1 = 1 -> a = 2.
- Answer: a = 2, b = -1.

Q9. Relation R on N defined by a = b². Check validity.

(i) (a,a) ∈ R: False. 2 is not equal to 2².
(ii) (a,b) implies (b,a): False. 4 = 2² (True), but 2 = 4² (False).
(iii) (a,b), (b,c) implies (a,c): False. 16 = 4² and 4 = 2², but 16 ≠ 2² (Transitive property fails).

Q12. A = {9, 10, 11, 12, 13}. f(n) = highest prime factor of n. Find Range.

Solution:
- f(9): Factors are 1, 3, 9. Highest prime = 3.
- f(10): Factors are 1, 2, 5, 10. Highest prime = 5.
- f(11): Prime itself. Highest prime = 11.
- f(12): Factors include 2, 3... Highest prime = 3.
- f(13): Prime itself. Highest prime = 13.
- Range: {3, 5, 11, 13}.