Relations and Functions — Class 11 Mathematics NCERT Solutions (Free)
Free step-by-step NCERT solutions for Class 11 Mathematics chapter "Relations and Functions" — 8 important questions with detailed answers for CBSE board exam preparation.
TL;DR: Free step-by-step NCERT solutions for Class 11 Mathematics chapter "Relations and Functions" — 8 important questions with detailed answers for CBSE bo…
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NCERT Solutions for Class 11 Mathematics — Relations and Functions. Step-by-step answers to all 8 textbook questions from this chapter, written for CBSE board preparation and free to use.
Relations and Functions — All 8 Questions Solved
Q1. Let A = {1, 2, 3} and B = {4, 5}. Define a relation R from A to B as R = {(x, y) : x < y}. Write R and determine its domain and range.
Step 1: Find all ordered pairs (x, y) where x ∈ A, y ∈ B, and x < y.
Check (1, 4): 1 < 4? Yes ✓
Check (1, 5): 1 < 5? Yes ✓
Check (2, 4): 2 < 4? Yes ✓
Check (2, 5): 2 < 5? Yes ✓
Check (3, 4): 3 < 4? Yes ✓
Check (3, 5): 3 < 5? Yes ✓
Step 2: Write the relation.
R = {(1, 4), (1, 5), (2, 4), (2, 5), (3, 4), (3, 5)}
Step 3: Find the domain (all first elements).
Domain(R) = {1, 2, 3}
Step 4: Find the range (all second elements).
Range(R) = {4, 5}
Final Answer: R = {(1,4), (1,5), (2,4), (2,5), (3,4), (3,5)}, Domain = {1, 2, 3}, Range = {4, 5}
Q2. Determine whether the relation R = {(1, 1), (1, 2), (2, 2), (2, 3), (3, 3)} on set A = {1, 2, 3} is reflexive, symmetric, and transitive.
Step 1: Check reflexivity.
A relation is reflexive if (a, a) ∈ R for all a ∈ A.
Check (1, 1): Yes ✓
Check (2, 2): Yes ✓
Check (3, 3): Yes ✓
R is reflexive.
Step 2: Check symmetry.
A relation is symmetric if (a, b) ∈ R ⟹ (b, a) ∈ R.
Check (1, 2): (1, 2) ∈ R, but (2, 1) ∉ R
R is not symmetric.
Step 3: Check transitivity.
A relation is transitive if (a, b) ∈ R and (b, c) ∈ R ⟹ (a, c) ∈ R.
Check (1, 1) and (1, 2): (1, 2) ∈ R ✓
Check (1, 2) and (2, 2): (1, 2) ∈ R ✓
Check (1, 2) and (2, 3): (1, 3) ∉ R
R is not transitive.
Final Answer: R is reflexive but not symmetric and not transitive.
Q3. Check whether f(x) = 2x + 3 is a function from ℝ to ℝ. If yes, find f(-2), f(0), and f(1).
Step 1: Verify f is a function.
For f to be a function, each x ∈ ℝ must have exactly one image in ℝ.
For any x ∈ ℝ, 2x + 3 is uniquely defined and is a real number.
Therefore, f is a function from ℝ to ℝ.
Step 2: Calculate f(-2).
f(-2) = 2(-2) + 3 = -4 + 3 = -1
Step 3: Calculate f(0).
f(0) = 2(0) + 3 = 0 + 3 = 3
Step 4: Calculate f(1).
f(1) = 2(1) + 3 = 2 + 3 = 5
Final Answer: Yes, f is a function. f(-2) = -1, f(0) = 3, f(1) = 5
Q4. For the function f(x) = x² - 2x + 3, find (i) f(-1), (ii) f(a + 1), (iii) the domain and range.
Step 1: Calculate f(-1).
f(-1) = (-1)² - 2(-1) + 3 = 1 + 2 + 3 = 6
Step 2: Calculate f(a + 1).
f(a + 1) = (a + 1)² - 2(a + 1) + 3
= a² + 2a + 1 - 2a - 2 + 3
= a² + 2
Step 3: Find the domain.
f(x) = x² - 2x + 3 is a polynomial, defined for all real numbers.
Domain = ℝ
Step 4: Find the range.
Rewrite f(x) by completing the square:
f(x) = x² - 2x + 3 = (x² - 2x + 1) + 2 = (x - 1)² + 2
Since (x - 1)² ≥ 0 for all x ∈ ℝ,
(x - 1)² + 2 ≥ 2
The minimum value is 2 (when x = 1).
Range = [2, ∞)
Final Answer: (i) f(-1) = 6, (ii) f(a+1) = a² + 2, (iii) Domain = ℝ, Range = [2, ∞)
Q5. Determine whether the function f(x) = x³ from ℝ to ℝ is one-one (injective) and onto (surjective).
Step 1: Check if f is one-one (injective).
f is one-one if f(a) = f(b) ⟹ a = b.
Assume f(a) = f(b):
a³ = b³
Taking cube root: a = b
Therefore, f is one-one.
Step 2: Check if f is onto (surjective).
f is onto if for every y ∈ ℝ, there exists x ∈ ℝ such that f(x) = y.
Let y ∈ ℝ be arbitrary.
We need x³ = y
This gives x = ∛y (cube root of y)
For any real y, ∛y is a real number.
Therefore, every y ∈ ℝ has a pre-image x = ∛y ∈ ℝ.
So f is onto.
Step 3: Conclusion.
Since f is both one-one and onto, f is bijective (one-one correspondence).
Final Answer: f(x) = x³ is both injective and surjective (hence bijective).
Q6. Given f(x) = 2x - 1 and g(x) = x + 3, find (i) (f ∘ g)(x), (ii) (g ∘ f)(x), (iii) (f ∘ g)(2).
Step 1: Find (f ∘ g)(x) = f(g(x)).
(f ∘ g)(x) = f(x + 3) = 2(x + 3) - 1 = 2x + 6 - 1 = 2x + 5
Step 2: Find (g ∘ f)(x) = g(f(x)).
(g ∘ f)(x) = g(2x - 1) = (2x - 1) + 3 = 2x + 2
Step 3: Find (f ∘ g)(2).
(f ∘ g)(2) = 2(2) + 5 = 4 + 5 = 9
Alternatively: g(2) = 2 + 3 = 5, then f(5) = 2(5) - 1 = 9
Step 4: Note that (f ∘ g)(x) ≠ (g ∘ f)(x).
Composition of functions is not commutative.
Final Answer: (i) (f ∘ g)(x) = 2x + 5, (ii) (g ∘ f)(x) = 2x + 2, (iii) (f ∘ g)(2) = 9
Q7. Find the inverse of the function f(x) = 3x - 5. Verify that f⁻¹(f(x)) = x and f(f⁻¹(x)) = x.
Step 1: Find the inverse f⁻¹(x).
Let y = f(x) = 3x - 5
Swap x and y: x = 3y - 5
Solve for y: 3y = x + 5
y = (x + 5)/3
Therefore, f⁻¹(x) = (x + 5)/3
Step 2: Verify f⁻¹(f(x)) = x.
f⁻¹(f(x)) = f⁻¹(3x - 5)
= ((3x - 5) + 5)/3
= 3x/3
= x ✓
Step 3: Verify f(f⁻¹(x)) = x.
f(f⁻¹(x)) = f((x + 5)/3)
= 3((x + 5)/3) - 5
= (x + 5) - 5
= x ✓
Step 4: Check that f is bijective (one-one and onto).
f(x) = 3x - 5 is linear with non-zero slope, so it is bijective.
Therefore, f⁻¹ exists.
Final Answer: f⁻¹(x) = (x + 5)/3; both f⁻¹(f(x)) = x and f(f⁻¹(x)) = x are verified.
Q8. A relation R is defined on ℤ (set of integers) by aRb if and only if a - b is divisible by 5. Show that R is an equivalence relation.
Step 1: Check reflexivity.
For any a ∈ ℤ, is a - a divisible by 5?
a - a = 0, and 0 is divisible by 5.
So aRa for all a ∈ ℤ.
R is reflexive.
Step 2: Check symmetry.
Assume aRb, so a - b is divisible by 5.
This means a - b = 5k for some k ∈ ℤ.
Then b - a = -(a - b) = -5k = 5(-k)
Since -k ∈ ℤ, b - a is also divisible by 5.
So bRa.
R is symmetric.
Step 3: Check transitivity.
Assume aRb and bRc.
Then a - b = 5m and b - c = 5n for some m, n ∈ ℤ.
Adding: (a - b) + (b - c) = 5m + 5n
a - c = 5(m + n)
Since m + n ∈ ℤ, a - c is divisible by 5.
So aRc.
R is transitive.
Step 4: Conclusion.
Since R is reflexive, symmetric, and transitive, R is an equivalence relation.
Final Answer: R is an equivalence relation.
Relations and Functions — Practice MCQs with Answers
Attempt these 20 multiple-choice questions after working through the solutions above. Each carries the correct option and the reasoning behind it, so a wrong answer tells you which idea to revisit.
MCQ 1. If A = {1, 2} and B = {a, b}, what is A × B?
- A. { (1,a), (1,b), (2,a), (2,b) }
- B. { (a,1), (b,1), (a,2), (b,2) }
- C. { (1,1), (a,a), (2,2), (b,b) }
- D. { (1,a), (2,b) }
Answer: A. { (1,a), (1,b), (2,a), (2,b) } — The Cartesian product A × B is the set of all ordered pairs (x, y) where x ∈ A and y ∈ B.
MCQ 2. If set A has 3 elements and set B has 2 elements, how many relations can be defined from A to B?
- A. 6
- B. 32
- C. 64
- D. 2^5
Answer: C. 64 — If a set A has 'm' elements and set B has 'n' elements, then the number of relations from A to B is 2^(m*n). Here, m=3, n=2, so 2^(3*2) = 2^6 = 64.
MCQ 3. The domain of the function f(x) = √(x - 3) is:
- A. (-∞, 3]
- B. (3, ∞)
- C. [3, ∞)
- D. (-∞, 3)
Answer: C. [3, ∞) — For the square root function √(y) to be defined, the expression inside the square root must be non-negative. So, x - 3 ≥ 0, which implies x ≥ 3.
MCQ 4. What is the range of the function f(x) = |x - 2|?
- A. (-∞, ∞)
- B. [0, ∞)
- C. (-∞, 0]
- D. (2, ∞)
Answer: B. [0, ∞) — The absolute value function always returns a non-negative value. Thus, |x - 2| ≥ 0 for all real x, so the range is [0, ∞).
MCQ 5. Which of the following relations is NOT a function from {1, 2, 3} to {a, b, c}?
- A. { (1,a), (2,b), (3,c) }
- B. { (1,a), (2,a), (3,a) }
- C. { (1,a), (1,b), (2,c) }
- D. { (1,b), (2,c), (3,a) }
Answer: C. { (1,a), (1,b), (2,c) } — A relation is a function if every element in the domain has exactly one image in the codomain. In option C, the element '1' from the domain has two images, 'a' and 'b', violating the definition of a function.
MCQ 6. Find the domain of the function f(x) = 1 / (x^2 - 4).
- A. R
- B. R - {2}
- C. R - {-2, 2}
- D. R - {0}
Answer: C. R - {-2, 2} — For a rational function, the denominator cannot be zero. So, x^2 - 4 ≠ 0, which means (x - 2)(x + 2) ≠ 0. Therefore, x ≠ 2 and x ≠ -2.
MCQ 7. The range of the function f(x) = x^2 + 1 is:
- A. (-∞, ∞)
- B. [1, ∞)
- C. (0, ∞)
- D. (-∞, 1]
Answer: B. [1, ∞) — The square of any real number x^2 is always non-negative (x^2 ≥ 0). Adding 1 to x^2 means x^2 + 1 ≥ 1. Thus, the minimum value is 1, and it can go up to infinity.
MCQ 8. What is the value of ⌊3.7⌋ + ⌈-2.3⌉? (⌊x⌋ is the greatest integer less than or equal to x, ⌈x⌉ is the smallest integer greater than or equal to x)
- A. 1
- B. 2
- C. 0
- D. -1
Answer: A. 1 — ⌊3.7⌋ = 3, as 3 is the greatest integer less than or equal to 3.7. ⌈-2.3⌉ = -2, as -2 is the smallest integer greater than or equal to -2.3. So, 3 + (-2) = 1.
MCQ 9. If f(x) = x + 1 and g(x) = x^2, then (f + g)(x) is:
- A. x^3 + x^2
- B. x^2 + x + 1
- C. x^2 + 1
- D. (x+1)x^2
Answer: B. x^2 + x + 1 — (f + g)(x) is defined as f(x) + g(x). So, (x + 1) + x^2 = x^2 + x + 1.
MCQ 10. For an identity function f(x) = x, what is its range?
- A. {0}
- B. R
- C. [0, ∞)
- D. Z (integers)
Answer: B. R — An identity function f(x) = x maps every real number to itself. Therefore, its range is the set of all real numbers, R.
MCQ 11. The domain of the function f(x) = √(x - 1) / (x - 4) is:
- A. [1, ∞)
- B. [1, 4) U (4, ∞)
- C. (1, 4) U (4, ∞)
- D. (-∞, 1] U (4, ∞)
Answer: B. [1, 4) U (4, ∞) — For √(x - 1) to be defined, x - 1 ≥ 0, which means x ≥ 1. For 1/(x - 4) to be defined, x - 4 ≠ 0, which means x ≠ 4. Combining these, the domain is [1, ∞) excluding x = 4, so [1, 4) U (4, ∞).
MCQ 12. What is the range of the signum function, sgn(x)?
- A. { -1, 0, 1 }
- B. R
- C. [-1, 1]
- D. { -1, 1 }
Answer: A. { -1, 0, 1 } — The signum function is defined as sgn(x) = 1 if x > 0, -1 if x < 0, and 0 if x = 0. Therefore, its range is the set {-1, 0, 1}.
MCQ 13. If f(x) = 5 for all x ∈ R, then f is a:
- A. Identity function
- B. Modulus function
- C. Constant function
- D. Polynomial function of degree 1
Answer: C. Constant function — A function that assigns the same constant value to every element in its domain is called a constant function.
MCQ 14. Which of the following is an odd function?
- A. f(x) = x^2
- B. f(x) = cos(x)
- C. f(x) = x^3 - x
- D. f(x) = |x|
Answer: C. f(x) = x^3 - x — A function f(x) is odd if f(-x) = -f(x). For f(x) = x^3 - x, f(-x) = (-x)^3 - (-x) = -x^3 + x = -(x^3 - x) = -f(x).
MCQ 15. The range of the function f(x) = x / (x - 1) is:
- A. R
- B. R - {1}
- C. R - {0}
- D. R - {-1}
Answer: B. R - {1} — Let y = x / (x - 1). Solving for x, we get x = y / (y - 1). For x to be real, the denominator (y - 1) cannot be zero, which means y ≠ 1. Thus, the range is R - {1}.
MCQ 16. Let R be a relation on N (natural numbers) defined by x + 2y = 8. What is the domain of R?
- A. {2, 4, 6}
- B. {1, 2, 3}
- C. {2, 3, 4}
- D. {2, 4, 6, 8}
Answer: A. {2, 4, 6} — Given x + 2y = 8, where x, y ∈ N. If y=1, x=6; if y=2, x=4; if y=3, x=2. If y=4, x=0 which is not a natural number. So the relation is {(6,1), (4,2), (2,3)}. The domain is the set of first elements {2, 4, 6}.
MCQ 17. A relation R from a non-empty set A to a non-empty set B is a subset of:
- A. A
- B. B
- C. A ∩ B
- D. A × B
Answer: D. A × B — By definition, a relation from set A to set B is any subset of the Cartesian product A × B.
MCQ 18. The domain of the function f(x) = 1 / √(x^2 - 9) is:
- A. (-3, 3)
- B. (-∞, -3) U (3, ∞)
- C. [-3, 3]
- D. (-∞, -3] U [3, ∞)
Answer: B. (-∞, -3) U (3, ∞) — For the function to be defined, x^2 - 9 must be strictly greater than 0 (because it's in the denominator and under a square root). So, x^2 - 9 > 0 implies x^2 > 9, which means x > 3 or x < -3.
MCQ 19. If set A has m elements and set B has n elements, how many distinct functions can be mapped from A to B?
- A. m * n
- B. m^n
- C. n^m
- D. n!
Answer: C. n^m — For each of the m elements in set A, there are n choices in set B to map to. Since there are m such elements, the total number of functions is n * n * ... (m times) = n^m.
MCQ 20. The range of the function f(x) = |x| + 5 is:
- A. [0, ∞)
- B. [5, ∞)
- C. (-∞, ∞)
- D. (-5, ∞)
Answer: B. [5, ∞) — The absolute value |x| is always non-negative, i.e., |x| ≥ 0. Adding 5 to it, we get |x| + 5 ≥ 5. So the minimum value is 5, and it extends to infinity.
How to Use These Solutions
Attempt each question yourself first and only then compare with the worked answer. Marks in the CBSE board exam are awarded for the METHOD as much as the final result, so reproduce the steps rather than memorising the last line. Where a solution states a law, a formula or a definition, learn that wording — examiners look for it.
Related practice for this chapter: chapter MCQs, previous-year questions, revision notes and sample papers.