Relations and Functions — Class 12 Mathematics NCERT Solutions (Free)
Free step-by-step NCERT solutions for Class 12 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 12 Mathematics chapter "Relations and Functions" — 8 important questions with detailed answers for CBSE bo…
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NCERT Solutions for Class 12 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. Check whether the relation R = {(a, b) : a ≤ b²} on the set ℕ is reflexive, symmetric, and transitive.
We need to check three properties for R = {(a, b) : a ≤ b²} on ℕ.
REFLEXIVE: For R to be reflexive, (a, a) ∈ R for all a ∈ ℕ.
This means a ≤ a².
For a = 1: 1 ≤ 1² = 1 ✓
For a = 2: 2 ≤ 2² = 4 ✓
For all a ≥ 1: a ≤ a² is true (since a² - a = a(a-1) ≥ 0).
So R is REFLEXIVE.
SYMMETRIC: For R to be symmetric, if (a, b) ∈ R then (b, a) ∈ R.
If a ≤ b², does this imply b ≤ a²?
Counter-example: (2, 3) ∈ R since 2 ≤ 3² = 9 ✓
But (3, 2) ∉ R since 3 ≤ 2² = 4 is false.
So R is NOT SYMMETRIC.
TRANSITIVE: For R to be transitive, if (a, b) ∈ R and (b, c) ∈ R, then (a, c) ∈ R.
If a ≤ b² and b ≤ c², does this imply a ≤ c²?
From b ≤ c², we get b² ≤ c⁴.
Since a ≤ b², we have a ≤ b² ≤ c⁴.
But this doesn't guarantee a ≤ c².
Counter-example: (2, 3) and (3, 5) are in R since 2 ≤ 9 and 3 ≤ 25.
For (2, 5): 2 ≤ 25 is true, so (2, 5) ∈ R.
Actually, checking more carefully: if a ≤ b² and b ≤ c², then a ≤ b² but we need a ≤ c².
Since b ≤ c² does not guarantee b² ≤ c², transitivity fails.
Counter-example: a = 6, b = 3, c = 2. Then 6 ≤ 9 (true) and 3 ≤ 4 (true) but 6 ≤ 4 (false).
So R is NOT TRANSITIVE.
CONCLUSION: R is reflexive only.
Q2. Let f: ℝ → ℝ be defined by f(x) = x + 1 and g: ℝ → ℝ be defined by g(x) = x². Find (f ∘ g) and (g ∘ f) and check if they are equal.
Given: f(x) = x + 1 and g(x) = x²
FINDING (f ∘ g):
(f ∘ g)(x) = f(g(x)) = f(x²) = x² + 1
FINDING (g ∘ f):
(g ∘ f)(x) = g(f(x)) = g(x + 1) = (x + 1)²
Expanding: (x + 1)² = x² + 2x + 1
COMPARISON:
(f ∘ g)(x) = x² + 1
(g ∘ f)(x) = x² + 2x + 1
Clearly (f ∘ g)(x) ≠ (g ∘ f)(x) for all x ∈ ℝ
(except at x = 0 where both equal 1).
CONCLUSION: (f ∘ g) and (g ∘ f) are NOT equal. Function composition is not commutative.
Q3. Determine whether the function f(x) = 2x/(x² + 1) defined on ℝ is one-one (injective).
To check if f(x) = 2x/(x² + 1) is one-one, we need to verify if f(a) = f(b) implies a = b.
Assume f(a) = f(b):
2a/(a² + 1) = 2b/(b² + 1)
Cross-multiplying:
2a(b² + 1) = 2b(a² + 1)
a(b² + 1) = b(a² + 1)
ab² + a = ba² + b
ab² - ba² = b - a
ab(b - a) = b - a
(ab - 1)(b - a) = 0
So either ab = 1 or a = b.
If ab = 1, then b = 1/a.
Let's check with a = 2, b = 1/2:
f(2) = 2(2)/(4 + 1) = 4/5
f(1/2) = 2(1/2)/(1/4 + 1) = 1/(5/4) = 4/5
So f(2) = f(1/2) but 2 ≠ 1/2.
CONCLUSION: f is NOT one-one (not injective) because distinct elements can have the same image.
Q4. If f(x) = √(x - 1) and g(x) = x², find the domain of (f ∘ g).
We need to find the domain of (f ∘ g)(x) = f(g(x)) = f(x²) = √(x² - 1).
For f(g(x)) to be defined:
1. g(x) must be defined (this is true for all x ∈ ℝ since g(x) = x² is defined everywhere)
2. g(x) must be in the domain of f
Domain of f is {x : x - 1 ≥ 0}, i.e., x ≥ 1
So we need g(x) ≥ 1
x² ≥ 1
x ≥ 1 or x ≤ -1
Also, for (f ∘ g)(x) = √(x² - 1) to be defined:
x² - 1 ≥ 0
x² ≥ 1
x ≥ 1 or x ≤ -1
CONCLUSION: Domain of (f ∘ g) = (-∞, -1] ∪ [1, ∞)
Q5. Show that f(x) = 3x + 5 is a bijection from ℝ to ℝ and find its inverse.
To show f: ℝ → ℝ defined by f(x) = 3x + 5 is a bijection, we need to prove it is both injective and surjective.
INJECTIVITY:
Let f(a) = f(b)
3a + 5 = 3b + 5
3a = 3b
a = b
So f is injective (one-one).
SURJECTIVITY:
For any y ∈ ℝ, we need to find x ∈ ℝ such that f(x) = y.
3x + 5 = y
3x = y - 5
x = (y - 5)/3
Since y is arbitrary and (y - 5)/3 is always a real number, f is surjective (onto).
So f is a bijection.
FINDING THE INVERSE:
Let y = f(x) = 3x + 5
Solving for x:
y = 3x + 5
3x = y - 5
x = (y - 5)/3
Therefore, f⁻¹(x) = (x - 5)/3
VERIFICATION:
f(f⁻¹(x)) = f((x - 5)/3) = 3((x - 5)/3) + 5 = (x - 5) + 5 = x ✓
f⁻¹(f(x)) = f⁻¹(3x + 5) = ((3x + 5) - 5)/3 = 3x/3 = x ✓
Q6. Verify whether the function f: {1, 2, 3} → {4, 5, 6} defined by f = {(1, 4), (2, 5), (3, 6)} is a bijection.
Given f: {1, 2, 3} → {4, 5, 6} with f = {(1, 4), (2, 5), (3, 6)}
CHECKING INJECTIVE:
Domain elements: 1, 2, 3 have images 4, 5, 6 respectively.
All images are different, so distinct elements map to distinct elements.
f is INJECTIVE.
CHECKING SURJECTIVE:
Codomain: {4, 5, 6}
Range (set of actual outputs): {4, 5, 6}
Every element in the codomain is the image of some domain element.
f is SURJECTIVE.
CONCLUSION: Since f is both injective and surjective, f is a BIJECTION.
The inverse function is:
f⁻¹: {4, 5, 6} → {1, 2, 3}
f⁻¹ = {(4, 1), (5, 2), (6, 3)}
Q7. Let R be a relation on ℤ defined by R = {(a, b) : 2 divides (a - b)}. Prove that R is an equivalence relation.
R = {(a, b) : 2 | (a - b)} on the set ℤ. We prove R is an equivalence relation by verifying reflexive, symmetric, and transitive properties.
REFLEXIVE:
For any a ∈ ℤ, we need (a, a) ∈ R.
This means 2 | (a - a) = 0.
Since 0 = 2 × 0, we have 2 | 0.
So (a, a) ∈ R for all a ∈ ℤ. R is REFLEXIVE.
SYMMETRIC:
Let (a, b) ∈ R. Then 2 | (a - b).
So a - b = 2k for some k ∈ ℤ.
Then b - a = -2k = 2(-k).
Since -k ∈ ℤ, we have 2 | (b - a).
So (b, a) ∈ R. R is SYMMETRIC.
TRANSITIVE:
Let (a, b) ∈ R and (b, c) ∈ R.
Then 2 | (a - b) and 2 | (b - c).
So a - b = 2m and b - c = 2n for some m, n ∈ ℤ.
Adding: (a - b) + (b - c) = 2m + 2n
a - c = 2(m + n)
Since m + n ∈ ℤ, we have 2 | (a - c).
So (a, c) ∈ R. R is TRANSITIVE.
CONCLUSION: R is an equivalence relation.
Q8. If f(x) = sin(x) and g(x) = x/π are functions from ℝ to ℝ, find the domain of f∘g and compute (f ∘ g)(π).
Given: f(x) = sin(x), g(x) = x/π
FINDING (f ∘ g):
(f ∘ g)(x) = f(g(x)) = f(x/π) = sin(x/π)
FINDING THE DOMAIN:
g(x) = x/π is defined for all x ∈ ℝ.
f(y) = sin(y) is defined for all y ∈ ℝ.
Since the range of g is ℝ (all real numbers), and f is defined on all of ℝ, the domain of (f ∘ g) is ℝ.
Domain of (f ∘ g) = ℝ
COMPUTING (f ∘ g)(π):
(f ∘ g)(π) = sin(π/π) = sin(1)
where 1 is in radians.
sin(1) ≈ 0.8414710... (using standard tables or calculator)
CONCLUSION: Domain of (f ∘ g) is ℝ and (f ∘ g)(π) = sin(1) ≈ 0.841
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. Let A = {1, 2, 3}. Which of the following relations on A is reflexive?
- A. R1 = {(1,1), (2,2)}
- B. R2 = {(1,1), (2,2), (3,3), (1,2)}
- C. R3 = {(1,1), (2,3), (3,2)}
- D. R4 = {(1,2), (2,3), (3,1)}
Answer: B. R2 = {(1,1), (2,2), (3,3), (1,2)} — A relation R on set A is reflexive if for every element 'a' in A, (a,a) belongs to R. R2 contains (1,1), (2,2), and (3,3), satisfying the condition for A = {1,2,3}.
MCQ 2. A function f: N → N defined by f(x) = 2x is:
- A. One-to-one and onto
- B. One-to-one but not onto
- C. Onto but not one-to-one
- D. Neither one-to-one nor onto
Answer: B. One-to-one but not onto — For f(x1) = f(x2) => 2x1 = 2x2 => x1 = x2, so f is one-to-one. For surjectivity, there is no natural number x such that f(x) = 1 (since 2x will always be even), so f is not onto.
MCQ 3. Let R be a relation on the set A = {1, 2, 3, 4, 5} defined by R = {(a,b) : |a-b| is even}. Then R is:
- A. Reflexive but not symmetric
- B. Symmetric but not transitive
- C. Transitive but not reflexive
- D. An equivalence relation
Answer: D. An equivalence relation — R is reflexive since |a-a| = 0, which is even. R is symmetric since if |a-b| is even, then |b-a| is also even. R is transitive because if |a-b| and |b-c| are even, then a-b and b-c have the same parity, implying (a-b)+(b-c) = a-c is even, so |a-c| is even. Thus, R is an equivalence relation.
MCQ 4. If f(x) = x^2 and g(x) = sin(x), then (gof)(x) is:
- A. sin(x^2)
- B. (sin x)^2
- C. x^2 sin(x)
- D. sin(x^2) - x^2
Answer: A. sin(x^2) — The composition (gof)(x) means g(f(x)). Substituting f(x) = x^2 into g(x) = sin(x) gives g(x^2) = sin(x^2).
MCQ 5. If a set A has 3 elements, then the number of equivalence relations on A is:
- A. 1
- B. 3
- C. 5
- D. 8
Answer: C. 5 — The number of equivalence relations on a set with n elements is given by the nth Bell number. For n=3, the Bell number B3 is 5.
MCQ 6. The inverse of the function f(x) = 3x - 2 is:
- A. g(x) = (x+2)/3
- B. g(x) = 3x + 2
- C. g(x) = (x-2)/3
- D. g(x) = 1/(3x-2)
Answer: A. g(x) = (x+2)/3 — To find the inverse, let y = f(x) = 3x - 2. Swap x and y: x = 3y - 2. Solving for y gives y = (x+2)/3, so f^(-1)(x) = (x+2)/3.
MCQ 7. A function f: R → R defined by f(x) = |x| is:
- A. One-to-one and onto
- B. One-to-one but not onto
- C. Onto but not one-to-one
- D. Neither one-to-one nor onto
Answer: D. Neither one-to-one nor onto — It's not one-to-one because f(1) = 1 and f(-1) = 1. It's not onto because the range is [0, ∞), but the codomain is R, so negative numbers have no pre-image.
MCQ 8. If f: R → R is given by f(x) = x^3 + 5, then f^(-1)(x) is:
- A. (x-5)^(1/3)
- B. (x+5)^(1/3)
- C. 1/(x^3+5)
- D. x^(1/3) - 5
Answer: A. (x-5)^(1/3) — Let y = x^3 + 5. Swap x and y: x = y^3 + 5. Solving for y, we get y^3 = x - 5, so y = (x-5)^(1/3). Thus, f^(-1)(x) = (x-5)^(1/3).
MCQ 9. Let L be the set of all lines in a plane and R be a relation in L defined as R = {(L1, L2) : L1 is perpendicular to L2}. Which of the following is true about R?
- A. R is reflexive
- B. R is symmetric
- C. R is transitive
- D. R is an equivalence relation
Answer: B. R is symmetric — R is symmetric because if L1 is perpendicular to L2, then L2 is perpendicular to L1. It is not reflexive (a line is not perpendicular to itself) nor transitive (if L1 ⊥ L2 and L2 ⊥ L3, then L1 || L3).
MCQ 10. Let * be a binary operation on Q defined by a * b = ab/2. The identity element for * is:
- A. 0
- B. 1
- C. 2
- D. 1/2
Answer: C. 2 — An element e is the identity element if a * e = a for all a in Q. So, ae/2 = a, which implies ae = 2a. For a ≠ 0, e = 2. If a=0, 0*e = 0, so 0=0, which holds for e=2.
MCQ 11. If A = {1, 2, 3} and B = {a, b}, then the number of onto functions from A to B is:
- A. 2
- B. 6
- C. 8
- D. 0
Answer: B. 6 — The total number of functions from A to B is 2^3 = 8. The functions that are not onto are those where all elements map to 'a' (1 function) or all map to 'b' (1 function). So, 8 - 2 = 6 onto functions.
MCQ 12. Let A = {1, 2, 3, 4, 5, 6} and R be a relation on A defined by R = {(x,y) : y = x+1}. The range of R is:
- A. {1, 2, 3, 4, 5, 6}
- B. {1, 2, 3, 4, 5}
- C. {2, 3, 4, 5, 6}
- D. {1, 3, 5}
Answer: C. {2, 3, 4, 5, 6} — For each x in A, y = x+1. The pairs in R are (1,2), (2,3), (3,4), (4,5), (5,6). The element (6,7) is not in R as 7 is not in A. The range is the set of all second elements, which is {2, 3, 4, 5, 6}.
MCQ 13. The function f: R → R defined by f(x) = x^2 + 1 is:
- A. Bijective
- B. One-to-one but not onto
- C. Onto but not one-to-one
- D. Neither one-to-one nor onto
Answer: D. Neither one-to-one nor onto — It's not one-to-one because f(1) = 2 and f(-1) = 2. It's not onto because the range is [1, ∞), but the codomain is R, meaning numbers like 0 have no pre-image.
MCQ 14. A relation R on set A is transitive if for all a,b,c in A:
- A. (a,b) in R implies (b,a) in R
- B. (a,a) in R
- C. (a,b) in R and (b,c) in R implies (a,c) in R
- D. (a,b) in R
Answer: C. (a,b) in R and (b,c) in R implies (a,c) in R — This is the standard definition of a transitive relation: if an element 'a' is related to 'b' and 'b' is related to 'c', then 'a' must be related to 'c'.
MCQ 15. If f(x) = sqrt(x-1) for x ≥ 1, then the domain of f^(-1)(x) is:
- A. [1, ∞)
- B. [0, ∞)
- C. (-∞, ∞)
- D. [2, ∞)
Answer: B. [0, ∞) — The domain of f^(-1)(x) is the range of f(x). For f(x) = sqrt(x-1) with x ≥ 1, the minimum value of x-1 is 0, so sqrt(x-1) ≥ 0. The function can take any non-negative real value, so the range of f(x) is [0, ∞).
MCQ 16. Let * be a binary operation on Z defined by a * b = a + b - 4. The inverse of 3 is:
- A. -1
- B. 1
- C. 5
- D. 4
Answer: C. 5 — First, find the identity element 'e': a * e = a => a + e - 4 = a => e = 4. To find the inverse of 3, let it be x. Then 3 * x = e => 3 + x - 4 = 4 => x - 1 = 4 => x = 5.
MCQ 17. Let f: R → R be defined by f(x) = x^2. The element 4 in the codomain has pre-images:
- A. 2
- B. -2
- C. 2 and -2
- D. None
Answer: C. 2 and -2 — To find the pre-images of 4, we set f(x) = 4, so x^2 = 4. This yields x = 2 and x = -2. Both are real numbers, so they are valid pre-images.
MCQ 18. Let R be the relation in the set Z of integers defined by R = {(x,y) : x - y is an even integer}. What is the equivalence class of 0?
- A. {0}
- B. Z
- C. {..., -2, 0, 2, ...} (Even integers)
- D. {..., -1, 1, 3, ...} (Odd integers)
Answer: C. {..., -2, 0, 2, ...} (Even integers) — The equivalence class of 0 consists of all integers x such that (x,0) is in R. This means x - 0 is an even integer, which implies x must be an even integer. So, the equivalence class of 0 is the set of all even integers.
MCQ 19. Let f: R → R be defined by f(x) = x^3. Then f is:
- A. One-to-one but not onto
- B. Onto but not one-to-one
- C. Neither one-to-one nor onto
- D. One-to-one and onto
Answer: D. One-to-one and onto — For injectivity, x1^3 = x2^3 => x1 = x2. For surjectivity, for any y in R, x = y^(1/3) is also in R. Thus, f is both one-to-one and onto, hence bijective.
MCQ 20. If f: R → R is given by f(x) = 2x + 3 and g: R → R is given by g(x) = (x-3)/2, then (fog)(x) is:
- A. 4x + 3
- B. 2x
- C. x
- D. x - 3/2
Answer: C. x — (fog)(x) = f(g(x)) = f((x-3)/2). Substituting (x-3)/2 into f(x) gives 2((x-3)/2) + 3 = (x-3) + 3 = x. This shows that g is the inverse of f.
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.