RD Sharma Solutions for Class 11 Maths Chapter 3 Functions

Mathematics forms the backbone of logical reasoning and analytical skills, especially in higher classes. For Class 11 students, mastering the concept of functions is essential, as it lays the groundwork for advanced topics in calculus and algebra. RD Sharma Solutions for Class 11 Maths Chapter 3 Functions offer clear, step-by-step explanations of the fundamental principles, types, and applications of functions, making this chapter vital for exam success.

A function is a special type of relation that links each element of a set (domain) to a unique element in another set (co-domain). In this chapter, students learn to define, identify, and represent functions, distinguish them from relations, and explore key concepts like domain, co-domain, range, and types of functions. Whether you’re looking for RD Sharma solutions PDF download or help with domain and range problems, these solutions are designed to support your learning and exam preparation.

The RD Sharma Solutions for Class 11 Maths Chapter 3 Functions are designed to make these abstract concepts accessible and easy to understand. Each solution is presented in a step-by-step manner, guiding students through definitions, solved examples, and a variety of exercises. Whether you’re learning about domains, ranges, types of functions (like polynomial, linear, quadratic, modulus, signum, greatest integer, and more), or exploring operations such as addition, subtraction, multiplication, and division of functions, these solutions provide clear explanations and logical reasoning at every step.

With the help of RD Sharma Class 11 solutions, students can easily navigate the complexities of functions, ensuring thorough preparation for school exams and competitive entrance tests. These solutions are aligned with the latest CBSE Class 11 syllabus and are an invaluable resource for effective self-study and revision.

Download RD Sharma Solutions for Class 11 Maths Chapter 3 Functions PDF Here

The RD Sharma Class 11 Chapter 3 Solution PDF includes comprehensive solutions, worked examples, and extra questions to help you master the topic of functions and related mathematical concepts. Click here to download the RD Sharma Solutions Class 11 Chapter 3 PDF.

Q1. List the relation R on A={1,2}, B={3,4} defined by “aRb if a<b.”

Solution:

1. Form A×B: {(1,3),(1,4),(2,3),(2,4)}.
2. Check each pair for a<b:
   • (1,3): 1<3 ✓
   • (1,4): 1<4 ✓
   • (2,3): 2<3 ✓
   • (2,4): 2<4 ✓
3. R = {(1,3),(1,4),(2,3),(2,4)}.

Q2. For A={1,2,3}, let R={(x,y): x=y}. List R and state if it’s reflexive, symmetric, transitive.

Solution:

1. R = {(1,1),(2,2),(3,3)}.
2. Reflexive? All (a,a) present ✓.
3. Symmetric? If (x,y)∈R ⇒ x=y ⇒ (y,x)=(x,x)∈R ✓.
4. Transitive? (x,y),(y,z) ⇒ x=y=z ⇒ (x,z)=(x,x)∈R ✓.
5. R is reflexive, symmetric, transitive (the identity relation).

Q3. Let A={1,2,3}, R = {(x,y): x<y}. Find domain and range.

Solution:

1. List A×A: check x<y:
   R = {(1,2),(1,3),(2,3)}.
2. Domain = {x | ∃y: (x,y)∈R} = {1,2}.
3. Range = {y | ∃x: (x,y)∈R} = {2,3}.
4. Domain={1,2}, Range={2,3}.

Q4. Given R = {(a,b),(a,c),(b,a),(c,a)} on A={a,b,c}, find domain and range.

Solution:

1. Domain: all first components = {a,b,c}.
2. Range: all second components = {b,c,a} = {a,b,c}.
3. Domain=Range={a,b,c}.

Q5. If |A|=m and |B|=n, how many relations from A to B exist?

Solution:

1. |A×B| = m·n.
2. Each subset of A×B is a relation.
3. Number of subsets = 2^(m·n).
4. Answer: 2^(m·n).

Q6. How many reflexive relations exist on a set A of size n?

Solution:

1. A×A has n² pairs.
2. Reflexive requires all (a,a) (n of them).
3. Other n²–n pairs free to include/exclude ⇒ 2^(n²–n) relations.
4. Answer: 2^(n²–n).

Q7. For A={1,2,3}, how many relations are there on A?

Solution:

1. |A×A| = 3×3 = 9.
2. Total relations = 2^9 = 512.
3. Answer: 512.

Q8. Determine if R = {(1,1),(2,2),(3,3),(1,2),(2,1)} on A={1,2,3} is symmetric, transitive.

Solution:

1. Symmetric?
   • (1,2)∈R ⇒ (2,1)∈R ✓
   • No other off-diagonal to check ⇒ symmetric ✓.
2. Transitive?
   • (1,2),(2,1) ⇒ need (1,1)∈R ✓
   • All other combinations of (x,y),(y,z) checked ✓.
3. R is symmetric and transitive (also reflexive on 1,2,3 except (3,3) present, so reflexive ✓).

Q9. On integers Z define R={(x,y): x–y is even}. Show R is equivalence and find two classes.

Solution:

1. Reflexive? x–x=0 even ✓.
2. Symmetric? x–y even ⇒ y–x=–(x–y) even ✓.
3. Transitive? x–y, y–z even ⇒ (x–z)=(x–y)+(y–z) even ✓.
4. Equivalence classes:
   • [0] = {...,–4,–2,0,2,4,...} (even ints)
   • [1] = {...,–3,–1,1,3,5,...} (odd ints)

Q10. Relation “is sibling of” on people. Check reflexive, symmetric, transitive.

Solution:

1. Reflexive? One isn’t sibling of oneself ⇒ no.
2. Symmetric? If A sibling of B ⇒ B sibling of A ✓.
3. Transitive? A sibling of B & B sibling of C ⇒ A sibling of C? Not necessarily.
4. It’s only symmetric.

Q11. R on students where xRy if x and y are in the same class. Is R an equivalence?

Solution:

1. Reflexive: same student ⇒ same class ✓.
2. Symmetric: if x,y same class ⇒ y,x same class ✓.
3. Transitive: x,y and y,z same class ⇒ x,z same class ✓.
4. R is an equivalence relation.

Q12. Given R: A→B and S: B→C, find S∘R.

Example: A={1,2}, B={a,b}, C={x,y}. R={(1,a),(2,b)}, S={(a,x),(b,y)}.
Solution:

1. Compose: (1→a→x),(2→b→y).
2. S∘R = {(1,x),(2,y)}.

Q13. For R={(1,2),(2,3),(1,3)} on A={1,2,3}, find R⁻¹ (the inverse).

Solution:

1. Swap each pair:
   R⁻¹ = {(2,1),(3,2),(3,1)}.

Q14. Determine if R = {(1,4),(2,5),(3,6)} from A={1,2,3} to B={4,5,6} is a function.

Solution:

1. Each a∈A appears exactly once as first element ⇒ yes.
2. R is a function (one–one correspondence).

Q15. Given f: A→B, f(x)=2x on A={1,2,3}, B={2,4,6}, show f is a relation and function.

Solution:

1. Relation: f ⊆ A×B = {(1,2),(2,4),(3,6)}.
2. Each x∈A maps to exactly one y ⇒ function.

Q16. How many antisymmetric relations on A={1,2}?

Solution:

1. A×A has 4 pairs: (1,1),(2,2),(1,2),(2,1).
2. For antisymmetry: if (1,2) and (2,1) both in R, then 1=2 impossible ⇒ can’t include both.
3. Each diagonal must or may be chosen freely (2 bits), and for the off-diagonal pair choose either, neither, but not both (3 choices).
4. Total = 2²×3 = 4×3 = 12.

Q17. Check if R={(1,1),(2,2),(1,2)} on A={1,2} is antisymmetric.

Solution:

1. Off-diagonals: (1,2)∈R but (2,1)∉R ⇒ no violation.
2. R is antisymmetric.

Q18. Is the “divides” relation on positive integers an equivalence? (xRy if x|y)

Solution:

1. Reflexive: x|x ✓.
2. Symmetric? If x|y does not imply y|x ⇒ no.
3. Transitive: x|y and y|z ⇒ x|z ✓.
4. Only reflexive & transitive (a partial order, not equivalence).

Q19. Partition induced by R in Q9 (even/odd). List blocks.

Solution:

1. Blocks = equivalence classes:
   • Even block = {...,–2,0,2,4,...}
   • Odd block = {...,–3,–1,1,3,5,...}.

Q20. Parent–child relation on family. Check reflexivity, etc.

Solution:

1. Reflexive? One isn’t own parent ⇒ no.
2. Symmetric? Parent of A≠child of A ⇒ no.
3. Transitive? A parent of B & B parent of C ⇒ A grandparent of C, not parent ⇒ no.
4. It’s neither reflexive, symmetric, nor transitive.

Q21. On Z, let R: xRy if x²=y². Is R an equivalence?

Solution:

1. Reflexive: x²=x² ✓.
2. Symmetric: x²=y² ⇒ y²=x² ✓.
3. Transitive: x²=y² & y²=z² ⇒ x²=z² ✓.
4. Yes, equivalence.
5. Classes: [0]={0}, [1]={1,–1}, [2]={2,–2}, …

Q22. On Z, R: xRy if x+y is odd. Check properties.

Solution:

1. Reflexive? x+x = 2x is even ⇒ no.
2. Symmetric? x+y odd ⇒ y+x odd ✓.
3. Transitive? x+y odd & y+z odd ⇒ (x+y)+(y+z)=2y + x+z even ⇒ x+z even ⇒ xRz false ⇒ no.
4. Only symmetric.

Q23. For R={(1,2),(2,3)} on A={1,2,3}, check transitivity.

Solution:

1. (1,2),(2,3)∈R ⇒ need (1,3) in R.
2. (1,3)∉R ⇒ not transitive.

Q24. List all relations on A={0,1} that are reflexive.

Solution:

1. Must include (0,0) and (1,1).
2. Off-diagonals (0,1),(1,0) either: include both, one, or neither ⇒ 2 choices each ⇒ 4 total.
3. Reflexive relations = 4.

Q25. If R={(a,b): a,b ∈{1,2,3} and a+b=4}, list R.

Solution:

1. Find pairs s.t. sum=4: (1,3),(2,2),(3,1).
2. R = {(1,3),(2,2),(3,1)}.