NCERT Exemplar Class 11 Maths Chapter 2 Relations and Functions are part of NCERT Exemplar Class 11 Maths. Here we have given NCERT Exemplar Class 11 Maths Chapter 2 Relations and Functions.

## NCERT Exemplar Class 11 Maths Chapter 2 Relations and Functions

**Short Answer Type Questions**

**Q1. If A = {-1, 2, 3 } and B = {1, 3}, then determine**

**(i) AxB (ii) BxC (c) BxB (iv) AxA
Sol: **We have A = {-1,2,3} and B = {1,3}

(i) A x B = {(-1, 1), (-1, 3), (2, 1), (2, 3), (3,1), (3, 3)}

(ii) BxA = {( 1, -1), (1, 2), (1,3), (3,-1), (3,2), (3, 3)}

(iii) BxB= {(1,1), (1,3), (3,1), (3, 3)}

(iv) A xA = {(-1, -1), (-1, 2), (-1, 3), (2, -1), (2, 2), (2, 3), (3, -1), (3, 2), (3,3)}

**Q2. If P = {x : x < 3, x e N}, Q= {x : x≤2,x ∈ W}. Find (P∪ Q) x (P∩ Q), where W is the set of whole numbers.**

**Sol:** We have, P={x: x<3,x ∈ N} = {1,2}

And Q = {x :x≤ 2,x∈ W] = {0,1,2}

P∪Q= {0, 1,2} and P ∩ Q= {1,2}

(P ∪ Q) x (P ∩ Q) = {0,1, 2} x {1,2}

= {(0,1), (0, 2), (1,1), (1,2), (2,1), (2, 2)}

**Q3. lfA={x:x∈ W,x < 2}, 5 = {x : x∈N, 1 <.x < 5}, C= {3, 5}. Find**

**(i) Ax(B∩Q) (ii) Ax(B∪C)**

**Sol: **We have, A = {x :x∈ W,x< 2} = {0, 1};

B = {x : x ∈ N, 1 <x< 5} = {2, 3,4}; and C= {3, 5}

**(i)** B∩ C = {3}

A x (B ∩ C) = {0, 1} x {3} = {(0, 3), (1, 3)}

**(ii)** (B ∪ C) ={2,3,4, 5}

A x (B ∪ C) = {0, 1} x {2, 3,4, 5}

= {(0,2), (0,3), (0,4), (0,5), (1,2), (1,3), (1,4), (1, 5)}

**Q4. In each of the following cases, find a and b. (2a + b, a – b) = (8, 3) (ii) {a/4, a – 2b) = (0, 6 + b)**

**Sol: **(i) We have, (2a + b,a-b) = (8,3)

=> 2a + b = 8 and a – b = 3

On solving, we get a = 11/3 and b = 2/3

**Q5. Given A = {1,2,3,4, 5}, S= {(x,y) :x∈ A,y∈ A}.Find the ordered pairs which satisfy the conditions given below**

**x+y = 5 (ii) x+y<5 (iii) x+y>8 **

** Sol:** We have, A = {1,2, 3,4, 5}, S= {(x,y) : x ∈ A,y∈ A}

**(i)** The set of ordered pairs satisfying x + y= 5 is {(1,4), (2,3), (3,2), (4,1)}

**(ii)** The set of ordered pairs satisfying x+y < 5 is {(1,1), (1,2), (1,3), (2, 1), (2,2), (3,1)}

**(iii)** The set of ordered pairs satisfying x +y > 8 is {(4, 5), (5,4), (5, 5)}.

**Q6. Given R = {(x,y) : x,y ∈ W, x ^{2} + y^{2} = 25}. Find the domain and range of R**

**Sol:**We have, R = {(x,y):x,y∈ W, x

^{2}+ y

^{2}= 25}

= {(0,5), (3,4), (4, 3), (5,0)}

Domain of R = Set of first element of ordered pairs in R = {0,3,4, 5}

Range of R = Set of second element of ordered pairs in R = {5,4, 3, 0}

**Q7. If R _{1} = {(x, y)| y = 2x + 7, where x∈ R and -5 ≤ x ≤ 5} is a relation. Then find the domain and range of R_{1}.**

**Sol:**We have, R

_{1}= {(x, y)|y = 2x + 7, where x∈ R and -5 ≤x ≤ 5}

Domain of R

_{1}= {-5 ≤ x ≤ 5, x ∈ R} = [-5, 5]

x ∈ [-5, 5]

=> 2x ∈ [-10,10]

=>2x + 7∈ [-3, 17]

Range is [-3, 17]

**Q8. If R _{2} = {(x, y) | x and y are integers and x^{2} +y^{2} = 64} is a relation. Then find R_{2
}Sol:** We have, R

_{2}= {(x, y) | x and y are integers and x

^{2}+ y

^{2}– 64}

Clearly, x

^{2}= 0 and y

^{2}= 64 or x

^{2}= 64 andy

^{2}= 0

x = 0 and y = ±8

or x = ±8 and y = 0

R

_{2}= {(0, 8), (0, -8), (8,0), (-8,0)}

**Q9. If R _{3} = {(x, |x|) | x is a real number} is a relation. Then find domain and range**

**Sol:**We have, R

_{3}= {(x, |x)) | x is real number}

Clearly, domain of R

_{3}= R

Now, x ∈ R and |x| ≥ 0 .

Range of R

_{3}is [0,∞)

**Q10. Is the given relation a function? Give reasons for your answer.**

**(i) h={(4,6), (3,9), (-11,6), (3,11)}**

**(ii) f = {(x, x) | x is a real number}**

**(iii) g = {(n, 1 In)| nis a positive integer}**

**(iv) s= {(n, n ^{2}) | n is a positive integer}**

**(v) t= {(x, 3) | x is a real number}**

**Sol: (i)**We have, h = {(4,6),(3,9), (-11,6), (3,11)}.

Since pre-image 3 has two images 9 and 11, it is not a function.

**(ii)**We have, f = {(x, x) | x is a real number}

Since every element in the domain has unique image, it is a function.

(iii) We have, g= {(n, 1/n) | nis a positive integer}

For n, it is a positive integer and 1/n is unique and distinct. Therefore,every element in the domain has unique image. So, it is a function.

**(iii)**We have, s = {(n, n

^{2}) | n is a positive integer}

Since the square of any positive integer is unique, every element in the domain has unique image. Hence, ibis a function.

**(iv)**We have, t = {(x, 3)| x is a real number}.

Since every element in the domain has the image 3, it is a constant function.

**Q11. If f and g are real functions defined byf( x) = x ^{2} + 7 and g(x) = 3x + 5, find each of the following
**

**Q12. Let f and g be real functions defined by f(x) = 2x+ 1 and g(x) = 4x – 7.**

**(i) For what real numbers x,f(x)= g(x)?**

**(ii) For what real numbers x,f (x) < g(x)?**

**Sol: **We have,f(x) = 2x + 1 and g(x) = 4x-7

**(i)** Now f (x) = g(x)

=> 2x+l=4x-7

=> 2x = 8 =>x = 4

**(ii)** f (x) < g(x)

=> 2x + 1 < 4x – 7

=> 8 < 2x

=> x > 4

**Q13. If f and g are two real valued ftmctions defined as f(x) = 2x + 1, g(x) = x ^{2} + 1, then find.
**

**Q14. Express the following functions as set of ordered pairs and determine their range.**

**f:X->R,f{x) = x ^{3} + 1, where X= {-1,0, 3, 9, 7} **

**Sol:**We have, f:X→ R,flx) = x

^{3}+ 1.

Where X = {-1, 0, 3, 9, 7}

Now f (-l) = (-l)

^{3}+1 =-l + 1 =0

f(0) = (0)

^{3}+l=0+l = l

f(3) = (3)

^{3}+ 1 = 27 + 1 = 28

f(9) = (9)

^{3}+ 1 = 729 + 1 = 730

f(7) = (7)

^{3}+ 1 = 343 + 1 = 344

f= {(-1, 0), (0, 1), (3, 28), (9, 730), (7, 344)}

Range of f= {0, 1, 28, 730, 344}

**Q15. Find the values of x for which the functions f(x) = 3x ^{2} -1 and g(x) = 3+ x are equal.**

**Sol:**f(x) = g(x)

=> 3x

^{2}-l=3+x => 3x

^{2}-x-4 = 0 => (3x – 4)(x+ 1) – 0

x= -1,4/3

**Q16. Is g = {(1, 1), (2, 3), (3, 5), (4, 7)} a function? Justify. If this is described by the relation, g(x) = x +, then what values should be assigned to and ?**

**Sol:**We have, g = {(1, 1), (2, 3), (3, 5), (4,7)}

Since, every element has unique image under g. So, g is a function.

Now, g(x) = x + For (1,1), g(l) = a(l) + P

=> l = + (i)

For (2, 3), g(2) = (2) +

=> 3 = 2 + (ii)

On solving Eqs. (i) and (ii), we get = 2, = -l

f(x) = 2x-1

Also, (3, 5) and (4, 7) satisfy the above function.

**Q17. Find the domain of each of the following functions given by
**

**Q18. Find the range of the following functions given by**

**Q19. Redefine the function f(x) = |x-2| + |2+x| , -3 ≤x ≤3**

**Q21. Let f (x) = √x and g(x) = xbe two functions defined in the domain R ^{+} ∪ {0}. Find**

**(i) (f+g)(x)**

**(ii) (f-g)(x)**

**(iii) (fg)(x)**

**(iv) f/g(x)**

**Q23. If f( x)= y = ax-b/ cx-a then prove that f (y) = x
**

**Objective Type Questions**

**Q24. Let n(A) = m, and n(B) = n. Then the total number of non-empty relations that can be defined from A to B is**

**(a) m ^{n} **

**(b) n**

^{m}– 1**(c) mn – 1**

**(d) 2**

^{mn}– 1**Sol:** (d) We have, n(A) = m and n(B) = n

n(A xB) = n(A). n(B) = mn

Total number of relation from A to B = Number of subsets of AxB = 2^{mn
}So, total number of non-empty relations = 2^{mn} – 1

**Q25. If [x] ^{2} – 5[x] + 6 = 0, where [. ] denote the greatest integer function, then**

**(a) x ∈ [3,4]**

**(b) x∈ (2, 3]**

**(c) x∈ [2, 3]**

**(d) x ∈ [2, 4)**

**Sol:**(d) We have [x]

^{2}– 5[x] + 6 = 0 => [(x – 3)([x] – 2) = 0

=> [x] = 2,3 .

For [x] = 2, x ∈ [2, 3)

For [x] = 3, x ∈ [3,4)

x ∈ [2, 3) u [3,4)

Or x ∈ [2,4)

**Q29. If fx) ax+ b, where a and b are integers,f(-1) = -5 and f(3) – 3, then a and b are equal to **

**(a) a = -3, b =-1**

**(b) a = 2, b =-3**

**(c) a = 0, b = 2**

**(d) a = 2, b = 3**

**Fill in the Blanks Type Questions**

**Q36. Let f and g be two real functions given by f= {(0, 1), (2,0), (3,.-4), (4,2), (5, 1)} **

** g= {(1,0), (2,2), (3,-1), (4,4), (5, 3)} then the domain of f x g is given by________ .**

**Sol: **We have, f = {(0, 1), (2, 0), (3, -4), (4, 2), (5,1)} and g= {(1, 0), (2, 2), (3, 1), (4,4), (5, 3)}

Domain of f = {0,2, 3, 4, 5}

And Domain of g= {1, 2, 3,4, 5}

Domain of (f x g) = (Domain of f) ∩ (Domain of g) = {2, 3,4, 5}

**Matching Column Type Questions**

**Q37. Let f= {(2,4), (5,6), (8, -1), (10, -3)} andg = {(2, 5), (7,1), (8,4), (10,13), (11, 5)} be two real functions. Then match the following:
**

**True/False Type Questions**

**Q38. The ordered pair (5,2) belongs to the relation R ={(x,y): y = x – 5, x,y∈Z}**

**Sol:** False

We have, R = {(x, y): y = x – 5, x, y ∈ Z}

When x = 5, then y = 5-5=0 Hence, (5, 2) does not belong to R.

**Q39. If P = {1, 2}, then P x P x P = {(1, 1,1), (2,2, 2), (1, 2,2), (2,1, 1)}**

**Sol:**False

We have, P = {1, 2} and n(P) = 2

n(P xPxP) = n(P) x n(P) x n(P) = 2 x 2 x 2

= 8 But given P x P x P has 4 elements.

**Q40. If A= {1,2, 3}, 5= {3,4} and C= {4, 5, 6}, then (A x B) ∪ (A x C) = {(1, 3), (1, 4), (1, 5), (1, 6), (2, 3), (2, 4), (2, 5), (2, 6), (3, 3), (3,4), (3, 5), (3,6)}.**

**Sol:** True

We have.4 = {1,2, 3}, 5= {3,4} andC= {4,5,6}

AxB= {(1, 3), (1,4), (2, 3), (2,4), (3, 3), (3,4)}

And A x C = {(1,4), (1, 5), (1, 6), (2,4), (2, 5), (2, 6), (3,4), (3, 5), (3, 6)}

(A x B)∪(A xC)= {(1, 3), (1, 4), (1, 5), (1, 6), (2, 3), (2, 4), (2, 5), (2, 6), (3,3), (3,4), (3, 5), (3,6)}

**Q42. If Ax B= {(a, x), (a, y), (b, x), (b, y)}, thenM = {a, b},B= {x, y}.**

**Sol:** True

We have, AxB= {{a, x), {a, y), (b, x), {b, y)}

A = Set of first element of ordered pairs in A x B = {a, b}

B = Set of second element of ordered pairs in A x B = {x, y}

## NCERT Exemplar Class 11 Maths Solutions

- Chapter 1 Sets
- Chapter 2 Relations and Functions
- Chapter 3 Trigonometric Functions
- Chapter 4 Principle of Mathematical Induction
- Chapter 5 Complex Numbers and Quadratic Equations
- Chapter 6 Linear Inequalities
- Chapter 7 Permutations and Combinations
- Chapter 8 Binomial Theorem
- Chapter 9 Sequence and Series
- Chapter 10 Straight Lines
- Chapter 11 Conic Sections
- Chapter 12 Introduction to Three-Dimensional Geometry
- Chapter 13 Limits and Derivatives
- Chapter 14 Mathematical Reasoning
- Chapter 15 Statistics
- Chapter 16 Probability

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