In this chapter, we provide NCERT Exemplar Problems Solutions for Class 11 Maths Chapter 2 Relations and Functions for English medium students, Which will very helpful for every student in their exams. Students can download the latest NCERT Exemplar Problems Solutions for Class 11 Maths Chapter 2 Relations and Functions pdf, free NCERT Exemplar Problems Solutions for Class 11 Maths Chapter 2 Relations and Functions book pdf download. Now you will get step by step solution to each question.

Textbook | NCERT |

Class | Class 11 |

Subject | Maths |

Chapter | Chapter 2 |

Chapter Name | Relations and Functions |

Category | NCERT Exemplar |

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

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) AxASol: **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}

**All Chapter NCERT Exemplar Problems Solutions For Class 11 Maths**

—————————————————————————–**All Subject NCERT Exemplar Problems Solutions For Class 11**

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