# CBSE Solutions for Class 11 Maths

#### Select CBSE Solutions for class 10 Subject & Chapters Wise :

Set of odd natural numbers divisible by 2 isn’t a null set

Answer :

False

Set of even prime numbers is an null set

Answer :

False

{x:is a natural numbers, < 3 and > 11 } is an finite set

Answer :

True

{z:is a point common to any two parallel lines} is an finite set

Answer :

True

The set of months of a leap year is a infinite set

Answer :

False

{0,1, 2, 3, 4 ...} is a infinite set

Answer :

True

{1, 2, 3 ... 99} is an finite set

Answer :

True

The set of positive integers greater than 9 is an finite set

Answer :

False

The set of lines which are parallel to the x-axis  is an finite set

Answer :

False

The set of letters in the English alphabet is a infinite set

Answer :

False

The set of natural numbers under 200 which are multiple of 7 is infinite set

Answer :

False

The set of animals living on the earth is a infinite set

Answer :

False

The set of circles passing through the origin (0, 0) is a infinite set

Answer :

True

The set A = {-2, -3}; B = {xis solution of x2 + 5+ 6 = 0} are not equal sets

Answer :

False

The set P = {xis a letter in the word FOLLOW}; Q = {yis a letter in the word WOLF} are  equal sets

Answer :

True

{2, 3, 4} ⊄ {1, 2, 3, 4, 5}

Answer :

False

{abc}⊂  {bcd}

Answer :

False

{xis a student of Class X of your school} ⊂  {xstudent of your school}

Answer :

True

{xis a square in the plane} ⊂  {xis a rectangle in the same plane}

Answer :

True

{pis a triangle in a plane}⊄  {pis a rectangle in the plane}

Answer :

True

Find the smallest set X such that X∪{1, 2}={1, 2, 3, 5, 9}.

Answer :

We have to find the smallest set X such that X∪{1, 2}={1, 2, 3, 5, 9}.

The union of the two sets X & Y is the set of all those elements that belong to X or to Y or to both X & Y.

Thus, X must be {3, 5, 9}.

Let U = {1, 2, 3, 4, 5, 6, 7, 8, 9}, P = {2, 4, 6, 8} and Q = {2, 3, 5, 7}. Verify that
(i) (A∪B)'=A'∩B'
(ii) (A∩B)'=A'∪B'.

Answer :

Given:
U = {1, 2, 3, 4, 5, 6, 7, 8, 9}, P = {2, 4, 6, 8} and Q = {2, 3, 5, 7}
We have to verify:

(i) (P∪Q)'=P'∩Q'
LHS
P∪Q ={2,3,4,5,6,7,8} (P∪B )'={1,9}

RHS
P'={1,3,5,7,9} Q'={1,4,6,8,9} P'∩Q'={1,9}

LHS = RHS
Hence proved.

(ii) (P∩Q)'=P'∪Q'
LHS
P∩Q={2} (P∩Q)'={1,3,4,5,6,7,8,9}

RHS
P'={1,3,5,7,9} Q'={1,4,6,8,9} P'∪Q'={1,3,4,5,6,7,8,9}

LHS = RHS
Hence proved.

Let U = {1, 2, 3, 4, 5, 6, 7, 8, 9}, P = {1, 2, 3, 4}, Q= {2, 4, 6, 8} and R = {3, 4, 5, 6}. Find
(i) P'
(ii) Q'
(iii) (P∩R)'
(iv) (P∪Q)'
(v) (P')'
(vi) (Q−R)'

Answer :

Given:
U = {1, 2, 3, 4, 5, 6, 7, 8, 9}, P = {1, 2, 3, 4}, Q= {2, 4, 6, 8} and R = {3, 4, 5, 6}
(i) P' = {5, 6, 7, 8, 9}
(ii) Q' = {1, 3, 5, 7, 9}
(iii) (P∩R)' = {1, 2, 5, 6, 7, 8, 9}
(iv) (P∪Q)' = {5, 7, 9}
(v) (P')' = {1, 2, 3, 4} = A
(vi) (Q−R)' = {1, 3, 4, 5, 6, 7, 9}

Let P = {3, 6, 12, 15, 18, 21}, Q = {4, 8, 12, 16, 20}, R = {2, 4, 6, 8, 10, 12, 14, 16} and S = {5, 10, 15, 20}. Find:
(i) P−Q
(ii) P−R
(iii) P−S
(iv) Q−P
(v) R−P
(vi) S−P
(vii) Q−R
(viii) Q−S

Answer :

Given:
P = {3, 6, 12, 15, 18, 21}, Q = {4, 8, 12, 16, 20}, R = {2, 4, 6, 8, 10, 12, 14, 16} and S = {5, 10, 15, 20}
(i) P−Q  = {3, 6, 15, 18, 21}
(ii) P−R = {3, 15, 18, 21}
(iii) P−S = {3, 6, 12, 18, 21}
(iv) Q−P = {4, 8, 16, 20}
(v) R−P = {2, 4, 8, 10, 14, 16}
(vi) S−P = {5, 10, 20}
(vii) Q−R = {20}
(viii) Q−S = {4, 8, 12, 16}

Let P={x:x∈N}, B={x:x−2n, n∈N}, C={x:x=2n−1, n∈N} and D = {x : x is a prime natural number}. Find:
(i) P∩Q
(ii) P∩R
(iii) P∩S
(iv) Q∩R
(v) Q∩S
(vi) R∩S

Answer :

P={x:x∈N}={1,2,3,...} Q={x:x−2n, n∈N}={2,4,6,8,...} R={x:x=2n−1, n∈N}={1,3,5,7,...} S = {x:x is a prime natural number.} = {2, 3, 5, 7,...}
(i) P∩Q = Q
(ii) P∩R = R
(iii) P∩S = S
(iv) B∩R = ϕ
(v) B∩S = {2}
(vi) R∩S = S−{2}

If P = {1, 2, 3, 4, 5}, Q = {4, 5, 6, 7, 8}, R = {7, 8, 9, 10, 11} and S = {10, 11, 12, 13, 14}, find:
(i) P∪Q
(ii) P∪R
(iii) Q∪R
(iv) Q∪S
(v) P∪Q∪R
(vi) P∪Q∪S
(vii) Q∪R∪S
(viii) P∩Q∪R
(ix) (P∩Q)∩(Q∩R)
(x) (P∪S)∩(Q∪R)

Answer :

Given:
P = {1, 2, 3, 4, 5}, Q = {4, 5, 6, 7, 8}, R = {7, 8, 9, 10, 11} and S = {10, 11, 12, 13, 14}
(i) P∪Q = {1, 2, 3, 4, 5, 6, 7, 8}
(ii) P∪R = {1, 2, 3, 4, 5, 7, 8, 9, 10, 11}
(iii) Q∪R = {4, 5, 6, 7, 8, 9, 10, 11}
(iv) Q∪S = {4, 5, 6, 7, 8, 10, 11, 12, 13, 14}
(v)  P∪Q∪R = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11}
(vi) P∪Q∪S = {1, 2, 3, 4, 5, 6, 7, 8, 10, 11, 12, 13, 14}
(vii)  Q∪R∪S = {4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14}
(viii) P∩(Q∪R) = {4, 5}
(ix) (P∩Q)∩(Q∩R) = ϕ
(x) (P∪S)∩(Q∪R) = {4, 5, 10, 11}

If X and Y are two sets such that XY, then find:
(i) X∩Y
(ii) XY

Answer :

From the Venn diagrams given below, we can clearly say that if X and Y are two sets such that XY, then
(i) Form the given Venn diagram, we can see that  X∩Y = X
(ii) Form the given Venn diagram, we can see that  XY = Y

If P = {1, 2, 3, 4, 5}, Q = {4, 5, 6, 7, 8}, R = {7, 8, 9, 10, 11} and S = {10, 11, 12, 13, 14}, find:

 1 Q∪S A {1, 2, 3, 4, 5, 6, 7, 8, 10, 11, 12, 13, 14} 2 P∪Q∪R B {4, 5, 6, 7, 8, 10, 11, 12, 13, 14} 3 P∪Q∪S C {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11}

Answer :

1-B, 2-C, 3-A

sets in Roster form

 1 {x ∈ R : x > x}. A {17, 26, 35, 44, 53, 62, 71, 80} 2 {x : x is a prime number which is a divisor of 60} B Φ 3 {x : x is a two digit number such that the sum of its digits is 8} C {T, R, I, G, O, N, M, E, Y} 4 The set of all letters in the word 'Trigonometry' D {2, 3, 5}

Answer :

1-B, 2-D, 3-A, 4-C

sets in Roster form

 1 {x : x is a letter before e in the English alphabet} A {1, 2, 3, 4} 2 {x ∈ N : x2 < 25} B {11, 13, 17, 19} 3 {x ∈ N : x is a prime number, 10 < x < 20} C {a, b, c, d} 4 {x ∈ N : x = 2n, n ∈ N} D {2, 4, 6, 8, 10,...}

Answer :

1-C, 2-A, 3-B, 4-D

If P = {1, 2, 3, 4, 5}, Q = {4, 5, 6, 7, 8}, R = {7, 8, 9, 10, 11} and S = {10, 11, 12, 13, 14}, find:

 1 P∪Q A {1, 2, 3, 4, 5, 6, 7, 8} 2 P∪R B {4, 5, 6, 7, 8, 9, 10, 11} 3 Q∪R C {1, 2, 3, 4, 5, 7, 8, 9, 10, 11}

Answer :

1-A, 2-C, 3-B

sets in set-builder form

 1 A = {1, 2, 3, 4, 5, 6} A {x:x∈N, 9

Answer :

1-B, 2-C, 3-D, 4-A

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The GSEB Books for class 10 are designed as per the syllabus followed Gujarat Secondary and Higher Secondary Education Board provides key detailed, and a through solutions to all the questions relating to the GSEB textbooks.

The purpose is to provide help to the students with their homework, preparing for the examinations and personal learning. These books are very helpful for the preparation of examination.

For more details about the GSEB books for Class 10, you can access the PDF which is as in the above given links for the same.