CBSE Solutions for Class 11 English

GSEB std 10 science solution for Gujarati check Subject Chapters Wise::

Write the following sets in roster form:

W = {xis a two-digit natural number such that the sum of its digits is 4}

Hide | Show

જવાબ : {13, 22, 31, 40}


Write the following sets in roster form:

X = {xis a prime number which is divisor of 10}.

Hide | Show

જવાબ : {2, 5}


Write the following sets in roster form:

V = {xis a natural number less than 3}.

Hide | Show

જવાબ : {1,2}


Write the following sets in roster form:

U = {xis an integer and –1 < < 1}.

Hide | Show

જવાબ : { -1, 0, 1}


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

Hide | Show

જવાબ : False


Set of even prime numbers is an null set

Hide | Show

જવાબ : False


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

Hide | Show

જવાબ : True


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

Hide | Show

જવાબ : True


If A ⊄ B and ∉ B, then ∉ A

Hide | Show

જવાબ : False


If ∈ A and A ⊂   B, then ∈ B

Hide | Show

જવાબ : True


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

Hide | Show

જવાબ : False


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

Hide | Show

જવાબ : True


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

Hide | Show

જવાબ : True


If A ⊄ B and B ⊄ C, then A ⊄ C

Hide | Show

જવાબ : False


If A ⊄ B and B ⊂ C, then A ⊂ C

Hide | Show

જવાબ :  False


If A ⊄ B and B ∈ C, then A ∈ C

Hide | Show

જવાબ : False


If ∈ A and A ∈ B, then ∈ B

Hide | Show

જવાબ : False


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

Hide | Show

જવાબ : False


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

Hide | Show

જવાબ : False


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

Hide | Show

જવાબ : False


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

Hide | Show

જવાબ : False


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

Hide | Show

જવાબ : False


{xis an even natural number less than 6} ⊄ {xis a natural number which divides 36}

Hide | Show

જવાબ : False


{a} ∈ (123)

Hide | Show

જવાબ : False


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

Hide | Show

જવાબ : True


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

Hide | Show

જવાબ : False


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

Hide | Show

જવાબ : True


{2, 3, 4} ⊄ {1, 2, 3, 4, 5}

Hide | Show

જવાબ : False


{abc}⊂  {bcd}

Hide | Show

જવાબ : False


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

Hide | Show

જવાબ : True


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

Hide | Show

જવાબ : True


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

Hide | Show

જવાબ : True


{xis an equilateral triangle in a plane} ⊄  {xis a triangle in the same plane}

Hide | Show

જવાબ : False


{yis an odd natural number} ⊄  {yis an integer}

Hide | Show

જવાબ : False


{ab} ⊂ {bca}

Hide | Show

જવાબ : True


{ae} ⊄  {pis a vowel in the English alphabet}

Hide | Show

જવાબ : False


{1, 2, 3} ⊄ {1, 3, 5}

Hide | Show

જવાબ : True


{p} ⊄  {p, qs}

Hide | Show

જવાબ : False


If A = {1, 3, 5, 7} and B = {2, 3, 5, 7}, Then A ∪ B is _________

Hide | Show

જવાબ : {1, 2, 3, 5, 7}


The sets A and B are having elements 10 and 8,n(A ∩ B) =1, the  n(A ∪B) is _______

Hide | Show

જવાબ : 17


D={x: x is positive integer less than 1000 and have distinct digits and even} , n(D)=?

Hide | Show

જવાબ : 373


C={x: x is positive integer less than 1000 and have distinct digits} , n(C)=?

Hide | Show

જવાબ : 738


A set Z contains 2 elements, and then the number of elements in the Power set of Z will be ______

Hide | Show

જવાબ : 4


The set Z={x: x2 - 4=0,x is a rational number) is an______set

Hide | Show

જવાબ : Finite


U={x: x is positive integer less than 1000 and divisible by 7}  , n(U) =?

Hide | Show

જવાબ : 142


V={x: x is positive integer less than 1000 and divisible by 7 but not by 11} , n(V)=?

Hide | Show

જવાબ : 130


P={x: x is positive integer less than 1000 and divisible by 7 and  11} , n(P)=?

Hide | Show

જવાબ : 12


Q={x: x is positive integer less than 1000 and divisible by either 7 or  11} , n(Q)=?

Hide | Show

જવાબ : 220


A={x: x is positive integer less than 1000 and divisible by exactly one of 7 or 11} , n(A)=?

Hide | Show

જવાબ : 208


B={x: x is positive integer less than 1000 and divisible by neither 7 nor 11} , n(B)=?

Hide | Show

જવાબ : 779


Is it true that for any sets X and Z, P(X) P(Z)=P (XZ)? Justify your answer.

Hide | Show

જવાબ : False.

Let MP(X)P(Z)

 MP(X) or  MP(Z)

MX or MZ

M(XZ)

MP(X∩Z) 

P(X)P(Z) P(XZ)     ...(1)

Again, let MP(XZ)But MP(X) or MP(Z)       [For example let X={2,5} and Z={1,3,4} and take M={1,2,3,4}]So, MP(X)P(Z)Thus, P(XZ) is not necessarily a subset of P(X)P(Z).

Question 20:

Show that for any sets X and Z,
(i) X=(X∩Z)(X−Z)Z
(ii) X(Z−X)=XZ

ANSWER:

(i)

RHS=(X∩Z)(X−Z)

RHS=(X∩Z)(X∩Z')

RHS=[(X∩Z)X]∩((X∩Z)X')

RHS=X∩[(XZ')∩(ZZ')]

RHS=X∩[(XZ')∩U]

RHS=X∩(XZ')

RHS=X=LHS
(ii)
LHS=X(Z−X)

LHS=X(Z∩X')

LHS=(XZ)∩(XX')

LHS=(XZ)∩U

LHS=XZ=RHS


For any two sets of X and Z, prove that:
(i) X'Z=UXZ
(ii) Z'X'XZ

Hide | Show

જવાબ :

  1. Let aX.

aUaX'Z      ( U=X'Z)aZ       (aX')Hence, XZ.

  1. Let aX.

aX'aZ'     (Z'X')

aZ

Hence, XZ.


Using properties of sets, show that for any two sets X and Z,
(XZ)∩(X∩Z')=X

Hide | Show

જવાબ : LHS=(XZ)(X∩Z')

LHS={(XZ)∩X}{(XZ)∩Z'}

LHS={(XZ)∩X}{(XZ)∩Z'}

LHS=X{(XZ)∩Z'}

LHS=X{(X∩Z')(Z∩Z')}     ( B∩B=ϕ)

LHS=X(X∩Z')

LHS=A=RHS


If X and Z are sets, then prove that X−Z, X∩Z and X−Z are pair wise disjoint.

Hide | Show

જવાબ : (i) ( X−Z) and ( X∩Z)

Let a∈X−Z⇒a∈X and a∉Z

⇒a∉X∩Z

Hence, (X−Z) and  X∩Z are disjoint sets.

(ii) ( Z−X) and ( X∩Z)

Let a∈Z−X

⇒a∈Z and a∉X⇒a∉X∩Z

Hence, (Z−X) and  X∩Z are disjoint sets.

(iii) (X−Z) and (Z−X)

(X−Z)={x:x∈X and x∉Z}

(Z−X)={x:x∈Z and x∉X}

Hence, (X−Z) and  (Z−X) are disjoint sets.


For any two sets X and Z, prove that: X∩Z=ϕ ⇒X⊆Z'

Hide | Show

જવાબ : Let a∈X⇒a∉Z          (∵X∩Z=ϕ)

⇒a∈Z'

Thus, a∈X and a∈Z' ⇒ X⊆Z'


Find sets X, Y and Z such that X∩Y, X∩Z and Y∩Z are non-empty sets and X∩Y∩Z=ϕ.

Hide | Show

જવાબ : Let us consider the following sets,

X = {5, 6, 10 }
Y = {6,8,9}
Z = {9,10,11}

Clearly, X∩B={6} Y∩Z={9},  X∩Z={10}  and X∩Y∩Z = ϕ It means that, X∩Y, Y∩Z and X∩Z are non empty sets and X∩Y∩Z = ϕ


For any two sets, prove that:
(i) X∪(X∩Z)=X
(ii) X∩(X∪Z)=X

Hide | Show

જવાબ : (i)

LHS = X∪(X∩Z)⇒LHS=(X∪X)∩(X∪Z)     

⇒LHS=X∩(X∪Z)                (∵X⊂X∪Z)

⇒LHS=X = RHS
(ii)

LHS=X∩(X∪Z)

⇒LHS=(X∩X)∪(X∩Z)     

⇒LHS=X∪(X∩Z) 

⇒LHS=X = RHS


For three sets X, Y and Z, show that
(i) X∩Y=X∩Z need not imply Y = Z.
(ii) X⊂Y⇒X−Y⊂Z−X

Hide | Show

જવાબ : (i) Let X = {2, 4, 5, 6},  Y = {6, 7, 8, 9} and Z = {6, 10, 11, 12,13}

So, X∩Y={6} and X∩Z={6}Hence, X∩Y=X∩Z but Y≠Z

(ii)

Let a∈Z−Y      ...(1)

⇒a∈Z and a∉Y⇒a∈Z and a∉X       [∵A⊂B]

⇒a∈Z−X        ...(2)

From (1) and (2),

 we get Z−Y⊂Z−X


For any two sets X and Z, show that the following statements are equivalent:
(i) X⊂Z
(ii) X−Z=ϕ
(iii) X∪Z=Z
(iv) X∩Z=X

Hide | Show

જવાબ : We have that the following statements are equivalent:
(i) X⊂Z
(ii) X−Z=ϕ
(iii) X∪Z=Z
(iv) X∩Z=X

Proof:
Let X⊂Z

Let a be an arbitary element of (A−B). 

Now,a∈(X−Z)

⇒a∈X & A∉Z          (Which is contradictory) 

Also,∵X⊂Z⇒X−Z⊆ϕ                 ...(1) 

We know that null sets are the subsets of every set.∴ϕ ⊆ X−Z     ...(2)

From (1) & (2), we get,(X−Z)=ϕ

∴(i)=(ii)

Now, we have,(X−Z)=ϕ

That means that there is no element in X that does not belong to Z.

Now, X∪Z=Z∴(ii)=(iii) 

We have,X∪Z=Z⇒X⊂Z⇒X∩Z=X∴(iii)=(iv)

We have, X∩Z=X

It should be possible if X⊂Z.

Now,X⊂Z∴ (iv)=(i)

We have,(i)=(ii)=(iii)=(iv) Therefore, we can say that all statements are equivalent.


For any two sets X and Y, prove that

(i) Y ⊂ X ∪ Y                          (ii) X ∩ Y ⊂ X                          (iii) X ⊂ Y ⇒ X ∩ Y = X

Hide | Show

જવાબ : (i) For all x ∈ Y

⇒ x ∈ X or x ∈ Y

⇒ x ∈ X ∪ Y            (Definition of union of sets)

⇒ Y ⊂ X ∪ Y

(ii) For all x ∈ X ∩ Y

⇒ x ∈ X and x ∈ Y              (Definition of intersection of sets)

⇒ x ∈ X

⇒ X ∩ Y ⊂ X

(iii) Let X ⊂ Y. We need to prove X ∩ Y = X.

For all x ∈ X

⇒ x ∈ X and x ∈ Y          (X ⊂ Y)

⇒ x ∈ X ∩ Y 

⇒ X ⊂ X ∩ Y    
  
Also, X ∩ Y ⊂ X

Thus, X ⊂ X ∩ Y and X ∩ Y ⊂ X

⇒ X ∩ Y = X         [Proved in (ii)]

∴ X ⊂ Y ⇒ X ∩ Y = X


If U = {2, 3, 5, 7, 9} is the universal set and X = {3, 7}, Y = {2, 5, 7, 9}, then prove that:
(i) (X∪Y)'=X'∩Y'
(ii) (X∩Y)'=X'∪Y'.

Hide | Show

જવાબ : Given:
U = {2, 3, 5, 7, 9}
X = {3, 7}
Y = {2, 5, 7, 9}

To prove :
(i) (X∪Y)'=X'∩Y'
(ii) (X∩Y)'=X'∪Y'

Proof :

(i) LHS:
(X∪Y)={2,3,5,7,9} (X∪Y)'=ϕ

RHS:
X'={2,5,9} Y'={3} X'∩Y'=ϕ

LHS=RHS 

∴ (X∪Y)'=X'∩Y'

(ii) LHS:
(X∩Y)={7} (X∩Y)'={2,3,5,9}

RHS:
X'={2,5,9} Y'={3} X'∪Y'={2,3,5,9}

LHS = RHS

∴ (X∩Y)'=X'∪Y'


Let X = {1, 2, 4, 5} Y = {2, 3, 5, 6} Z = {4, 5, 6, 7}. Verify the following identities:
(i) X∪(Y∩Z)=(X∪Y)∩(X∪Z)
(ii) X∩(Y∪Z)=(X∩Y)∪(X∩Z)
(iii) X∩(Y−Z)=(X∩Y)−(X∩Z)
(iv) X−(Y∪Z)=X(X−Y)∩(X−Z)
(v) X−(Y∩Z)=(X−Y)∪(X−Z)
(vi) X∩(YΔZ)=(X∩Y)Δ(X∩Z).

Hide | Show

જવાબ : Given:
X = {1, 2, 4, 5}, Y = {2, 3, 5, 6} and Z = {4, 5, 6, 7}
We have to verify the following identities:
(i) X∪(B∩C)=(X∪B)∩(A∪C)

LHS
(Y∩C)={5,6} X∪(Y∩C)={1,2,4,5,6}

RHS
(X∪Y)={1,2,3,4,5,6} (X∪Z)={1,2,4,5,6,7} (X∪Y)∩(X∪Z)={1,2,4,5,6}

LHS = RHS
∴ X∪(B∩C)=(X∪B)∩(X∪C)

(ii) A∩(B∪C)=(A∩B)∪(A∩C)

LHS
(B∪C)={2,3,4,5,6,7}A∩(B∪C)={2,4,5}

RHS
A∩B={2,5}A∩C={4,5}(A∩B)∪(A∩C)={2,4,5}

LHS = RHS
∴ A∩(B∪C)=(A∩B)∪(A∩C)

 (iii) A∩(B−C)=(A∩B)−(A∩C)
 
LHS
(B−C) ={2,3}A∩(B−C)={2}

RHS
(A∩B)={2,5}(A∩C)={4,5}(A∩B)−(A∩C)={2}

LHS = RHS
∴ A∩(B−C)=(A∩B)−(A∩C)

 (iv) A−(B∪C)=(A−B)∩(A−C)

LHS
(B∪C)={2,3,4,5,6,7}A−(B∪C)={1}

RHS
(A−B)={1,4}(A−C)={1,2}(A−B)∩(A−C)={1}

LHS = RHS
∴ A−(B∪C)=(A−B)∩(A−C)

(v) A−(B∩C)=(A−B)∪(A−C)

LHS
(B∩C)={5,6}A−(B∩C)={1,2,4}

RHS
(A−B)={1,4}(A−C)={1,2}(A−B)∪(A−C)={1,2,4}

LHS = RHS
∴ A−(B∩C)=(A−B)∪(A−C)


(vi) A∩(BΔC)=(A∩B)Δ(A∩C)

LHS
(BΔC)=(B−C)∪(C−B)(B−C)={2,3} (C−B)={4,7} (B−C)∪(C−B)={2,3,4,7}

⇒(BΔC)={2,3,4,7} A∩(BΔC)={2,4}
RHS
(A∩B)={2,5} (A∩C)={4,5} (A∩B)Δ(A∩C)={(A∩B)−(A∩C)}∪{(A∩C)−(A∩B)} (A∩B)−(A∩C)={2} (A∩C)−(A∩B)={4} {(A∩B)−(A∩C)}∪{(A∩C)−(A∩B)}={2,4}

⇒(A∩B)Δ(A∩C)={2,4} 

LHS = RHS
∴ A∩(BΔC)=(A∩B)Δ(A∩C)A∩B∆C=A∩B∆A∩C


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

Hide | Show

જવાબ : 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'.

Hide | Show

જવાબ : 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)'

 

Hide | Show

જવાબ : 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

Hide | Show

જવાબ : 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

Hide | Show

જવાબ : 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)

Hide | Show

જવાબ : 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

 

Hide | Show

જવાબ : 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


There are No Content Availble For this Chapter

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:

1

Q−R

A

{4, 8, 12, 16}

2

Q−S

B

{5, 10, 20}

3

S - P

C

{20}

Hide | Show

જવાબ :

1-C, 2-A, 3-B

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:

1

Q−P

A

{5, 10, 20}

2

R−P

B

{4, 8, 16, 20}

3

S−P

C

{2, 4, 8, 10, 14, 16}

Hide | Show

જવાબ :

1-B, 2-C, 3-A

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:

1

P−Q

A

{4, 8, 16, 20}

2

P−R

B

{3, 6, 15, 18, 21}

3

P−S

C

{3, 15, 18, 21}

4

Q-P

D

{3, 6, 12, 18, 21}

Hide | Show

જવાબ :

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

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

QRS

A

{4, 5, 10, 11}

2

P∩QR

B

ϕ

3

 (P∩Q)∩(Q∩R)

C

{4, 5}

4

 (PS)∩(QR)

D

{4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14}

Hide | Show

જવાબ :

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

  1. sets in set-builder form

1

E = {0};

A

{x:x=2n, nN} 

2

{1, 4, 9, 16, ..., 100}

B

{5n:nN, 1≤n≤4}

3

{2, 4, 6, 8 .....}

C

{x2:xN, 1≤n≤10}

4

{5, 25, 125 625}

D

{x:x=0}

Hide | Show

જવાબ :

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

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

QS

A

{1, 2, 3, 4, 5, 6, 7, 8, 10, 11, 12, 13, 14}

2

PQR

B

{4, 5, 6, 7, 8, 10, 11, 12, 13, 14}

3

PQS

C

{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11}

Hide | Show

જવાબ :

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}
 

Hide | Show

જવાબ :

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 = 2nn  N}

D

{2, 4, 6, 8, 10,...}

Hide | Show

જવાબ :

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

PQ

A

{1, 2, 3, 4, 5, 6, 7, 8}

2

PR

B

{4, 5, 6, 7, 8, 9, 10, 11}

3

QR

C

{1, 2, 3, 4, 5, 7, 8, 9, 10, 11}

Hide | Show

જવાબ :

1-A, 2-C, 3-B

sets in set-builder form

1

 A = {1, 2, 3, 4, 5, 6}

A

{x:xN, 9<x<16}

2

B={1, ½ , 1/3 , ¼ , 1/5, ...}

B

{x:xN, x<7}

3

C = {0, 3, 6, 9, 12, ...}

C

{x: x=1n, xN}

4

D = {10, 11, 12, 13, 14, 15}

D

{x:x=3n, nZ+}

Hide | Show

જવાબ :

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

Download PDF

Take a Test

Choose your Test :

Sets

Math

Browse & Download CBSE Books For Class 11 All Subjects

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.