Math | Easy Education

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

Write the following sets in roster form:

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

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જવાબ : {13, 22, 31, 40}

Write the following sets in roster form:

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

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જવાબ : {2, 5}

Write the following sets in roster form:

V = {x: x is a natural number less than 3}.

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જવાબ : {1,2}

Write the following sets in roster form:

U = {x: x is an integer and –1 < x < 1}.

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જવાબ : { -1, 0, 1}

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

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જવાબ : False

Set of even prime numbers is an null set

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જવાબ : False

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

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જવાબ : True

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

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જવાબ : True

If A ⊄ B and x ∉ B, then x ∉ A

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જવાબ : False

If x ∈ A and A ⊂ B, then x ∈ B

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જવાબ : True

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

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જવાબ : False

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

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જવાબ : True

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

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જવાબ : True

If A ⊄ B and B ⊄ C, then A ⊄ C

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જવાબ : False

If A ⊄ B and B ⊂ C, then A ⊂ C

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જવાબ : False

If A ⊄ B and B ∈ C, then A ∈ C

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જવાબ : False

If x ∈ A and A ∈ B, then x ∈ B

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જવાબ : False

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

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જવાબ : False

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

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જવાબ : False

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

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જવાબ : False

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

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જવાબ : False

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

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જવાબ : False

{x: x is an even natural number less than 6} ⊄ {x: x is a natural number which divides 36}

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જવાબ : False

{a} ∈ (1, 2, 3)

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જવાબ : False

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

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જવાબ : True

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

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જવાબ : False

The set P = {x: x is a letter in the word FOLLOW}; Q = {y: y is a letter in the word WOLF} are equal sets

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જવાબ : True

{2, 3, 4} ⊄ {1, 2, 3, 4, 5}

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જવાબ : False

{a, b, c}⊂ {b, c, d}

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જવાબ : False

{x: x is a student of Class X of your school} ⊂ {x: x student of your school}

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જવાબ : True

{x: x is a square in the plane} ⊂ {x: x is a rectangle in the same plane}

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જવાબ : True

{p: p is a triangle in a plane}⊄ {p: p is a rectangle in the plane}

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જવાબ : True

{x: x is an equilateral triangle in a plane} ⊄ {x: x is a triangle in the same plane}

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જવાબ : False

{y: y is an odd natural number} ⊄ {y: y is an integer}

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જવાબ : False

{a, b} ⊂ {b, c, a}

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જવાબ : True

{a, e} ⊄ {p: p is a vowel in the English alphabet}

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જવાબ : False

{1, 2, 3} ⊄ {1, 3, 5}

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જવાબ : True

{p} ⊄ {p, q, s}

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જવાબ : False

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

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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 _______

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જવાબ : 17

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

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જવાબ : 373

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

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જવાબ : 738

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

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જવાબ : 4

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

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જવાબ : Finite

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

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જવાબ : 142

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

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જવાબ : 130

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

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જવાબ : 12

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

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જવાબ : 220

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

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જવાબ : 208

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

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જવાબ : 779

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

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જવાબ : False.

Let M∈P(X)∪P(Z)

⇒ M∈P(X) or M∈P(Z)

⇒M⊂X or M⊂Z

⇒M⊂(X∪Z)

⇒M∈P(X∩Z)

∴P(X)∪P(Z) ⊂P(X∪Z) ...(1)

Again, let M∈P(X∪Z)But M∉P(X) or M∉P(Z) [For example let X={2,5} and Z={1,3,4} and take M={1,2,3,4}]So, M∉P(X)∪P(Z)Thus, P(X∪Z) 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)=X∪Z

ANSWER:

(i)

RHS=(X∩Z)∪(X−Z)

⇒RHS=(X∩Z)∪(X∩Z')

⇒RHS=[(X∩Z)∪X]∩((X∩Z)∪X')

⇒RHS=X∩[(X∪Z')∩(Z∪Z')]

⇒RHS=X∩[(X∪Z')∩U]

⇒RHS=X∩(X∪Z')

⇒RHS=X=LHS
(ii)
LHS=X∪(Z−X)

⇒LHS=X∪(Z∩X')

⇒LHS=(X∪Z)∩(X∪X')

⇒LHS=(X∪Z)∩U

⇒LHS=X∪Z=RHS

For any two sets of X and Z, prove that:
(i) X'∪Z=U⇒X⊂Z
(ii) Z'⊂X'⇒X⊂Z

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જવાબ :

Let a∈X.

⇒a∈U⇒a∈X'∪Z (∵ U=X'∪Z)⇒a∈Z (∵a∉X')Hence, X⊂Z.

Let a∈X.

⇒a∉X'⇒a∉Z' (∵Z'⊂X')

⇒a∈Z

Hence, X⊂Z.

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

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જવાબ : LHS=(X∪Z)∪(X∩Z')

⇒LHS={(X∪Z)∩X}∪{(X∪Z)∩Z'}

⇒LHS=X∪{(X∪Z)∩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.

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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'

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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.

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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

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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

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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

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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

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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'.

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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).

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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}.

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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'.

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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)'

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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

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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

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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)

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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 X⊂Y, then find:
(i) X∩Y
(ii) X∪Y

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

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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}

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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}

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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}

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

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જવાબ :

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

sets in set-builder form

1	E = {0};	A	{x:x=2n, n∈N}
2	{1, 4, 9, 16, ..., 100}	B	{5n:n∈N, 1≤n≤4}
3	{2, 4, 6, 8 .....}	C	{x2:x∈N, 1≤n≤10}
4	{5, 25, 125 625}	D	{x:x=0}

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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	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}

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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}

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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 = 2n, n ∈ N}	D	{2, 4, 6, 8, 10,...}

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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	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}

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જવાબ :

1-A, 2-C, 3-B

sets in set-builder form

1	A = {1, 2, 3, 4, 5, 6}	A	{x:x∈N, 9<x<16}
2	B={1, ½ , 1/3 , ¼ , 1/5, ...}	B	{x:x∈N, x<7}
3	C = {0, 3, 6, 9, 12, ...}	C	{x: x=1n, x∈N}
4	D = {10, 11, 12, 13, 14, 15}	D	{x:x=3n, n∈Z+}

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જવાબ :

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

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