Set Relation and Function

6 )

Let X = {(x, y, z)/x, y, z ∈ N, x + y + z = 10, x < y < z } and Y = {(x, y, z)/x, y, z ∈ N, y = |x – z|} then X ∩ Y is equal to

{(2, 3, 5)}

{(1, 4, 5)}

{(5, 1, 4)}

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

View Solution

Answer :

X = {(x, y, z)/x, y, z ∈ N, x + y + z = 10, x < y < z }

Y = {(x, y, z)/x, y, z ∈ N, y = |x – z|}

Hence

X = {(1, 2, 7), (1, 3, 6), (1, 4, 5), (2, 3, 5)}

elements of X which belong to Y are (1, 4, 5) and (2, 3, 5)

Hence

X ∩ Y = {(1, 4, 5), (2, 3, 5)}.

7 )

In a certain town 30% families own a scooter and 40% on a car 50% own neither a scooter nor a car 2000 families own both a scooter and car consider the following statements in this regard

(1) 20% families own both scooter and car

(2) 35% families own either a car or a scooter

(3) 10000 families live in town.

Which of the above statements are correct?

2 and 3

1, 2 and 3

1 and 2

1 and 3

View Solution

Answer :

Let

X families live in the town

A = {families with scooter)

B = {families with car)

n(A) = (30 / 100)x, n(B) = (40 / 100)x

n(A ∩ B) = {families own both) = 2000

n(A ∪ B)' = {families with neither of it}

as n(A ∪ B) = 1 – n(A ∩ B)' = (50 / 100)x

∵ n(A ∪ B) = n(A) + n(B) – n(A ∩ B)

hence

(50x / 100) = (30x / 100) + (40x / 100) – n(A ∩ B)

∴ n(A ∩ B) = (20x / 100)

Hence

(20x / 100) = 2000

∴ x = 10000

∴ No. of families = 10000

Hence

n(A ∩ B) = (20 / 100)x = (20 / 100) × 10000 = 2000

i.e. 20% families own both car and scooter.

8 )

Let A = {θ: tan θ + sec θ = √2 sec θ} and B = {θ: sec θ – tan θ = √2 tan θ} be two sets then.

A = B

A âŠ‚ B

A â‰ B

B âŠ‚ A

View Solution

Answer :

given: tan θ + sec θ = √2 sec θ

∴ tan θ = (√2 – 1) sec θ

∴ [(sin θ) / (cos θ)] = (√2 – 1) [1 / (cos θ)]

∴ sin θ = (√2 – 1) (1)

also

sec θ – tan θ = √2 tan θ

∴ sec θ = (√2 + 1) tan θ

∴ [1 / (cos θ)] = (√2 + 1) [(sin θ) / (cos θ)]

∴ [1 / (√2 + 1)] = sin θ

i.e. sin θ = [(√2 – 1) / {(√2 + 1) (√2 – 1)}]

= [(√2 – 1) / (2 – 1)]

= √2 – 1 (2)

from (1) & (2), A = B.

9 )

Suppose sets A_i(i = 1, 2,... 60) each set having 12 elements and set B_j(j = 1, 2, 3….n) each set having 4 elements let ⁶⁰U_i=1 A_i = ⁿU_j=1 B_j = C and each element of C belongs to exactly 20 of A_i’s exactly 18 of B_j’s then n is equal to

120

135

162

177

View Solution

Answer :

A_i(i = 1, 2,... 60)

B_j(j = 1, 2, 3….n)

also ⁶⁰U_i=1 A_i = ⁿU_j=1 B_j = C

A_i → each set has 12 elements

B_j → each set has 4 elements.

∴ n(c) = n[⁶⁰U_i=1 A_i] = (1 / 20)(12 × 60) = 36

n(c) = n[ⁿU_j=1 B_j] = (1 / 18)(n × 4) = (4n / 18)

∴ (4n / 18) = 36

∴ n = 9 × 18

∴ n = 162.

10 )

If set A is empty set then n [P[P[P(A)]]] = ...........

View Solution

Answer :

A is empty set hence power set will contain 2°= 1 element

∴ n(P(A)) = 1

∴ n[P[P(A)]] = 2

n[P[P[P(A)]]] = 2² = 4

Hence

n[P[P[P(A)]]] = 4.

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