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(c) 6

n[(A × B) ∪ (A × C)] = n[ A × (B ∪ C)] 

= n(A) × n(B ∪ C) = 2 × 3 = 6

A = {set of students in 11^th standard} 

B = {set of sections in 11sup>th standard} 

R : A ➝ B ⇒ x related to y

⇒ Every students in eleventh Standard must in one section of the eleventh standard. 

⇒ It is a function. 

Inverse relation cannot be a function since every section of eleventh standard cannot be related to one student in eleventh standard.

n(A × A) = 16 ⇒ n( A) = 4 

S ={(-1, 0), (-1, 1), (0, 2), (1, 2)}

n(A ∪ B) = 10; n(A ∩ B) = 3 

n(A ∆ B) = 10 – 3 = 7 and n(P(A ∆ B)) = 27 = 128

n(A) = 3 ⇒ set A contains 3 elements 

n(B) = 2 ⇒ set B contains 2 elements – 

We are given (x, 1), (y, 2), (z, 1) are elements in A × B ⇒ A = {x, y, z} and B = {1, 2}

f(-4) = -(-4) + 4 = 8 

f(1) = 1 – 1² = 0 

f(-2) = (-2)² – (-2) = 4 + 2 = 6 

f(7) = 0 

f(0) = 0

(b) 17²

n (A ∩ B) = 17 

So n [(A × B) ∩ (B × A)] 

= n(A ∩ B) × n(B ∩ A) 

= 17 × 17 = 17²

A × A = 16 elements = 4 × 4 

⇒ A has 4 elements 

∴ A = {0, 1, 2, 3}

(b) A × A

When A ⊂ B, (A × B) ∩ (B × A) = A × A

Draw the graph of y = x² 

To get y = (x – 1)² we have to shift the curve 1 unit to the right.

Then we have to draw the curve y = 3(x – 1)² and finally we have to draw y = 3(x – 1)² + 5

f(-3) = (-3)² – 3 – 5 = 9 – 8 = 1 

f(5) = (5)² + 3(5) – 2 = 25 + 15 – 2 = 38 

f(2) = 4 – 3 = 1 

f(-1) = (-1)² + (-1) – 5 = 1 – 6 = -5 

f(0) = 0 – 3 = -3

Answer is (c) 512

Number of relations = 2^{n^2} = 2^{3^2} = 2⁹ = 512

Given that: f(x) = 1 + 3 cos 2x 

We know that -1 ≤ cos 2x ≤ 1 

⇒ -3 ≤ 3 cos 2x ≤ 3 ⇒ -3 + 1 ≤ 1 + 3 cos 2x ≤ 3 + 1 

⇒ -2 ≤ 1 + 3 cos 2x ≤ 4 ⇒ -2 ≤ f(x) ≤ 4 

Hence the range of f = [-2, 4]

(d) [0, ∞), [0, ∞)

(d) (-∞, 1]

f: R ➝ R defined by 

f(x) = 1 – |x| 

For example, 

f(1) = 1 – 1 = 0 

f(8) = 1 – 8 = -1 

f(-9) = 1 – 9 = -8 

f(-0.2) = 1 – 0.2 = 0.8 

so range = (-∞, 1]

Answer is (d) [0, 9]

(b) Onto

For x = π/4, x = 3π/4, f(x) = 1/√2

So it is not one-to-one 

So it is an onto function

(b) Neither an odd function nor an even function

(i) (c, c) 

(ii) (c, a) 

(iii) nothing 

(iv) (c, c) and (c, a)

(a) Range of R is {0, -1, 1}

A= {0, -1, 1, 2} 

|x² + y²| ≤ 2 

The values of x and y can be 0, -1 or 1 

So range = {0, -1, 1}

P = {set of all triangles in a plane} 

aRb ⇒ a similar to b 

(a) aRa ⇒ every triangle is similar to itself 

∴ aRa is reflexive 

(b) aRb ⇒ if a is similar to b ⇒ b is also similar to a. 

⇒ It is symmetric 

(c) aRb ⇒ bRc ⇒ aRc 

a is similar to b and b is similar to c 

⇒ a is similar to a

⇒ It is transitive 

∴ R is an equivalence relation

Let f + g = k 

Now k ⊙ h = k(h(x))

= (f + g((h(x)) 

= f[h(x)] + g [h(x)] 

= foh + goh 

(i.e.,)(f + g)(o)h = foh + goh

fo(g + h) is also a function

Let H = set of people speaking Hindi;

E = set of people speaking English.

Then n(H∪E) = 400; n(H) = 250; n(E) = 200

∴ n(H∩E) = n(H) = n(E)-n(H∪E)

= 250 + 200 – 400 = 50

Hence, 50 people can speak both Hindi and English.

Let H be the set of people who speak Hindi, and E be the set of people who speak English
∴ n(H ∪ E) = 400, n(H) = 250, n(E) = 200,  n(H ∩ E) = ?
We know that: n(H ∪ E) = n(H) + n(E) – n(H ∩ E)
∴ 400 = 250 + 200 – n(H ∩ E)
⇒ 400 = 450 – n(H ∩ E) ⇒ n(H ∩ E) = 450 – 400
∴ n(H ∩ E) = 50
Thus, 50 people can speak both Hindi and English.

It is given that: 

n(X) = 40, n(X ∪ Y) = 60,

n(X ∩ Y) = 10 

We know that:

n(X ∪ Y) = n(X) + n(Y) – n(X ∩ Y) 

∴ 60 = 40 + n(Y) – 10

∴ n(Y) = 60 – (40 – 10) = 30

Thus, the set Y has 30 elements.

A' is the set of all equilateral triangles.

A’ = set of equilateral triangles.

(i) A∪A’ = U

(ii) (ii) φ’∩A = A

(iii) A∩A’ = φ

(iv) U’∩A = φ.

Given n(X∪Y) = 50; n(X) = 28; n(Y) = 32

But n(X∪Y) = n(x) + n(y)-n(X∩Y)

∴ n(X∩Y) = 28 + 32-50 = 10

B. 25

Given:

Total number of students are 60

Students who play cricket and tennis are 25 and 20 respectively

Students who play both the games are 10

To find: number of students who play neither

Let S be the total number of students, C and T be the number of students who play cricket and tennis respectively

n(S) = 60, n(C) = 25, n(T) = 20, n(C ∩ T) = 10

Number of students who play either of them = n(C ∪ T)

= n(C) + n(T) – n(C ∩ T)

= 25 + 20 – 10

= 35

Number of student who play neither

= Total – n(C ∪ T)

= 60 – 35

= 25

Hence, there are 25 students who play neither cricket nor tennis.

Let C be the set of students who play cricket and T be the set of students who play tennis.

n(U) = 60, n(C) = 25, n(T) = 20, and n(C ∩ T) = 10

n(C ∪ T) = n(C) + n(T) – n(C n T) = 25 + 20 – 10 = 35

Given n(X) = 40; n(X∪Y) = 60; n(X∩Y) = 10

n(Y) = n(X∪Y) + n(X∩Y)-n(X)

= 60 + 10 – 40 = 30

According to the question,

Total number of students = 60

Students who play cricket = 25

Students who play tennis = 20

Students who play both the games = 10

To find: number of students who play neither

Let the total number of students = S

Let the number of students who play cricket = C

Let the number of students who play tennis = T

n(S) = 60, n(C) = 25, n(T) = 20, n(C ∩ T) = 10

So, Number of students who play either of them,

n(C ∪ T) = n(C) + n(T) – n(C ∩ T)

= 25 + 20 – 10

= 35

Hence, Number of student who play neither = Total – n(C ∪ T)

= 60 – 35

= 25

Therefore, there are 25 students who play neither cricket nor tennis.

the number of students who study all the three subjects.

Answers is 20

Let C = set of people who like coffee 

T = set of people who like tea 

Then n(C∪T) = 70; n(C) = 37; n(T) = 52 n(C∩T) 

= n(C) + n(J)-n(C∪T) 

= 37 + 52-70 =19

According to the question,

Total number of students = n(U) = 200

Number of students who study Mathematics = n(M) = 120

Number of students who study Physics = n(P) = 90

Number of students who study Chemistry = n(C) = 70

Number of students who study Mathematics and Physics = n(M ∩ P) = 40

Number of students who study Mathematics and Chemistry = n(M ∩ C) = 50

Number of students who study Physics and Chemistry = n(P ∩ C) = 30

Number of students who study none of them = 20

Let the total number of students = U

Let the number of students who study Mathematics = M

Let the number of students who study Physics = P

Let the number of students who study Chemistry = C

number of students who study all the three subjects n(M ∩ P ∩ C)

Number of students who play either of them = n(P ∪ M ∪ C)

n(P ∪ M ∪ C) = Total – none of them

= 200 – 20

= 180 …(i)

Number of students who play either of them = n(P ∪ M ∪ C)

n(P ∪ M ∪ C) = n(C) + n(P) + n(M) – n(M ∩ P) – n(M ∩ C) – n(P ∩ C) + n(P ∩ M ∩ C)

= 120 + 90 + 70 – 40 – 30 – 50 + n(P ∩ M ∩ C)

= 160 + n(P ∩ M ∩ C) …(ii)

From equation (i) and (ii), we get,

160 + n(P ∩ M ∩ C) = 180

⇒ n(P ∩ M ∩ C) = 180 – 160

⇒ n(P ∩ M ∩ C) = 20

Therefore, there are 20 students who study all the three subjects.

Let total number of people n(P) = 50 

A number of people who drink Tea n(T) = 30. 

A number of people who drink coffee n(C). 

n(T–C) = 14 

i. how may drink tea and coffee both. 

We can see that T is disjoint union of n(T–C) and n (T ∩ C). 

(If A and B are disjoint then n (A ∪ B) = n(A) + n(B)) 

∴ T = n(T–C) ∪ n (T ∩ C). 

⇒ n(T) = n(T–C) + n (T ∩ C). 

⇒ 30 = 14 + n (T ∩ C). 

⇒ n(T ∩ C) = 16. 

16 People drink both coffee and tea. 

ii. how many drink coffee but not tea. 

We know 

n (P) = n(T) + n(C) – n (T ∩ C) 

Substituting the values we get 

50 = 30+n(C) – 16 

n(C) = 36. 

We can see that T is disjoint union of n(C–T) and n (T ∩ C). 

(If A and B are disjoint then n (A ∪ B) = n(A) + n(B)) 

∴ C = n(C–T) ∪ n (T ∩ C). 

⇒ n(C) = n(C–T) + n (T ∩ C). 

⇒ 36 = n(C–T) + 16. 

⇒ n(C–T) = 20. 

20 People drink coffee but not tea.

(i) n (B)

As we know,

n (A ∪ B) = n (A) + n (B) – n (A ∩ B)

By substituting the values we get

42 = 20 + n (B) – 4

42 = 16 + n (B)

n (B) = 26

∴ n (B) = 26

(ii) n (A – B)

As we know,

n (A – B) = n (A ∪ B) – n (B)

By substituting the values we get

n (A – B) = 42 – 26

= 16

∴ n (A – B) = 16

(iii) n (B – A)

As we know,

n (B – A) = n (B) – n (A ∩ B)

By substituting the values we get

n (B – A) = 26 – 4

= 22

∴ n (B – A) = 22

Total number of People n(P) = 60. 

n(H) =25. 

n(T) = 26. 

n(I) = 26. 

n (H ∩ I) = 9 

n (H ∩ T) = 11 

n (T ∩ I) = 8 

n (H ∩T ∩ I) = 3 

The people who read at least one newspaper would be n(H or I or T) = n(H∪I∪T) 

We know, 

n(H∪I∪T) = n(H)+n(I)+n(T) – n (H ∩ I)– n (H ∩ T)– n (T ∩ I)+ n (H ∩T ∩ I) 

n(H∪I∪T) = 25+26+26–9–11–8+3 

n(H∪I∪T) = 52. 

There are 52 people who read at least one newspaper.