Maths

54. Given, P (A) = P (B) where P is power set

Let x$\cancel{\in }$A.

Then, {x} $\subset$P (A)$\subset$ P (B)

i.e., x$\cancel{\in }$ B

A$\subset$ B

Similarly, B $\subset$A

A = B

53. Given, A $\subset$B.

Let x$\cancel{\in }$ C - B then x $\cancel{\in }$C but X$B$ .

However, A $\subset$B, elements of B should have elements of A

i.e., X$B$ 

So, x $\cancel{\in }$C A i.e., x$\cancel{\in }$ C but X$A$

C - B $\subset$C-  A

52. (i) Let A-  B≠$?$ , our assumption.

i.e., x$\in$ A But x≠ B where x is an element.

But as A B, the above condition of assumption is wrong if A $\subset$B then A -  B =$?$

(ii) Let x$\in$ A.

As A - B =$?$ we can say that x$\in$ B because if $x\cancel{\in }B,$ A - B ≠$?$

if A - B =$?$ then A $\subset$B.

(iii) We know that,

B $\subset$A$\cup$ B always true

Let x$\in$A$\cup$B i.e., x$\in$ A or x$\in$ B.

As A $\subset$B,

If x $\in$A then x$\in$ B, all elements of A are among the elements of B

So, (A$\cup$B) = B

(iv) We know that,

(A$\cap$ B) $\subset$A as A$\subset$ B.

Let x $\in$A then x $\in$B

So, x $\in$(A$\cap$ B)

i.e., A$\subset$ (A$\cap$ B)

So, A = (A $\cap$B)

Hence, A$\subset$ B

A - B = $?$.

A$\cup$ B = B

A $\cap$B = A.

i.e., the 4 conditions are equivalent.

51. Let x$\in$b

As B$\subset$ A$\cup$ B we can write

Let x $\in$A $\cup$B.

as A $\cup$B = A $\cup$C.

x$\in$ A$\cup$ C.

i.e., x$\in$ A or x $\in$C

when x$\in$ A, and x$\in$ B,

x$\in$ A$\cap$ B

But A$\cap$ B = A $\cap$C

So, x$\in$ A $\cap$C

i.e., x$\in$ A and x$\in$ C

x $\in$C

when, x$\in$ C

as x $\in$B and x $\in$C

So, B $\subset$C

Similarly, C $\subset$B

So, B = C

50. (i) False Let A = {a}, a $\in$A then B = {a}, b} I e,  $a\cancel{\in B.}$

(ii) False. Let A. = {a}, if A $\subset$B, B = {a, b} and B $\in$C I e, C = {a, b}, c} I e, A = {a}. 

(iii) True. Let x$\in$A, if A$\subset$ B then x$\in$B and if B$\subset$ C, x$\in$ C I e, elements of A are also elements of C. A$\subset$ C.

(iv) False. Let A = {a} and B = {b} then A$\not\subset$$B$ . Let C = {a, c} then B$\not\subset$$\subset$ but a $\in$A and a c, i.e., A$\subset$ C.

(v) False. Let A = {a} and B = {b} so, A$\not\subset$$B$ i.e., A$B$ .

(vi) True. Let A$\subset$ B such that y$\in$ B i.e., y A But x$\in$$B$ and suppose x$\in$A.

Then by above definition,  A $\subset$B i.e., x$\in$ B and x$\in$ A which is not the case

49. A = {x: xR and x satisfy x² - 8x + 12 = 0}

So, x² - 8x + 12 = 0

x² - 6x 2x + 12 = 0

x (x- 6) 2 (x -6) = 0

(x -6) (x- 2) = 0

x = 6, 2.

So, A = {2, 6) B = {2, 4, 6} C = {2, 4, 6, 8, ….}

D = {6}

D⊂ A ⊂B ⊂C

i e, D ⊂A, D⊂ B, D⊂ C, A⊂ B, A ⊂C and B⊂ C.

48. Let A and B be set of people who speaks Frenchand Spanish respectively. Then,

n (A) = 50, people speak French

n (B) = 20, people speak Spanish.

n (A $\cap$B) = 10, speaks both French& Spanish.

So, number of people who speaks at least one of these two languages

= n (A $\cup$B)

= n (A) + n (B) n (A $\cap$B)

= 50 + 20 10

= 60

47. Let C and T be the set of people who likes cricket and tennis respectively. Then,

n (C) = 40, people who likes cricket

n (C $\cup$T) = 65, people who likes either cricket or tennis

and n (C $\cap$T) = 10, people who likes both

So, n (C $\cup$T) = n (C) + n (T) n (C $\cap$T)

65 = 40 + n (T) 10

n (T) = 35

So, 35 people likes tennis

And number of people who likes tennis only and not cricket

(TC) = n (T) n (C $\cap$T) = 35 10 = 25.

46. Let A and B but the set of people who likes coffee and tea.

Then, n (A) = 37. no. of people who like coffee

n (B) = 52, no. of people who like tea.

As each person likes at least one of the two drink,

n (A $\cup$B) = 70.

So using, n (A $\cup$B) = n (A) + n (B) n (A $\cap$B)

n (A $\cap$B) = n (A) + n (B) n (A $\cup$B)

= 37 + 52 70

= 89 70

= 19

So, 19 people likes both coffee and tea.

45. Given, n (X) = 40

n (Y $\cup$Y) = 60.

n (X $\cap$Y) = 10.

n (Y) =?

Using, n (X $\cup$Y) = n (X) + n (Y) n (X $\cap$Y)

60 = 40 + n (y) 10.

n (Y) = 60 - 40 + 10

n (Y) = 30.