Class 11th

41. Given, n (X)= 17

n (Y) = 23

n (X $\cup$Y)= 38.

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

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

= 17 + 23 38.

= 2

40.  (i) A $\cup$A$\text{'}$ = U

(ii) $?$∩ A = U ∩ A = A.

(iii) A ∩ A$\text{'}$ = $?$ .

(iv) U$\text{'}$ ∩ A = $?$ ∩ A = $?$ .

39. A$\text{'}$ = U – A

= Set of all triangle in a plane - Set of all triangle with at least the angle different from 60°.

= Set of all triangle with each angle 60°.

A? = set of all equilateral triangle.

38. (i) (A$\cup$ B)$\text{'}$ = U – (A $\cup$B)

(ii) A$\text{'}$ ∩ B$\text{'}$ = (AU B)$\text{'}$ = U – (AU B)

(iii) (A ∩ B)$\text{'}$ = U – (A ∩ B)

(iv) A$\text{'}$ U B$\text{'}$ = (A ∩ B)$\text{'}$ = U – (A ∩ B)$\text{'}$.

37. (i) L.H.S = (A$\cup$ B)$\text{'}$ = U – (A $\cup$B)

= {1,2,3,4,5,6,7,8,9} – [ {2,4,6,8)$\cup$ {2,3,5,7}]

= {1,2,3,4,5,6,7,8,9} – {2,3,4,5,6,7,8}

= {1,9}

R.H.S. = A$\text{'}$ ∩ B$\text{'}$ = [U – A] ∩ [U$\cup$ B]

= $\left[\left\{1,2,3,4,5,6,7,8,9\right\}–\{2,4,6,8\}\right]$ ∩ $\left[\left\{1,2,3,4,5,6,7,8,9\right\}–\{2,3,5,7\}\right]$

= {1,3,5,7,9} ∩ {1,4,6,8,9}

= {1,9}

? L.H.S. = R.H.S.

(A B) = A ∩ B.

(ii) L.H.S. = (A ∩ B)$\text{'}$ = U – (A ∩ B)

= {1,2,3,4,5,6,7,8,9} – [ {2,4,6,8} ∩ {2,3,5,7}]

= {1,2,3,4,5,6,7,8,9} – {2}

= {1,3,4,5,6,7,8,9}

R.H.S. = A$\text{'}$ $\cup$B$\text{'}$

= [U – A] $\cup$ [U – B]

= [ {1,2,3,4,5,6,7,8,9} – {2,4,6,8}] [ {1,2,3,4,5,6,7,8,9} – {2,3,5,7}]

= {1,3,5,7,9}$\cup$ {1,4,6,8,9}

= {1,3,4,5,6,7,8,9}

? L.H.S. = R.H.S.

(A ∩ B)$\text{'}$ = A$\text{'}$$\cup$ B$\text{'}$.

3. (i) {x : x is an odd natural number}

(ii) {x : x is an even natural number}

(iii) {x : x is not a multiple of 3}

(iv) {x : x is a positive composite number and x = 1}

(v) {x : x is a natural number not divisible by 3 and 5}.

(vi) {x : x is not a perfect square}

(vii) {x : x is not a perfect cube}

(viii) We have, x + 5 = 8.

x = 8 – 5 = 3

x = 3

? {x : x ≠ 3, x? N}

(ix )We have,

2x + 5 = 9

2x = 9 – 5

2x = 4

x = 2

? {x : x? N and x ≠ 2}

(x) {x : x < 7} = {1,2,3,4,5,6}

(xi) We have,

2x + 1 >10

2x >10 – 1

x > $\frac{9}{2}$

?  $\left\{x:x\in x<\frac{9}{2}\right\}$ = {1,2,3,4}

35. (i) A$\text{'}$ = U – A = {a, b, c, d, e, f, g, h} – {a, b, c}

= {d, e, f, g, h}

(ii) B$\text{'}$ = U – B = {a, b, c, d, e, f, g, h} – {d, e, f, g}

= {a, b, c, h}.

(iii) C$\text{'}$ = U – C = {a, b, c, d, e, f, g, h} – {a, c, e, g}.

= {b, d, f, h}

(iv) D$\text{'}$ = U – D = {a, b, c, d, e, f, g, h} – {f, g, h, a}

= {b, c, d, e}

34. (i) A$\text{'}$ = U – A = {1,2,3,4,5,6,7,8,9} – {1,2,3,4}

= {5,6,7,8,9}

(ii) B$\text{'}$ = U – B = {1,2,3,4,5,6,7,8,9} – {2,4,6,8}

= {1,3,5,7,9}.

(iii) (A $\cup$C)$\text{'}$ = A$\text{'}$ ∩ C$\text{'}$

= {5,6,7,8,9} ∩ [U – C] [? (i)]

= {5,6,7,8,9} ∩ [ {1,2,3,4,5,6,1,8,9} – {3,4,5,6}]

= {5,6,7,8,9} ∩ {1,2,7,8,9}

= {7,8,9}

(iv) A$\cup$ B)$\text{'}$ = A$\text{'}$ ∩ B$\text{'}$ [By demorgan's law]

= {5,6,7,8,9} ∩ {1,3,5,7,9} [? (i) and (ii)]

= {5,7,9}.

(v) (A$\text{'}$)$\text{'}$ = U – A$\text{'}$ = {1,2,3,4,5,6,7,8,9} – {5,6,7,8,9} [? (1)]

= {1,2,3,4} = A

(A$\text{'}$)$\text{'}$ = A.

(vi) (B – C)$\text{'}$ = U – (B – C) = {1,2,3,4,5,6,7,8,9} – [ {2, 4, 6, 8} – {3, 4, 5, 6}]

= {1,2,3,4,5,6,7,8,9} – {2,8}

= {1,3,4, 5, 6,7,9}.

19. Given, x-coordinate of R = 4

Let R divides line segment joining points P(2, –3, 4) and Q(8, 0,10) internally in the ratio k : 1. Then coordinate of R is

$\left(\frac{k(8)+1(2)}{k+1},\frac{k(0)+1(-3)}{k+1},\frac{k(10)+1(4)}{k+1}\right)$

= $\left(\frac{8k+2}{k+1},\frac{-3}{k+1},\frac{10k+4}{k+1}\right)$

Then,

$\frac{8k+2}{k + 1}$ = 4

=> $8k+2=4\left(k+1\right)$

=> $8k+2=4k+4$

=> $8k-4k=4-2$

=> $4k=2$

=> $k=\frac{2}{4}$

=> $k=\frac{1}{2}$

Hence, $y=$ $\frac{-3}{k + 1}$

= $\frac{-3}{\frac{1}{2} + 1}$

= $-3÷\frac{1+2}{2}$

= $-\mathrm{3*} \frac{2}{3}$

= $-2$

And,

z = $\frac{10k + 4}{k + 1}$

= $\frac{\left(10\frac{1}{2}\right) + 4}{\frac{1}{2} + 1}$

= $\left(5+4\right)÷\frac{1+2}{2}$

= $9\frac{2}{3}$

= 6

Therefore, coordinates of R is (4, –2, 6).