36. Taking the set of natural numbers as the universal set, write down the complements of the following sets:

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

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

(iii) {x : x is a positive multiple of 3}

(iv) { x : x is a prime number }

(v) {x : x is a natural number divisible by 3 and 5}

(vi) {x : x is a perfect square }

(vii) { x : x is a perfect cube}

(viii) {x : x + 5 = 8 }

(ix) { x : 2x + 5 = 9}

(x) {x : x $\ge$ 7} (xi) { x : x $\in$ N and 2x + 1 > 10 }

Given    n = 2^x. 3^y. 5^z . (i)

$y+z=5\&\frac{1}{y}+\frac{1}{z}=\frac{5}{6}$

On solving we get y = 3, z = 2

So, n = 2^x. 3³. 5²

So that no. of odd divisor = (3 + 1) (2 + 1) = 12

Hence no. of divisors including 1 = 12

Let A = {a, b, c}, B = {1, 2, 3, 4, 5} n (A × B) = 15

x = number of one-one functions from A to B.

${=}^5{C}_3.3!=60$

y = number of one-one functions for A to (A × B)

${=}^{15}{C}_3.3!=15\times 14\times 13=2730$

$\frac{Y}{x}=\frac{2730}{60}\Rightarrow 2y=91x$

2cos $\left(\frac{x^2+x}{6}\right)=4^x+4^{-x}$

$-2\le LHS\le 2LHS=2\&RHS=2\Rightarrow x=0only\to thenLHS=2also$

RHS $\ge$ 2

66. Given series is 1× 2× 3 + 2× 3 ×4 + 3× 4 ×5 + … to n term

a_n = (n^th term of A. P. 1, 2, 3, …) ´× (n^th terms of A. P. 2, 3, 4) ×

i e, a = 1, d = 2- 1 = 1i e, a = 2, d = 3- 2 = 1

(nth term of A. P. 3, 4, 5)

i e, a = 3, d = 3 -4 = 1.

= [1 + (n -1) 1] ×[2 + (n -1):1]× [3 + (n- 1) 1]

= (1 + n -1)×(2 + n -1