6. Match each of the set on the left in the roster form with the same set on the right described in set-builder form:

(i) {1, 2, 3, 6}                             (a) {x : x is a prime number and a divisor of 6}

(ii) {2, 3}                                   (b) {x : x is an odd natural number less than 10}

(iii) {M,A,T,H,E,I,C,S}                 (c) {x : x is natural number and divisor of 6}

(iv) {1, 3, 5, 7, 9}                       (d) {x : x is a letter of the word MATHEMATICS}.

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