(3x + 4y)²    [we have (a+b)² = a²+ 2ab + b²]
(3x + 4y)² = (3x)²+2(3x)(4y)+(4y)²   put [a = 3x, b = 4y]
= 9x² + 24xy + 16y²

6.34 \(\times\) 10⁴

(i) \(\frac{1}{\sqrt{50}}=\frac{1}{\sqrt{50}}\times\frac{\sqrt{50}}{\sqrt{50}}=\frac{\sqrt{50}}{\sqrt{50}}=\frac{\sqrt{5\times5\times2}}{5\times5\times2}=\frac{ ̶5̶\sqrt{2}}{ ̶5̶\times5\times2}=\frac{\sqrt{2}}{10}\)
(ii) \(\frac{5}{3\sqrt{5}}=\frac{5}{3\sqrt{5}}\times\frac{\sqrt{5}}{\sqrt{5}}=\frac{ ̶5̶\sqrt{5}}{3\times ̶5̶}=\frac{\sqrt{5}}{3}\)
(iii) \(\frac{\sqrt{75}}{\sqrt{18}}=\frac{\sqrt{3\times5\times5}}{\sqrt{3\times2\times3}}=\frac{5\sqrt{3}\times\sqrt{2}}{3\sqrt{2}\times\sqrt{2}}=\frac{5\sqrt{6}}{3\times2}=\frac{5\sqrt{6}}{6}\)
(iv) \(\frac{3\sqrt{5}}{\sqrt{6}}=\frac{3\sqrt{5}\times\sqrt{6}}{\sqrt{6}\times\sqrt{6}}=\frac{ ̶3̶\sqrt{30}}{ ̶6̶}=\frac{\sqrt{30}}{2}\)

\({ 5 }^{ \frac { 1 }{ 2 } }\)

Associative Property of intersection of sets
A⋂(B⋂C) = (A⋂B)⋂C
B⋂C = {√3, √5, 6, 13} ∩ {√2, √3, √5, 9} = {√3,√5}
A⋂( B⋂C) = {-11,√2,√5,7} ⋂ {√3,√5} = {√5}...(1)
A⋂B = {-11,√2,√5,7}√{√3,√5,6,13} = {√5}
(A∩B)∩C = {√5} ∩ {√2,√3,√5,9} = {√5} ...(2)
From (1) and (2), it is verified that A∩(B∩C) = (A∩B)∩C

Now,  \((B\cap C)=\left\{ \frac { 1 }{ 4 } ,2,\frac { 5 }{ 2 } \right\} \)
\(A\cap (B\cap C)=\left\{ \frac { 1 }{ 4 } ,2 \right\} \)    .......(1)
Then, \(A\cap B=\left\{ 0,\frac { 1 }{ 4 } ,\frac { 3 }{ 4 } ,2 \right\} \)
\((A\cap B)\cap C=\left\{ \frac { 1 }{ 4 } ,2 \right\} \)        .......(2)
From (1) and (2), it is verified that
\((A\cap B)\cap C=A\cap (B\cap C)\).

Y-X

R- S = {l, m, n, o, p} - {j, l, n, q} = {m, o, p}
S-R = {j, l, n, q} - {l, m, n, o, p} = {j, q}
R\(\Delta\)S = (R - S) \(\cup \) (S-R)
= { m, o, p} - {j, q} = {j, m, o, p, q}
R\(\cup \)S = {l, m, n, o, p} \(\cup \) {j, l, n, q}
= {l, m, n, o, p ,j, q}
R\(\cap \)S = {l, m, n, o, p} \(\cap \) {j, l, n, q}
= { l, n}
R\(\Delta\)S = (R\(\cup \)S)-(R\(\cap \)S)
= {l, m, n, o, p, j, q} - {l, n} = {m, o, p, j, q}

 A′\(\cup \) B′ = {a, c, e, g} \(\cup \) {b, c, f, g}
{a, b, c, e, f, g}

A { m, a, t, h, e, i, c, s}
B = {g, e, o, m, t, r, y}
A\(\cup\)B = {m, a, t, h, e, c, s} \(\cup\) {g, e, o, m, t, r, y}
= {a, c, e, g, h, i, m, o, r, s, t, y}
A\(\cap\)B = {m, a, t, h, e, i, c, s} \(\cap\) {g, e, o, m, t, r, y}
= {e, m, t}
A-B = {m, a, t, h, e, i, c, s} - {g, e, o, m, t, r, y}
= {a, c, h, i, s}
B-A = {g, e, o, m, t, r, y} - { m,a, t, h, e, i, c, s}
= {g, o, r, y}

{2, 3, 4, 6, 7, 8, 9, 10, 11}

n(X) = 10
The number of subsets of X = n[P(X)]
= 2^m
= 2¹⁰ = 1024
number of proper subsets of X = 2^m-1
= 1024 - 1
= 1023

P(E)= {{ }}

The set A is a null set

X & Y are equivalent sets and also equal sets equal sets

S = The set of English consonants.

Q = The set of Prime numbers between 5 and 31

{P, A, R, L, E, O, G, M}

not a set

When y = 1
f(y) = 6y-3y² + 3
f(1) = 6(1) - 3(1)²+ 3
= 6 - 3+ 3 = 6

n(A) = 36, n(B) = 10, n(A∪B) = 40, n(A′) = 27
(i) n(U) = n(A) + n(A′) = 36 + 27 = 63
(ii) n(A∩B) = n(A) + n(B) – n(A∪B) = 36 + 10 – 40 = 46 – 40 = 6.

A′

A= {a,b}
Subsets of A are \(\phi \),{a}, {b} and {a, b}.

P = {–3, –2, –1, 0}, The prime factors of 210 are 2, 3, 5 and 7 and so, Q = {2, 3, 5, 7}
n(P) = 4 and n(Q) = 4. Therefore P and Q are equivalent sets.

PRINCIPAL
B = {P, R, I, N, C, A, L}