We have \(n(A \cup B)=6,n(A\cap B)=2\) and \(n(A \triangle B)=4.\)
So, \(n((A\cup B)\times(A\cap B)\times(A \triangle B))=n(A \cup B)\times n(A\cap B)\times n(A\triangle B)= 6 \times 2 \times 4 = 48.\)

Given \(\frac { x+2 }{ x+3 } =\frac { x+4 }{ 2x+3 } \)
⇒ (x + 2)(2x + 3) = (x + 4) (x + 3)
⇒ 2x² + 3x + 4x + 6 = x² + 3x + 4x +12

⇒  x² = 6
⇒   x = ± \(\sqrt { 6 } \)
The roots are \(\sqrt { 6 } \) and -\(\sqrt { 6 } \)

Let a, b, c are people in Chennai
Reflexivity: "a" is a friend of "a" \(\Rightarrow\)  a R    a \(\Rightarrow\) R is not reflexive.
Symmetric: a is friend of b \(\Rightarrow\) b is the friend of a.
\(\therefore\) aRb \(\Rightarrow\) bRa \(\Rightarrow\) R is symmetric
Transitive: a is the friend of b and b is the friend of c \(\Rightarrow\) a need not be the friend of c.
\(\therefore\) aRb \(\Rightarrow\) bRc \(\neq \) aRc \(\Rightarrow\) R is not transitive
Hence, the relation "friendship" is not equivalent.

n(A\(\times\) A) = 16. \(\Rightarrow\) n(A) = 4
Given S = {a,b) \(\in \) A\(\times\)A : a
\(\therefore\) A = {0, 1, 2, -1}
(A\(\times\)A) = {(0, 0)(0, 1)(0, 2)(0, -1),(1, 1) (1, 2)(1, -1)(2, 0)(2, 1)(2, 2)(2, -1)(-1, 0)(-1, 1)(-1, 2)(-1, -1)}
Now, S = {(0, 1)(0, 2)(-1, 0)(-1, 1)(-1, 2)}
\(\therefore\) The remaining elements of S are (0, 2)(-1, 0)(-1, 1)(1, 2)(0, 1) (1, 2)(0, 2)

Let r be the radius of the spherical tank
 Then, Volume of the spherical tank = \(\frac { 32\pi }{ 3 } \)
⇒ \(\frac { 4 }{ 3 } { \pi r }^{ 3 }=\frac { 32\pi }{ 3 } \)
⇒ 4r³ = 32
⇒  r³ = \(\frac { 32 }{ 4 } \) = 8
⇒ r³ = 2³
⇒  r = 2
∴ Radius of the spherical tank is 2 unit

Let p = {-1, 1}
\(\Rightarrow\) P = {x \(\in \) R : x is a root of the equation x²-1 = 0}

\(\left( \left[ (256)^{ \frac { -1 }{ 2 } } \right] ^{ \frac { -1 }{ 4 } } \right) ^{ 3 }\)  =  \((256)^{ \frac { -1 }{ 2 } \times \frac { -1 }{ 4 } \times 3 }\)   \([\because \frac { { a }^{ m } }{ { a }^{ n } } ={ a }^{ m-n }]\)
=  \((256)^{ \frac { 3 }{ 8 } }=({ 2 }^{ 8 })^{ \frac { 3 }{ 8 } }={ 2 }^{ 8\times \frac { 3 }{ 8 } }={ 2 }^{ 3 }=8\)

\(\frac { (x-2) }{ (x+4) } -\frac { 5 }{ x+3 } \ge 0\)
\(\frac { (x-2)(x+3)-5(x+4) }{ (x+4)(x+3) } \ge 0\)
\(\frac { { x }^{ 2 }+x-6-5x-20 }{ (x+4)(x+3) } \ge 0\)
(i.e)\(\frac { { x }^{ 2 }-4x-26 }{ (x+4)(x+3) } \ge 0\)
x+4 = 0 ⇒ x = -4 ; x + 3 = 0 ⇒ x = -3
Plotting the points -4, -3 on number line and taking limits \((-∞,-4)(-4,-3),(-3,∞)\)

The solution for the inequality\(\frac { x-2 }{ x+4 } \ge \frac { 5 }{ x+3 } \)are the intervals \((-∞,-4)\) and (-4, -3)

Method 1:
One-to-one: Let f(x) = f(y). Then 2x - 3 = 2y - 3; this implies that x = y. That is, f(x) = f(y) implies that x = y. Thus f is one-to-one.
Onto: Let y \(\in\) R. Let x \(={y+3 \over 2}.\)  Then \(f(x)=2\left( {y+3\over2} \right)-3=y.\)  Thus f is onto. This also can be proved by saying the following statement. The range of f is R (how?) which is equal to the co-domain and hence f is onto.
Inverse: Let y = 2x - 3. Then y + 3 = 2x and hence \(x={y+3\over 2}.\)  Thus \({f}^{-1}(y)={y+3\over2}\). By replacing y as x, we get \({f}^{-1}(x)={x+3\over 2}.\)
Method 2:
Let y = 2x − 3. Then \(x=\frac{y+3}{2}\). Let \(g(y)=\frac{y+3}{2}\)
Now
\((g \circ f)(x)=g(f(x))=g(2 x-3)=\frac{(2 x-3)+3}{2}=x .\)
\((f \circ g)(y)=f(g(y))=f\left(\frac{y+3}{2}\right)=2\left(\frac{y+3}{2}\right)-3=y\)
Thus, \(g \circ f=I_{X} \text { and } f \circ g=I_{Y}\)
This implies that f and g are bijections and inverses to each other. Hence f is a bijection and \(f^{-1}(y)=\frac{y+3}{2}\). Replacing y by x we get \(f^{-1}(x)=\frac{x+3}{2}\)

Given inequality is \({x^3(x-1)\over x-2}>0\)
The critical numbers are 0, 1, 2
The possible intervals are (-∞, 0) (0, 1) (1, 2) (2, ∞)

The given inequality  \({x^2(x-1)\over x-2}>0\) is satisfied by the intervals (0, 1) and (2, ∞)
∴ Solution set is (0, 1)∪(2, ∞)
∴ Solution set is \((0,1)\bigcup(2, \infty)\)

Given equation is x⁴+ 1
x⁴+1 = (x²)²+1²   [ \(\because\) a² + b² = (a + b)²-2b where a = x², b = 1 ]
= (x² + 1)²- 2x²(1)
= (x² + 1)²-\((-\sqrt{2}x)^2\)
= (x² + 1 + \(\sqrt{2}\) x) (x²+ 1 -\(\sqrt{2}\)x)    [ \(\because\)  a²- b² = (a + b) (a - b)]

Let x be the number of litres of 30% acid solution
∴ Total mixture = (600 + x) litres
30% of x + 12% of 600 > 5% of (600 + x)
\(⇒\ {30x\over 100}+{12\over 100}\times600>{15\over 100}(600+x)\)
⇒30x + 7200 > 9000 + 15x [Multiplying by 100]
⇒ 30x + 7200 - 15x > 9000 [Subtracting 15x]
⇒ 15x + 7200 > 900
⇒ 15x > 1800
⇒ x > 120      ....(1)
Also, 30% of x + 12% of 600 < \(18\over 100\)(600 + x)
\({30x\over 100}+{12\over 100}\times600<{18\over 100}(600 +x)\)
⇒ 30x + 7200 < 18 (600 + x) [Multiplying by 100]
⇒ 30x + 7200 < 10, 800 + 18x
⇒ 12x + 7200 < 10, 800 [Subtracting 18x]
⇒ 12x < 10, 800 - 7200 [Subtracting 7200]
⇒ 2x < 3600
⇒ \(x<{3600\over12}\)
⇒ x < 300    ....(2)
From (1) and (2), 120 < x < 300
The number of litres of the 30% acid solution will have to be greater than 120 litres and less than 300 litres.

Given inequality is \({1\over 5}|10x-2|<1\)
⇒ |10x - 2| < 5
This means, -5 < 10 x - 2 < 5.
⇒ -5 + 2 < 10 x < 5 + 2
⇒ -3 < 10x <7

\(⇒\ -{3\over 10}< x < {7\over 10}\)
∴Solution set is \(\left[ \frac { -3 }{ 10 } ,\frac { 7 }{ 10 } \right] \)

Given A(x) = 30,000 + 0.04x
S(x) = 25,000 + 0.05x
∴ (A + S)(x) = 30,000 + 0.04x + 25,000 + 0.05x
= 55,000+0.09x
Given x = Rs. 1,50,00,000
Then (A+S)(x) = 55000 + 0.09(1,50,00,000)
= 55000 + 1,350,000
= 1,405,000
Hence total family income = Rs. 14,05,000

Let \(\frac{x}{(x+3)(x-4)}=\frac{A}{x+3}+\frac{B}{x-4}\) where A and B are constants.
Then, \(\frac{x}{(x+3)(x-4)}=\frac{A(x-4)+B(x+3)}{(x+3)(x-4)}\) which gives x = A(x - 4) + B(x + 3)
When x = 4, we have B = \(\frac{4}{7}\)
When x = -3 we have A = \(\frac{3}{7}\)
Hence, \(\frac{x}{(x+3)(x-4)}=\frac{3}{7(x+3)}+\frac{4}{7(x-4)}\)

Clearly x ∈ A ∩ B
⇒ x ∈ A and x ∈ B
⇒ x = 3n and x = 4n, n ∈ Z
⇒ x is a multiple of 3 and x is a multiple of 4
⇒ x is a multiple of 3 and 4 both
⇒ x is a multiple of 12
⇒ x = 12n, n ∈ Z
Hence A ∩ B= {x : x = 12n, n∈Z}

Let f(x) = \(\frac { 1 }{ 1-2sinx } \)
When the denominator is 0,
1-2 sin x = 0
\(\Rightarrow\) 1 = 2 sin x
\(\Rightarrow sin\quad x=\frac { 1 }{ 2 } \)
\(\Rightarrow sin\quad x=sin\frac { \pi }{ 6 } \)
\(\Rightarrow x=n\pi +{ (-1) }^{ n }\frac { \pi }{ 6 } n\in Z\)   \(\left[ \because sin\quad x=sin\alpha \Rightarrow x=n\pi +{ (-1) }^{ n }\alpha \quad n\in Z \right] \)
Domain of f(x) is R - \(\left( n\pi +{ (-1) }^{ n }\frac { \pi }{ 6 } \right) ,n\in Z\)

LHS = \(log\frac { { a }^{ 2 } }{ bc } +log\frac { b^{ 2 } }{ ca } +log\frac { c^{ 2 } }{ ab } \)
= \(log\left( \frac { { a }^{ 2 } }{ bc } .\frac { b^{ 2 } }{ ca } .\frac { c^{ 2 } }{ ab } \right) \)
= \(log\left( \frac { { a }^{ 2 }b^{ 2 }c^{ 2 } }{ { a }^{ 2 }b^{ 2 }c^{ 2 } } \right) \)
= log 1 = 0 =RHS
Hence proved

Given a²+ b² = 7ab
Adding 2ab both sides we get,
a²+b²+2ab = 7ab + 2ab
⇒ (a+b)² = 9ab
⇒ \(\frac { (a+b)^{ 2 } }{ 9 } \)  = ab
⇒  \(\left( \frac { a+b^{ 2 } }{ 9 } \right) ^{ 2 }\)  = ab
Taking square root,we get
\(\frac { a+b }{ 3 } \)  = ab
\(log\left( \frac { a+b }{ 3 } \right) =log(ab)^{ \frac { 1 }{ 2 } }\)
= \(\frac { 1 }{ 2 } \) log (ab)
log \(\left( \frac { a+b }{ 3 } \right) \) = \(\frac { 1 }{ 2 } \) [log a + log b]
Hence proved.

Given quadratic equation is
5^2x-5^x+3 + 125 = 5x
\(\Rightarrow\) (5^x)²-5^x 5³- 5^x + 125 = 0
\(\Rightarrow\) (5^x)²-125 5^x- 5^x + 125 = 0
\(\Rightarrow\) (5^x)²-126.5^x+125=0
Let 5^x=y
\(\Rightarrow\)y²-126y + 125 = 0
\(\Rightarrow\)(y-1)(y-125) = 0
\(\Rightarrow\)    y = 1 or 125
\(\Rightarrow\)   5^x = 1 or 125
Case (i) When 5^x = 1 \(\Rightarrow\) 5⁰ \(\Rightarrow\) x = 0
Case (ii) When 5^x = 125 \(\Rightarrow\) 5^x = 5³\(\Rightarrow\) x = 3
\(\therefore\) The roots are 0, 3.

Let \(f(x)={5x-160\over 9}\)
Given \(y={5x\over 9}-{160\over 9}\)
\(⇒\ \ y={5x-160\over 9}\)
Then 9y = 5x - 160 ⇒ 5x = 9y + 160 ⇒ \(x={9y+160\over5}\)
Let \(g(y)={9y+160\over 5}\)
Now gof(x) = \(g[f(x)=g\left(5x-160\over 6\right)\)
\(={9\left(5x-160\over 9\right)+160\over 5}={5x-160+160\over 5}={5x\over 5}=x\)
and fog(y) = f(g(y)) =\(f\left(9y+160\over 5\right)={5\left(9y+160\over5\right)-160\over 5}={9y+160-160\over 9}=y\)
Thus gof = I_x and fog = I_y
This implies that f and g are bijections and inverses to each other
\(f^{-1}(y)={9y+160\over5}\)
Replacing y by x, we get f^-1 (x) = \(\frac { 9x+160 }{ 5 } \)​​​​​

\(\text { Let } \log _{\sqrt{2}} 512=x\)
\(\text { Then }(\sqrt{2})^{x}=2^{9}\)
\(\Rightarrow 2^{\frac{x}{2}}=2^{9} \Rightarrow x / 2=9 \Rightarrow x=18\)

\((A \times B) \cap(A \times C) =A \times(B \cap C) \)
\(n[(A \times B) \cap(A \times C)] =n(A) \times n(B \cap C) \)
\(\Rightarrow 8 =n(A) \times 2 \Rightarrow n(A)=4 \)

\(n(A \cap B)=1\)

It is onto not one-one

\(\text { Since } \sin 30^{\circ}=\frac{1}{2}\)
\(\sin 150^{\circ}=\frac{1}{2}\)