(i) \(\phi\) \(\subset\) B as \(\phi\) is a subset of every set.
(ii) A \(\not \subset\) B as 3 \(\in\) A and 3 \(\notin\) B
(iii) A \(\subset\) C as 1, 3 \(\in\) A also belongs to C
(iv) B \(\subset\) C as each element of B is also an element of C.

Here \(3{ x }^{ 2 }-4x+\frac { 20 }{ 3 } =0\)
Comparing the given quadratic equation with ax² + bx + C = 0, we have
a=3,b=-4 and c=\(\frac { 20 }{ 3 } \)
∴  \(x=\frac { -(-4)\pm \sqrt { (-4)^{ 2 }-4\times 3\times \frac { 20 }{ 3 } } }{ 2\times 3 } \)
= \(\frac { 4\pm \sqrt { 16-80 } }{ 6 } =\frac { 4\pm \sqrt { -64 } }{ 6 } \)
= \(\frac { 4\pm 8\sqrt { -1 } }{ 6 } =\frac { 4\pm 8i }{ 6 } =\frac { 2\pm 4i }{ 3 } \)
Thus x = \(\frac { 2+4i }{ 3 } \)and x=\(\frac { 2-4i }{ 3 } \)

(i) To convert degree measures into radians, we multiply by \({\pi\over 180}\)
\(\therefore 25^o=(25\times {\pi\over 180})^c=({5\pi\over 36})^c\)
(ii) - 47° 30' = -\((47{30\over 60})^c=-({95\over 2})^o\)
\(=-({95\over 2}\times{\pi\over 180})^c=-({19\pi\over 72})^c\)
(iii) 240^o
We know that 180° = π radian
=\(({240\times {\pi\over 180}})^C=({4\pi\over 3})^C\)
(iv) 520^o
We know that 180° = π radian
\(=({520\times {\pi\over180}})^c=({26\pi\over 9})^C\)

Let A(- 2, 3, 5), B(1, 2, 3) and C(7, 0, -1) be three given points.
Then PQ = \(\sqrt { { \left( 1+2 \right) }^{ 2 }+{ \left( 2-3 \right) }^{ 2 }+{ \left( 3-5 \right) }^{ 2 } } \)
= \(\sqrt { 9+1+4 }\)
\( =\sqrt { 14 } \)
QR = \(\sqrt { { \left( 7-1 \right) }^{ 2 }+{ \left( 0-2 \right) }^{ 2 }+{ \left( -1-3 \right) }^{ 2 } } \)
\(\sqrt { 36+4+16 } \)
\(=\sqrt { 56 }\)
\( =2\sqrt { 14 } \)
PR = \(\sqrt { { \left( 7+2 \right) }^{ 2 }+{ \left( 0-3 \right) }^{ 2 }+{ \left( -1-5 \right) }^{ 2 } } \)
\(\sqrt { 81+9+36 } \)
\(=\sqrt { 126 } \)
\(=3\sqrt { 14 } \)
\(\text { Here, } \mathrm{PQ}+\mathrm{QR}=\sqrt{14}+2 \sqrt{14}=3 \sqrt{14}=\mathrm{PR}\)
\(\text { Hence, points } \mathrm{P}(-2,3,5), \mathrm{Q}(1,2,3), \text { and } \mathrm{R}(7,0,-1) \text { are collinear. }\)

2/7

We have ,mean
\(\bar { (x) } =\frac { { x }_{ 1 }+{ x }_{ 2 }+{ x }_{ 3 }+...+{ x }_{ n } }{ n } \)
Now, mean of \( a{ x }_{ 1 },a{ x }_{ 2 },...,a{ x }_{ n }\)
\(=\frac { { ax }_{ 1 }+{ ax }_{ 2 }+...+a{ x }_{ n } }{ n } \)
\(=\frac { a({ x }_{ 1 }+{ x }_{ 2 }+{ x }_{ 3 }+...+{ x }_{ n } }{ n }\)
\(=a\bar { x } \quad [using\quad eq.(i)]\)
Also, we have variance.
\({ \sigma }^{ 2 }=\frac { \sum { ({ x }_{ i }-\bar { x } { ) }^{ 2 } } }{ n } ....(ii)\) 
\(\therefore Variance\ of\ a{ x }_{ 1 },a{ x }_{ 2 },a{ x }_{ 3 },...,a{ x }_{ n }=\frac { \sum { (a{ x }_{ 1 }-a\bar { x } { ) }^{ 2 } } }{ n } \)
\(=\frac { { a }^{ 2 }({ x }_{ 1 }-\bar { x } { ) }^{ 2 }+{ a }^{ 2 }({ x }_{ 2 }-\bar { x } { ) }^{ 2 }+...+{ a }^{ 2 }({ x }_{ n }-\bar { x } { ) }^{ 2 } }{ n } \)

\(=\frac { { a }^{ 2 }\sum { ({ x }_{ 1 }-\bar { x } { ) }^{ 2 } } }{ n } ={ a }^{ 2 }{ \sigma }^{ 2 }\)

R = {(1,3),(2,3),(3,3),(4,3)}
∴ Domain(R) = {1,2,3,4} and range(R) = {3}

Given lines are
3x – 2y = 5 … (1)
and 3x + 2y = 5 … (2)
Let (h, k) is any point, whose distances from the lines (1) and (2) are equal. Therefore
\(\frac{|3 h-2 k-5|}{\sqrt{9+4}}=\frac{|3 h+2 k-5|}{\sqrt{9+4}} \text { or }|3 h-2 k-5|=|3 h+2 k-5|\)
which gives 3h – 2k – 5 = 3h + 2k – 5 or – (3h – 2k – 5) = 3h + 2k – 5.
Solving these two relations we get k = 0 or h \(\frac{5}{3}\)Thus, the point (h, k) satisfies the equations y = 0 or x =\(\frac{5}{3}\) which represent straight lines. Hence, path of the point equidistant from the lines (1) and (2) is a straight line.

Given that \(\text { A.M. }=\frac{a+b}{2}=10 \) ..(1)
and \(\text { G.M. }=\sqrt{a b}=8\) ..(2)
From (1) and (2), we get
a + b = 20 ... (3)
ab = 64 ... (4)
Putting the value of a and b from (3), (4) in the identity (a – b)² = (a + b)² – 4ab,
we get
(a – b)² = 400 – 256 = 144
or a – b = ± 12... (5)
Solving (3) and (5), we obtain
a = 4, b = 16 or a = 16, b = 4
Thus, the numbers a and b are 4, 16 or 16, 4 respectively.

We have 5(2x-7) -3 (2x + 3) \(\le\)0 and 2x +19 \(\le\) 6x +47
From inequality (i),we get
5(2x-7) -3 (2x + 3) \(\le\)0
\(\Rightarrow\)  10x - 35 - 6x - 9 \(\le\) 0 and -4x \(\le\) 28
\(\Rightarrow\) -4x -44 \(\le\) 0  and x \(\ge\) -7 [adding 44 on both sides]
\(\Rightarrow\) 4x\(\le\)44 and x \(\ge\) -7
\(\Rightarrow\) x\(\le\)11  and x\(\ge\) -7 [dividing both sides by 4]
\(\therefore\) The solution set is (- \(\infty\), 11].
From inequality (ii), we get
\(2 x+19 \leq 6 x+47\)
\(\Rightarrow \quad 2 x+19-2 x \leq 6 x+47-2 x\)  [subtracting 2x from both sides
\(\Rightarrow 19 \leq 4 x+47 \)
\(\Rightarrow 19-47 \leq 4 x+47-47\)  [subtracting 47 from both sides]
\(\Rightarrow \quad-28 \leq 4 x \text { or } 4 x \geq-28\)
\(\Rightarrow \frac{4 x}{4} \geq \frac{-28}{4}\) [dividing both sides by 4]
\(\Rightarrow x \geq-7\)
\(\therefore \text { The solution set is }[-7, \infty) \text { . }\)
Now, let us draw the graphs of the solutions of both inequalities on number line.

It can be seen that the values of x, which are common to both are lying in the interval [-7,11].
Hence, the solution set of given system of inequations is [- 7, 11] and this can be represented graphically on the number line as

Here A = {1, 2, 3, 5} and B = {4, 6, 9}, x ∈ A,y ∈ B
\(\therefore\)x - y = (1 - 4), (1 - 6), (1 - 9), (2 - 4),
(2- 6), (2 - 9), (3-4), (3-6),
(3- 9), (5- 4), (5- 6), (5- 9)
x - Y = -3, -5, -8, -2, -4, -7, -1, -3,-6,1, -1,-4
\(\therefore\)R = {(1, 4), (1, 6), (2, 9), (3, 4), (3, 6), (5, 4), (5, 6)}

Let POQ be the parabolic reflector which is 20 cm in diameter and 5 cm deep.

Then, PQ = 20 cm and OR = 5 cm, where R is the midpoint of PQ. We take OX as X-axis and OY as Y-axis. The equation of parabola may be taken as 4ax. Since, the point P(5, 10) lies on the parabola.
\({ 10 }^{ 2 }=4a(5)\Rightarrow a=5\)
Therefore, the coordinate of the focus are (a, o) i.e. (5, 0). Hence, the focus is the midpoint of the given diameter.

0 do not comes in hundred's place, so only 4 numbers are posiible in hundred's place. Since, the digits are repeated, so remaining two digits, 5 numbers are possible in each place.
\(\therefore \) Total number of ways = 4 x 5 x 5 = 100

Let z = \(\sqrt { 3 } \)+i= r (cos\(\theta\) + isin\(\theta\))
Now, equate real and imaginary parts of this equation and solve it for r and \(\theta\)
\(2\left( cos\frac { \pi }{ 6 } +isin\frac { \pi }{ 6 } \right) ;2,\frac { \pi }{ 6 } \)

(i) Here f(x)=2x-\(\frac{3}{4}\)
∴ \(f^{'}(x)=\frac{d}{dx}(2x-\frac{3}{4})\)
=\(2\frac{d}{dx}(x)-\frac{d}{dx}(\frac{3}{4})\)
=\(2\times 1 -0=2\).
(ii) Here f(x)=(5x³ + 3x - 1) (x - 1)
∴ \(f^{'}(x)=\frac{d}{dx}[(5x^{3}+3x-1)(x-1)]\)
=\((5x^{3}+3x-1)\frac{d}{dx}(x-1)+(x-1)\frac{d}{dx}(5x^{3}+3x-1)\)
=\(5x^{3}+3x-1)\times 1 + (x-1)(15x^{2}+3)\)
=\(5x^{3}+3x-1+15x^{3}+3x-15x^{2}-3\)
=\(20x^{3}-15x^{2}+6x-4\).
(iii) Here f(x)=\(x^{-3} (5+3x)\)
ஃ \(f^{'}(x)=\frac{d}{dx}[x^{-3}(5+3x)]\)
=\(x^{-3}\frac{d}{dx}(5+3x)+(5+3x)\frac{d}{dx}(x^{-3}) \)
=\(x^{-3}\times 3+(5+3x)\times -3x^{-4}\)
=\(\frac{3}{x^{3}}-\frac{3}{x^{4}}(5+3x)\)
=\(\frac{3}{x^{3}}[1-\frac{5+3x}{x}]=\frac{3}{x^{3}}[\frac{x-5-3x}{x}]\)
=\(\frac{-3}{x^4}(5+2x)\).
(iv) Here \(f(x)=x^{5}(3-6x^{-9})\)
∴ \(f^{'}(x)=\frac{d}{dx}[x^{5}(3-6x^{-9})] \)
=\(x^{5}\frac{d}{dx}(3-6x^{-9})+(3-6x^{-9})\frac{d}{dx}(x^{5})\)
=\(x^{5}(54x^{-10})+(3-6x^{-9})\times 5x^{4}\)
=\(54x^{-5}+15x^{4}-30x^{-5}\)
=\(24x^{-5}+15x^{4}\)
=\(\frac{24}{x^{5}}+15x^{4}\).
(v) Here f(x)=x^-4(3-4x^-5)
∴ \(f^{'}(x)=\frac{d}{dx}[x^{-4}(3-4x^{-5})]\)
=\(x^{-4}\frac{d}{dx}(3-4x^{-5})+(3-4x^{-5})\frac{d}{dx}(x^{-4})\)
=\(x^{-4}(20x^{-6})+(3-4x^{-5})(-4x^{-5})\)
=\(20x^{-10}-12x^{-5}+16x^{-10}\)
=\(36x^{-10}-12x^{-5}=\frac{36}{x^{10}}-\frac{12}{x^{5}}\)
(vi) Here, f(x)=\(\frac{2}{x+1}-\frac{x^{2}}{3x-1}\)
ஃ \(f^{'}(x)=\frac{d}{dx}[\frac{2}{x+1}-\frac{x^{2}}{3x-1}]\)
=\(\frac{d}{dx}(\frac{2}{x+1})-\frac{d}{dx}(\frac{x^{2}}{3x-1})\)
=\(\frac{(x+1)\frac{d}{dx}(2)-2\frac{d}{dx}(x+1)}{(x+1)^{2}}\)-\(\frac{(3x-1)\frac{d}{dx}(x^{2}-x^{2}\frac{d}{dx}(3x-1)}{(3x-1)^{2}}\)
=\(\frac{(x+1)\times 0 -2 \times 1}{(x+1)^{2}}-\frac{(3x-1)(2x)-x^{2}\times 3}{(3x-1)^{2}}\)
=\(\frac{-2}{(x+1)^{2}}-\frac{6x^{2}-2x-3x^{2}}{(3x-1)^{2}}\)
=\(\frac{-2}{(x+1)^{2}}-\frac{3x^{2}-2x}{(3x-1)^{2}}\)

(a) Here a = 2, and r = \(\frac { 2\sqrt { 2 } }{ 2 } =\sqrt { 2 } \).
Let nth term of given sequence be 128 is,
a_n = 128.
We know that a_n = arⁿ-1
\(\therefore \)  128 =2\(\times \)\(\left( \sqrt { 2 } \right) ^{ n-1 }\)\(\Rightarrow \)64 = \(\left( \sqrt { 2 } \right) ^{ n-1 }\)
\(\Rightarrow \)\(\left( \sqrt { 2 } \right) ^{ 12 }\)=\(\left( \sqrt { 2 } \right) ^{ n-1 }\) \(\Rightarrow \) n-1=12
\(\Rightarrow \)  n=13
Thus, 13^thterm of the given sequence is 128.
(b) Here a =\(\sqrt { 3 } \)and r =\(\frac { 3 }{ \sqrt { 3 } } \)=\(\sqrt { 3 } \)
Let n^th term of given sequence be 729 i.e
a_n =729
We know that a_n = ar^n-1
\(\therefore \)729=\(\sqrt { 3 } \)x\(\left( \sqrt { 3 } \right) ^{ n-1 }\)\(\Rightarrow \)\(\left( \sqrt { 3 } \right) ^{ 12 }\)=\(\left( \sqrt { 3 } \right) ^{ n}\)
\(\Rightarrow \) n=12 
Thus, the 12th term of the given sequence is 729.
(c) Here a=\(\frac { 1 }{ 3 } \)and r= \(\frac { \frac { 1 }{ 9 } }{ \frac { 1 }{ 3 } } \)
Let n^th term of given sequence be\( \frac { 1 }{ 19683 } \)
i.e a_n\( \frac { 1 }{ 19683 } \)
We know that a_n = ar^n-l.
\(\therefore \)\( \frac { 1 }{ 19683 } \)\( \frac { 1 }{ 19683 } \)=\(\frac { 1 }{ 3 } \)x \(\left( \frac { 1 }{ 3 } \right) ^{ n-1 }\)\(\Rightarrow \)\(\left( \frac { 1 }{ 3 } \right) ^{ 9 }\)=\(\left( \frac { 1 }{ 3 } \right) ^{ n }\)
\(\Rightarrow \)n =9
Thus, the 9th term of the given sequence is
\( \frac { 1 }{ 19683 } \).

Here tan x =\(-{5\over12}\)
cot x=\({1\over tan \ x}={-12\over 5}\)
Now sec²x = 1 + tan ²x
\(\Rightarrow sec^2 x=1+({-5\over 12})^2\)
\(\Rightarrow sec \ x={169\over 144}\)
\(\Rightarrow sec x=\pm{13\over 12}\)
But x lies in second quadrant.
\(\therefore \ sec \ x={-13\over 12}\)
\(cos \ x={1\over sec \ x}={-12\over 13}\)
Also sin² x + cos² x = 1
\(\Rightarrow sin^2 x+({-12\over 13})^2=1\)
\(\Rightarrow sin ^2=1-{144\over 169}\)
\(\Rightarrow sin ^2 \ x ={25\over 169}\)
\(\Rightarrow sin \ x=\pm{5\over13}\)
But x lies in second quadrant.
\(\therefore sin x={5\over13}\)
\(cosec \ x={1\over sin \ x}={13\over 5}\)

\(\frac{N}{2}=\frac{100}{2}=50\)
∴ Median class is 35.5-40.5
∴ Median =\(35.5+\frac { 50-37 }{ 26 } \times 5=35.5+2.5=38\)
M.D. about median
\(=\frac { 1 }{ N } \sum _{ i=1 }^{ n }{ { f }_{ i }\left| { x }_{ i }-M \right| } =\frac { 1 }{ 100 } \times 735=7.35\)

General term in \(\left( \frac { 3 }{ 2 } x^{ 2 }-\frac { 1 }{ 3x } \right) ^{ 9 }\)
\(T_{ r+1 }={ ^{ 9 }C }_{ r }\left( \frac { 3 }{ 2 } \right) ^{ 9-r }\left( -\frac { 1 }{ 3 } \right) ^{ r }x^{ 18-3r }\)
So, general form in the expansion of
\((1+x+{ 2x }^{ 3 })\left( \frac { 3 }{ 2 } x^{ 2 }-\frac { 1 }{ 3x } \right) ^{ 9 }\)
\(={ ^{ 9 }C }_{ r }\left( \frac { 3 }{ 2 } \right) ^{ 9-r }\left( -\frac { 1 }{ 3 } \right) ^{ r }x^{ 18-3r }+{ ^{ 9 }C }_{ r }\left( \frac { 3 }{ 2 } \right) ^{ 9-r }\left( -\frac { 1 }{ 3 } \right) ^{ r }x^{ 19-3r }+2{ ^{ 9 }C }_{ r }\left( \frac { 3 }{ 2 } \right) ^{ 9-r }\left( -\frac { 1 }{ 3 } \right) ^{ r }x^{ 21-3r }\)
For the term independent of x
18-3r=0, 19-3r=0 and 21-3r=0
\(\Rightarrow \ r=6,\quad \frac { 19 }{ 3 } and\quad 7\)
Thus,  posible value of r are 6 and 7
Term independent of x
\(={ ^{ 9 }C }_{ r }\left( \frac { 3 }{ 2 } \right) ^{ 9-6 }\left( -\frac { 1 }{ 3 } \right) ^{ 6 }+2.^{ 9 }C_{ 7 }\left( \frac { 3 }{ 2 } \right) ^{ 9-7 }\left( -\frac { 1 }{ 3 } \right) ^{ 7 }\)
\(=\frac { 84 }{ 8 } .\frac { 1 }{ 3^{ 3 } } -\frac { 36 }{ 4 } .\frac { 2 }{ 3^{ 5 } } =\frac { 17 }{ 54 } \)