\(Let\quad u=\log { (1+{ x }^{ 2 } } )\quad and\quad v=\tan ^{ -1 }{ x } \)
\( \Rightarrow \frac { du }{ dx } =\frac { 1 }{ 1+{ x }^{ 2 } } .\frac { d }{ dx } (1+{ x }^{ 2 })\quad \frac { dv }{ dx } =\frac { 1 }{ 1+{ x }^{ 2 } } \)
\(\frac { du }{ dx } =\frac { 2x }{ 1+{ x }^{ 2 } } \)

\(\therefore \frac { du }{ dv } =2x\)

\(=x^2-4 x+5 \)
\(=\int \frac{1}{\sqrt{x^2-4 x+5}} d x\)
\(=\left(x^2-4 x\right)+5\)
Completing the square on x
\(=\left(x^2-4 x+2^2\right)-2^2+5 \)
\(=(x-2)^2+1 \)
\(=(x-2)^2+1^2\)
\(\therefore \int \frac{1}{\sqrt{x^2-4 x+5}} d x =\int \frac{1}{\sqrt{(x-2)^2+1^2}} d x\)
\(=\int \frac{1}{\sqrt{u^2+a^2}} d u, u=x-2, a=1\)
\(=\log \left|u+\sqrt{u^2+a^2}\right|+c\)
\(=\log \left|(x-2)+\sqrt{x^2-4 x+5}\right|+c\)

\(\int { { e }^{ 8-7x } } dx=\frac { { e }^{ 8-7x } }{ (-7) } +c\)

Given vector is \(\hat{j}\)
The direction ratio of \(\hat{j}\) are 0, 1, 0.
\(x=\sqrt{x^2+y^2+z^2}=\sqrt{0+1^2+0}=1\)
Hence, its direction cosines are \({0\over1},{1\over 1},{0\over1}\Rightarrow 0,1,0.\)

Let x be the marks obtained by Ravi in the third test.
Then, \(\frac { 70+75+x }{ 3 } \ge 60\)
⇒ 145 + x ≥ 180  ⇒ x ≥ 180 - 145
⇒ x ≥ 35
Thus, Ravi must obtain a minimum of 35 marks to get an average of at least 60 marks.
Note : A minimum of 35 marks.
⇒  Marks greater than or equal to 35.

Clearly A = {x: x = 4n + 1, n = 2, 3, 4, 5} = {9, 13, 17, 21}.
Hence n(A) = 4. This implies that \(n(\mathscr{P}(A))=2^4=16.\)

Using binomial theorem, we have
(1 + ax)ⁿ-1 + nC₁(ax) + nC₂(ax)²+.......+aⁿxⁿ
= \(1+nax+{{n(n-1)}\over{2}}a^2x^2+....a^nx^n\)
Given (1 + ax)ⁿ = 1 + 12x + 64x² +....
Conparing the Co-efficient of x and x², we get
n a = 12
and \({n(n-1)\over 2}a^2=64\)
\((n-1).{na.a\over2}=64\Rightarrow(n-1){(12)a\over2}=64\)
\((n-1)6a=64\Rightarrow(n-1)a={{64}\over{6}}\)    \(\left[ \because na=12\Rightarrow a={12\over n} \right]\)
\(\Rightarrow(n-1)\left( {12\over n} \right)={64 \over 6}\)
\(={n-1\over n}={ 64 \over 6\times 12}\Rightarrow{n-1\over n}={8\over 9}\)
\(\Rightarrow\)  9n - 9 = 8n
\(\Rightarrow\)  n = 9 and \(a=\frac { 12 }{ n } =\frac { 12 }{ 9 } =\frac { 4 }{ 3 } \)

Given that P (-3,1) lie on the locus of x² - 5x + ky = 0.
⇒ (-3)²-5 (-3)+k(1) = 0
⇒ 9 +15 + k = 0
⇒ k = -24
Also, it is given that (2, b) lie on the locus of x² - 5x + ky = 0.
⇒ 2² - 5 (2) + kb = 0
⇒ 4 - 10 - 24 (b) = 0
⇒ - 6 - 24b = 0
⇒ -24b = 6
⇒ b = \(\frac{-6}{24}=\frac{-1}{4}\)

LHS = sin x + sin 2x + sin 3x
= (sin x + sin 3x) + sin 2x
\(=2sin\left( \frac { x+3x }{ 2 } \right) cos\left( \frac { x-3x }{ 2 } \right) +sin2x\)
= 2sin 2x.cos(-x) + sin 2x
= 2sin 2x + cos x + sin 2x
= sin 2x(1 + 2cos x) = RHS

Let I = \(\int { \frac { 1 }{ { x }^{ \frac { 1 }{ 2 } }+{ x }^{ \frac { 1 }{ 3 } } } } \)dx
Here, the exponents of x are \(\frac { 1 }{ 2 } \) and \(\frac { 1 }{ 3 } \) and the LCM of their denominator is 6.
So, to remove fractional exponents put x = t⁶ \(\Rightarrow\) dx = 6t⁵ dt.
I = \(\int { \frac { { 6t }^{ 5 } }{ { t }^{ 3 }+{ t }^{ 2 } } } dt=6\int { \frac { { t }^{ 5 }dt }{ { t }^{ 2 }\left( t+1 \right) } } =6\int { \frac { { t }^{ 3 } }{ t+1 } } dr=6\int { \frac { \left( { t }^{ 3 }+1 \right) -1 }{ t+1 } } dt\)
= \(\int { \frac { \left( t+1 \right) ^{ 3 }-3t\left( t+1 \right) -1 }{ t+1 } } dt=6\) \(\int { \left[ \frac { \left( t+1 \right) ^{ 3 } }{ t+1 } -\frac { 3t(t+1) }{ t+1 } -\frac { 1 }{ t+1 } \right] dt } \)
= \(6\left[ f\left( t+1 \right) ^{ 2 }-3t-\frac { 1 }{ t+1 } dt \right] 6\left[ f{ t }^{ 2 }+2t+1-3t-\frac { 1 }{ t+1 } dt \right] =6\left( f{ t }^{ 2 }-t+1-\frac { 1 }{ t+1 } \right) dt\)
= \(6\left( \frac { { t }^{ 3 } }{ 3 } -\frac { { t }^{ 2 } }{ 2 } +t-log|t+1| \right) +c\)
= \(2.\sqrt { x } -3x^{ \frac { 1 }{ 3 } }+6x^{ \frac { 1 }{ 6 } }-6log|{ x }^{ \frac { 1 }{ 6 } }+1|+c\)

E₁: Student knows the answer
E₂: Student guesses the answer
A: event that the answer is correct
\(P({ E }_{ 1 })=\frac { 3 }{ 4 } \) and P(E₂) = \(\frac { 1 }{ 4 } \)
P(A/E₁) = 1 and P(A/E₂) = \(\frac { 1 }{ 4 } \)
\(\therefore\) By Bayes's theorem,
\(P({ E }_{ 1 }/A)=\frac { P({ E }_{ 1 })P(A/{ E }_{ 1 }) }{ P({ E }_{ 1 }).P(A/{ E }_{ 1 })+P({ E }_{ 2 }).P(A/{ E }_{ 2 }) } =\frac { \frac { 3 }{ 4 } \times 1 }{ \frac { 3 }{ 4 } \times 1+\frac { 1 }{ 4 } \times \frac { 1 }{ 4 } } =\frac { \frac { 3 }{ 4 } }{ \frac { 3 }{ 4 } +\frac { 1 }{ 16 } } =\frac { \frac { 3 }{ 4 } }{ \frac { 12+1 }{ 16 } } \)
\(P({ E }_{ 1 }/A)=\frac { 3 }{ 4 } \times \frac { 16 }{ 13 } =\frac { 12 }{ 13 } \)

Given \(\log { ({ x }^{ 2 }+{ y }^{ 2 }) } =2\tan ^{ -1 }{ \frac { y }{ x } , } \)
Differentiating both sides, with respect to 'x' we get
\(\frac { 1 }{ { x }^{ 2 }+{ y }^{ 2 } } .\frac { d }{ dx } ({ x }^{ 2 }+{ y }^{ 2 })=2.\frac { 1 }{ 1+{ \left( \frac { y }{ x } \right) }^{ 2 } } .\frac { d }{ dx } { \left( \frac { y }{ x } \right) }\)

\(\Rightarrow 2x+2y\frac { dy }{ dx } =2x\frac { dy }{ dx } -2y\\ \Rightarrow x+y\frac { dy }{ dx } =x\frac { dy }{ dx } -y\\ \Rightarrow \frac { dy }{ dx } (y-x)=-x-y\\ \Rightarrow \frac { dy }{ dx } =\frac { -(x+y) }{ y-x } =\frac { x+y }{ x-y } \quad\)
Hence Proved

Let A₁, A₂ and A₃ be the event of A, B, C becoming managers of the company respectively. Let X be the event that the office canteen will be improved.
Then, \(P\left(A_1\right)=\frac{5}{10}=0.5 \)
\(P\left(A_2\right)=\frac{3}{10}=0.3 \)
\(P\left(A_3\right)=\frac{2}{10}=0.2 \)
\(P\left(X / A_1\right)=0.4 \)
\(P\left(X / A_2\right)=0.5\)
\(P\left(X / A_3\right)=0.3\)
\(P\left(A_2 / X\right)=\frac{P\left(A_2\right) P\left(X / A_2\right)}{P\left(A_1\right) P\left(X / A_1\right)+P\left(A_2\right) P\left(X / A_2\right)}+P\left(A_3\right) P\left(X / A_3\right)\)
\(=\frac{0.3(0.5)}{0.5(0.4)+0.3(0.5)+0.2(0.3)} \)
\(=\frac{0.15}{0.2+0.15+0.06} \)
\(=\frac{0.15}{0.41}=\frac{15}{41}\)

\(f(x)=\frac{x\lfloor x\rfloor}{\sin |x|} = \begin{cases}\frac{-x}{\sin (-x)} & \text { if }-1<x<0 \\ \frac{x \cdot 0}{\sin x} & \text { if } 0<x<1\end{cases} \)
\(= \begin{cases}\frac{x}{\sin x} & \text { if }-1<x<0 \\ 0 & \text { if } 0<x<1\end{cases} \)
Therefore, \(lim_{x\rightarrow 0^-}f(x)=+1\)
\(lim_{x\rightarrow 0^+}f(x)=0\).
Hence the limit does not exist.

Let the position vector of the vertices of the \(\triangle\) ABC be \(\overrightarrow{a},\overrightarrow{b}\) and \(\overrightarrow{c}\) respectively.
\(\therefore \overrightarrow{OA}=\overrightarrow{a},\overrightarrow{OB}=\overrightarrow{b},\overrightarrow{OC}=\overrightarrow{c}.\)
Since G is the centroid of \(\triangle\) ABC, we have
\(\Rightarrow \overrightarrow{OG}={\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}\over3}\) \(\Rightarrow 3\overrightarrow{OG}=\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}\)
Now,LHS \(=\overrightarrow{GA}+\overrightarrow{GB}+\overrightarrow{GC}\)
\(=\overrightarrow{OA}-\overrightarrow{OG}+\overrightarrow{OB}-\overrightarrow{OG}+\overrightarrow{OC}-\overrightarrow{OG}=(\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC})\)\(-3\overrightarrow{OG}\)
\(=(\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC})-(\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}=\overrightarrow{0}=RHS\)
Hence proved.

We have |B| = \(\begin{vmatrix} 2(a+b+c) & 2(a+b+c) &2(a+b+c) \\ c+a & a+b &b+c \\ a+b & b+c & c+a \end{vmatrix}\)\((R_1 \rightarrow R_1+R_2+R_3)\)
= 2\(\begin{vmatrix} a+b+c & a+b+c & a+b+c \\ c+a & a+b &b+c \\ a+b & b+c & c+a \end{vmatrix}\)
= 2\(\begin{vmatrix} a+b+c & a+b+c & a+b+c \\ -b & -c &-a \\ -c & -a & -b \end{vmatrix}\)\((R_2 \rightarrow R_2-R_1andR_3\rightarrow R_3-R_1)\)

= 2\(\begin{vmatrix}a &b &c \\ -b & -c & -a \\ -c & -a & -b \end{vmatrix}\)\((R_1 \rightarrow R_1+R_2+R_3)\)
= 2(-1)² \(\begin{vmatrix}a &b &c \\b &c & a \\ c & a & b \end{vmatrix}\)
= 2| A |.

The given inequality is 3x+2y≤12.
Draw the graph of the line  3x+2y≤12.
Table of values satisfying the equation
3x+2y≤12

Putting (0, 0) in the given inequation, we have 3\(\times\)0+2\(\times\)0≤12 which is true
∴ Half plane of 3x+2y≤12 is towards origin
Also, the given inequality is 1
Draw the graph of the line x = 1.
 
Putting (0, 0) in the given inequation, we have 0 ≥1 which is false.
∴ Half plane of x≥1 is away from origin.
The given inequality is y 
Putting (0, 0) in the given inequation, we have 0 ≥ 2 which is false.
∴ Half plane of is always from origin.

Let (x)be any point lying in the equation x + y = 4
x₁+y₁=4....(i)
Distance of the point (x₁, y₁) from the equation 4x + 3y = 10
\(\frac { 4{ x }_{ 1 }+3{ y }_{ 1 }-10 }{ \sqrt { ({ 4 })^{ 2 }+(3)^{ 2 } } } \)=1
\(\left| \frac { 4{ x }_{ 1 }+3{ y }_{ 1 }-10 }{ 5 } \right| \)=1
4x₁+3y₁-10=±5
Taking 4x₁+3y₁-10=5
⇒ 4x₁+3y₁=15    ........(ii)
From equation (i) we get y₁=4-x₁
Putting the value of y1 in equation (ii) we get
4x₁+3(4-x₁)=15
⇒ 4x₁+12-3x₁=15
⇒ x₁+12=15
x₁=3 and y₁=4-3=1
So, the required point is (3, 1)
Now taking(-) sign, we have
4x₁+3y₁-10=-5
⇒ 4x₁+3y₁=5.....(iii)
From equation (i) we get y1=4-x1
⇒ 4x₁+3(4-x₁)=5
⇒ 4x₁+12-3x₁=5
⇒ x₁=5-12=-7
and y₁=4-(-7)=11
So, the required point is (- 7, 11)
Hence, the required points on the given line are (3, 1) and (-7, 11).

LHS = cos A cos 2A cos 2²A cos2³A ... cos 2^n-1A
\(=\frac{1}{2sinA}2sin\ A\ cos\ A\ cos\ 2A\ cos2^2\ Acos2^3A....cos2^{n-1}A\)
\(=\frac{1}{2sin\ A}sin\ 2A\ cos\ 2A\ cos2^2A\ cos\ 2^3A....cos\ 2^{n-1}A\)
\(=\frac{1}{2^2\ sinA}sin\ 4A\ cos\ 2^2A\ cos\ 2^3A...cos2^{n-1}A\)
Continuing the process, we get
\(=\frac{sin2^nA}{2^nsinA}\)

In a pack of 52 cards, there are four aces
3 aces will be selected from 4 cards and remaining 2 cards will be selected from rest of 48 cards.
∴ Number of ways of selecting 3 aces from 4 aces = 4C₃
Number of ways of selecting 2 cards out of 48 cards = 48C₂
∴ Required number of ways = 4C₃ \(\times \) 48C₂
= \(4\times \frac { 48\times 47 }{ 2\times 1 } \)     [∵ nC_r = 1C_n-r]
= \(2\times 48\times 47\)
= 4512

Draw graph of the cost as x goes from 0 to 50 copies.

Since x and y represent the number of copies and the cost of photocopying,
\(y=1.50x,\ 0\le x\le 10\)   [Given]
For x > 10,
y = 10 (1.50) + (x - 10) ...(1)
[ First 10 copies 1 Re and the remaining (x - 10) ]
= 15 + x - 10
= x + 5
\(\begin{matrix} \therefore \quad y \\ \therefore \quad y \end{matrix}=\begin{cases} x+5\ if\quad x>10 \\ 1.50x,\quad 0\le x\le 10 \end{cases}\)
\(x+5 \ if \ x>10\)
(ii) Find the cost of making 40 copies
Cost of making 40 copies which is greater than 10
y = x + 5
when x = 40
y = 40 + 5
y = Rs. 5

(i) Find the linear relationship between C and F.
By the given data
x₁ (100°C)  y₁(212°F)
x₂(0°C) y₂(32°F)
Using two point form, the linear relationship between C and F is
\(\frac { y-{ y }_{ 1 } }{ { y }_{ 2 }-{ y }_{ 1 } } =\frac { x-{ x }_{ 1 } }{ { x }_{ 2 }-{ x }_{ 1 } } \)
\(\frac { y-212 }{ 32-212 } =\frac { x-100 }{ 0-100 } \)
\(\Rightarrow \quad \frac { y-212 }{ -180 } =\frac { x-100 }{ -100 } \)
\(\Rightarrow \quad \frac { y-212 }{ 9 } =\frac { x-100 }{ 5 } =\frac { 5 }{ 9 } (y-212)=x-100\)
\(\Rightarrow \quad x=\frac { 5 }{ 9 } (y-212)+100\quad \Rightarrow \quad x=\frac { 5 }{ 9 } y-\frac { 5 }{ 9 } \times 212+100\)
\(\Rightarrow \quad x=\frac { 5 }{ 9 } y-118+100\quad \Rightarrow x=\frac { 5 }{ 9 } y-18\)
\(\Rightarrow \quad x=\frac { 5 }{ 9 } (y-32) \Rightarrow C=\frac { 5 }{ 9 } (F-32)\quad .....(1)\)
[\(\because \)  x represents Celsius and y represents Fahrenheit]
Which is the required relationship between C and F.
(ii) Find the value of C for 98.6°F and
Find C when F = 98.6° F
Substituting F = 98.6° in (1) we get,
C = \(\frac{5}{9}(98.6-32)=\frac{5}{9}(66.6)=\frac{333}{9}=37°\)
(iii) Find the value of F for 38°C.
Substituting C = 38° in (1) we get,
38=\(\frac{5}{9}(F-32)\)
⇒ \(\frac{342}{5}+32\) = F
⇒ F = 100.4°C

Let = 6 + 66 + 666 + ... upto n terms
= 6 (I + 11 + 111+ ....) upto n terms
\(={6\over9}(9+99+999+ ...)\) upto n terms
\(={63\over 6}[(10 -1) + (10^2-1) + (10^3 -1) + ...]\) upto n terms
\(={6\over 9}[(10+ 10^2 + 10^3+ ...) - (1+ 1+1...)]\) upto n terms​​​​​​​
\(={6\over9}\left[{ 10(10^n-1)\over 10-1}n\right]\)[In a G.P with a = 10 r = 10, \(S_n={(r^n-1)\over r-1}\)]
\(={6\over9}\left[{ 10(10^n-1)\over 10-1}n\right]={6\over9}\left[ 10(10^n-1)-9n\over9\right]\)
\({ S }_{ n }=\frac { 6 }{ 81 } \left[ 10\left( { 10 }^{ n }-1 \right) -9n \right] \)

LHS = cos (A + B) cos C - cos (B + C) cos A
= (cos A cos B - sin A sin B) cos C - cos A (cos B cos C - sin B sin C)
= cos A cos B cos C - sin A sin B cos C - cos A cos B cos C + cos A sin B sin C
= cos A sin B sin C - sin A sin B cos C
= sin B (cos A sin C - sin A cos C)
= sin B (sin C cos A - cos C sin A)
= sin B sin (C - A)
= RHS
Hence proved.

(i) \(y=-x^{ \left( \frac { 1 }{ 3 } \right) }\)

\(Let \ y=-x^{ ^{ \frac { 1 }{ 3 } } }\)
\(Then \ y=-x^{ ^{ \frac { 1 }{ 3 } } }\) is the reflection of the graph of \(y=x^{ ^{ \frac { 1 }{ 3 } } }\)  about the x-axis.
(ii) \(y=x^{1\over 3}+1\)

Let \(y=x^{ ^{ \frac { 1 }{ 3 } } }\)
Then \(y=x^{ ^{ \frac { 1 }{ 3 } } }+1\) is the x graph of \(y=x^{ ^{ \frac { 1 }{ 3 } } }\) shifts to the upward for one unit
(iii) \(y=x^{1\over 3}-1\)

Let \(y=x^{1\over 3}\)
Then \(y=x^{1\over 3}\)-1 is the graph of \(x^{ ^{ \frac { 1 }{ 3 } } }\) shifts to the downward for one unit.
(iv)  \(y=(x+1)^{1\over 3}\)

\(y=(x+1)^{1\over 3}\) causes the graph of \({x}^{\frac{1}{3}}\), shifts to the left for one unit.

Given P(A) = \(\frac { 1 }{ 2 } \), P(B) = \(\frac { 7 }{ 12 } \) and P(\(\bar { A } \cup \bar { B } \)) = \(\frac { 1 }{ 4 } \).
Now, \(P(\bar { A } \cup \bar { B } )=P(\overline { A\cap B } )=1-P(A\cap B)\)
\(\Rightarrow \frac { 1 }{ 4 } =1-P(A\cap B)\quad \Rightarrow P(A\cap B)=1-\frac { 1 }{ 4 } =\frac { 3 }{ 4 } \)
Now P(A) \(\times\) P(B) = \(\frac { 1 }{ 2 } \times \frac { 7 }{ 12 } =\frac { 7 }{ 24 } \)
\(\therefore P(A\cap B)\neq P(A)\times P(B)\)
Thus, A and B are not independent.

\(Given,\quad \lim _{ x\rightarrow 2 }{ \frac { { x }^{ n }-{ 2 }^{ n } }{ x-2 } } =80\)
\( \Rightarrow n.{ 2 }^{ n-1 }=80\)
By trial method, put n 5
\(\Rightarrow 5({ 2 }^{ 5-1 })=80\)
\(\Rightarrow 5({ 2 }^{ 4 })=80\)
\( \Rightarrow 5(16)=80\)
\( \Rightarrow 80=80\)
\(\therefore n=5\)

\(\int {1\over \sqrt{1-25 x^2}} dx=\int {1\over \sqrt{1-(5x)^2}}dx={1\over5}sin^{-1}(5x)+c\)

We have, y = \({1\over x}=x^{-1}\)
y' = -1x^-2 = -\({1\over x^2}\)
y'' = (-1)(-2)x^-3 \(={(-1)^22 !\over x^3}\)
and y"' = (-1)(-2)(-3)x^-4 \(={(-1)^33!\over x^4}\)

Limit of f(x) as x approaches 2 is the nature of (x) on both sides of 2.
It is independent ofthe nature of f(x) at x = 2.
Therefore we can not conclude anything about \(lim_{x \rightarrow 2}f(x) from f(2)=4\)

When atleast 3 girls are included, then
Number of ways = ⁴C₃ \(\times\)⁷C₂ + ⁴C₄ \(\times\)⁷C_I
\(=4\times {7\times6\over2\times1}+1\times7=84+7=91ways\)
Hence the required number of ways are 91 ways

Let the G.P. be a, ar, ar² ,...
Given T₁ = 3₁ T_n = 96 and S_n = 189
T_n = 96 ⇒ a.r^n-1 = 96
Also, S_n = 189 ⇒ \(\frac{a(1-r^{n})}{1-r}=189\)
⇒ \(\frac{a-a.r^{n-1}.r}{1-r}=189\)
⇒ \(\frac{3-(96)r}{1-r}=189\)       [∵a = 3 and ar^n-1 = 96]
⇒ 3 - 96r = 186(1-r)
⇒ 189-189r = 3 - 96r
⇒ 189 - 3 = 189r - 96r
⇒ 186 = 93r
⇒ r = 2
Substituting r = 2 in (1) we get,
a.r^n-1 = 96 ⇒ 3.2^n-1= 96
⇒ 2^n-1= 32 ⇒ 2^n-1 = 2⁵
⇒ n-1 = 5
⇒ n = 6

(A\(\times\) B) = {(1,4) (1,5) (1,6) (1,7) (2,4) (2,5) (2,6) (2,7) (3,4) (3,5) (3,6) (3,7)}
(B\(\times\)A) = {(4,1) (4,2) (4,3) (5,1) (5,2) (5,3) (6,1) (6,2) (6,3) (7,1) (7,2) (7,3)}
LHS = (A\(\times\)B)\(\cap \)(B\(\times\)A) = { }....(1)
(A\(\cap \)B) = { }, (B\(\cap \)A) = { }
\(\therefore\) RHS = (A\(\cap \)B) \(\times\) (B\(\cap \)A) = { }.....(2)
From (1) and (2), LHS = RHS

\(n(S) =2^{14}=16 \)
\(A =\left\{\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}|,| \begin{array}{ll} 0 & 1 \\ 1 & 0 \end{array}|,| \begin{array}{ll} 1 & 1 \\ 0 & 1 \end{array}|,| \begin{array}{ll} 1 & 0 \\ 1 & 1 \end{array}|,| \begin{array}{ll} 0 & 1 \\ 1 & 1 \end{array}|,| \begin{array}{ll} 1 & 1 \\ 1 & 0 \end{array} \mid\right\} \)
\(n(A) =6 \)
\(P(A) =\frac{6}{16}=\frac{3}{8} \)

\(\int \frac{\sec x}{\sqrt{\cos 2 x}} d x =\int \frac{\sec x}{\sqrt{\cos ^{2} x-\sin ^{2} x}} d x \)
\(=\int \frac{\sec x}{\sqrt{\cos ^{2} x\left(1-\frac{\left.\sin ^{2} x\right)}{\cos ^{2} x}\right)} d x} \)
\(=\int \frac{\sec x}{\cos x \sqrt{1-\tan ^{2} x}} d x \)
\(=\int \frac{1}{\sqrt{1-\tan ^{2} x}} \times \sec ^{2} x d x \)
\(=\int \frac{1}{\sqrt{1-u^{2}}} \times d u, u=\tan x \)
\(=\sin ^{-1} u+c \)
\(=\sin ^{-1}(\tan x)+c \)

\(y=f\left(x^{2}+2\right) \)
\(\frac{d y}{d x} =f^{\prime}\left(x^{2}+2\right)(2 x) \)
\(\text { At } x =1, \frac{d y}{d x}=f^{\prime}(1+2)(2)=f^{\prime}(3)(2) \)
\(=5(2)=10 \)

\(\lim _{x \rightarrow \infty}\left(\frac{x^{2}+5 x+3}{x^{2}+x+3}\right)^{x} =\lim _{x \rightarrow \infty}\left(\frac{x^{2}+5 x}{x^{2}+x}\right)^{x} \)
\(=\lim _{x \rightarrow \infty}\left(\frac{1+\frac{5}{x}}{1+1 / x}\right)^{x}=\frac{e^{5}}{e}=e^{4} \)

\(A=\left[\begin{array}{ll} a_{11} & a_{12} \\ a_{21} & a_{22} \end{array}\right] \)
\(a_{11}=\frac{1}{2}(3-2)=\frac{1}{2} ; a_{12}=\frac{1}{2}(3-4)=\frac{-1}{2} \)
\(a_{21}=\frac{1}{2}(3(2)-2)=\frac{4}{2}=2 ; a_{22}=\frac{1}{2}(6-4)=\frac{2}{2}=1 \)
\(\therefore A=\left[\begin{array}{ll} \frac{1}{2} & -\frac{1}{2} \\ 2 & 1 \end{array}\right] \)

\(x^{2}+a x+c=0 \)
\(x^{2}+d x+b=0 \)
\(8 \& 2 \text { are the roots }\)\(\text { 3. } 3 \text { are the roots }\)
\(\therefore a=-10 ; c=16 \quad d=-6, \quad b=9\)
\(x^{2}+a x+b =0 \)
\(x^{2}-10 x+9 =0 \)
\(\Rightarrow(x-1)(x-9) =0 \)
\(\therefore x =1 \text { (or) } 9 \)

\(\sqrt{2+2 \cos 4 \theta} =\sqrt{2+2\left(2 \cos ^{2} 2 \theta-1\right)} \)
\(=\sqrt{4 \cos ^{2} 2 \theta} \)
\(=2 \cos 2 \theta \)
\(\sqrt{2+\sqrt{2+2 \cos 4 \theta}} =\sqrt{2+2 \cos 2 \theta} \)
\(=\sqrt{2+2\left(2 \cos ^{2} \theta-1\right)} \)
\(=\sqrt{4 \cos ^{2} \theta} \)
\(=\pm 2 \cos \theta \)
\(\Rightarrow \pi<2 \theta<\frac{3 \pi}{2} \)
\(\Rightarrow \frac{\pi}{2}<\theta<\frac{3 \pi}{4} \text { in II quadrant. } \)
\(\therefore \sqrt{2+\sqrt{2+2 \cos 4 \theta}}=-2 \cos \theta\)

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

\(|A|=\left|\begin{array}{ccc} 2 & -3 & 5 \\ 6 & 0 & 4 \\ 1 & 5 & -7 \end{array}\right|\)
\(=2(0-20)+3(-42-4)+5(30-0)\)
\(=2(-20)+3(-46)+5(30)\)
\(=-40-138+150\)
\(=-28 \neq 0\)
\(|A| \neq 0\)
\(\therefore\) A is non singular