Let A and B the events of drawing a black card in first draw and second draw respectively.
In the first draw, there are 26 black cards out of 52 cards.
\(\therefore P(A)=\frac { 26 }{ 52 } =\frac { 1 }{ 2 } \)
In the second draw, there are 25 black cards out of 51 cards.
\(\therefore P(B/A)=\frac { 25 }{ 51 } \)
\(\therefore P(A\cap B)=P(A).P(B/A)=\frac { 1 }{ 2 } \times \frac { 25 }{ 51 } =\frac { 25 }{ 102 } \) 

\(\lim _{ x\rightarrow 1 }{ \frac { 1+(x-1{ ) }^{ 2 } }{ 1+{ x }^{ 2 } } } =\frac { 1+0 }{ 1+{ 1 }^{ 2 } } =\frac { 1 }{ 2 } \)

\(\int \frac{1}{(2-3 x)^4} d x=\int(2-3 x)^{-4} d x \)
\(=-\frac{1}{3} \frac{(2-3 x)^{-4+1}}{-4+1}\)
\(=\frac{1}{9} \cdot \frac{1}{(2-3 x)^3}+c \)

\(\int { { e }^{ x } } dx\) = e^x + c

\(lim_{x\rightarrow3}(4-x)\)

At x = 3, the value of the curve on y-axis is 1.
\(\therefore lim_{x\rightarrow3}(4-x)=1\)

\(\overrightarrow{a}.\overrightarrow{b}\)= (\(\hat{i}-\hat{j}+5\hat{k}\) ).(\(3\hat{i}-2\hat{k}\)) = (1)(3) + (1)(0) + (5)(2)
= 3 - 10 = -7

(3x +y) (x - 3y) = 0 \(\Rightarrow\) 3x²- 8xy - 3y² = 0

Given a, b, c are in A.P
\(\Rightarrow b=\frac { a+c }{ 2 } \)
RHS = 4[b² - ac]
\(=4\left[ { \left( \frac { a+c }{ 2 } \right) }^{ 2 }-ac \right] =4\left[ { \left( \frac { a+c }{ 4 } \right) }^{ 2 }-ac \right] \)
\(=4\left[ \frac { { \left( a+c \right) }^{ 2 }-4ac }{ 4 } \right] ={ a }^{ 2 }+{ c }^{ 2 }+2ac-4ac\)
= a² + c² - 2ac
= (a - c)² = LHS
Hence proved.

Given nP₄ = 20 \(\times\)nP₂.

\(\Rightarrow\) (n-2) (n-3) = 20 \(\Rightarrow\) n²-5n+6 = 20
\(\Rightarrow\) n²-5n-14 = 0
\(\Rightarrow\) (n-7)(n+2) = 0
\(\Rightarrow\) n = 7 or -2 \(\Rightarrow\) n = 7

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.

30⁰
30⁰ = 30 \(\times\) \(\frac { \pi }{ 180 } =\frac { \pi }{ 6 } \)

The set of all positive roots of the equation (x-1)(x+1)(x²-1)=0.
Let B = { the set of positive roots of the equation (x-1)(x+10(x²-1)=0}
\(\Rightarrow\) x = 1, -1
B = {1}.

\({ x }^{ 2 }+2xy+{ y }^{ 3 }=42\)
Differentiating both sides with respect to 'x' we get,
\(2x+2\left[ x.\frac { dy }{ dx } +y(1) \right] +3{ y }^{ 2 }\frac { dy }{ dx } =0 \Rightarrow 2x+2x\frac { dy }{ dx } +2y+3{ y }^{ 2 }\frac { dy }{ dx } =0\)
\(\Rightarrow \frac { dy }{ dx } (2x+3{ y }^{ 2 })=-2x-2y \Rightarrow \frac { dy }{ dx } =\frac { -2\left( x+y \right) }{ 2x+3{ y }^{ 2 } } \)

\( \int \frac{\sin ^2 x}{1+\cos x} d x =\int \frac{1-\cos ^2 x}{1+\cos x} d x \)
\(=\int \frac{(1-\cos x)(1+\cos x)}{(1+\cos x)} d x \)
\(=\int(1-\cos x) d x \)
\(=x-\sin x+c \)

\(y={log \ x \over e^x }=e^{-x}.log \ x\)
\({dy\over dx}=e^{-x}({1\over x})+log \ x(e^{-x})(-1)\)
\(=e^{-x}[{1\over x}-log \ x]\).

\(lim_{x\rightarrow 0}{e^{ax}-e^{bx}\over x}\)\(=lim_{x\rightarrow 0}{e^{ax}-1-e^{bx}+1\over x}=lim_{x\rightarrow 0}{e^{ax}-1-(e^{bx}-1)\over x}\)
\([lim_{ax\rightarrow 0}{e^{ax}-1\over ax}\times a]-[lim_{bx\rightarrow0}{e^{bx}-1\over bx}\times b]\)
\(=log \ a -log \ b=log({a\over b})\)

Let f(x) = sin²x - 2 cos²x - cos x
f(-x) = f(x) [since sin(-x) = -sin x and cos(-x) = cos x]
Thus f(x) is even.

Let the G.P. be a, ar, ar², .. with |r|<1 ....(1)
Given a + ar + ar²+...= 23 ...(2)
and a²+(ar)²+(ar²)²+....= 69    ..(3)
\((1)\Rightarrow{a\over1-r}=23\Rightarrow a=23(1-r)\)
\((2)\Rightarrow{a^2\over 1-r^2}=69\)
\(\Rightarrow\) a² = 39 (1-r²)
\(\Rightarrow\) 23²(1-r)² = 69(1-r²)    [from (3)]
\(\Rightarrow\) 23² (1-r)² = 3(1+r)(1-r)
\(\Rightarrow\) 23(1-r) = 3(1+r)
\(\Rightarrow\) 23-23r = 3 + 3r
\(\Rightarrow\)  \(20=26r\Rightarrow r={20\over60}={10\over13}.\)
From (3) \(a=23\left(1-{{10}\over{13}} \right)=23\left({3\over 13} \right)={69\over 13}\)
\(\therefore\)The G.P is \(\frac { 69 }{ 13 } ,\frac { 69 }{ 13 } \left( \frac { 10 }{ 13 } \right) ,\frac { 69 }{ 13 } \left( \frac { 10 }{ 13 } \right) ^{ 2 },..\)

Given pair of lines is ax² + 2hxy + by² = 0
Since the lines passes through the origin, let the equation of one of the line by y = mx.
Slope of the other line is Km (given)
\(\therefore\) Equation of the second line is y = Kmx.
\(\therefore\) m₁ = m and m₂ = m
We know that sum of the slopes
m₁ + m₂ = \(-\frac{2h}{b}\)
\(\therefore\) m + Km = \(-\frac{2h}{b}\)
\(\Rightarrow\)   m (1 + K) = \(-\frac{2h}{b}\) \(\Rightarrow\) m = \(\frac{-2h}{b(1+K)}\)
Also product of the slopes
m₁ m₂ = \(\frac{a}{b}\) \(\Rightarrow\) m.Km = \(\frac{a}{b}\)
\(\Rightarrow\)  Km² = \(\frac{a}{b}\)
\(\Rightarrow\)   m² = \(\frac{a}{Kb}\)
  From (1), \({ m }^{ 2 }={ \left[ -\frac { 2h }{ b\left( 1+K \right) } \right] }^{ 2 }=\frac { 4{ h }^{ 2 } }{ { b }^{ 2 }{ \left( 1+K \right) }^{ 2 } } \)
Equation (2) and (3) we get,
 \(\frac { a }{ Kb } =\frac { { 4h }^{ 2 } }{ { b }^{ 2 }{ \left( 1+K \right) }^{ 2 } } \)
 \(\Rightarrow \frac { a }{ K } =\frac { { 4h }^{ 2 } }{ b{ \left( 1+K \right) }^{ 2 } } \)
\(\Rightarrow\) 4Kh² = ab (1 + K)² Hence proved.

The family of equations of straight lines is of the form (a₁ x + b₁ y + c₁) + \(\lambda\)(a₂ x + b₂ y + c₂) = 0
(i) That is (4x - y + 3) + \(\lambda\)(5x + 2y + 7) = 0....(1)
Since the required equation passes through the point (-1, 2), the point satisfies equation (1),
\(\therefore\)  (-4 - 2 + 3) + \(\lambda\)(-5 + 4 + 7) = 0
\(\Rightarrow\) (-3) + \(\lambda\)(6) = 0
\(\Rightarrow\) 6\(\lambda\) = 3
\(\Rightarrow\)\(\lambda\) = 2
Substituting \(\lambda\) = 2 in (1) we get,
(4x - y + 3) + 2(5x + 2y + 7) = 0
\(\Rightarrow\)4x - y + 3 + 10x + 4y + 14​​​​​​​ = 0
\(\Rightarrow\) 14x + 3y + 17​​​​​​​ = 0
(ii) Any line perpendicular to x - 2y + 1 = 0 will be of the form 2x + y + k = 0
Since this line passes through (-1,-1)
2 (-1) -1 + k = 0
\(\Rightarrow\) 2 - 1 + k = 0
\(\Rightarrow\) -3 + k = 0
\(\Rightarrow\) k = 3
\(\therefore\)  Equation of the required line is 2x + y + 3 = 0
(ii) Parallel to x - y + 5 = 0....(1)
Given lines are   4x - y + 3 = 0 and...(2)
5x + 2y + 7.....(3)
\(\begin{matrix} (2)\times 2\Rightarrow 8x-2y+6=0 \\(3)\Rightarrow \ \ \ \ \ 58x+2y+7=0 \\ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_\_ \_ \_ \_ \_ \_ \_ \_ \_\_ \_ \_ \_\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 13x\ \ \ +13=0 \end{matrix}\\\Rightarrow x=-1\)
Substituting x = -1 in (2) we get,
-4x - y + 3 = 0
\(\Rightarrow\) -1 - y = 0
\(\Rightarrow\) -1 = +y
\(\Rightarrow\) y = -1
\(\therefore\) (-1, -1) is the point of intersection of the given lines.
Slope of the required line is \(m=-\frac { Co-efficient\quad of\quad x }{ Co-efficient\quad of\quad y } =\frac { -1 }{ -1 } =1\)
\(\therefore\) Equation of the line passing through (-1, -1) with slope 1 is
y + 1 = 1(x + 1)        [ y - y₁ = m(x - x₁) ]
\(\Rightarrow\) y + 1 = x + 1  
\(\Rightarrow\)  x - y = 0 
\(\Rightarrow\) x - y = 0
(iii) Any line perpendicular to x - 2y + 1 = 0 will be of the form 2x + y + k = 0
Since this line passes through (-1, -1)
2 (-1) -1 + k = 0
\(\Rightarrow\) 2 - 1 + k = 0
\(\Rightarrow\) -3 + k = 0
\(\Rightarrow\) k = 3
\(\therefore\)  Equation of the required line is 2x + y + 3 = 0

(i) Since f, g, h are functions from R \(\rightarrow\) R,
(f + g)o h: R \(\rightarrow\) R and f o h + g o h: R\(\rightarrow\) R. For any x \(\in \) R,
[(f + g)oh](x) = (f + g)(h(x) - f(h(x)) + g(h(x)) = f o h(x) + g o h(x)
(f + g)o h = f o h + g o h
(ii) Also fo(g + h) = f[(g + h)(x)] for any x \(\in \)R
= f[g(x) + h(x)] = f(g(x) + f(h(x)) = f o g(x) + f o h(x)
\(\therefore\) fo(g + h) = fog(x) + f o h(x).

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.

Given f is differentiable
\(\therefore f^{\prime}\left(1^{-}\right)=f^{\prime}\left(1^{+}\right)=1 \)
\(f^{\prime}\left(1^{-}\right) =\lim _{x \rightarrow 1^{-}} \frac{f(x)-f(1)}{x+1}=\lim _{x \rightarrow 1^{-}} \frac{\left(a x^{2}-b\right)-(a-b)}{x+1} \)
\(=\lim _{x \rightarrow 1^{-}} \frac{a x^{2}-b-a+b}{x+1}=\lim _{x \rightarrow 1^{-}} \frac{a\left(x^{2}-1\right)}{x+1} \)
\(=\lim _{x \rightarrow 1^{-}}-a(x+1) \)
\(=a(1+1)=2 a \)
\(\therefore f^{\prime}\left(1^{+}\right) =\lim _{x \rightarrow 1^{+}} \frac{f(x)-f(1)}{x-1}=\lim _{x \rightarrow 1^{+}} \frac{\frac{1}{x}-1}{x-1} \)
\(=\lim _{x \rightarrow 1^{+}} \frac{1-x}{x(x-1)}=\lim _{x \rightarrow 1^{+}} \frac{-1}{x}=-1\)
\(\therefore 2 a =-1 \)
\(a =\frac{-1}{2} \)
\(\text { and } f(1)=1\)
\(a-b=1 \)
\(-1 / 2-1=b \)
\(b=-3 / 2 \)

\(\lim _{x \rightarrow \pi / 2} \frac{2 x-\pi}{\cos x} =\lim _{x \rightarrow \pi / 2} \frac{2 x-\pi}{\sin \left(\frac{\pi}{2}-x\right)} \)
\(=\lim _{\left(\frac{\pi}{2}-x\right) \rightarrow 0} \frac{-2\left(\frac{\pi}{2}-x\right)}{\sin \left(\frac{\pi}{2}-x\right)}=-2 \)

\(2 \vec{a}=\vec{b}+\vec{c} \Rightarrow \vec{a}+\vec{a}=\vec{b}+\vec{c} \Rightarrow \vec{a}-\vec{b}=\vec{c}-\vec{a} \)
\(\overrightarrow{O A}-\overrightarrow{O B}=\overrightarrow{O C}-\overrightarrow{O A} \Rightarrow \overrightarrow{B A}=\overrightarrow{A C} \)
\(\Rightarrow \vec{a}, \vec{b}, \vec{c} \text { are collinear }\)

\(-1 \leq x<0 \Rightarrow\lfloor x\rfloor=-1 ;\)
\(0 \leq y<1 \Rightarrow\lfloor y\rfloor=0 ; 1 \leq z<2 \Rightarrow\lfloor z\rfloor=1 \)
\(\therefore|\cdot A|=\left|\begin{array}{ccc} -1+1 & 0 & 1 \\ -1 & 0+1 & 1 \\ -1 & 0 & 1+1 \end{array}\right|=\left|\begin{array}{ccc} 0 & 0 & 1 \\ -1 & 1 & 1 \\ -1 & 0 & 2 \end{array}\right|=1(0+1)=1 \)
\(|A|=1=\lfloor z\rfloor \)

\(\text {Perimeter }=4+2 \sqrt{2}\)
\(a+a+\sqrt{2} =4+2 \sqrt{2} \)
\(2 a+\sqrt{2} a =4+2 \sqrt{2} \)
\(\therefore a =2 \)
\(\text {Equation of line is } \frac{x}{2}+\frac{y}{2}=1\)
\(x+y-2=0\)

\(\text { Number of points of intersection }={ }^{10} \mathrm{C}_{2}\)
\(=\frac{10 \times 9}{1 \times 2}=45\)

\(\frac{1}{\sqrt{1}+\sqrt{3}} =\frac{1}{\sqrt{3}+\sqrt{1}} \times \frac{\sqrt{3}-1}{\sqrt{3}-1}=\frac{\sqrt{3}-1}{2} \)
\(\frac{1}{\sqrt{3}+\sqrt{5}} =\frac{1}{\sqrt{5}+\sqrt{3}} \times \frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}-\sqrt{3}} \)
\(=\frac{\sqrt{5}-\sqrt{3}}{2} \)
\(\text { Sum to } \mathrm{n} \text { terms }=\frac{(\sqrt{3}-1)}{2}+\frac{(\sqrt{5}-\sqrt{3})}{2}+\ldots . .\left(\frac{\sqrt{2 n+1}-\sqrt{2 n-1}}{2}\right)\)
\(=\frac{\sqrt{2 n+1}-1}{2}\)

\(\text { Let } \log _{3} \frac{1}{81}=x \Rightarrow 3^{x}=\frac{1}{81}\)
\(\Rightarrow 3^{x}=3^{-4}\)
\(\Rightarrow x=-4 \)

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

Let I = \(\int { \frac { { x }^{ 3 }dx }{ { x }^{ 4 }+{ 3x }^{ 2 }+2 } } \) = \(\int { \frac { { x }^{ 2 }.xdx }{ \left( { x }^{ 2 } \right) ^{ 2 }+{ 3x }^{ 2 }+2 } } \)
Let t = x² \(\Rightarrow\) dt = 2x dx \(\Rightarrow\) \(\frac { dt }{ 2 } \) = x dx
\(\therefore\) I = \(\int { \frac { t.\frac { dt }{ 2 } }{ { t }^{ 2 }+3t+2 } =\frac { 1 }{ 2 } \int { \frac { tdt }{ { t }^{ 2 }+3t+2 } } }\)
Now t = A \(\frac { d }{ dx } \) (t² + 3t + 2) + B \(\Rightarrow\) t = A (2t  + 3)  + B
Equating the x - term constant terms we get,
1 = 2A \(\Rightarrow\) A = \(\frac { 1 }{ 2 } \) 0 = 3A + B \(\Rightarrow\) 0 = \(\frac { 3 }{ 2 } \) + B\(\Rightarrow\) B = \(-\frac { 3 }{ 2 } \);    \(\therefore\) t = \(\frac { 1 }{ 2 } \)(2t + 3) -\(\frac { 3 }{ 2 } \)
I = \(\int { \frac { tdt }{ { t }^{ 2 }+3t+2 } } =\frac { 1 }{ 2 } \left[ \int { \frac { 2t+3 }{ 2\left( { t }^{ 2 }+3t+2 \right) } dt-\frac { 3 }{ 2 } \int { \frac { dt }{ { t }^{ 2 }+3t+2 } } } \right] \)
\(\Rightarrow\) I = \(\frac { 1 }{ 4 } \left[ log\left| { t }^{ 2 }+3t+2 \right| -3\int { \frac { dt }{ { t }^{ 2 }+3t+\frac { 9 }{ 4 } -\frac { 9 }{ 4 } +2 } } \right] =\frac { 1 }{ 4 } \left[ log\left| { t }^{ 2 }+3t+2 \right| -3\int { \frac { dt }{ \left( t+\frac { 3 }{ 2 } \right) -\left( \frac { 1 }{ 2 } \right) ^{ 2 } } } \right] \)
Now I₁ = \(\int { \frac { dt }{ { t }^{ 2 }+3t+2 } } =\int { \frac { dt }{ { t }^{ 2 }+3t+\left( \frac { 3 }{ 2 } \right) ^{ 2 }-\left( \frac { 3 }{ 2 } \right) ^{ 2 }+2 } } \) = \(3\frac { 1 }{ 4 } \left[ log\left| { t }^{ 2 }+3t+2 \right| -3log\left| \frac { t+\frac { 3 }{ 2 } -\frac { 1 }{ 2 } }{ t+\frac { 3 }{ 2 } +\frac { 1 }{ 2 } } \right| \right] \)
I = \(\frac { 1 }{ 4 } log\left| { t }^{ 2 }+3t+2 \right| -3log\left| \frac { t+1 }{ t+2 } \right| +c\)
\(\therefore\) I = \(\frac { 1 }{ 4 } log\left| { x }^{ 4 }+3{ x }^{ 2 }+2 \right| -3log\left| \frac { { x }^{ 2 }+1 }{ { x }^{ 2 }+1 } \right| +c\) 

Consider the following events.
E_1: I purse is chosen
E2: II purse is chosen
A: Coin pulled out is silver
\(\Rightarrow \quad \therefore P(E_{ 1 })=P({ E }_{ 2 })=\frac { 1 }{ 2 }\)
There are 3 silver and 4 copper coins in I purse
\(\Rightarrow \quad P(A/{ E }_{ 1 })=\frac { 3 }{ 7 } \)
There are 4 silver and 3 copper coins in II Purse
\(\therefore\) P(A/E₁) =  \(\frac { 4 }{ 7 } \)
By the theorem of total probability
P(A) = P(E₁).P(A/E₁)+P(E₂).P(A/E₂) = \(\frac { 1 }{ 2 } \times \frac { 3 }{ 7 } +\frac { 1 }{ 2 } \times \frac { 4 }{ 7 } =\frac { 3 }{ 14 } +\frac { 4 }{ 17 } =\frac { 7 }{ 14 } =\frac { 1 }{ 2 } \)

\(f\left( x \right) =\begin{cases} x{ e }^{ -\left( \frac { 1 }{ \left| x \right| } +\frac { 1 }{ x } \right) }=x{ e }^{ -\frac { 2 }{ x } },\quad x>0 \\ \quad \quad \ 0,\quad \quad \quad \quad \quad \quad \quad x=0\quad \\ x{ e }^{ -\left( -\frac { 1 }{ x } +\frac { 1 }{ x } \right) }=x,\quad \quad \quad \quad x<0 \end{cases}\)
\(\therefore f\left( x \right) =\begin{cases} x{ e }^{ -\frac { 2 }{ x } }\quad x>0 \\ 0,\quad \quad \quad x=0 \\ x,\quad \quad x<0 \end{cases}\)
\( \therefore f^{ ' }\left( { 0 }^{ - } \right) =\lim _{ x\rightarrow { 0 }^{ + } }{ \frac { f\left( x \right) -f\left( 1 \right) }{ x-0 } } =\lim _{ x\rightarrow { 0 }^{ + } }{ \frac { x-0 }{ x-0 } } =\lim _{ x\rightarrow { 0 }^{ - } }{ \frac { x }{ x } } =1\quad \left[ \because f\left( x \right) =x\quad for\quad x<0\quad and\quad f\left( 0 \right) =0 \right] \)
\(\therefore f^{ ' }\left( { 0 }^{ + } \right) =\lim _{ x\rightarrow { 0 }^{ + } }{ \frac { f\left( x \right) -f\left( 0 \right) }{ x-0 } } =\lim _{ x\rightarrow { 0 }^{ + } }{ \frac { x{ e }^{ -\frac { 2 }{ x } }-0 }{ x } } =\lim _{ x\rightarrow { 0 }^{ + } }{ { e }^{ -\frac { 2 }{ x } } } =0\quad \left[ \because f\left( x \right) =x{ e }^{ -2x }\quad for\quad x>0\quad and\quad f\left( 0 \right) =0 \right] \)
\(\therefore f^{ ' }\left( { 0 }^{ - } \right) \neq f^{ ' }\left( { 0 }^{ + } \right) \)
\(\therefore f\left( x \right) \)is not differentiate at x = 0.

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, \(P\left(A_1\right)=\frac{5}{10}=0.5 \)
\(P\left(A_2\right)=\frac{3}{10}=0.3 \)
\(P\left(A_3\right)=\frac{2}{10}=0.2 \)
\(P\left(X / A_1\right)=0.4 \)
\(P\left(X / A_2\right)=0.5\)
\(P\left(X / A_3\right)=0.3\)
\(P\left(A_2 / X\right)=\frac{P\left(A_2\right) P\left(X / A_2\right)}{P\left(A_1\right) P\left(X / A_1\right)+P\left(A_2\right) P\left(X / A_2\right)}+P\left(A_3\right) P\left(X / A_3\right)\)
\(=\frac{0.3(0.5)}{0.5(0.4)+0.3(0.5)+0.2(0.3)} \)
\(=\frac{0.15}{0.2+0.15+0.06} \)
\(=\frac{0.15}{0.41}=\frac{15}{41}\)

Given \(\vec{a}, \vec{b}\) and \(\vec{c}\)are unit vectors.
\(\Rightarrow \left| \vec { a } \right| =\left| \vec { b } \right| =\left| \vec { c } \right| =1\)
\(\vec{a}.\vec{b}=\vec{a}.\vec {c}\)= 0, and angle between \(\vec{b}\) and \(\vec{c}\) is \(\frac{\pi}{3}\)
\(\vec{a}.\vec{b}=\vec{a}.\vec {c}\)= 0 \(\Rightarrow \vec{a}\) is \({ \bot }^{ r }\) to both\(\vec{b}\) and \(\vec{c}\)
\(\vec{a}\) is \({ \bot }^{ r }\) to \(\vec { b } \times \vec { c } \Rightarrow \vec { a } =\lambda (\vec { b } \times \vec { c } )\) for some scalar \(\lambda\).
\(\therefore { \left| \vec { a } \right| }^{ 2 }={ \lambda }^{ 2 }{ \left| \vec { b } \times \vec { c } \right| }^{ 2 }\)
\(\Rightarrow 1={ \lambda }^{ 2 }[{ \left| \vec { b } \right| }^{ 2 }{ \left| \vec { c } \right| }^{ 2 }-{ (\vec { b } \times \vec { c } ) }^{ 2 }]\left[ \therefore \left| \vec { a } \right| =1{ and\left| \vec { a } \times \vec { b } \right| }^{ 2 } \right] ={ \left| \vec { a } \times \vec { b } \right| }^{ 2 }={ \left| \vec { a } \right| }^{ 2 }{ \left| \vec { b } \right| }^{ 2 }-{ (\vec { a } .\vec { b } ) }^{ 2 }]\)
\(\Rightarrow 1={ \lambda }^{ 2 }[(1)(1)-{ \left| \vec { b } \right| }^{ 2 }{ \left| \vec { c } \right| }^{ 2 }{ cos }^{ 2 }\frac { \pi }{ 3 } ]\) [\(\therefore\)angle between \(\vec{a}\)and\(\vec{b}\) is\(\frac{\pi}{3}\) ]
\(\Rightarrow 1={ \lambda }^{ 2 }[1-{ cos }^{ 2 }\frac { \pi }{ 3 } ][\because \left| \vec { b } \right| \left| \vec { c } \right| =1]\)
\(\Rightarrow 1={ \lambda }^{ 2 }[1-\frac { 1 }{ 4 } ]\Rightarrow 1={ \lambda }^{ 2 }\left( \frac { 3 }{ 4 } \right) \)
\(\Rightarrow 1={ \lambda }^{ 2 }\frac { 4 }{ 3 } \Rightarrow 1={ \lambda }\pm \frac { 2 }{ \sqrt { 3 } } \)
Substituting \(\)\({ \lambda }\pm \frac { 2 }{ \sqrt { 3 } } \) in (1) we get,
\(\vec { a } =\pm \frac { 2 }{ \sqrt { 3 } } (\vec { b } \times \vec { c } )\)

Gift pack matrix
\(A=\left[\begin{array}{ccc} \text { Pack I } & \text { PII } & \text { PIII } \\ 100 & 200 & 250 \\ 100 & 100 & 250 \\ 50 & 100 & 150 \end{array}\right] \begin{aligned} &\text { Cashew } \\ &\text { raisins } \\ &\text { almonds } \end{aligned}\)
Given Cost of 50 gm of cashew is Rs. 50
\(\therefore\)1 gm of cashew is Rs. 1
Cost of 50 gm of raisin is Rs. 10
1 gm of raisin is Rs. 1/5
Cost of 50gm of almonds is Rs. 60
1 gm of raisin is Rs. 6/5
Cost matrix \(B=\left(1, \frac{1}{5}, \frac{6}{5}\right)\)
Cost of package is AB.
\(A B=\left(\begin{array}{lll} 1 & \frac{1}{5} & \frac{6}{5} \end{array}\right)\left[\begin{array}{ccc} 100 & 200 & 250 \\ 100 & 100 & 250 \\ 50 & 100 & 150 \end{array}\right]\)
\(=\left[\begin{array}{c} 100+20+60 \\ 200+20+120 \\ 250+50+180 \end{array}\right]=\left[\begin{array}{c} 180 \\ 340 \\ 480 \end{array}\right]\)
Pack I Cost Rs. 180
Pack II Cost Rs. 340
Pack III Cost Rs. 480

x³+5x²-16x-14 = (x+7)(x²+px-2)
equating co-eff of x
7p-2 = -16
7p = -16+2 =-14
⇒ p = -2
so the other factor is x²-2x-2
Solving x²-2x-2 = 0
⇒ x =\(\frac { 2\pm \sqrt { 4-4(1)(-2) } }{ 2(1) } \)
= \(\frac { 2\pm \sqrt { 4+8 } }{ 2 } =\frac { 2\pm 2\sqrt { 3 } }{ 2 } =1\pm \sqrt { 3 } \)
so the roots\(1\pm \sqrt { 3 } ,-7\)

Given equation of the line is \(x\cos\theta+y\sin\theta=p\) .
Let C (h, k) be the mid point of the given line AB where it meets the two axis atA (a, 0) andB (0, b).
Since (a, 0) lies on eq (i) then \(a\cos\theta+0=p^n\)
\(\Rightarrow a=\frac{p}{\cos\theta}\) .....(ii)
B (0, b) also lies on the eq (i) then 0 + \(b\sin\theta\) = p
\(\Rightarrow b=\frac{p}{\sin\theta}\)  ...(iii)
Since C(h, k) is the mid point of AB
\(\therefore h=\frac{0+a}{2}\Rightarrow a=2h\) and \(k=\frac{b+0}{2}\Rightarrow b=2k\)
Putting the values of a and b is eq (ii) and (iii) we get
\(2h=\frac{p}{\cos\theta}\Rightarrow\cos\theta=\frac{p}{2k}\) .....(iv)
and \(2k=\frac{p}{\sin\theta}\Rightarrow\sin\theta=\frac{p}{2k}\) ....(v)
Squaring and adding eq (iv) and (v) we get
\(\Rightarrow \cos^2\theta+\sin^2\theta=\frac{p^2}{4h^2}+\frac{p^2}{4k^2}\Rightarrow 1=\frac{p^2}{4h^2}+\frac{p^2}{4k^2}\)
So, the locus of the mid point is \(1=\frac{p^2}{4x^2}+\frac{p^2}{4y^2}\)
\(\Rightarrow \)  4x²y² = p² (x² + y²)
Hence, the value of the filter is 4x²y² = p² (x² + y²).

\(\sqrt [ 3 ]{ x^3+7 } ={(x^3+7)}^{{1\over 3}}\)
\(={\left[ x^3\left( 1+{7\over x^3} \right) \right]}^{{1\over 3}}\)   (\(\left |{7\over x^3}\right |<1\) as x is large)
\(=x{\left( 1+{7\over x} \right)}^{1\over 3}\)
\(=x\left( 1+{1\over 3} \times {7\over x^3}+{{{1\over3}\left( {1\over 3}-1 \right)}\over{2!}} {\left( {{7\over x^3}} \right)}^{2} +......\right)\)
\(=x\left( 1+{7\over 3}\times{1\over x^3}-{49\over 9}\times{1\over x^6}+...... \right)\)
\(=x+{7\over 3}\times{1\over x^2}-{49\over 9}\times{1\over x^5}+...\)
\(\sqrt [ 3 ]{ x^3+4 } ={(x^3+4)}^{1\over 3}\)
\(={\left[ x^3\left( 1+{4\over x^3} \right) \right]}^{1\over 3}\)
\(=x\left( 1+{4\over x^3} \right)^{1\over 3}\)
\(=x{\left( 1+{1\over3}\times{4\over x^3}+{{1\over 3}\left( {1\over3}-1 \right)\over{2!}} {\left( {4\over x^3} \right)}^{2}+... \right)}^{1\over3}\)
\(=x+{4\over 3}\times{1\over x^3}-{16\over 9}\times{1\over x^5}+...\)
Since x is large, \({1 \over x}\) is very small and hence higher powers of \({1 \over x}\) are negligible.
Thus \(\sqrt [ 3 ]{ x^3+7 } =x+{7\over 3}\times{1\over x^2}\) and \(\sqrt [ 3 ]{ x^3+4 } =x+{4\over3}\times{1\over x^3}.\) Therefore
\(\sqrt [ 3 ]{ x^3+7 } -\sqrt [ 3 ]{x^3+4 } =\left(x+{7\over 3}\times{1\over x^2} \right)-\left( x+{4\over 3}\times{1\over x^2} \right)={1\over x^2}\)

The number of 4-digit numbers that can be formed using the given 5 digits is ⁵P₄ = 120. We first find the sum of the digits in the unit place of all these 120 numbers. By filling the 1 in unit place, the remaining three places can be filled with remaining 4 digits in ⁴P₃ = 24 ways. This means, the number of 4-digit numbers having 1 in units place is ⁴P₃ = 24. Similarly, each of the digits 2, 4, 6, 8 appear 24 times in units place. An addition of all these digits gives the sum of all the unit digits of all 120 numbers. Therefore,
(⁴P₃ \(\times\)1) + (⁴P₃ \(\times\) 2) + (⁴P₃ \(\times\) 4) + (⁴P₃ \(\times\) 6) + (⁴P₃ \(\times\) 8)
= ⁴P₃ \(\times\) (1 + 2 + 4 + 6 + 8)
= ⁴P₃ \(\times\) (sum of the digits)
= ⁴P₃ \(\times\) 21
Similarly, we get the sum of the digits in 10^th place as ⁴P₃ \(\times\) 21. Since it is in 10^th place, its value is ⁴P₃ x 21 x 10. Similarly, the values of the sum of the digits in 100^th place and 1000^th place are ⁴P₃ \(\times\) 21 \(\times\) 100 and ⁴P₃ \(\times\) 21 \(\times\) 1000 respectively. Hence the sum of all the 4 digit numbers formed by using the digits 1, 2, 4, 6, 8 is (⁴P₃ \(\times\) 21) + (⁴P₃ \(\times\) 21 \(\times\) 10) + (⁴P₃ \(\times\) 21 \(\times\) 100) + (⁴P₃ \(\times\) 21\(\times\) 1000)
= ⁴P₃ (21\(\times\)1111)
= 24 \(\times\) 21 \(\times\) 1111
= 559944.

Using sine formula,
Let \(\frac { sinA }{ a } =\frac { sinB }{ b } =\frac { sinC }{ c } =k\)
sin A = ak, sin B = bk, sin C = ck
Given \(\frac { sin\quad A }{ sin\quad C } =\frac { sin\left( A-B \right) }{ sin\left( B-C \right) } \)
\(\Rightarrow \frac { sin\left( B+C \right) }{ sin\left( A+B \right) } =\frac { sin\left( A-B \right) }{ sin\left( B-C \right) } \)
⇒ Sin(B + C) sin(B - C) = sin(A + B) sin(A - B)
⇒ sin²B - sin²C = sin²A - sin²B
⇒ k²b²- k²c² = k²a² - k²b²
⇒ b²- c² = a²- b²
⇒ 2b² = a² + c²
⇒ a², b², c² are in A.P.
Hence proved

\(|A|=\left|\begin{array}{ccc} 0 & a-b & k \\ b-a & 0 & 5 \\ -k & -5 & 0 \end{array}\right|\)
\(=0-(a-b)[0+5 k]+k(-5(b-a)-0)\)
\(=(-a+b)(5 k)+k(-5 b+5 a)\)
\(=-5 k a+5 k b-5 k b+5 a k\)
= 0
\(|A|=0\)
\(\therefore\) A is singular