The given differential equation is homogeneous (verify).
Now, we rewrite the given equation in differential form \(\frac { dy }{ dx } =\frac { y+\sqrt { { x }^{ 2 }+{ y }^{ 2 } } }{ x } \)
Since the initial value of x is 1, we consider x > 0 and take x =\(\sqrt { { x }^{ 2 } } \)
We have \(\frac { dy }{ dx } =\frac { y }{ x } +\sqrt { 1+{ \left( \frac { y }{ x } \right) }^{ 2 } } \)
Let y = vx. Then, \(v+x\frac { dv }{ dx } =v+\sqrt { 1+{ v }^{ 2 } } \), which becomes \(x\frac { dv }{ dx } =\sqrt { 1+{ v }^{ 2 } } \)
By separating variables, we have \(\frac { dv }{ \sqrt { { v }^{ 2 }+1 } } =\frac { dx }{ x } \)
Upon integration, we get \(|v+\sqrt { { v }^{ 2 }+1 } |=log|x|+log|C|\ or\ v+\sqrt { { v }^{ 2 }+1 } =xC\)
Now, we replace v by \(\frac{y}{x}\), we get \(\frac { y }{ x } +\sqrt { \frac { { y }^{ 2 } }{ { x }^{ 2 } } +1 } =Cx\quad (or)\quad y+\sqrt { { x }^{ 2 }+{ y }^{ 2 } } ={ Cx }^{ 2 }\) gives the general solution of the given differential equation 
To determine the value of C, we use the condition that y = 0 when x = 1. So, we get C = 1.
Thus \(y=\sqrt { { x }^{ 2 }+{ y }^{ 2 } } ={ x }^{ 2 }\)is the particular solution of the given differential equation.

Equation of the given curve is y = x sin x
∴ slope = m = \(\left( \frac { dy }{ dx } \right) \)_{\(\left( \frac { \pi }{ 2 } ,\frac { \pi }{ 2 } \right) \)}
= \(\frac { \pi }{ 2 } cos\frac { \pi }{ 2 } +sin\frac { \pi }{ 2 } \)
= \(\frac { \pi }{ 2 } \) (0) + 1 = 1
∴ Equating of the tangent is y - y₁ = m (x - x₁)
⇒ y - \(\frac { \pi }{ 2 } \) = 1 \(\left( x-\frac { \pi }{ 2 } \right) \)
\(\Rightarrow \frac{2 y-\pi}{\not 2}=\frac{2 x-\pi}{\not 2}\)
Equating of the normal is y - y₁ = \(\frac{-1}{m}\) (x - x₁)
\(\Rightarrow \frac{2 y-\pi}{\not 2}=\frac{2 x-\pi}{\not 2}\)
⇒ 2x + 2y = 2π
⇒ x + y- π = 0

Given y = ax + \(\frac { b }{ x } \) .......(1)
Differentiating with respect to x
y' = ax - \(\frac { b }{ x ^2} \) ......(2)
Differentiating again with respect to x
\(y'' = \frac{-b(-2)}{x^3}= \frac{2b}{x^3} \)
\(Now, x^2y'' + xy'-y \)
\( = x^2 \times \frac{2b}{x^3}+x(a- \frac{b}{x^2})-(ax+\frac{b}{x}) \)
\(= 2\times (\frac{b}{x})+ax-(\frac{b}{x})-ax-(\frac{b}{x})\)
= 0
Hence, y = ax + b is the solution of the differential equation x²y"+xy'-y = 0.

From the graph the vertex is at (0, 0) and the parabola is open down
Equation of the parabola is x² = -4ay
(-20, -15) and (20, -15) lie on the parabola 
20² = -4a(-15)
\(4a=\frac { 400 }{ 15 } \)
x² =\(\frac { -80 }{ 3 } \) x y
Therefore equation is 3x² = -80y

Since A² = \(\left[ \begin{matrix} 4 & 3 \\ 2 & 5 \end{matrix} \right] \left[ \begin{matrix} 4 & 3 \\ 2 & 5 \end{matrix} \right] =\left[ \begin{matrix} 22 & 27 \\ 18 & 31 \end{matrix} \right] \).
A² + xA + yI₂ = O₂ ⇒ \(\left[ \begin{matrix} 22 & 27 \\ 18 & 31 \end{matrix} \right] +x\left[ \begin{matrix} 4 & 3 \\ 2 & 5 \end{matrix} \right] +y\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix} \right] =\left[ \begin{matrix} 0 & 0 \\ 0 & 0 \end{matrix} \right] \)
⇒ \(\left[ \begin{matrix} 22+4x+y & 27+3x \\ 18+2x & 31+5x+y \end{matrix} \right] =\left[ \begin{matrix} 0 & 0 \\ 0 & 0 \end{matrix} \right] \).
So, we get 22 + 4x + y = 0, 31 + 5x + y = 0, 27 + 3x = 0 and 18 + 2x = 0.
Hence x = −9 and y = 14. Then, we get  A² - 9A + 14I₂ = O₂.
Postmultiplying this equation by A^-1, we get A - 9I₂ + 14A^-1 = O₂. Hence, we get
A^-1 = \(\frac { 1 }{ 14 } \) (9I₂ - A) = \(\frac { 1 }{ 14 } \left( 9\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix} \right] -\left[ \begin{matrix} 4 & 3 \\ 2 & 5 \end{matrix} \right] \right) =\frac { 1 }{ 14 } \left[ \begin{matrix} 5 & -3 \\ -2 & 4 \end{matrix} \right] \).

given A = {Q\{1}}
A is defined on A by x*y = x+y-xy
Let x,y ≠ 1
∴ x*y = x + y - xy
Now to prove that x + y - xy ≠ 1
Let us assume that x + y - xy = 1
x+y-xy-1 = 0
(x-1)-y(x-1) = 0
(x-1)(1-y) = 0
x =1 or y = 1 which is a false [∵x, y ≠ 1]
∴ is a binary operation on A.
Commutative property:
Let x,y ∈A ⇒ x, y≠1
∴x*y = x+y-xy
and y*x = y+x-yx
⇒x+y = y*x∀x, y∈A
A has commutative property under *
Associative property:
Let x,y,z ∊A ⇒x,y,z≠1
Consider (x*y)*z = (x+y-xy)*z
= x +y~xy +z- (x +y-xy)z
= x +y-xy+ z-xz- yz + xyz
= x +y+z-xy-yz-zx +xyz ...(1)
= x +y+z - yz - x (y + z - yz)
= x +y +z-yz-zy-xz +xyz ..(2)
From (1) & (2), (x*y)*z = x*(y*z)
A has associative property under *.

The entries in the columns corresponding to p ↔ q and ( p ⟶ q) ∧ (q ⟶ p) are identical and hence they are equivalent

Given 2a = 20 cm \(\Rightarrow\) a = 10 cm;
2b = 10 cm \(\Rightarrow\)a = 5 cm
\(\therefore\) Equation of the ellipse is \(\frac { { x }^{ 2 } }{ { a }^{ 2 } } +\frac { { y }^{ 2 } }{ { b }^{ 2 } } =1\)
\(\Rightarrow \frac { { x }^{ 2 } }{ 100 } +\frac { { y }^{ 2 } }{ 25 } =1\Rightarrow \frac { { y }^{ 2 } }{ 25 } =1-\frac { { x }^{ 2 } }{ 100 } =\frac { 100-{ x }^{ 2 } }{ 100 } \)
\(\Rightarrow { y }^{ 2 }=\frac { 25 }{ 100 } (100-{ x }^{ 2 })\)

\(\therefore\) Required volume \(=2\pi \int _{ 0 }^{ 10 }{ { y }^{ 2 }dx } \)
\(=2\pi \int _{ 0 }^{ 10 }{ \frac { 25 }{ 100 } (100-{ x }^{ 2 })dx=\frac { 50\pi }{ 100 } \int _{ 0 }^{ 10 }{ (100-{ x }^{ 2 })dx } } \)
\(=\frac { \pi }{ 2 } { \left[ 100x-\frac { { x }^{ 3 } }{ 3 } \right] }_{ 0 }^{ 10 }=\frac { \pi }{ 2 } \left[ 100-\frac { 1000 }{ 3 } \right] \)
\(V=\frac { \pi }{ 2 } \left( \frac { 3000-1000 }{ 3 } \right) =\frac { \pi }{ 2 } \left( \frac { 2000 }{ 3 } \right) \)
\(=\frac { 1000\pi }{ 3 } \)

Let X denote the winning amount. The possible events of selection are 
(i) both balls are black, or
(ii) one white and one black or
(iii) both are white 
Therefore X is a random variable that can be defined as
X (both are black balls) = Rs. 2(20) = Rs. 40
X (one black and one white ball) = Rs. 20 − Rs. 10 = Rs. 10
X (both are white balls) = (Rs. 20) = - Rs. 20
Therefore X takes on the values 40,10 and −20
Total number of balls n = 12
Total number of ways of selecting 2 balls = \(\left( \begin{matrix} 12 \\ 2 \end{matrix} \right) =\frac { 12\times 11 }{ 1\times 2 } =66\)
Number of ways of selecting 2 black balls =  \(\left( \begin{matrix} 4 \\ 2 \end{matrix} \right) =6\) 
Number of ways of selecting one black ball and one white ball = \(\left( \begin{matrix} 8 \\ 1 \end{matrix} \right) \left( \begin{matrix} 4 \\ 1 \end{matrix} \right) =32\) 
Number of ways of selecting 2 white balls = \(\left( \begin{matrix} 8 \\ 2 \end{matrix} \right) =28\)

Probability mass function is

Mean :
\(E(X)\Sigma xf(x)=40.\left( \frac { 6 }{ 66 } \right) +10.\left( \frac { 32 }{ 66 } \right) +\left( -20 \right) .\left( \frac { 28 }{ 66 } \right) =\frac { 4000 }{ 11 } \)
That is expected winning amount is 0
 Variance :
\(\Sigma x^{ 2 }=\Sigma { x }^{ 2 }f(x)=40^{ 2 }.\left( \frac { 6 }{ 66 } \right) +10^{ 2 }.\left( \frac { 32 }{ 66 } \right) +\left( -20 \right) ^{ 2 }.\left( \frac { 28 }{ 66 } \right) =\frac { 4000 }{ 11 } \)
 (E(X )² = 0² = 0
This gives  \(V(X)=E({ X }^{ 2 })-\left( E(X))^{ 2 } \right) =\frac { 4000 }{ 11 } -0=\frac { 4000 }{ 11 } \)
Therefore E(X ) = 0 and \(V(x)=\frac { 4000 }{ 11 } \)

Equation of AB is \(\frac { y-1 }{ 2-1 } =\frac { x+1 }{ 3+1 } or\quad y=\frac { 1 }{ 4 } (x+5)\)
Equation of BC is \(\frac { y-5 }{ 2-5 } =\frac { x-0 }{ 3-0 } or\quad y=-x+5\)
Equation of AC is \(\frac { y-1 }{ 5-1 } =\frac { x+1 }{ 0+1 } or\quad y=4x+5\)
\(\therefore\) Area of \(\Delta\)ABC = Area DACO+ Area of OCBE − Area of DABE
\(=\int _{ -1 }^{ 0 }{ (4x+5)dx+\int _{ 0 }^{ 3 }{ (-x+5)dx-\frac { 1 }{ 4 } \int _{ -1 }^{ 3 }{ (x+5)dx } } } \)
\(\\ \\ \\ ={ \left[ \frac { { 4x }^{ 2 } }{ 2 } +5x \right] }_{ -1 }^{ 0 }+{ \left[ -\frac { { x }^{ 2 } }{ 2 } +5x \right] }_{ 0 }^{ 3 }-\frac { 1 }{ 4 } { \left[ \frac { { x }^{ 2 } }{ 2 } +5x \right] }_{ -1 }^{ 3 }\)
\(=0-(+2-5)+\left( -\frac { 9 }{ 2 } +15 \right) -0-\frac { 1 }{ 4 } \left[ \frac { 9 }{ 2 } +15 \right] +\frac { 1 }{ 4 } \left[ \frac { 1 }{ 2 } -5 \right] =\frac { 15 }{ 2 } \)

Since the given function is a probability density function

\(\int _{ -\infty }^{ \infty }{ f(x) } dx=1\) 
That is \(\int _{ -\infty }^{ 1 }{ f(x) } dx+\int _{ 1 }^{ 4 }{ f(x) } dx+\int _{ 4 }^{ \infty }{ f(x) } dx=1\) 
From the given information
\(\int _{ -\infty }^{ 1 }{ 0dx } +\int _{ 1 }^{ 4 }{ { Cx }^{ 2 }dx } +\int _{ 4 }^{ \infty }{ 0dx } =1\) 
\(0+C\left[ \cfrac { { x }^{ 3 } }{ 3 } \right] _{ 1 }^{ 4 }+0=1\Rightarrow C\left[ \cfrac { 64-1 }{ 3 } \right] =1\Rightarrow 21C\Rightarrow C=\cfrac { 1 }{ 21 } \) 
Therefore the probability density function is
\(f(x)= \begin{cases}C x^{2} & 1
Since f (x) is continuous, the probability that X is equal to any particular value is zero. Therefore when the random variable is continuous, either or both of the signs < by ≤ and > by ≥ can be interchanged. Thus
(i) P(1.5 < X < 3.5) = P(1.5 ≤ X< 3.5)= P(1.5 < X ≤3.5) = P(1.5 ≤X ≤ 3.5)
Therefore
\(P(1.5
 
= \(\cfrac { 1 }{ 21 } \left( \cfrac { { x }^{ 3 } }{ 3 } \right) =\cfrac { 1 }{ 21 } \left( \cfrac { \left( 3.5 \right) ^{ 3 }-\left( 1.5 \right) ^{ 3 } }{ 3 } \right) \) 
= \(\cfrac { 79 }{ 126 } \) 
(ii) \(P(X\le 2)=\int _{ -\infty }^{ 2 }{ f(x) } dx=\int _{ -\infty }^{ 1 }{ f(x)dx } +\int _{ 1 }^{ 2 }{ f(x)dx } \)
 
Therefore
\(P(X\le 2)=0+\cfrac { 1 }{ 21 } \int _{ 1 }^{ 2 }{ { x }^{ 2 }dx=\cfrac { 1 }{ 21 } \left( \cfrac { { x }^{ 3 } }{ 3 } \right) } ^{ 2 }_{ 1 }\) 
= \(\cfrac { 1 }{ 21 } \left( \cfrac { { 2 }^{ 3 }-{ 1 }^{ 3 } }{ 3 } \right) =\cfrac { 7 }{ 63 } \) 
(iii) \(P(3
 
= \(\cfrac { 1 }{ 21 } \int _{ 3 }^{ 4 }{ { x }^{ 2 }dx+0=\cfrac { 1 }{ 21 } \left( \cfrac { { x }^{ 3 } }{ 3 } \right) } _{ 3 }^{ 4 }\) 
= \(\cfrac { 1 }{ 21 } \left( \cfrac { { 4 }^{ 3 }-{ 3 }^{ 3 } }{ 3 } \right) =\cfrac { 37 }{ 63 } \)

X~ B(n, p)
Given mean np = 6
\(S.D=\sqrt { npq } =2\) 
\(\Rightarrow npq=4\) 
\( \rightarrow \frac { npq }{ np } =\frac { 4 }{ 6 } =\frac { 2 }{ 3 } \) 
\(\Rightarrow q=\frac { 2 }{ 3 } \) 
\(\Rightarrow 1-P=\frac { 2 }{ 3 } \) 
\(\Rightarrow 1-\frac { 2 }{ 3 } =P\) 
\(\therefore P=\frac { 1 }{ 3 } \)  
\(n\times \frac { 1 }{ 3 } =6\Rightarrow n=18\)
(i) The probability mass function
P(X = x) nC_x p^x (1 - p )^n-x,
X = 0,1,2, ... , n
\(\therefore P(X=x)=\ ^{18}{ C }_{ x }\left( \frac { 1 }{ 3 } \right) ^{ x }\left( \frac { 2 }{ 3 } \right) ^{ 18-x }\) 
x=0,1,2...,8
(ii) \(P(X=3)=\ ^{ 18}{C }_{ 3 }\left( \frac { 1 }{ 3 } \right) ^{ 3 }\left( \frac { 2 }{ 3 } \right) ^{ 18-3 }\) 
= \(^{ 18}{C }_{ 3 }\left( \frac { 1 }{ 3 } \right) ^{ 3 }\left( \frac { 2 }{ 3 } \right) ^{ 15 }\)
(iii) P(X ≥ 2)  
 P(X ≥ 2) 1 -P(X < 2)
= 1 - [P(X = 0) + P(X = 1)]
= \(1-\left[ ^{18}{ C }_{ 0 }\left( \frac { 1 }{ 3 } \right) ^{ 0 }\left( \frac { 2 }{ 3 } \right) ^{ 18 }+^{ 18}{C }_{ 1 }\left( \frac { 1 }{ 3 } \right) ^{ 1 }\left( \frac { 2 }{ 3 } \right) ^{ 17 } \right] \)
=  \(1-\left[ \left( \frac { 2 }{ 3 } \right) ^{ 18 }+6\left( \frac { 2 }{ 3 } \right) ^{ 17 } \right] \) 
= \(1-\left( \frac { 2 }{ 3 } \right) ^{ 17 }\left[ \frac { 2 }{ 3 } +6 \right] \) 
= \(1-\left( \frac { 2 }{ 3 } \right) ^{ 17 }\left( \frac { 20 }{ 3 } \right) \)
= \(1-\frac { 20 }{ 3 } \left( \frac { 2 }{ 3 } \right) ^{ 17 }\)

Given V(x, y) = e^x(x cos y - y sin y)
\(\frac { \partial V }{ \partial x } \) = e^x (cos y) +(x cos y - y sin y)e^x
= e^x (cos y + x cos y - y sin y)
\(\frac { { \partial }^{ 2 }V }{ \partial { x }^{ 2 } } \) = e^x(0 + cos y - 0) + (cos y +x cos y.- y sin y)e^x
= e^x(2 cos y + x cos y - y sin y) ... (1)
\(\frac { \partial V }{ \partial y } \) = e^x(-x sin y- y cos y- sin y)

\(\frac { { \partial }^{ 2 }V }{ \partial { y }^{ 2 } } \) = e^x (-x cos y - (-y sin y + cos y) - cos y)
= e^x(- x cos y + y sin y - cos y - cos y)
= e^x (- x cos y + y sin y - 2 cos y) ... (2)
(1)+(2)➝ 
\(\frac { { \partial }^{ 2 }V }{ \partial { x }^{ 2 } } \) + \(\frac { { \partial }^{ 2 }V }{ \partial { y }^{ 2 } } \) = e^x(2 cos y + x cos y - y sin y - x cos y + y sin y - 2 cos y]
= e^x (0) = 0
Hence proved

\(Let\ I=\int _{ 0 }^{ \pi }{ x\left[ { sin }^{ 2 }(sinx)+{ cos }^{ 2 }(cosx) \right] } dx\quad ...(1)\)
\(\left[ \because \int _{ 0 }^{ a }{ f(x)dx=\int _{ 0 }^{ a }{ f(a-x)dx } } \right] \)
\(I=\int _{ 0 }^{ \pi }{ (\pi -x) } [{ sin }^{ 2 }(sin(\pi -x))+{ cos }^{ 2 }(cos(\pi -x)]dx\)
\(=\int _{ 0 }^{ \pi }{ (\pi -x)[{ sin }^{ 2 }(sinx)+{ cos }^{ 2 }(-cosx)]dx } \)
\(=\int _{ 0 }^{ \pi }{ (\pi -x)[{ sin }^{ 2 }(sin\quad x)+{ cos }^{ 2 })(cosx)]dx } \)
\([\because cos(-x)=cos\quad x]\)
\(=\int _{ 0 }^{ \pi }{ \pi [{ sin }^{ 2 }(sinx)+{ cos }^{ 2 }(cosx)] } dx\)
\(-\int _{ 0 }^{ \pi }{ x[{ sin }^{ 2 }(sinx)+{ cos }^{ 2 }(cosx)] } dx\)
\(=\int _{ 0 }^{ \pi }{ \pi [{ sin }^{ 2 }(sinx)+{ cos }^{ 2 }(cosx)]dx-I } [from\quad (1)]\)
\(2I=\pi \int _{ 0 }^{ \pi }{ [{ sin }^{ 2 }(sinx)+{ cos }^{ 2 }(cosx)] } dx\)
\(I=\pi \int _{ 0 }^{ \pi /2 }{ \left[ { sin }^{ 2 }(sin\quad x)+{ cos }^{ 2 }(cos\quad x) \right] } dx\quad ...(2)\)
\(I=\pi \int _{ 0 }^{ \pi /2 }{ \left[ { sin }^{ 2 }(sin\frac { \pi }{ 2 } -x)+{ cos }^{ 2 }(cos\frac { \pi }{ 2 } -x) \right] dx } \)
\(I=\pi \int _{ 0 }^{ \pi /2 }{ [{ sin }^{ 2 }(cosx)+{ cos }^{ 2 }(sin\quad x)]dx] } ....(3)\)
Adding (2) and (3) we get,
\(2I=\pi \int _{ 0 }^{ \pi /2 }{ \left[ { sin }^{ 2 }(sinx)+{ cos }^{ 2 }(cosx) \right] dx } +{ sin }^{ 2 }(cosx)+{ cos }^{ 2 }(sin\quad x)]dx\)
\(I=\pi \int _{ 0 }^{ \pi /2 }{ (1+1)dx } \ [\because { sin }^{ 2 }\theta +{ cos }^{ 2 }\theta =1]\)
\(=\pi (2){ [x] }_{ 0 }^{ \frac { \pi }{ 2 } }=2\pi \left[ \frac { \pi }{ 2 } -0 \right] ={ \pi }^{ 2 }\ \)
\( \therefore I=\frac { { \pi }^{ 2 } }{ 2 } \)

Let I = \(\int ^{\pi}_{-\pi} \frac{cos ^2 x}{1+ a^x} dx\)-- (1) 
Using \(\int ^{b}_{a}\) f(x) dx =\(\int ^{b}_{a}\) f(a+b -x)dx  we get,
I = \(\int ^{\pi}_{-\pi} \frac{cos ^2 (\pi -\pi - x)}{1+ a^{\pi -\pi - x}} dx\)
= \(\int ^{\pi}_{-\pi} \frac{cos ^2 (-x)}{1+ a^{-x}} dx\)
= \(\int ^{\pi}_{-\pi} a^x (\frac{cos ^2 x }{a^x+ 1} )dx\) --- (2)
Adding (1) and (2) we get
2I =  \(\int ^{\pi}_{-\pi} \frac{cos ^2 x }{a^x+ 1}(a^x+ 1) dx\) =  \(\int ^{\pi}_{-\pi} cos ^2 x dx\)
= 2 \(x =\int ^{\pi}_{-\pi} cos ^2 x dx\)  (since cos² x is an even function)
Hence, I =  \(\int ^{\pi}_{0} (\frac{1 + cos2x }{2} )dx\) 
=  \(\frac {1}{2} [ x + \frac {sin 2x}{2}]^{x}_{0}\) 
= \(\frac {1}{2} [\pi]\)
= \(\frac {\pi}{2}\)

\(f(x)=\begin{cases} \begin{matrix} x+1 & -1\le x<0 \end{matrix} \\ \begin{matrix} -x+1 & 0\le x<1 \end{matrix} \\ \begin{matrix} 0 & otherwise \end{matrix} \end{cases}\)
(i) Distribution function
Case 1 : x < -1
F(x) = \(\int _{ -\infty }^{ x }{ f(u)du } \) = 0
Case 2 : -1 ≤ x < 0
\(\int _{ -\infty }^{ x }{ f(u)du } \) 
= \(\int _{ -\infty }^{ x }{ f(x) } dx=\left[ \frac { { x }^{ 2 } }{ 2 } +x \right] _{ -1 }\) 
= \(\left( \frac { {u }^{ 2 } }{ 2 } +u \right)=\frac{x^2}{2}+x -\left( \frac { 1 }{ 2 } +1 \right) \) 
= \(\frac { { x }^{ 2 } }{ 2 } +x+\frac { 1 }{ 2 } \) 
Case 3 : 0 ≤ x < 1,
 \(F(X)=\int _{ 0 }^{ x }{ (-x+1)dx } =\left[ -\frac { { x }^{ 2 } }{ 2 } +x \right] _{ 0 }^{ x }\) 
= \(\left( -\frac { { x }^{ 2 } }{ 2 } +x \right) -\left( 0 \right) =\frac { { x }^{ 2 } }{ 2 } +x\) 
When 1 ≤ x,
\(F(x)=\int _{ 1 }^{ x }{ f(x)dx } =\int _{ 1 }^{ x }{ 0dx } \)
= \(\therefore F(X)=\begin{cases} \begin{matrix} \frac { { x }^{ 2 } }{ 2 } +x+\frac { 1 }{ 2 } & -1\le x<0 \end{matrix} \\ \begin{matrix} -\frac { { x }^{ 2 } }{ 2 } +x & 0\le x<1 \end{matrix} \\ \begin{matrix} 0 & otherwise \end{matrix} \end{cases}\) 
(ii) p(0.5 ≤ X ≤ 0.5)
= \(\int _{ -0.5 }^{ 0.5 }{ f(x)dx } =\int _{ 0.5 }^{ 0 }{ f(x)dx+\int _{ 0 }^{ 0.5 }{ f(x)dx } } \)
= \(\int _{ -0.5 }^{ 0 }{ (x+1) } dx+\int _{ 0 }^{ 0.5 }{ \left( -x+1 \right) } dx\) 
= \(\left[ \frac { { x }^{ 2 } }{ 2 } +x \right] _{ -0.5 }^{ 0 }+\left[ \frac { -{ x }^{ 2 } }{ 2 } +x \right] _{ 0 }^{ 0.5 }\) 
= \(0-\left( \frac { { 0.5 }^{ 2 } }{ 2 } -0.5 \right) +\left( -\frac { \left( 0.5 \right) ^{ 2 } }{ 2 } +0.5 \right) -0\) 
= \(-\left( \frac { .25 }{ 2 } -0.5 \right) +\left( \frac { -0.25 }{ 2 } +0.5 \right) \) 
= \(\frac { .25 }{ 2 } +0.5-\frac { 0.25 }{ 2 } +0.5=0.25+1\)
= 0.75

We use the formula
\(\int _{ a }^{ b }{ f(x)dx } =\underset { n\rightarrow \infty }{ lim } \frac { b-a }{ n } \sum _{ r=1 }^{ n }{ f } \left( a+(b-a)\frac { r }{ n } \right) \)
Here f(x) = 2x²+3, a = 1 and b = 4
So, we get
\(f\left( a+(b-a)\frac { r }{ n } \right) =f\left( 1+(4-1)\frac { r }{ n } \right) =f\left( 1+\frac { 3r }{ n } \right) =2{ \left( 1+\frac { 3r }{ n } \right) }^{ 2 }+3=5+\frac { { 18r }^{ 2 } }{ { n }^{ 2 } } +\frac { 12r }{ n } \)Hence, we get
\(\int _{ 1 }^{ 4 }{ ({ 2x }^{ 2 }+3) } dx=\underset { n\rightarrow \infty }{ lim } \frac { 3 }{ n } \sum _{ r=1 }^{ n }{ \left( 5+\frac { { 18r }^{ 2 } }{ { n }^{ 2 } } +\frac { 12r }{ n } \right) } =\underset { n\rightarrow \infty }{ lim } \left[ \frac { 15 }{ n } \sum _{ r=1 }^{ n }{ 1+\frac { 54 }{ { n }^{ 3 } } \sum _{ r=1 }^{ n }{ { r }^{ 2 } } +\frac { 36 }{ { n }^{ 2 } } \sum _{ r=1 }^{ n }{ r } } \right] \)
\(=\underset { n\rightarrow \infty }{ lim } \left[ \frac { 15 }{ n } n+\frac { 54 }{ { n }^{ 3 } } ({ 1 }^{ 2 }+{ 2 }^{ 2 }+...+{ n }^{ 2 })+\frac { 36 }{ { n }^{ 2 } } (1+2+..+n) \right] \)
\(=\underset { n\rightarrow \infty }{ lim } \left[ 15+\frac { 54 }{ { n }^{ 3 } } \frac { n(n+1)(2n+1) }{ 6 } +\frac { 36 }{ { n }^{ 2 } } \frac { n(n+1) }{ 2 } \right] \)
\(=\underset { n\rightarrow \infty }{ lim } \left[ 15+9\left( 1+\frac { 1 }{ n } \right) \left( 2+\frac { 1 }{ n } \right) +18\left( 1+\frac { 1 }{ n } \right) \right] \)
= 15+9(1+ 0)(2 + 0) +18(1+ 0) = 51.

Let the price of low-grade steel be Rs. p per tonne. Then the price of high-grade steel is Rs. 2p per tonne.
The total receipt per day is given by \(R=px+py=px+2p\left( \frac { 40-5x }{ 10-x } \right) \). Hence the problem is to maximise R . Now, simplifying and differentiating R with respect to x , we get
\(R=p\left( \frac { 80-{ x }^{ 2 } }{ 10-x } \right) \)
\(\frac { dR }{ dx } =p\left( \frac { { x }^{ 2 }-20x+80 }{ (10-x)^{ 2 } } \right) \)
\(\frac { dR }{ dx } =-\frac { 40P }{ \left( 10-x \right) ^{ 3 } } \)
Now, \(\frac { dR }{ dx } =0\Rightarrow { x }^{ 2 }-20x+80=0\) and hence \(x=10\pm 2\sqrt { 5 } \) 
At \(x=10-2\sqrt { 5 } ,\frac { { d }^{ 2 }R }{ { dx }^{ 2 } } <0\) and hence R will be maximum. If x \(x=10-2\sqrt { 5 } \) then \(y=5-\sqrt { 5 } \)  
Therefore the steel plant must produce low-grade and high-grade steels respectively in tonnes per day are \(10-2\sqrt { 5 } \)  and \(5-5\sqrt { 5 } \)

Given f(x) = 4x³ + 3x²- 6x + 1
f'(x) = 12x² + 6x - 6
f"(x) = 24x + 6
f'(x) = 0
⇒12x² + 6x - 6 = 0
⇒ 2x² + x - 1 = 0
⇒ (x + 1)(2x - 1) = 0
\(\Rightarrow x=-1,\frac { 1 }{ 2 } \)
The critical numbers are -1, \(\frac { 1 }{ 2 } \)
The possible intervals of monotonicity are
\(\left( -\infty ,-1 \right) \left( -1,\frac { 1 }{ 2 } \right) \left( \frac { 1 }{ 2 } ,\infty \right) \)

∴ f(x) is strictly increasing in \(\left( -\infty ,-1 \right) \left( \frac { 1 }{ 2 } ,\infty \right) \) and strictly decreasing in \(\left( -1,\frac { 1 }{ 2 } \right) \)
f"(x) = 0
\(\Rightarrow 24x+6=0\Rightarrow 24x=-6\)
\(x=\frac { -6 }{ 24 } =\frac { -1 }{ 4 } \)
The possible intervals of concavity are \(\left( -\infty ,\frac { -1 }{ 4 } \right) \left( \frac { -1 }{ 4 } ,\infty \right) \) 

ஃf(x) concave down in \(\left( -\infty ,\frac { -1 }{ 4 } \right) \) and concave up in \(\left( \frac { -1 }{ 4 } ,\infty \right) \)
As f"(x) changes its sign when it passes through
\(x=\frac { -1 }{ 4 } \), the point of inflection is \(\left( -\frac { 1 }{ 4 } ,f\left( -\frac { 1 }{ 4 } \right) \right) \)
Now \(f\left( \frac { -1 }{ 4 } \right) =\left( -\frac { 1 }{ 4 } \right) ^{ 3 }+3\left( \frac { -1 }{ 4 } \right) ^{ 2 }-6\left( \frac { -1 }{ 4 } \right) +1\)
= \(4\left( \frac { -1 }{ 4 } \right) +\frac { 3 }{ 16 } +\frac { 6 }{ 4 } +1\)
= \(\frac { -1 }{ 16 } +\frac { 3 }{ 16 } +\frac { 3 }{ 22 } +1=\frac { 1 }{ 8 } +\frac { 3 }{ 2 } +1\)
= \(\frac { 1+12+8 }{ 8 } =\frac { 21 }{ 8 } \)
ஃ Point of  inflection. \(\left( \frac { -1 }{ 4 } ,\frac { 21 }{ 8 } \right) \)
Since f'(x) changes its sign from positive to negative at x = -1, it has a local maximum at x = -1.
ஃf (-1) = 4 (-1)³ + 3 (-1)² - 6(-1) + 1
Since f'(x) changes its sign from negative to positive at \(x=\frac { 1 }{ 2 } \) it has a local minimum at \(x=\frac { 1 }{ 2 } \)
\(\therefore f\left( \frac { 1 }{ 2 } \right) =4\left( \frac { 1 }{ 2 } \right) ^{ 3 }+3\left( \frac { 1 }{ 2 } \right) ^{ 2 }-6\left( \frac { 1 }{ 2 } \right) +1\)
= \(\frac { 4 }{ 8 } +\frac { 3 }{ 4 } -\frac { -3 }{ 2 } +1=\frac { 1 }{ 2 } +\frac { 3 }{ 4 } -\frac { 3 }{ 2 } +1\)
= \(\frac { 2+3-6+4 }{ 4 } =\frac { 3 }{ 4 } \)

Let x(t) denote the amount of salt in the tank at time t.
Its rate of change is
\(\frac{dx}{dt}\) = inflow rate - outflow rate
Now, 2 gram time 3 litres per minutes is inflow rate = 6 grams of salt. (3 x 2 = 6)
The out flow of salt is \(\frac{3}{50}\) times x = \(\frac{3x}{50}\)
\(\therefore \frac { dx }{ dt } =6-\frac { 3x }{ 50 } =\frac { 300-3x }{ 50 } \)
\(=-\frac { 3(x-100) }{ 50 } \)
\(\Rightarrow \frac { dx }{ x-100 } =-\frac { 3 }{ 50 } dt\)
\(\Rightarrow \int { \frac { dx }{ x-100 } =-\frac { 3 }{ 50 } \int { dt } } \)
\(\Rightarrow log(x-100)=-\frac { 3 }{ 50 } t+logC\)
\(\\ \Rightarrow log(x-100)-logC=-\frac { 3 }{ 50 } t\)
\(\Rightarrow log\left( \frac { x-100 }{ C } \right) =-\frac { 3 }{ 50 } t\)
\(\Rightarrow \frac { x-100 }{ C } ={ e }^{ -\frac { 3t }{ 50 } }\)
\(\Rightarrow x-100={ C }_{ e }-\frac { 3t }{ 50 } \quad ...(1)\)
When t = 0, x = 0
[Since initial water was pure without any salt]
\(\Rightarrow\) 0-100 = Ce⁰ 
\(\Rightarrow\) C = -100
(1) becomes x-100 = -100\(\\ \\ \\ \\ \\ \\ \\ { e }^{ -\frac { 3t }{ 50 } }\)
\(\Rightarrow\) x = 100-100\(\\ \\ \\ \\ \\ \\ \\ { e }^{ -\frac { 3t }{ 50 } }\)
\(\Rightarrow\) x = 100(1-\(\\ \\ \\ \\ \\ \\ \\ { e }^{ -\frac { 3t }{ 50 } }\))
Hence the amount of salt in the tank at time t is x = 100(1-\(\\ \\ \\ \\ \\ \\ \\ { e }^{ -\frac { 3t }{ 50 } }\))

\(​​​​​​\underset { x\rightarrow { 0 }^{ + } }{ lim } { x }^{ x }\)
This is an indeterminate of the form 0⁰
Let g(x) x^x
Taking logarithm, we get
\(log \ g(x)=log({ x }^{ 2 })=xlogx=\frac { log\quad x }{ \frac { 1 }{ x } } \)
\(\therefore ​​​​​​\underset { x\rightarrow { 0 }^{ + } }{ lim } log \ g(x)={ \left[ \frac { logx }{ \frac { 1 }{ x } } \right] }=\frac { \infty }{ \infty } \)
\(=\underset { x\rightarrow { 0 }^{ + } }{ lim } \left( \frac { \frac { 1 }{ x } }{ -\frac { 1 }{ { x }^{ 2 } } } \right) \) [by L' Hopital rule]
= \(\underset { x\rightarrow { 0 }^{ + } }{ lim } \frac { 1 }{ x } \times \frac { { x }^{ 2 } }{ 1 } =\underset { x\rightarrow { 0 }^{ + } }{ lim } -x\)
= 0
But \(\underset { x\rightarrow { 0 }^{ + } }{ lim } (log(g(x))=log(\underset { x\rightarrow { 0 }^{ + } }{ lim } (g(x))\)
ஃ \(\underset { x\rightarrow { 0 }^{ + } }{ lim } (log(g(x))=0\)
\(\Rightarrow { e }^{ log }(\underset { x\rightarrow { 0 }^{ + } }{ lim } log(g(x))={ e }^{ 0 }\)
\(\Rightarrow \underset { x\rightarrow { 0 }^{ + } }{ lim } g(x)=1\)
\(\therefore \underset { x\rightarrow { 0 }^{ + } }{ lim } { x }^{ x }=1\)

Given E = Ri + L \(\frac{di}{dt}\)
\(\frac { E }{ L } =\frac { Ri }{ L } +\frac { di }{ dt } \)
\(\Rightarrow \frac { Ri }{ L } +\frac { di }{ dt } =\frac { E }{ L } \)
This is a linear differential equation
\(Here\quad P=\frac { R }{ L } and\quad Q=\frac { E }{ L } \)
\(\therefore \int { pdt } =\int { \frac { R }{ L } dt } =\frac { R }{ L } t\)
\(\therefore I.F={ e }^{ \int { pdt } }={ e }^{ \frac { Rt }{ L } }\)
\(\therefore\) Solution is i\({ e }^{ \int { pdt } }=\int { Q{ e }^{ \int { pdt } }dt+C } \)
\(\Rightarrow i{ e }^{ \frac { Rt }{ L } }=\int { \frac { E }{ L } . } { e }^{ \frac { Rt }{ L } }dt+C\)
\(\therefore i{ e }^{ \frac { Rt }{ L } }=\frac { E }{ L } \frac { { e }^{ \frac { Rt }{ L } } }{ \frac { R }{ L } } dt+C\)
\(i=\frac { E }{ R } { e }^{ \frac { Rt }{ L } }+C\)
\(i=\frac { E }{ R } +c{ e }^{ -\frac { Rt }{ L } }\)
When E = 0,
\(i=0+c{ e }^{ -\frac { Rt }{ L } }\)
\(\Rightarrow i=c{ e }^{ -\frac { Rt }{ L } }\)

The given differential cquation may be written as
\(\left(1+x+x y^2\right) \frac{d y}{d x}+\left(y+y^3\right)=0  \)
\(\left(1+x+x y^2\right) \frac{d y}{d x}=-\left(y+y^3\right)  \)
\(\left(1+x+x y^2\right)=-1\left(y^2+1\right) \frac{d x}{d y}  \)
\(y\left(y^2+1\right) \frac{d x}{d y}+1+x\left(y^2+1\right)=0\)
Divided by \( y\left(y^2+1\right) ,\) 
\(\frac{d x}{d y}+\frac{1}{y\left(y^2+1\right)}+\frac{x\left(y^2+1\right)}{y\left(y^2+1\right)}   =0  \)
\(\frac{d x}{d y}+\frac{x}{y}  =-\frac{1}{y\left(y^2+1\right)}\)
This is the form of \( \frac{d x}{d y}+\mathrm{Px}=\mathrm{Q}  \) where \( \mathrm{P}=\frac{1}{y} and  \mathrm{Q}=\frac{-1}{y\left(1+y^2\right)}\)
\(\text { I.F }=e^{\int P d y}=e^{\int \frac{1}{y} d y}=e^{\log y}=y\)
So, the solution of the equation is given by
\(x \times \mathrm{I} . \mathrm{F}  =\int(Q \times I . F) d y+c  \)
\(x \times \mathrm{y}  =\int \frac{-1}{y\left(1+y^2\right)} \times y \times d y+c  \)
\(x \mathrm{y}  =\int \frac{-1}{1+y^2} d y+c=-\int \frac{1}{1+y^2} d y+c\)
xy = -tan^-1y + c
xy + tan^-1y = c
Which is the required solution.

Let x and y be the length and breadth of the rectangle.
ஃ Perimeter P = 2x + 2y
\(\Rightarrow 2y=P-2x\Rightarrow y=\frac { P-2x }{ 2 } \) 
Let \(f(x)=Area=xy=x\left( \frac { P-2x }{ 2 } \right) \)
\(f(x)=\frac { Px-{ 2x }^{ 2 } }{ 2 } \)
\(f'(x)=\frac { 1 }{ 2 } \left[ P-4x \right] \)
f'(x) = 0
\(\Rightarrow \frac { 1 }{ 2 } \left[ P-4x \right] =0\)
\(\Rightarrow P=4x\) 
\(\Rightarrow x=\frac { P }{ 4 } \)
∴The critical number is \(\frac { P }{ 4 } \) 
 Now,\(f''\left( \frac { P }{ 4 } \right) =-2<0\)
ஃf(x) is maximum when \(x=\frac { P }{ 4 } \)
When \(x=\frac { P }{ 4 } \)
\(\Rightarrow y=\frac { P-2\left( \frac { P }{ 4 } \right) }{ 2 } =\frac { P-\frac { P }{ 2 } }{ 2 } =\frac { P }{ 4 } \)
\(\therefore x=y=\frac { P }{ 4 } \)
ஃThe rectangle is a square when the area is maximum for a given perimeter.

The given equation can be rewritten as \(\frac { dy }{ dx } +\frac { (x\quad tan\quad x-1) }{ x } y=xcosx\)
This is a linear differential equation. Here \(P=\frac { (x\quad tan\quad x-1) }{ x } ;Q=xcosx\)
\(\int { Pdx } =\int { \frac { (xtanx-1) }{ x } } dx=-log|cosx|-log|x|=-log|xcosx|=log\frac { 1 }{ |xcosx| } \)
Thus, \(I.F.={ e }^{ \int { pdx } }={ e }^{ log\frac { 1 }{ |xcosx| } }=\frac { 1 }{ xcosx } \)
Hence the solution is \({ ye }^{ \int { Pdf } }=\int { Q{ e }^{ \int { Pdx } }dx+C } \)
i.e., \(y\frac { 1 }{ xcosx } =\int { (xcosx)\frac { 1 }{ xcosx } dx+C } \)
or \(\\ \\ \\ \\ y\frac { 1 }{ xcosx } =x+C\)
or y = x² cos x + Cx cos x is the required solution.

Givenf(x) = sin x + cos x, 0 f'(x) = cos x - sin x
f"(x) = sin x - cos x
\(\therefore\) f"(x) = 0
\(\Rightarrow\) sin x - cos X = 0
\(\Rightarrow\) -sin x = cosx
\(\Rightarrow\) sin (-x) = cos x
\(\Rightarrow\) \(x=\frac { 3\pi }{ 4 } ,\frac { 7\pi }{ 4 } ,2\pi \)
ஃ The possible intervals are \(\left( 0,\frac { 3\pi }{ 4 } \right) \left( \frac { 3\pi }{ 4 } ,\frac { 7\pi }{ 4 } \right) \left( \frac { 7\pi }{ 4 } ,2\pi \right) \) 

ஃf (x) is concave upward in \(\left( \frac { 3\pi }{ 4 } ,\frac { 7\pi }{ 4 } \right) \) and concave downward in \(\left( 0,\frac { 3\pi }{ 4 } \right) \left( \frac { 7\pi }{ 4 } ,2\pi \right) \) 
Since f" (x) changes its sign from negative to positive at \(\frac { 3\pi }{ 4 } \) and positive at \(\frac { 7\pi }{ 4 } \) f(x) has point of inflection at \(\left( \frac { 3\pi }{ 4 } ,f\left( \frac { 3\pi }{ 4 } \right) \right) \) and 

\(\left( \frac { 7\pi }{ 4 } ,f\left( \frac { 7\pi }{ 4 } \right) \right) \) 
\(\therefore f\left( \frac { 3\pi }{ 4 } \right) =sin\frac { 3\pi }{ 4 } -cos\frac { \pi }{ 4 } \) 
= \(sin\left( \pi -\frac { \pi }{ 4 } \right) +cos\left( \frac { \pi }{ 4 } \right) \) 
= \(\frac { 1 }{ \sqrt { 2 } } -\frac { 1 }{ \sqrt { 2 } } =0\) 
ஃ The points of inflection are \(\left( \frac { 3\pi }{ 4 } ,0 \right) \) and \(\left( \frac { 7\pi }{ 4 } ,0 \right) \)

Let AB be the position of the ladder at any time t such that OA = x and OB = y
Then OA² + OB² = AB²
⇒ x² + y² = 17²
Given \(\frac { dx }{ dt } \) = 5 and x = 8
When x = 8, 8² + y² = 17²
 ⇒ y² = 289 - 64 = 225
 ⇒ y = 15
Differentiating (1) with respect to 't' we get,
\(2x\frac { dx }{ dt } +2y\frac { dy }{ dt } =0\)
⇒ 8(5) + 15 \(\frac { dy }{ dt } \) = 0 [∵ x = 8, \(\frac { dx }{ dt } \) = 15, y = 15]
⇒ 40 + 15\(\frac { dy }{ dt } \) = 0
⇒ \(\frac { dy }{ dt } =\frac { -40 }{ 15 } =\frac { -8 }{ 3 } \) m/sec
∴ The rate of top of the ladder moving down the wall is \(\frac{-8}{3}\) m/sec
(ii) The ladder, the wall and the floor forms a right angled triangle.
 Area = \(\frac12\)xy
Differentiating with respect to 't' we get,
\(\frac { dA }{ dt } =\frac { 1 }{ 2 } \left[ x\frac { dy }{ dx } +y\frac { dx }{ dt } \right] \)
\(=\frac { 1 }{ 2 } \left[ 8\left( -\frac { 8 }{ 3 } \right) +15(5) \right] \)
= \(\frac { 1 }{ 2 } \left[ \frac { -64 }{ 3 } +75 \right] =\frac { 1 }{ 2 } \)
= \(\frac { 1 }{ 2 } \left[ \frac { -64+225 }{ 3 } \right] =\frac { 1 }{ 2 } \left( \frac { 161 }{ 3 } \right) \)
= \(\frac { dA }{ dt } \) = 26.83 sq.m/sec

Given s (t) = 16t², height = 400 ft.
⇒ t² = \(\frac { 400 }{ 16 } =\frac { 100 }{ 4 } \)
t² = 25
t = 5 sec
(ii) Average velocity = \(\frac { ds }{ dt } \) = 32 t
When t = 2 sec
Average in the last
2 sec = \(\frac { V \ at \ t=3+V \ at \ t=5 }{ 2 } \)
= \(\frac { 32(3)+32(5) }{ 2 } \)
= \(\frac { 96+160 }{ 2 } =\frac { 256 }{ 2 } \)
= 128 f/sec
(iii) Instantaneous Velocity
=\(\frac { ds }{ dt } \) = 32t
When t = 5 sec
Velocity = \(\frac { ds }{ dt } \) = 32(5)
= 160 ft/sec

 y² - 4y - 8x + 12 = 0
y²-4y = 8x-12
Adding 4 both sides, we get,
y - 4y + 4 = 8x - 12 + 4 = 8x - 8
⇒ (y - 2)² = 8(x - 1)
This is a right open parabola and latus
rectum is 4a = 8 ⇒ a = 2.
(a) Vertex is (1, 2) ⇒ h = 1, k = 2
(b) focus is (h + a, 0 + k)
⇒ (1 + 2, 0 + 2)
⇒ (3, 2)
(c) Equation of directrix is x = h - a
⇒ x = 1-2
⇒ x = -1
(d) Length of latus rectum is 4a = 8 units.

3x + 3y − z = 11, 2x − y + 2z = 9, 4x + 3y + 2z = 25.
Δ = \(\left| \begin{matrix} 3 & 3 & -1 \\ 2 & -1 & 2 \\ 4 & 3 & 2 \end{matrix} \right| \)
= \(3\left| \begin{matrix} -1 & 2 \\ 3 & 2 \end{matrix} \right| -3\left| \begin{matrix} 2 & 2 \\ 4 & 2 \end{matrix} \right| -1\left| \begin{matrix} 2 & -1 \\ 4 & 3 \end{matrix} \right| \)
= 3(-2-6)-3(4-8)-1(6+4)·
= 3(- 8) - 3(- 4) - 1(10)
= - 24 + 12 - 10 = - 22
Δ₂ = \(\left| \begin{matrix} 11 & 3 & -1 \\ 9 & -1 & 2 \\ 25 & 3 & 2 \end{matrix} \right| \)
= \(11\left| \begin{matrix} -1 & 2 \\ 3 & 2 \end{matrix} \right| -3\left| \begin{matrix} 9 & 2 \\ 25 & 2 \end{matrix} \right| -1\left| \begin{matrix} 9 & -1 \\ 25 & 3 \end{matrix} \right| \)
= 11(-2-6)-3(18-50)-1(27+25)
= 11(-8)-3(-32)-1(52)
= -88+96-52 = -44
Δ₂ = \(\left| \begin{matrix} 3 & 11 & -1 \\ 2 & 9 & 2 \\ 4 & 25 & 2 \end{matrix} \right| \)
= \(3\left| \begin{matrix} 9 & 2 \\ 25 & 2 \end{matrix} \right| -11\left| \begin{matrix} 2 & 2 \\ 4 & 2 \end{matrix} \right| -1\left| 2\begin{matrix} 2 & 9 \\ 4 & 25 \end{matrix} \right| \)
= 3(18 - 50) -11(4 - 8) - 1(50- 36)
= 3(- 32) - 11(- 4) - 1(14)
= -96+44-14 = - 66
Δ₃ = \(\left| \begin{matrix} 3 & 3 & 11 \\ 2 & -1 & 9 \\ 4 & 3 & 25 \end{matrix} \right| \)
\(3\left| \begin{matrix} -1 & 9 \\ 3 & 25 \end{matrix} \right| -3\left| \begin{matrix} 2 & 9 \\ 4 & 25 \end{matrix} \right| -11\left| \begin{matrix} 2 & -1 \\ 4 & 3 \end{matrix} \right| \)
= 3(-25-27)-3(50-36)+ 11(6+4)
= 3(- 52) - 3(14) + 11(10)
= -156 - 42 + 110= - 88
∴ x = \(\frac { { \triangle }_{ 1 } }{ \triangle } =\frac { -44 }{ -22 } \) = 2
y = \(\frac { { \triangle }_{ 2 } }{ \triangle } =\frac { -66 }{ -22 } \) = 3
z = \(\frac { { \triangle }_{ 3 } }{ \triangle } =\frac { -88 }{ -22 } \) = 4
∴ Solution set is {2, 3, 4}

Assume that the system px + by + cz = 0, ax + qy + cz = 0, ax + by + rz = 0 has a non-trivial solution.
So, we have \(\left| \begin{matrix} p & b & c \\ a & q & c \\ a & b & r \end{matrix} \right| \) = 0, Applying R₂ ➝ R₂ - R₁ and R₃ ➝ R₃ - R₁ in the above equation,
we get \(\left| \begin{matrix} p & b & c \\ a-p & q-b & c \\ a-p & b & r-c \end{matrix} \right| \) = 0. That is,  \(\left| \begin{matrix} p & b & c \\ -\left( p-a \right) & q-b & c \\ -\left( p-a \right) & b & r-c \end{matrix} \right| \) = 0.
Since p ≠ a, q ≠ b, r ≠ c, we get (p - a)(q - b)(r - c) \(\left| \begin{matrix} \frac { p }{ p-a } & \frac { b }{ q-b } & \frac { c }{ r-c } \\ -1 & 1 & 0 \\ -1 & 0 & 1 \end{matrix} \right| \) = 0.
So, we have \(\left| \begin{matrix} \frac { p }{ p-a } & \frac { b }{ q-b } & \frac { c }{ r-c } \\ -1 & 1 & 0 \\ -1 & 0 & 1 \end{matrix} \right| \) = 0.
Expanding the determinant, we get \(\frac { p }{ p-a } +\frac { b }{ q-b } +\frac { c }{ r-c } \) = 0.
That is, \(\frac { p }{ p-a } +\frac { q-\left( q-b \right) }{ q-b } +\frac { r-\left( r-c \right) }{ r-c } \) = 0
⇒ \(\frac { p }{ p-a } +\frac { b }{ q-b } +\frac { c }{ r-c } \) = 2.

Comparing the two given lines with
\(\vec { r } =\vec { a } +t\vec { b } ,\vec { r } =\vec { c } +s\vec { d } \)
we have, \(\vec { a } =-\hat { i } -3\hat { j } -5\hat { k } ,\vec { b } =3\hat { i } +5\hat { j } +7\hat { k } ,\vec { c } =2\hat { i } +4\hat { j } +6\hat { k } \)  and \(\vec { d } =\hat { i } +4\hat { j } +7\hat { k } \)
We know that the two given lines are coplar, if \((\vec { c } -\vec { a } ).(\vec { b } \times \vec { d } )\) = 0
Here, \(\vec { b } \times \vec { d } \left| \begin{matrix} \hat { i } & \hat { j } & \hat { k } \\ 3 & 5 & 7 \\ 1 & 4 & 7 \end{matrix} \right| =7\hat { i } -14\hat { j } +7\hat { k } \)  and  \(\vec { c } -\vec { a } =3\hat { i } +7\hat { j } +11\hat { k } \)
Then, \((\vec { c } -\vec { a } ).(\vec { b } \times \vec { d } )=(3\hat { i } +7\hat { j } +11\hat { k } )(7\hat { i } -14\hat { j } +7\hat { k } )=0\)
Therefore the two given lines are coplanar. Then we find the non parametric form of vector equation of the plane containing the two given coplanar lines. We know that the plane containing the two given coplanar lines is
\((\vec { r } -\vec { a } ).(\vec { b } \times \vec { d } )\)= 0
which implies that \((\vec { r } -(-\hat { i } -3\hat { j } -5\hat { k } )).(7\hat { i } -14\hat { j } +7\hat { k } )\) = 0. 
Thus, the required non-parametric vector equation of the plane containing the two given coplanar lines is
\(\vec { r } .(\hat { i } -2\hat { j } +\hat { k } )\) = 0.

Let \({ z }^{ 3 }+8i=0\)
\(\Rightarrow\) z³ = -8i
= \(8(-i)=8\left( cos\left( -\frac { \pi }{ 2 } +2k\pi \right) isin\left( -\frac { \pi }{ 2 } +2k\pi \right) \right) \),k\(\in Z\)
\(z=\sqrt [ 3 ]{ 8 } \left( cos\left( \frac { -\pi +4k\pi }{ 6 } \right) +isin\left( \frac { -\pi +4k\pi }{ 6 } \right) \right) \)
Taking k = 0, 1, 2 we get,
k = 0, \(z=2\left( cos\left( -\frac { \pi }{ 6 } \right) +isin\left( -\frac { \pi }{ 6 } \right) \right) =2\left( -\frac { 1 }{ 2 } -i\frac { \sqrt { 3 } }{ 2 } \right) =2\left( \frac { \sqrt { 3 } }{ 2 } -i\frac { 1 }{ 2 } \right) \)
k = 1, \(z=2\left( cos\left( \frac { \pi }{ 2 } \right) +isin\left( \frac { \pi }{ 2 } \right) \right) =2=\left( 0+i \right) =0+2i=2i\)
k = 2,\(z=2\left( xcos\left( \frac { 7\pi }{ 6 } \right) +isim\left( \frac { 7\pi }{ 6 } \right) \right) =2\left( cos\left( \pi +\frac { \pi }{ 6 } \right) \right) +isin\left( \pi +\frac { \pi }{ 6 } \right) \)
= \(2\left( -cos\left( \frac { \pi }{ 6 } \right) -isin\left( \frac { \pi }{ 6 } \right) \right) =2\left( -\frac { \sqrt { 3 } }{ 2 } -i\frac { 1 }{ 2 } \right) =-\sqrt { 3 } -i\)
The values of z are \(\sqrt { 3 } -i,2i\) and \(-\sqrt { 3 } -i\)

Let the cross section of the satellite dish be an right open parabola.

Its equation is y² = 4ax
Since focus is placed 1.2 m from the vertex OA = 1.2 m and BC = 2.5 m since the width of the dish is 5 m.
From the diagram, a = 1.2 m
∴ y² = 4(1.2)x
(a) ⇒ y² = 4.8x   ...(1)
(b) Since (x₁, 2.5) lines on (1)(2.5)² = 4.8(x₁)
x₁ = \(\frac { 2.5\times 2.5 }{ 4.8 } \)
x₁ = 1.3 m
∴ Depth of the satellite dish at the vertex is 1.3 m.

Here the number of unknowns is 3.
The matrix form of the system is AX = B, where A = \(\left[ \begin{matrix} 1 & 2 & 1 \\ 1 & 1 & \lambda \\ 1 & 3 & -5 \end{matrix} \right] \), X = \(\left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] \), B = \(\left[ \begin{matrix} 7 \\ \mu \\ 5 \end{matrix} \right] \).
Applying elementary row operations on the augmented matrix [A | B], we get
[A | B] = \(\left[ \begin{matrix} 1 & 2 & 1 \\ 1 & 1 & \lambda \\ 1 & 3 & -5 \end{matrix}|\begin{matrix} 7 \\ \mu \\ 5 \end{matrix} \right] \overset { { R }_{ 2 }\leftrightarrow { R }_{ 3 } }{ \longrightarrow } \left[ \begin{matrix} 1 & 2 & 1 \\ 1 & 3 & -5 \\ 1 & 1 & \lambda \end{matrix}|\begin{matrix} 7 \\ 5 \\ \mu \end{matrix} \right] \overset { \begin{matrix} { R }_{ 2 }\longrightarrow { R }_{ 2 }-{ R }_{ 1 }, \\ { R }_{ 3 }\longrightarrow { R }_{ 3 }-{ R }_{ 1 } \end{matrix} }{ \longrightarrow } \left[ \begin{matrix} 1 & 2 & 1 \\ 0 & 1 & -6 \\ 0 & -1 & \lambda -1 \end{matrix}|\begin{matrix} 7 \\ -2 \\ \mu -7 \end{matrix} \right] \overset { { R }_{ 3 }\longrightarrow { R }_{ 3 }+{ R }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} 1 & 2 & 1 \\ 0 & 1 & -6 \\ 0 & 0 & \lambda -7 \end{matrix}|\begin{matrix} 7 \\ -2 \\ \mu -9 \end{matrix} \right] \).
(i) If λ =7 and μ \(\neq\) 9, then ρ(A) = 2 and ρ([A | B]) = 3. So ρ(A) ≠ ρ([A | B]) Hence the given system is inconsistent and has no solution.
(ii) If λ ≠ 7 and μ is any real number, then ρ(A) = 3 and ρ([A | B]) = 3.
So, ρ(A) = ρ([A | B]) = 3 = Number of unknown. Hence the given system is consistent and has a unique solution.
(iii) If λ = 7 and μ = 9, then ρ(A) = 2 and ρ([A | B]) = 2.
So, ρ(A) = ρ([A | B]) = 2 < Number of unknown. Hence the given system is consistent and has infinite number of solutions.

Now, \(tan^{ -1 }\left( \frac { x-1 }{ x-2 } \right) +tan^{ -1 }\left( \frac { x+1 }{ x+2 } \right) =tan^{ -1 }\left[ \frac { \frac { x-1 }{ x-2 } +\frac { x+1 }{ x+2 } }{ 1-\frac { x-1 }{ x-2 } \left( \frac { x+1 }{ x+2 } \right) } \right] =\frac { \pi }{ 4 } \)
Thus, \(\frac { \frac { x-1 }{ x-2 } +\frac { x+1 }{ x+2 } }{ 1-\frac { x-1 }{ x-2 } \left( \frac { x+1 }{ x+2 } \right) } \) = 1, which on simplification gives 2x²−4 = −3
Thus, x² = \(\frac{1}{2}\)gives x = \(\pm \frac { 1 }{ \sqrt { 2 } } \)

Let P(x) = ax²+ bx + c
Given P(-3) = 21
[∵ P(x) ÷ x + 3, the remainder is 21]
⇒ a(-3)2 + b(-3) + c = 21
⇒ 9a - 3b + c = 21
Also, P(5) = 61
⇒ a(5)2 + b(5) + c = 61
[using remainder theorem]
⇒ 25a +5b + c = 61..........(2)
and P(1) = 9
⇒ a(1)2 + b(1) + c = 9
⇒ a + b + c = 9 ............(3)
Reducing the augment matrix to an equivalent row-echelon form using elementary row operations, we get
\(\left[ \begin{matrix} 9 & - & 1 \\ 25 & 5 & 1 \\ -1 & 1 & 1 \end{matrix}|\begin{matrix} 21 \\ 61 \\ 9 \end{matrix} \right] \overset { { R }_{ 1 }\leftrightarrow { R }_{ 3 } }{ \longrightarrow } \left[ \begin{matrix} 1 & 1 & 1 \\ 25 & 5 & 1 \\ 9 & -3 & 1 \end{matrix}|\begin{matrix} 9 \\ 61 \\ 21 \end{matrix} \right] \)
\(\overset { { R }_{ 3 }\rightarrow { R }_{ 3 }-9{ R }_{ 1 }\\ { R }_{ 2 }\rightarrow { R }_{ 2 }-25{ R }_{ 1 } }{ \longrightarrow } \left[ \begin{matrix} 1 & 1 & 1 \\ 0 & -20 & -24 \\ 0 & -12 & -8 \end{matrix}|\begin{matrix} 9 \\ -164 \\ -60 \end{matrix} \right] \)
\(\overset { { R }_{ 2 }\rightarrow { R }_{ 2 }\div \\ { R }_{ 3 }\rightarrow { R }_{ 3 }\div 4 }{ \longrightarrow } \left[ \begin{matrix} 1 & 1 & 1 \\ 0 & -5 & -6 \\ 0 & -3 & -2 \end{matrix}|\begin{matrix} 9 \\ -41 \\ -15 \end{matrix} \right] \)
\(\overset { { R }_{ 3 }\rightarrow { R }_{ 3 }-\frac { 3 }{ 5 } { R }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} 1 & 1 & 1 \\ 0 & -5 & -6 \\ 0 & 0 & \frac { 8 }{ 5 } \end{matrix}|\begin{matrix} 9 \\ -41 \\ \frac { 48 }{ 5 } \end{matrix} \right] \)
\(\overset { { R }_{ 3 }\rightarrow 5{ R }_{ 3 } }{ \longrightarrow } \left[ \begin{matrix} 1 & 1 & 1 \\ 0 & -5 & -6 \\ 0 & 0 & 8 \end{matrix}|\begin{matrix} 9 \\ -41 \\ 48 \end{matrix} \right] \)
Writing the equivalent equations from the row-echelon matrix we get,
a + b + c = 9 ............(1)
5b + 6c = 41 ................(2)
-8c = -48
⇒ c = 6
Substituting c = 6 
⇒  5b + 36 = 41
⇒ 5b = 5  
b = 1
Substituting b = 1, c = 6
a + 1 + 6 = 9
⇒ a + 7 = 9
⇒ a = 9 - 7
⇒ a = 2
∴ a = 2, b = 1, and c = 6

x = 1 
y = x² -5x + 6
y = 1 -5 + 6 = 2
x = 2
y = 4 -10 + 6 = 0
x = 0 
y = 6
x = 3
y = 9 -15 + 6 = 12
x = -1 
y = 1 + 5 + 6 = 12
x = 4
y = 16 - 20 + 6 = 2

(1, 2), (0, 6), (-1, 12)
P(x) = (x²-5x + 6) (x²-5x+16)
= x⁴- 5x³+16x²-5x+25x²- 80x + 6x²- 30x + 96 = 0
x⁴-10x³+ 47x² -110x + 96 = 0
It has two sign changes
\(\therefore\) it has two positive real roots
P(-x) = x⁴-10x³+ 47x² -110x + 96
It has no sign changes, no negative real roots
y = x²- 5x + 16

Given z = x + iy and arg\(\left( \frac { z-i }{ z+2 } \right) =\frac { \pi }{ 4 } \)
⇒ arg(z-i) - arg(z+2) = \(\frac { \pi }{ 4 } \)
⇒ arg(x + iy-i) - arg(x+iy+2) = \(\frac { \pi }{ 4 } \)
⇒ arg(x+i(y-1)-arg((x+2)+iy) = \(\frac { \pi }{ 4 } \)
⇒ \(tan^{ -1 }\left( \frac { y-1 }{ x } \right) -tan^{ -1 }\left( \frac { y }{ x+2 } \right) \) = \(\frac { \pi }{ 4 } \)
⇒ \(tan^{ -1 }\left( \frac { \frac { y-1 }{ x } -\frac { y }{ x+2 } }{ 1+\frac { y-1 }{ x } .\frac { y }{ x+2 } } \right) \)
= \(\frac { \pi }{ 4 } \)\(\left[ \because tan^{ -1 }x-tan^{ -1 }y=tan^{ -1 }\left( \frac { x-y }{ 1+xy } \right) \right] \)
\(\Rightarrow \frac{\left(\frac{(x+2)(y-1)- x y}{\not {x (\not x+\not2)}}\right)}{\left(\frac{x(x+2)+y(y-1)}{\not x(\not x+\not 2)}\right)}=\tan \frac{\pi}{4}=1\)
⇒ \(\frac { (x+2)(y-1)-xy }{ x(x+2)+y(y-1) } \) = 1
⇒ -x + 2y-2 = x²+ 2x + y²-y
⇒ x² + 2x + y²-y + x-2y + 2 = 0
⇒ x² + y²+3x-3y + 2 = 0
Hence proved.

Let the prices per unit for the commodities A, B and C be Rs. x, Rs. y and Rs. z.
By the given data,
2x - 4y + 5z = 15000
3x + y - 2z = 1000
-x + 3y + z = 4000
The matrix form of the system of equations is
\(\left[ \begin{matrix} 2 & -4 & 5 \\ 3 & 1 & -2 \\ -1 & 3 & 12 \end{matrix} \right] \left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] =\left[ \begin{matrix} 15000 \\ 1000 \\ 4000 \end{matrix} \right] \)
AX = B where A =\(\left[ \begin{matrix} 2 & -4 & 5 \\ 3 & 1 & -2 \\ -1 & 3 & 1 \end{matrix} \right] ,X=\left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] \)
and B =\(\left[ \begin{matrix} 15000 \\ 1000 \\ 4000 \end{matrix} \right] \)
⇒ X = A^-1B
|A| = \(\left| \begin{matrix} 2 & -4 & 5 \\ 3 & 1 & -2 \\ -1 & 3 & 1 \end{matrix} \right| \)
= \(2\left| \begin{matrix} 1 & -2 \\ 3 & 1 \end{matrix} \right| +4\left| \begin{matrix} 3 & -2 \\ -1 & 1 \end{matrix} \right| +5\left| \begin{matrix} 3 & 1 \\ -1 & 3 \end{matrix} \right| \)
= 2 (1 + 6) + 4 (3 - 2) + 5 (9 + 1)
= 2 (7) + 4 (1) + 5(10) = 14 + 4 + 50 = 68.
adj A = \(\left[ \begin{matrix} +\left| \begin{matrix} 1 & -2 \\ 3 & 1 \end{matrix} \right| & -\left| \begin{matrix} 3 & -2 \\ -1 & 1 \end{matrix} \right| & +\left| \begin{matrix} 3 & 1 \\ -1 & 3 \end{matrix} \right| \\ -\left| \begin{matrix} -4 & 5 \\ 3 & 1 \end{matrix} \right| & +\left| \begin{matrix} 2 & 5 \\ -1 & 1 \end{matrix} \right| & -\left| \begin{matrix} 2 & -4 \\ -1 & 3 \end{matrix} \right| \\ +\left| \begin{matrix} -4 & 5 \\ 1 & -2 \end{matrix} \right| & -\left| \begin{matrix} 2 & 5 \\ 3 & -2 \end{matrix} \right| & +\left| \begin{matrix} 2 & -4 \\ 3 & 1 \end{matrix} \right| \end{matrix} \right] ^{ T }\)
= \(\left[ \begin{matrix} +(1+6) & -(3-2) & +(9+1) \\ -(-4-15) & +(2+5) & -(6-4) \\ +(8-5) & -(4-15) & +(2+12) \end{matrix} \right] \)
= \(\left[ \begin{matrix} 7 & -1 & 10 \\ 19 & 7 & -2 \\ 3 & 19 & 14 \end{matrix} \right] ^{ T }=\left[ \begin{matrix} 7 & 19 & 3 \\ -1 & 7 & 19 \\ 10 & -2 & 14 \end{matrix} \right] \)
∴ A^-1 = \(\frac { 1 }{ |A| } adj=\frac { 1 }{ 68 } \left[ \begin{matrix} 7 & 19 & 3 \\ -1 & 7 & 19 \\ 10 & -2 & 14 \end{matrix} \right] \)
∴ X = A^-1B = \(\frac { 1 }{ 68 } \left[ \begin{matrix} 7 & 19 & 3 \\ -1 & 7 & 19 \\ 10 & -2 & 14 \end{matrix} \right] \left[ \begin{matrix} 15000 \\ 1000 \\ 4000 \end{matrix} \right] \)
= \(\frac { 1 }{ 68 } \left[ \begin{matrix} 105000+19000+12000 \\ -15000+7000+76000 \\ 150000-2000+56000 \end{matrix} \right] \)
= \(\frac { 1 }{ 68 } \left[ \begin{matrix} 136000 \\ 68000 \\ 204000 \end{matrix} \right] =\left[ \begin{matrix} 2000 \\ 1000 \\ 3000 \end{matrix} \right] \)
∴ x = 2000, y = 1000, z = 3000
Hence the prices per unit of the commodities A, B and C are Rs. 2000, Rs. 1000 and Rs. 3000 respectively.

By the definition of a conic  \(\frac{SP}{PM}\)= e or SP² = e²PM²
Then, (x−2)² + (y−3)² = \(\frac { 1 }{ 4 } \) (x-7)² 
3x²+ 4y²−2x − 24y + 3 = 0
\({ 3\left( x-\frac { 1 }{ 3 } \right) }^{ 2 }+4(y-3)^{ 2 }=3\left( \frac { 1 }{ 9 } \right) +4\times 9-3=\frac { 100 }{ 3 } \)
\(\frac { { \left( x-\frac { 1 }{ 3 } \right) }^{ 2 } }{ \frac { 100 }{ 9 } } +\frac { (y-3{ ) }^{ 2 } }{ \frac { 100 }{ 12 } } \) = 1 which is in the standard form.
Therefore, the length of major axis = 2a = 2\(\sqrt { \frac { 100 }{ 9 } = } \frac { 20 }{ 3 } \) and
the length of minor axis = 2b = 2\(\sqrt { \frac { 100 }{ 12 } = } \frac { 10 }{ \sqrt { 3 } } \).

Since the coefficients of the equation are all rational numbers, 2+i and 3-\(\sqrt{2}\) are roots, we get 2-i and 3+\(\sqrt{2}\) are also roots of the given equation. Thus (x-(2+i)), (x-(2-i)), (x-(3-\(\sqrt{2}\))) and (x-(3+\(\sqrt{2}\))) are factors. Thus their product.
((x-(2+i))(x-(2-i))(x-(3-\(\sqrt{2}\)))(x-(3+\(\sqrt{2}\))) is a factor of the given polynomial equation.
That is, (x²-4x+5)(x²-6x+7) is a factor. Dividing the given polynomial equation by this factor, we get the other factor as (x²-3x-4) which implies that 4 and −1 are the other two roots. Thus
2+i, 2-i, 3+\(\sqrt{2}\), 3-\(\sqrt{2}\), -1, and 4 are the roots of the given polynomial equation.

Given
\(f(x)=\frac { { x }^{ 2 }+6x-4 }{ 3x-6 } \)
\(\underset { x\rightarrow { 2 }^{ + } }{ lim } \frac { { x }^{ 2 }+6x-4 }{ 3x-6 } =\underset { h\rightarrow 0^{ + } }{ lim } \frac { (2+h)^{ 2 }+6(2+h)-4 }{ 3(2+h)-6 } \) 
= \(\underset { h\rightarrow 0^{ + } }{ lim } \frac { (2+h)^{ 2 }+6(2+h)-4 }{ 6+3h-6 } \)
= -∞
ஃ x = 2 is the vertical asymptote,
Also

\(\therefore y=\frac { 1 }{ 3 } x+\frac { 8 }{ 3 } \) is the slanting asymptote.

If \(\vec { r } =x\hat { i } +y\hat { j } +z\hat { k } \) is the position vector of an arbitrary point (x, y, z) on the plane, then the given equation can be written as \((x\hat { i } +y\hat { j } +z\hat { k } ).(3\hat { i } -4\hat { j } +3\hat { k } )=-8\) or \((x\hat { i } +y\hat { j } +z\hat { k } ).(-3\hat { i } +4\hat { j } -3\hat { k } )=8\). 
That is, \(\hat { r } .(-3\hat { i } +4\hat { j } -3\hat { k } )=8\) which is the vector equation of the given plane in standard form.