Given \(\frac { dy }{ dx } =-\left( \frac { y+3 }{ x+2 } \right) \)
⇒ \(\frac { dy }{ y+3 } =-\frac { dx }{ x+2 } \)
⇒ \(\int { \frac { dy }{ y+3 } } =-\int { \frac { dx }{ x+2 } } \)
⇒ log(y + 3) = -log (x + 2) + log c
⇒ log(y + 3) + log (x + 2) = log c
⇒ log(x + 2) (y + 3) = log c
⇒ (x+2)(y+3) = c ..(1)
Since the curve passes through (0, 0)
(0 + 2)(0 + 3) = c ⇒ c = 6
(1) becomes,
(x + 2)(y + 3) = 6
⇒ xy + 3x + 2y + 6 = 6
⇒ xy + 3x + 2y = 0

Given s is a non-empty set and x 0 y = x, x,y \(\in \) s y₀x = y
x₀u ≠ y₀x ⇒ is not commutative
Now, x₀(y₀z) = x₀y = x
and (x₀y) 0 z = x₀z = x
x₀(y₀z) = (y₀z) 0 z
0 is associative.

Given sides are y = 2x + 1.....(1)
y = 3x + 1...(2)
x = 4...(3)
Solving (1) & (2), x = 0, y = 1
Solving (2) & (3), x = 4, y = 13
Solving (1) & (3), x = 4, y = 9

∴ Required area \(\int _{ 0 }^{ 4 }{ (3x+1) } dx-\int _{ 0 }^{ 4 }{ (2x+1) } dx\)
\({ =\left( \frac { { 3x }^{ 2 } }{ 2 } +x \right) }_{ 0 }^{ 4 }-{ \left( \frac { { 2x }^{ 2 } }{ 2 } +x \right) }_{ 0 }^{ 4 }\)
\(=\left( \frac { 48 }{ 2 } +4 \right) -(16+4)=28-20\)
Area = 8 sq. units

Given w(x, y, z) = \(\frac { 1 }{ \sqrt { { x }^{ 2 }+{ y }^{ 2 }+{ z }^{ 2 } } }\)
= (x² + y² + z²) -^\(\frac12\)
\(\frac { \partial w }{ \partial x } =\frac { -1 }{ 2 } ({ x^{ 2 }+{ y }^{ 2 }+{ z }^{ 2 }) }^{ -\frac { 3 }{ 2 } }(2x)\)
= (-x² + y² + z²) -^\(\frac12\)
\(\frac { { \partial }^{ 2 }w }{ \partial { x }^{ 2 } } =\frac { \partial }{ \partial x } \left( \frac { \partial w }{ \partial x } \right) \)
= -[x\(\left( \frac { -3 }{ 2 } \right) \)( x² + y² +z²)^\(-\frac32\)
\((\not 2 x)+\left(x^{2}+y^{2}+z^{2}\right)^{\frac{3}{2}}\)
= (x² + y² + z²)^-\(\frac52\) [-3x² + x² +y + z²]
= - (x² + y² + z²)^-\(\frac52\) [y² + z² - 2x²] ....(1)
\(\frac { \partial w }{ \partial y } =\frac { -1 }{ 2 } ({ x^{ 2 }+{ y }^{ 2 }+{ z }^{ 2 }) }^{ -\frac { 3 }{ 2 } }(2y)\)
= -y(x² +y² + z²)-^\(\frac32\) 
\(\frac { { \partial }^{ 2 }w }{ \partial { y }^{ 2 } } =\frac { \partial }{ \partial y } \left( \frac { \partial w }{ \partial y } \right) \)
\(=-\left[y\left(\frac{-3}{\not 2}\right)\left(x^{2}+y^{2}+z^{2}\right)^{\frac{5}{2}}(\not 2 y)+\left(x^{2}+y^{2}+z^{2}\right)^{\frac{3}{2}}(1)\right]\)
= -(x² + y² + z²)-^{\(\frac52 \)} 
= -(x² + y² + z²)-^{\(\frac52 \)} [3y² + x² + y² + z²]
= -(x² + y² + z²)-^{\(\frac52 \)} [x² + y² + 2z²] ....(2)
Now \(\frac { \partial w }{ \partial z } =\frac { -1 }{ 2 } ({ x^{ 2 }+{ y }^{ 2 }+{ z }^{ 2 }) }^{ -\frac { 3 }{ 2 } }(2z)\)
= -z(x² + y² + z²)-^{\(\frac32 \) 
∴} \(\frac { { \partial }^{ 2 }w }{ \partial { z }^{ 2 } } \) = -(x² + y² + z²)-^{\(\frac52 \)} [x² + y² - 2z²] ...(3)
(1)+(2)+(3)⟶
^∴ \(\frac { { \partial }^{ 2 }w }{ \partial { x }^{ 2 } } +\frac { { \partial }^{ 2 }w }{ \partial { y }^{ 2 } } +\frac { { \partial }^{ 2 }w }{ \partial { y }^{ 2 } } \) ⁼ -(x² + y² + z²)-^{\(\frac52 \)} [y² + z² - 2x² + x² + z² - 2y + x²+ y-2z²]
= -(x² + y² + z²)-^{\(\frac52 \)}(0) = 0
Hence proved 

Given \(\begin{cases} \begin{matrix} k & 200\le x\le 600 \end{matrix} \\ \begin{matrix} 0 & otherwise \end{matrix} \end{cases}\) 
(i) Since f{x) is a probability density function
\(\int _{ -\infty }^{ \infty }{ f(x)d=1\Rightarrow \int _{ 200 }^{ 600 }{ kda } =1 } \) 
\(\Rightarrow k[x]_{ 200 }^{ 600 }=1\Rightarrow k(600-200)=1\) 
400 k = 1
\(\Rightarrow k=\frac { 1 }{ 400 } \) 
(ii) The distribution function  
= \(\int _{ -\infty }^{ x }{ f(u) } du\) 
Case 1: x < 200
\(F(x) =\int _{ -\infty }^{ u }{ du } =0\) 
Case 1: x < 200 ≤ x ≤  600
\(\int _{ -\infty }^{ x }{ f(u) } du\)
\(F(x)=\int _{ -\infty }^{ 200 }{ f(u)du } =+\int _{ 200 }^{ x }{ f(u)du } \)  
= \( =\frac { 1 }{ 400 }(x-200) =\frac { x }{ 400 } =\frac { 1 }{ 2 } \) 
Case 3: x > 600
\(F(x) =\int _{ -\infty }^{ u }{ du } =0\) 
\(f(x)= \begin{cases}0, & x<200 \\ \frac{x}{400}-\frac{1}{2}, & 200 \leq x \leq 600 \\ 0, & x>600\end{cases}\)
(iii) P(300 < x < 500)
= \(\int _{ 300 }^{ 500 }{ kdx=\frac { 1 }{ 400 } \left[ x \right] _{ 300 }^{ 500 } } \) 
= \(\frac { 1 }{ 400 } \left[ 500-300 \right] =\frac { 200 }{ 400 } =\frac { 1 }{ 2 } \)

\(\Rightarrow \frac { dy }{ dx } =\frac { { y }^{ 2 }-2xy }{ { x }^{ 2 }-2xy } \)  ...(1)
This is a homogeneous differential equation
\(\therefore put\ y=vx\Rightarrow \frac { dy }{ dx } =v+x\frac { dv }{ dx } \)
\(\therefore\) (1) becomes,
\(v+x\frac { dv }{ dx } =\frac { { v }^{ 2 }{ x }^{ 2 }-2xvx }{ { x }^{ 2 }-2xvx } \)

\(\Rightarrow x\frac { dv }{ dx } =\frac { { v }^{ 2 }-2v }{ 1-2v } -v\)
\(=\frac { { v }^{ 2 }-2v-v+2{ v }^{ 2 } }{ 1-2v } \)
\(=\frac { { 3v }^{ 2 }-3v }{ 1-2v } \)
Separating the variables we get,
\(\frac { 1-2v }{ 3{ v }^{ 2 }-3v } dv=\frac { dx }{ x } \)
\(\frac { 6v-3 }{ 3{ v }^{ 2 }-3v } dv=-3\frac { dx }{ x } \)
Integrating on both sides,
\( \int \frac{(2 v-1) d v}{\left(v^2-v\right)} =\int-3 \frac{d x}{x} \)
\(\log \left(v^2-v\right) =-3 \log x+\log \mathrm{c} \)
\(\log \left(v^2-v\right)+\log \left(x^3\right) =\log \mathrm{c} \)
\(\left(v^2-v\right)\left(x^3\right) =\mathrm{c} \)
\(\left(\frac{y^2}{x^2}-\frac{y}{x}\right) x^3 =c \)
\(x y^2-x^2 \mathrm{y} =\mathrm{c}\)

Since the beacon makes one revolution (360°) in 10 sec.

At the point of observation, let 0 be the angle of deviation of the beacon of light from OA.
Let AB x km,when \(\angle\)AOB = 0
\(\frac { d\theta }{ dt } =\frac { 2\pi }{ 10 } \)
\(=\frac { \pi }{ 5 } \) rad /sec.
Let AB = x
Then, tan θ \( =\frac { x }{ 5 } \)
x = 5 tan θ
We know that velocity = \(\frac { dx }{ dt } \)
Differentiating (1) with respect to 't' we get,
\(\frac { dx }{ dt } =5{ sec }^{ 2 }\theta ,\frac { d\theta }{ dt } \)
\(=5({ sec }^{ 2 }{ 45 }^{ o })\left( \frac { \pi }{ 5 } \right) \)
\(=5({ \sqrt { 2 } })^{ 2 }\left( \frac { \pi }{ 5 } \right) \)
\(=5(2)\left( \frac { \pi }{ 5 } \right) \)
= 2π km/sec.

\(\frac { { (y-2) }^{ 2 } }{ 25 } -\frac { ({ x+1) }^{ 2 } }{ 16 } =1\)
Given equatlon is \(\frac { { (y-2) }^{ 2 } }{ 25 } -\frac { ({ x+1) }^{ 2 } }{ 16 } =1\)
Thi is an equation of the hyperbola where transverse axis is parallel to y-axis.
∴ a² = 25, b² = 16
⇒ c² - a² + b² = 25 +16 = 41
⇒ c = \(\sqrt { 41 } \)
\(e =\sqrt { 1-\frac { { b }^{ 2 } }{ { a }^{ 2 } } } =\sqrt { 1+\frac { 16 }{ 25 } } =\sqrt { \frac { 41 }{ 25 } } =\frac { \sqrt { 41 } }{ 5 } \)
(a) Center is (-1, 2) ⇒ h = -1, k = 2
(b) Foci are (h, k + c), (h, k- c)
⇒ (-1, 2 + \(\sqrt { 41 } \)), (-1, 2 - \(\sqrt { 41 } \))
(c) Vertices are (h, k + a), (h, k-a)
⇒ (-1, 2 + 5), (-1, 2 - 5)
= (-1, 7), (-1, -3)
(d) Equation of directrices are
\(y-2=\pm \frac { 5 }{ \frac { \sqrt { 41 } }{ 5 } } \)
\(\Rightarrow y-2=\pm \frac { 25 }{ \sqrt { 41 } } \)
\(\Rightarrow y=2+\frac { 25 }{ \sqrt { 41 } } \) and 
\( y=2-\frac { 25 }{ \sqrt { 41 } } \)

arg(1+i) + arg(1-i) = arg[(1+i) (1-i)]
LHS = arg (1+i) + arg(1-i)
1+i = \(\sqrt { 2 } \left( \frac { 1 }{ \sqrt { 2 } } +\frac { i }{ \sqrt { 2 } } \right) \)
= \(\sqrt { 2 } \left( cos\frac { \pi }{ 4 } +isin\frac { \pi }{ 4 } \right) \)
∴ arg (1+i) = π/4
-1+i =\(\sqrt { 2 } \left( \frac { -1 }{ \sqrt { 2 } } +\frac { i }{ \sqrt { 2 } } \right) \)
= \(\sqrt { 2 } \left( cos3\frac { \pi }{ 4 } +isin3\frac { \pi }{ 4 } \right) \)
∴ (-1+i) = 3\(\frac { \pi }{ 4 } \)
∴ LHS = \(\frac { \pi }{ 4 } +\frac { 3\pi }{ 4 } =\frac { 4\pi }{ 4 } =\pi \)
RHS = arg[(1+i) (-1+i)]
= arg[-1-i + i + i²]
= (-1-i + i-1) = arg(-2)
= arg(2) - (1) = 2 arg(-1)
= 2 (cos π + isin π) = π
∴ LHS = RHS

Given F₁P + F₂P = 8
By the focal property of ellipse
F₁P + F₂P = 2a
∴ 2a = 8 ⇒ a = 4
and distance between the foci = F₁F₂ = 6
2ae = 6 ⇒ ae = 3
∴ 4(e) = 3 ⇒ e \(\frac34\)

∴ b² = a²(1- e²)
= \(16\left( { 1-\left( \frac { 3 }{ 4 } \right) }^{ 2 } \right) =16\left( 1-\frac { 9 }{ 10 } \right) =16\left( \frac { 7 }{ 16 } \right) =7\)
∴ The path traced by him is an ellipse and its equation is \(\frac { { x }^{ 2 } }{ { a }^{ 2 } } +\frac { { y }^{ 2 } }{ { b }^{ 2 } } =1\)
⇒ \(\frac { { x }^{ 2 } }{ 16 } -\frac { { y }^{ 2 } }{ 7 } \) = 1

The domain of sin^-1x is [-1, 1]
\(\therefore\) f(x) = sin^-1(2x - 3) is defined for all x, satisfying
\(-1\le 2x-3\le 1\)
\(\Rightarrow 3-1\le 2x\le 1+3\)
\(\Rightarrow 2\le 2x\le 4\Rightarrow 1\le x\le 2\Rightarrow x\epsilon \left[ 1,2 \right] \)
\(\therefore\) Domain of f(x) = sin^-1(2x - 3) is [1, 2].
(ii) The domain of f(x) is [-1, 1] and that of cosx is R
\(\therefore\) Domain of f(x) = sin^-1x + cos x is
\(\left[ -1,1 \right] \cap R=\left[ -1,1 \right] \)

Given \(\overset { \rightarrow }{ AB } =\overset { \rightarrow }{ \alpha } \), \(\overset { \rightarrow }{ AD } =\overset { \rightarrow }{ \beta } \) and \(\overset { \rightarrow }{ AC } =2\overset { \rightarrow }{ \alpha } +3\overset { \rightarrow }{ \beta } \)
Area of the quadrilateral ABCD 
∴ = are of ∆ ABC + area of ∆ ACD
\(=\frac { 1 }{ 2 } \left| \overset { \rightarrow }{ AB } \times \overset { \rightarrow }{ AC } \right| +\frac { 1 }{ 2 } \left| \overset { \rightarrow }{ AC } \times \overset { \rightarrow }{ AD } \right| \)
\(=\frac { 1 }{ 2 } \left| \overset { \rightarrow }{ \alpha } \times \left( 2\overset { \rightarrow }{ \alpha } +3\overset { \rightarrow }{ \beta } \right) \right| +\frac { 1 }{ 2 } \left| \left( 2\overset { \rightarrow }{ \alpha } +3\overset { \rightarrow }{ \beta } \right) \times \overset { \rightarrow }{ \beta } \right| \)
\(=\frac { 1 }{ 2 } \left| 2\left( \overset { \rightarrow }{ \alpha } \times \overset { \rightarrow }{ \alpha } \right) +3\left( \overset { \rightarrow }{ \alpha } \times \overset { \rightarrow }{ \beta } \right) \right| +\frac { 1 }{ 2 } \left| 2\left( \overset { \rightarrow }{ \alpha } \times \overset { \rightarrow }{ \beta } \right) +3\left( \overset { \rightarrow }{ \beta } \times \overset { \rightarrow }{ \beta } \right) \right| \)
\(=\frac { 1 }{ 2 } \left| 3\left( \overset { \rightarrow }{ \alpha } \times \overset { \rightarrow }{ \beta } \right) \right| +\frac { 1 }{ 2 } \left| 2\left( \overset { \rightarrow }{ \alpha } \times \overset { \rightarrow }{ \beta } \right) \right| \quad \quad \quad \left[ \because \overset { \rightarrow }{ \alpha } \times \overset { \rightarrow }{ \alpha } =\overset { \rightarrow }{ \beta } \times \overset { \rightarrow }{ \beta } =0 \right] \)
\(=\left( \frac { 3 }{ 2 } +\frac { 2 }{ 2 } \right) \left( \overset { \rightarrow }{ \alpha } \times \overset { \rightarrow }{ \beta } \right) =\left( \frac { 5 }{ 2 } \right) \left| \overset { \rightarrow }{ \alpha } \times \overset { \rightarrow }{ \beta } \right| \quad \quad (1)\)
Now, Area of the parallelogram with \(\overset { \rightarrow }{ AB } \) and \(\overset { \rightarrow }{ AD } \) as
adjacent sides = \(\left| \overset { \rightarrow }{ AB } \times \overset { \rightarrow }{ AD } \right| =\left| \overset { \rightarrow }{ \alpha } \times \overset { \rightarrow }{ \beta } \right| .... (2)\)
From (1) & (2), \(\frac { 5 }{ 2 } \left| \overset { \rightarrow }{ \alpha } \times \overset { \rightarrow }{ \beta } \right| =\lambda \left| \overset { \rightarrow }{ \alpha } \times \overset { \rightarrow }{ \beta } \right| \) [Given]
\(\lambda =\frac { 5 }{ 2 } \)

Augmented matrix [A|B] is \(\left[ \begin{matrix} -2 & 1 & 1 \\ 1 & -2 & 1 \\ 1 & 1 & -2 \end{matrix}|\begin{matrix} a \\ b \\ c \end{matrix} \right] \)
[A|B]\(\overset { { R }_{ 1 }\leftrightarrow { R }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} 1 & -2 & 1 \\ -2 & 1 & 1 \\ 1 & 1 & -2 \end{matrix}|\begin{matrix} b \\ a \\ c \end{matrix} \right] \)
\(\overset { { R }_{ 2 }\rightarrow { R }_{ 2 }+2{ R }_{ 1 }\\ { R }_{ 3 }\rightarrow { R }_{ 3 }+{ R }_{ 1 } }{ \longrightarrow } \left[ \begin{matrix} 1 & -2 & 1 \\ 0 & -3 & 3 \\ 0 & 3 & -3 \end{matrix}|\begin{matrix} b \\ a+2b \\ c-b \end{matrix} \right] \)
\(\overset { { R }_{ 3 }\rightarrow { R }_{ 3 }+{ R }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} 1 & -2 & 1 \\ 0 & -3 & 3 \\ 0 & 0 & 0 \end{matrix}|\begin{matrix} b \\ c+2b \\ a+b+c \end{matrix} \right] \)
Here \(\rho\) (A) = 2
The given system is consistent only when \(\rho\)([A|B]) = 2\(\rho\)([A|B]) = 2 only if a + b + c = 0 Hence proved.

We have to find \(1^{\frac{1}{4}}\). Let z⁴ = \(1^{\frac{1}{4}}\). Then z⁴ = 1.
In polar form, the equation z = 1 can be written as
\(z^4=cos\left( 0+2k\pi \right) +isin\left( 0+2k\pi \right) ={ e }^{ i2k\pi }\), k = 0, 1, 2,...
Therefore,\({ \left( z \right) }^{ \frac { 1 }{ 4 } }=cos\left( \frac { 2k\pi }{ 4 } \right) +isin\left( \frac { 2k\pi }{ 4 } \right) ={ e }^{ i\frac { 2k\pi }{ 4 } }\), k=0,1,2,3.
Taking k = 0, 1, 2, 3, we get
k = 0, z = cos 0 + isin 0 = 1
k = 1, \(z=cos\left( \frac { \pi }{ 2 } \right) +isin\left( \frac { \pi }{ 2 } \right) =i\)
k = 2, \(z=cos\pi +isin\pi =-1\)
k = 3, \(z=cos\frac { 3\pi }{ 2 } +isin\frac { 3\pi }{ 2 } =-cos\frac { \pi }{ 2 } -isin\frac { \pi }{ 2 } =-i\)
Fourth roots of unity are 1, i, −1, −i \(\Rightarrow\) 1, \(\omega \), \({ \omega }^{ 2 }\) and \({ \omega }^{ 3 }\), where \(\omega ={ e }^{ i\frac { 2\pi }{ 4 } }=i\)

Now, \(tan^{ -1 }\left( \frac { d }{ 1+{ a }_{ 1 }{ a }_{ 2 } } \right) =tan^{ -1 }\frac { { a }_{ n }-{ a }_{ 1 } }{ 1+{ a }_{ 1 }{ a }_{ n } } =tan^{ -1 }{ a }_{ 2 }-tan^{ -1 }{ a }_{ 1 }\)
Similarly, \(tan^{ -1 }\left( \frac { d }{ 1+{ a }_{ 1 }{ a }_{ 2 } } \right) =tan^{ -1 }\left( \frac { { a }_{ n }-{ a }_{ 1 } }{ 1+{ a }_{ 1 }{ a }_{ n } } \right) =tan^{ -1 }{ a }_{ 3 }-tan^{ -1 }{ a }_{ 2 }\)
Continuing inductively, we get
\(tan^{ -1 }\left( \frac { d }{ 1+{ a }_{ n }{ a }_{ n-1 } } \right) =tan^{ -1 }\left( \frac { { a }_{ n }-{ a }_{ n-1 } }{ 1+{ a }_{ n }{ a }_{ n-1 } } \right) =tan^{ -1 }{ a }_{ n }-tan^{ -1 }{ a }_{ n-1 }\)
Adding vertically, we get
\(tan^{ -1 }\left( \frac { d }{ 1+{ a }_{ 1 }{ a }_{ 2 } } \right) +tan^{ -1 }\left( \frac { d }{ 1+{ a }_{ 2 }{ a }_{ 3 } } \right) +....+tan^{ -1 }\left( \frac { d }{ 1+{ a }_{ n }{ a }_{ n-1 } } \right) tan[tan^{ -1 }{ a }_{ n }-{ tan }^{ -1 }{ a }_{ 1 }]\\ \)
\(tan\left[ tan^{ -1 }\left( \frac { d }{ 1+{ a }_{ 1 }{ a }_{ 2 } } \right) +tan^{ -1 }\left( \frac { d }{ 1+{ a }_{ 2 }{ a }_{ 3 } } \right) +...+tan^{ -1 }\left( \frac { d }{ 1+{ a }_{ n }{ a }_{ n-1 } } \right) \right] =tan\left[ tan^{ -1 }{ a }_{ n }-tan^{ -1 }a_{ 1 } \right] \)\(=\left[ tan^{ -1 }\left( \frac { { a }_{ n }-{ a }_{ 1 } }{ 1+{ a }_{ 1 }{ a }_{ n } } \right) \right] =\frac { { a }_{ n }-{ a }_{ 1 } }{ 1+{ a }_{ 1 }{ a }_{ n } } \)

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

Let the roots be a−d, a, a+d.
Then the sum of the roots is 3a which is equal to 6 from the given equation.
Thus 3a = 6 and hence a = 2.
The product of the roots is a³− ad² which is equal to −24 from the given equation.
Substituting the value of a, we get 8−2d² = −24 and hence d = ±4.
If we take d = 4 we get −2, 2, 6 as roots and if we take d = −4, we get 6, 2, −2 as roots (same roots given in reverse order) of the equation.

Clearly 1 is the identity element of G
w² . a^-1 = e ⇒ w² . a^-1 = a^-1 = w
Since w² . w = w³ = 1
Inverse of w² is w.

Let I = \(\int _{ 1 }^{ 2 }{ \frac { 3x }{ { 9x }^{ 2 }-1 } dx } \) ⇒ IA³| = \(\left| I \right| \)
Put t = 9x² - 1 ⇒ dt = 18x dx
\(\frac{d t}{6}=3 x d x\)

∴ \(\int _{ 8 }^{ 35 }{ \frac { dt }{ 6t } } \)
= \(\frac { 1 }{ 6 } { \left[ log \ t \right] }_{ 8 }^{ 35 }\)
= \(\frac { 1 }{ 6 } [log35-log8]\)
= \(\frac { 1 }{ 6 } \left[ log\left( \frac { 35 }{ 8 } \right) \right] \)

x =\(\sqrt { t } \)
V = \(\frac { dx }{ dt } =\frac { 1 }{ 2 } t^{ -\frac { 1 }{ 2 } }\) ..(1)
Acceleration = \(\frac { { d }^{ 2 }x }{ { dt }^{ 2 } } =\frac { 1 }{ 2 } \left( -\frac { 1 }{ 2 } t^{ -\frac { 3 }{ 2 } } \right) =\frac { -t^{ -\frac { 3 }{ 2 } } }{ 4 } \)
∴ Acceleration is negative
Acceleration = \(-\frac { 1 }{ 4 } \left( { t }^{ -\frac { 1 }{ 2 } } \right) ^{ 3 }\)
= \(-2\left( \frac { 1 }{ 2 } t^{ -\frac { 1 }{ 2 } } \right) ^{ 3 }\) = 2V³    [using (1)]
Hence, acceleration is negative proportional to the cube of the velocity.

Given
\(f(x)=\cfrac { 4-x }{ 6 } \) 
\(f(x)=\cfrac { 4-1 }{ 6 } =\cfrac { 3 }{ 6 } =\cfrac { 1 }{ 2 } \) 
\(f(2)=\cfrac { 4-2 }{ 6 } =\cfrac { 2 }{ 6 } =\cfrac { 1 }{ 3 } \) 
\(f(3)=\cfrac { 4-3 }{ 6 } =\cfrac { 1 }{ 6 } \) 
ஃ The probability mass function is

Mean \(E(X)=\Sigma xf(x)\) 
= \(1\left( \cfrac { 1 }{ 2 } \right) +2\left( \cfrac { 1 }{ 3 } \right) +3\left( \cfrac { 1 }{ 6 } \right) \) 
= \(\cfrac { 1 }{ 2 } +\cfrac { 2 }{ 3 } +\cfrac { 3 }{ 6 } =\cfrac { 3+4+6 }{ 6 } \) 
= \(\cfrac { 10 }{ 6 } =\cfrac { 5 }{ 3 } =1.67\) 
\(E({ x }^{ 2 })=\Sigma { x }^{ 2 }f(x)\) 
= \({ 1 }^{ 2 }\left( \cfrac { 1 }{ 2 } \right) +{ 2 }^{ 2 }\left( \cfrac { 1 }{ 3 } \right) +3^{ 2 }\left( \cfrac { 1 }{ 6 } \right) \) 
= \(\cfrac { 1 }{ 2 } +\cfrac { 4 }{ 9 } +\cfrac { 9 }{ 6 } =\cfrac { 3+8+9 }{ 6 } \) 
= \(\cfrac { 20 }{ 6 } =\cfrac { 10 }{ 3 } =3.33\)
var(X) E(X²) - [E(X)]²
= \(\cfrac { 10 }{ 3 } -\left( \cfrac { 5 }{ 3 } \right) ^{ 2 }=\cfrac { 10 }{ 3 } -\cfrac { 25 }{ 9 } \) 
= \(\cfrac { 30-25 }{ 9 } =\cfrac { 5 }{ 9 } =0.54\) 

x = 3 and dx = 0.02
When x = 3 and dx = 0.02,
df = (6 + 3) (0.02)
= 9(0.02) = 0.18

Given that x² + y² = r², r∈R      ...(1)
The given equation contains exactly one arbitrary constant.
So, we have to differentiate the given equation once. Differentiate (1) with respect to x, we get
2x +2y\(\frac{dy}{dx}\) = 0  which implies \(\frac{dy}{dx}\) = \(-\frac{x}{y}\)
Thus, x² + y² = r² satisfies the differential equation \(\frac { dy }{ dx } \) = -\(\frac { x }{ y } \).
Hence, x² + y² = r² is a solution of the differential equation  \(\frac { dy }{ dx } \) = -\(\frac { x }{ y } \).

Let P(x, y) be the movable point.
By focal property of ellipse, PA + PB = 2a
∴ 2a = 10 ⇒ a = 5
Since focus is (4, 0), ae = 4 ⇒ 5e = 4 ⇒ e = \(\frac45\)
Also b² = a²(1 - e²) = 25\(\left( 1-\frac { 16 }{ 25 } \right) =25\left( \frac { 9 }{ 25 } \right) \) = 9
Equation of ellipse is \(\frac { { x }^{ 2 } }{ { a }^{ 2 } } +\frac { { y }^{ 2 } }{ { b }^{ 2 } } =1\)
\(\frac { { x }^{ 2 } }{ 25 } +\frac { { y }^{ 2 } }{ 9 } =1\)

Let ∝, β are the roots of the equation
Sum of the roots \(\alpha +\beta =\frac { -b }{ a } \)
 \(=\frac { [-(a-2)] }{ 1 } =a-2\)
Product of the roots \(=\alpha \beta =\frac { c }{ a } \)
\(=\frac { -(a+1) }{ 1 } =-(a+1)\)
we have \({ \alpha }^{ 2 }{ \beta }^{ 2 }=({ \alpha +\beta ) }^{ 2 }-2\alpha \beta \)
\(={ (a-2) }^{ 2 }+2(a+1)\)
\(={ a }^{ 2 }-4a+4+2a+2\)
\(=(a-1{ ) }^{ 2 }+5\)
Thus \(\\ { \alpha }^{ 2 }+{ \beta }^{ 2 }\) is least if a = 1

Agumented matrix
[A|B] \(\left[ \begin{matrix} 2 & 5 \\ 6 & 15 \end{matrix}|\begin{matrix} 7 \\ 13 \end{matrix} \right] \overset { { R }_{ 2 }\rightarrow { R }_{ 2 }-3{ R }_{ 1 } }{ \longrightarrow } \left[ \begin{matrix} 2 & 5 \\ 0 & 0 \end{matrix}|\begin{matrix} 7 \\ -8 \end{matrix} \right] \)
Here \(\rho\) (A) = 2 and \(\rho\)([A|B]) = 3
∴ \(\rho\) (a) ≠ \(\rho\) ([AIB])
Hence the system is inconsistent.

Let \(\vec { a } =-6\hat { i } +14\hat { j } +10\hat { k } \), \(\vec { b } =14\hat { i } -10\hat { j } -6\hat { k } \) and \(\vec { c } =2\hat { i } +4\hat { j } -2\hat { k } \)
Volume of the parallelepiped having \(\vec { a } ,\vec { b } \) and \(\vec { c } \) as its co-terminus edges is \(\vec { a } .(\vec { b } \times \vec { c } )\).
∴ \(\vec { a } .(\vec { b } \times \vec { c } )=\left| \begin{matrix} -6 & 14 & 10 \\ 14 & -10 & -6 \\ 2 & 4 & -2 \end{matrix} \right| \)
= \(-6\left| \begin{matrix} -10 & -6 \\ 4 & -2 \end{matrix} \right| -14\left| \begin{matrix} 14 & -6 \\ 2 & -2 \end{matrix} \right| +10\left| \begin{matrix} 14 & -10 \\ 2 & 4 \end{matrix} \right| \)
= -6(20 + 24) - 14(-28 + 12) + 10(56 + 20)
= -6(44) -14(-16) + 10(76)
= -264 + 224 + 760 = 720.
∴ Volume of the required parallelepiped = 720 cubic units.

(a) Let a, b \(\in \) 2.
The a * b = a + b + z = b + a + z = b * a
(z, *) has commutative property.
(b) Let a, b, c \(\in \) G
(a * b) * c = (a+b+2) * c = a + b + 2 + c + 2
= a + b + c + 4
∴ z is associate under *.

u = x; v = e^-2x
u' = 1; \({ v }_{ 1 }=\frac { { e }^{ -2x } }{ -2 } \)
\({ v }_{ 2 }=\frac { { e }^{ -2x } }{ 4 } \)
Bernoulli's formula
\(\int { uv\ dx={ uv }^{ (1) }-{ uv }^{ (2) } } \)
\(\therefore \int _{ 0 }^{ 1 }{ { xe }^{ -2x } } dx={ \left[ x\left( \frac { { e }^{ -2x } }{ -2 } \right) -1\left( \frac { { e }^{ -2x } }{ 4 } \right) \right] }_{ 0 }^{ 1 }\)
= \(-{ \left[ \frac { x }{ 2 } { e }^{ -2x }+\frac { { e }^{ -2x } }{ 4 } \right] }_{ 0 }^{ 1 }\)
= \({ -e }^{ -2x }{ \left[ \frac { x }{ 2 } +\frac { 1 }{ 4 } \right] }_{ 0 }^{ 1 }\)
= \({ -e }^{ -2 }\left( \frac { 1 }{ 2 } +\frac { 1 }{ 4 } \right) +{ e }^{ 0 }\left( 0+\frac { 1 }{ 4 } \right) \)
= \(-{ e }^{ -2 }\left( \frac { 3 }{ 4 } \right) +\frac { 1 }{ 4 } =\frac { 1 }{ 4 } (1-{ 3e }^{ -2 })\)

By the mean value theorem we have, there exists 'c'∈(2, 7) such that,
\(\frac { f(7)-f(2) }{ 7-2 } \) = f'(c) ≤ 29
Hence, f(7) ≤ 5× 29 +17 = 162
Therefore, the maximum value of f (7) is 162.

LHS = \({ tan }^{ -1 }\sqrt { x } =\frac { 1 }{ 2 } .2{ tan }^{ -1 }\left( \sqrt { x } \right) \)
= \(\frac { 1 }{ 2 } .\left( 2{ tan }^{ -1 }\left( \sqrt { x } \right) \right) =\frac { 1 }{ 2 } { cos }^{ -1 }\left( \cfrac { 1-\left( \sqrt { x } \right) ^{ 2 } }{ 1+\left( \sqrt { x } \right) ^{ 2 } } \right) \)
\(\left[ \because {2 tan }^{ -1 }x={ cos }^{ -1 }\left( \frac { 1-{ x }^{ 2 } }{ 1+{ x }^{ 2 } } \right) \right] \)
= \(\frac { 1 }{ 2 } { cos }^{ -1 }\left( \frac { 1-x }{ 1+x } \right) \)
= RHS
Hence proved .

Let \(\overset { \rightarrow }{ r } =x\overset { \wedge }{ i } +y\overset { \wedge }{ j } +z\overset { \wedge }{ k } \)
\(\therefore x\overset { \wedge }{ i } +y\overset { \wedge }{ j } +z\overset { \wedge }{ k } =\left( s-2t \right) \overset { \wedge }{ i } +\left( 3-t \right) \overset { \wedge }{ j } +\left( 2s+t \right) \overset { \wedge }{ k } \)
Equating the co-efficients of like components both sides,
We get, x = s - 2t
y = 3 - t
z = 2s + t
Eliminating x and t using determinates we get
\(\left| \begin{matrix} x \\ y-3 \\ z \end{matrix}\begin{matrix} 1 \\ 0 \\ 2 \end{matrix}\begin{matrix} -2 \\ -1 \\ 1 \end{matrix} \right| =0\)
⇒ x (0+2) -1(y - 3 + z) -2 (2y - 6 - 0) = 0
⇒ 2x - y + 3 - z- 4y + 12 = 0
⇒ 2x - 5y - z + 15 = 0

AB =\(\left[ \begin{matrix} 2 & 1 \\ 5 & 3 \end{matrix} \right] \left[ \begin{matrix} 4 & 5 \\ 3 & 4 \end{matrix} \right] =\left[ \begin{matrix} 8+3 & 10+4 \\ 20+9 & 25+12 \end{matrix} \right] \)
=\(\left[ \begin{matrix} 11 & 14 \\ 29 & 37 \end{matrix} \right]\)
|AB| =\(\left[ \begin{matrix} 11 & 14 \\ 29 & 37 \end{matrix} \right]\)
= 407 - 406 = 1 ≠ 0
(AB)^-1 = \(\frac { 1 }{ |AB| } \) adj(AB)
= \(\frac { 1 }{ 1 } \left[ \begin{matrix} 11 & 14 \\ 29 & 37 \end{matrix} \right] \)
= \(\left[ \begin{matrix} 37 & -14 \\ -29 & 11 \end{matrix} \right] \) ....(1)
|A| =\(\left[ \begin{matrix} 2 & 1 \\ 5 & 3 \end{matrix} \right] \) = 6 - 5 =1
|B| =\(\left[ \begin{matrix} 4 & 5 \\ 3 & 4 \end{matrix} \right] \) = 16 - 15 = 1
B^-1 = \(\frac { 1 }{ |B| } adjB=\left[ \begin{matrix} 4 & -5 \\ -3 & 4 \end{matrix} \right] \)
A^-1 = \(\frac { 1 }{ |A| } adjA=\left[ \begin{matrix} 3 & -1 \\ -5 & 2 \end{matrix} \right] \)
∴ B^-1A^-1 =\(\left[ \begin{matrix} 4 & -5 \\ -3 & 4 \end{matrix} \right] \left[ \begin{matrix} 3 & -1 \\ -5 & 2 \end{matrix} \right] \)
=\(\\ \left[ \begin{matrix} 12+25 & -4-10 \\ -9-20 & 3+8 \end{matrix} \right] \)
=\(\left[ \begin{matrix} 37 & -14 \\ -29 & 11 \end{matrix} \right] \)....(2)
From (1) and (2), (AB)^-1 = B^-1A^-1

The given equation can be written as
\({ 2 }^{ z }\left( 1+\frac { 1 }{ 2 } +\frac { 1 }{ 4 } \right) ={ 7 }^{ x }\left( 1+\frac { 1 }{ 7 } +\frac { 1 }{ 49 } \right) \)
\(\Rightarrow { 2 }^{ x }\left( \frac { 8+4+2 }{ 8 } \right) ={ 7 }^{ x }\left( \frac { 49+7+1 }{ 49 } \right) \)
\(\Rightarrow { 2 }^{ x }\left( \frac { 7 }{ 4 } \right) ={ 7 }^{ x }\left( \frac { 57 }{ 49 } \right) \Rightarrow \frac { 7 }{ 4 } \times \frac { 49 }{ 57 } =\frac { { 7 }^{ x } }{ { 2 }^{ x } } \)
\(\Rightarrow \frac { 7 }{ 4 } \times \frac { 49 }{ 57 } ={ \left( \frac { 7 }{ 4 } \right) }^{ x }\Rightarrow \frac { { 7 }^{ 3 } }{ 4\times 57 } ={ \left( \frac { 7 }{ 4 } \right) }^{ x }\)
\(\Rightarrow xlog\left( \frac { 7 }{ 4 } \right) =3log\ 7-log4-log57\)
\(\Rightarrow x=\frac { 3log7-log4-log57 }{ log\left( \frac { 7 }{ 2 } \right) } \)