Given n = 6, \(p=\frac { 1 }{ 3 } \), k = 3
\(P(X=k)=\left( \begin{matrix} n \\ k \end{matrix} \right) { p }^{ k }\left( 1-p \right) ^{ n-k },\) 
n = 0,1,2, ... n
\(\therefore P(X=k)=\left( \begin{matrix} n \\ k \end{matrix} \right) { p }^{ k }(1-p)^{ n-k }\) 
n = 0,1,2, ... n
\(P(X=3)=\left( \begin{matrix} 6 \\ 3 \end{matrix} \right) \left( \cfrac { 1 }{ 3 } \right) ^{ 3 }\left( 1-p \right) ^{ 6-3 }\) 
= \(\left( \begin{matrix} 6 \\ 3 \end{matrix} \right) \left( \cfrac { 1 }{ 3 } \right) ^{ 3 }\left( \cfrac { 2 }{ 3 } \right) ^{ 2 }\) 

\(P(X=3)=\frac { 160 }{ 729 } \) 

\({ I }_{ n }=\int _{ 0 }^{ \pi /2 }{ { cos }^{ 7 }xdx=\frac { n-1 }{ n } { I }_{ n-2 },n\ge 2 } \)
\(\therefore { I }_{ 7 }=\int _{ 0 }^{ \pi /2 }{ { cos }^{ 7 }xdx=\frac { 6 }{ 7 } \times \frac { 4 }{ 5 } \times \frac { 2 }{ 3 } \times 1 } =\frac { 16 }{ 35 } \)

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

The given differential equation is 
\({ \left( \frac { d^2y }{ dx^2 } \right) }^{ 3\times2 }= { 1+\left( \frac { dy }{ dx } \right) } \)
squaring both sides, we get
\({ \left( \frac { dy }{ dx } \right) }^{ 6 }=1+\left( \frac { dy }{ dx } \right) \)
In this equation, the highest order derivative is 2 and its power is 6.
∴ Order 2, degree 6.

Equation of the circle is x² + y² - x + 2y - 3 = 0
Here 2g = -1 ⇒ g = \(\frac { -1 }{ 2 } \)
2f = 2 ⇒ f = 1 and c = -3
Centre is (-g, -f) = \(\left( \frac { 1 }{ 2 } ,-1 \right) \)
and r = \(\sqrt { { g }^{ 2 }+{ f }^{ 2 }-c } \)
= \(\sqrt { \frac { 1 }{ 4 } +1+3 } \)
= \(\sqrt { \frac { 1 }{ 4 } +4 } =\sqrt { \frac { 1+16 }{ 2 } } \)
r = \(\sqrt { \frac { 17 }{ 2 } } \) units.

Since the given three vectors are coplanar, we have \(\left| \begin{matrix} 2 & -1 & 3 \\ 3 & 2 & 1 \\ 1 & m & 4 \end{matrix} \right| \) = 0 ⇒ m = -3

We compute \(\left| 6-8i \right| =\sqrt { { 6 }^{ 2 }+\left( -8 \right) ^{ 2 } } =10\)
and applying the formula for square root, we get
\(\sqrt { 6-8i } =\pm \left( \sqrt { \frac { 10+6 }{ 2 } } -i\sqrt { \frac { 10-6 }{ 2 } } \right) \) (\(\therefore\) b is negative\( \frac{b}{|b|}=-1 \))
= \(\pm \left( \sqrt { 8 } +i\sqrt { 2 } \right) \)
= \(\pm \left( 2\sqrt { 2 } -i\sqrt { 2 } \right) \)

Given adj (A) =\(\left[ \begin{matrix} 0 & -2 & 0 \\ 6 & 2 & -6 \\ -3 & 0 & 6 \end{matrix} \right] \)
We know that A^-1 = ±\(\frac { 1 }{ \sqrt { |adjA| } } \) (adj A) ...............(1)
|adj A| = 0 + 2\(\left| \begin{matrix} 6 & -6 \\ -3 & 6 \end{matrix} \right| \) + 0
[Expanded along R₁]
= 2(36-18) = 2(18) = 36
∴ A^-1 = \(\pm \frac { 1 }{ \sqrt { 36 } } \left[ \begin{matrix} 0 & -2 & 0 \\ 6 & 2 & -6 \\ -3 & 0 & 6 \end{matrix} \right] \)
= \(\pm \frac { 1 }{ 6 } \left[ \begin{matrix} 0 & -2 & 0 \\ 6 & 2 & -6 \\ -3 & 0 & 6 \end{matrix} \right] \).

Here Δ = b²−4ac = 0 for equal roots. This implies 4(k + 2)² = 4(9)k. This implies k = 4 or 1.

\(cos\left( { cos }^{ -1 }\left( \frac { 4 }{ 5 } \right) +{ sin }^{ -1 }\left( \frac { 4 }{ 5 } \right) \right) \)
\({ cos }^{ -1 }\left( \frac { 4 }{ 5 } \right) =\theta \Rightarrow \frac { 4 }{ 5 } =cos\theta \)
Also \({ sin }^{ -1 }\left( \frac { 4 }{ 5 } \right) ={ sin }^{ -1 }\left( cos\theta \right) \) [using (1)]
= \({ sim }^{ -1 }\left( sin\left( \frac { \pi }{ 2 } -\theta \right) \right) \) \(\left[ \because cos\theta =sin\left( \frac { \pi }{ 2 } -\theta \right) \right] \)
= \(\frac { \pi }{ 2 } -\theta \)
\(\therefore cos\left( cos^{ i1 }\left( \frac { 4 }{ 5 } \right) +sin^{ -1 }\left( \frac { 4 }{ 5 } \right) \right) \)
= \(cos\left( \theta +\frac { \pi }{ 2 } -\theta \right) \)[using (1) & (2)]
= \(cos\frac { \pi }{ 2 } \)
= 0

Given u (x, y) = \(\frac { { x }^{ 2 }+{ y }^{ 2 } }{ \sqrt { x+y } } \)
\(u({ \lambda }x,{ \lambda }y)=\frac { { \lambda }^{ 2 }{ x }^{ 2 }+{ { \lambda } }^{ 2 }{ y }^{ 2 } }{ \sqrt { { \lambda }x+{ \lambda }y } } \)
= \(\frac { { { \lambda } }^{ 2 }({ x }^{ 2 }+{ y }^{ 2 }) }{ \sqrt { { \lambda } } (\sqrt { x+y } ) } \)
= \({ { \lambda } }^{ 2-\frac { 1 }{ 2 } }u(x,y)\)
= \({ { \lambda } }^{ \frac { 3 }{ 2 } }u(x,y)\)
∴ u (x, y) is a homogeneous function of degree \(\frac32\)
∴ By Euler's theorem,
\(x\frac { \partial u }{ \partial x } +y\frac { \partial u }{ \partial y } \) = n.u ≍ \(x\frac { \partial u }{ \partial x } +y\frac { \partial u }{ \partial y } =\frac { 3 }{ 2 } u\)
Hence, proved.

\(\int _{ 0 }^{ 3 }{ (3{ x }^{ 2 }-4x+5) } dx=\int _{ 0 }^{ 3 }{ 3{ x }^{ 2 } } dx-\int _{ 0 }^{ 3 }{ 4x } dx+\int _{ 0 }^{ 3 }{ 5 } dx\)
\(=3\int _{ 0 }^{ 3 }{ { x }^{ 2 } } dx-4\int _{ 0 }^{ 3 }{ c } dx+5\int _{ 0 }^{ 3 }{ dx } \)
\(=3{ \left[ \frac { { x }^{ 3 } }{ 3 } \right] }_{ 0 }^{ 3 }-4{ \left[ \frac { { x }^{ 2 } }{ 2 } \right] }_{ 0 }^{ 3 }+5{ \left[ x \right] }_{ 0 }^{ 3 }\)
= (27 − 0) − 2(9 − 0) + 5(3− 0)
= 27 −18 +15 = 24.

Let X be the random variable of getting apples Given 5 mangoes and 4 apples are in an urn 
= {0, 1,2,3}
The sample space consists of ⁹C₃ = 84
X = 0, X (3 mangoes) = ⁵C₃ = 10
X = 1, X (2 mangoes and 1 apples) = ⁵C₂ x ⁴C₁ = 40
X = 2, X (1 mangoes and 2 apples) = ⁵C₁ x ⁴C₂ = 30
X = 3, X (apples) = ⁴C₃ = 4

Let f(x) = sin x ⇒ f(0) = sin 0 = 0
f^I(x) = cos x ⇒ f^I(0) = cos 0 = 1
f^Il(x) = - sin x ⇒ f^II(0) = -sin 0 = 0
f^IlI(x) = - cos x ⇒ f^III(0) = -cos 0 = 0
f^IV(x) = + sin x ⇒ f^IV(0) = sin 0 = 0
f^V(x) = cos x ⇒ f^v(0) = cos 0 = 0
f^VI(x) = - sin x ⇒ f^vI(0) = - sin 0 = 0
f^VII(x) = - cos x ⇒ f^vII(0) = - cos 0 = 0
∴ Maclaurins' series expansion
\(f(x)=f(0)+\frac { { f }^{ 1 }(0) }{ 1! } x+\frac { { f }^{ II }(0) }{ 2! } { x }^{ 2 }+\frac { { f }^{ III }(0) }{ 3! } { x }^{ 3 }+\frac { { f }^{ IV }(0) }{ 4! } { x }^{ 4 }+.....\)
\(\therefore sinx=0+\frac { 1 }{ 1! } x+0-\frac { 1 }{ 3! } { x }^{ 3 }+0-\frac { 1 }{ 5! } { x }^{ 5 }+0-\frac { 1 }{ 7! } { x }^{ 7 }+........\)
\(x-\frac { { x }^{ 3 } }{ 3! } -\frac { { x }^{ 5 } }{ 5! } -\frac { { x }^{ 7 } }{ 7! } +...\)

If \(\hat { b } =\frac { 2\hat { i } +2\hat { j } -\hat { k } }{ \left| \hat { 2i } +2\hat { j } -\hat { k } \right| } =\frac { 1 }{ 3 } (2\hat { i } +2\hat { j } -\hat { k } )\). Therefore from the definition of direction cosines of \(\hat { b } \), we have
\(cos\alpha =\frac { 2 }{ 3 } ,cos\beta =\frac { 2 }{ 3 } ,cos\gamma =-\frac { 1 }{ 3 } \)
where α, β, \(\gamma \) are the angles made by \(\hat { b } \) with the positive x -axis, positive y -axis, and positive
z -axis, respectively. As the angle between the given straight line with the coordinate axes are same as the angles made by \(\hat { b } \) with the coordinate axes, we have \(\alpha ={ cos }^{ -1 }\left( \frac { 2 }{ 3 } \right) ,\beta ={ cos }^{ -1 }\left( \frac { 2 }{ 3 } \right) ,\gamma ={ cos }^{ -1 }\left( \frac { -1 }{ 3 } \right) \) respectively.

\({ tan }^{ -1 }(\frac { 2 }{ 11 }) +{ tan }^{ -1 }(\frac { 7 }{ 24 }) ={ tan }^{ -1 }(\frac { 1 }{ 2 } )\)
\(LHS={ tan }^{ -1 }\left( \frac { 2 }{ 11 } \right) +{ tan }^{ -1 }\left( \frac { 7 }{ 24 } \right) \)
= \({ tan }^{ -1 }\left( \frac { \frac { 2 }{ 11 } +\frac { 7 }{ 24 } }{ 1-\left( \frac { 2 }{ 11 } \right) \left( \frac { 7 }{ 24 } \right) } \right) \) \(\left[ \because { tan }^{ -1 }(x)+{ tan }^{ -1 }y={ tan }^{ -1 }\left( \frac { x+y }{ 1-xy } \right) \right] \)
= \({ tan }^{ -1 }\left( \frac { \frac { 48+77 }{ 11\times 24 } }{ \frac { 264-14 }{ 264 } } \right) ={ tan }^{ -1 }\left( \frac { \frac { 125 }{ 264 } }{ \frac { 250 }{ 264 } } \right) \)

= RHS
Hence proved.

2x+5y = -2, x+2y = -3
The matrix form of the system is
\(\left( \begin{matrix} 2 & 5 \\ 1 & 2 \end{matrix} \right) \left( \begin{matrix} x \\ y \end{matrix} \right) =\left( \begin{matrix} -2 \\ -3 \end{matrix} \right) \)
⇒ AX = B where
A =\(\left( \begin{matrix} 2 & 5 \\ 1 & 2 \end{matrix} \right) ,B=\left( \begin{matrix} -2 \\ -3 \end{matrix} \right) \)
X =\(\left( \begin{matrix} x \\ y \end{matrix} \right) \)
⇒ = A^-1B
|A| = \(\left| \begin{matrix} 2 & 5 \\ 1 & 2 \end{matrix} \right| \)= 4 - 5 = -1 ≠ 0
∴ A^-1 =\(\frac { 1 }{ |A| } adjA=\frac { 1 }{ -1 } \left[ \begin{matrix} 2 & -5 \\ -1 & 2 \end{matrix} \right] \)
=\(\left[ \begin{matrix} -2 & 5 \\ 1 & -2 \end{matrix} \right] \)
∴ X = A^-1B =\(\left[ \begin{matrix} -2 & 5 \\ 1 & -2 \end{matrix} \right] \left[ \begin{matrix} -2 \\ -3 \end{matrix} \right] \)
= \(\left[ \begin{matrix} 4-15 \\ -2+6 \end{matrix} \right] =\left[ \begin{matrix} -11 \\ 4 \end{matrix} \right] \)
∴ Solution set is x = -11, y = 4

Given z³+2\(\overline { z } \) = 0
⇒ z³ = -2\(\overline { z } \)
Taking modulus, |z³| = |-2\(\overline { z } \)|
⇒ |z|³ = 2|z|
⇒ |z| [|z|2-2] = 0
|z| = 0 or |z|²-2 = 0
⇒ |z| = 0
⇒ z = 0 is a solution ....(1)
|z²| = 2
⇒ |z²| = 2 ⇒ (z\(\overline { z } \))² = 2
⇒ z\(\overline { z } \) = \(\sqrt { 2 } \Rightarrow \overline { z } \frac { \sqrt { 2 } }{ z } \)
Given z + 2\(\overline { z } \) = 0
⇒ z³+2\(\frac { \sqrt { 2 } }{ z } \) = 0
⇒ z⁴+2\(\sqrt { 2 } \) = 0
If has 4 non-zero solutions
Hence from (1) and (2), z³+2\(\overline { z } \) has 5 solutions.

Given vertex A(5, -2) and focus S(2, -2) and the focal distance
AS = a = 3  
Parabola is open left and symmetric about the line parallel to x -axis.
Then, the equation of the required parabola is
(y + 2)² = −4(3)(x − 5)
y² + 4y + 4 = −12x + 60
y² + 4y +12x − 56 = 0  

Given 2i + 3 is a root
∴ Its conjugate 3 - 2i is also a root of the polynomial equation.
∴ Sum of the roots 3 + 2i + 3 - 2i = 6
Product of the roots = (3 + 2i) (3 - 2i)
= 3² + 2² = 9+ 4 = 13
∴ The polynomial equation of minimum degree with rational co-efficients is
x² - x (sum of the roots) + product of the roots = 0
⇒x² -x (6) + 13 = 0
⇒ x²- 6x + 13 = 0

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 *.

Given z(x, y) = x³ - 3x²y³ ; x = se^t; y = se^-t
\(\frac { \partial z }{ \partial s } \) = 3x² - 6xy³; \(\frac { \partial z }{ \partial y } \) = -9x²y²
∴ \(\frac { \partial z }{ \partial s } \) 3s²e^2t - 6 se^t s³ e^-3t
\(\frac { \partial z }{ \partial y } \) = 9s²e^2t.s²e^-2t
\(\frac { \partial z }{ \partial y } \) = -9s⁴
\(\frac { \partial z }{ \partial s} \) = e^t; \(\frac { \partial y }{ \partial s} \) = e^-t
\(\frac { \partial x }{ \partial t} \) = se^t; \(\frac { \partial y }{ \partial s} \) = -se^-t 
By chain rule;
\(\therefore \frac { \partial z }{ \partial s } =\frac { \partial z }{ \partial x } .\frac { dx }{ ds } +\frac { \partial u }{ \partial y } .\frac { dy }{ dx } \)
= (3s²e^2t - 6 s⁴ e^-2t) (e^t) + (-9s⁴) (e^-t)
= 3s²e^3t - 6 s⁴ e^-t -9s⁴ e^-t
\(\frac { \partial z }{ \partial s } \) = 3s²e^3t - 15 s⁴ e^-t
By chain rule;
\(\frac { \partial z }{ \partial t } =\frac { \partial z }{ \partial x } .\frac { dx }{ dt } +\frac { \partial z }{ \partial y } .\frac { dy }{ dt } \)
= (3s²e^2t - 6 s⁴e-^2t) (se^t) + (-9 s⁴) (-se^-t)
= 3s³e^3t - 6 s⁵e^-t + 9 s⁵ e^-t
= 3s³ e^3t + 3 s⁵e^-t
= 3s³(e^3t + s²e^-t)

Equation of the given curve is 2+x-x²+y = 0

\(\Rightarrow\)y = x²-x-2

\(\therefore\) Required area= \(\int _{ -3 }^{ -1 }{ ydx } +\int _{ -1 }^{ 2 }{ -y } dx+\int _{ 2 }^{ 3 }{ ydx } \)
\(\\ =\int _{ -3 }^{ -1 }{ \left( { x }^{ 2 }-x-2 \right) } dx\int _{ -1 }^{ 2 }{ (2+x-{ x }^{ 2 })dx } +\int _{ 2 }^{ 3 }{ ({ x }^{ 2 }-x-2) } dx\)
\(={ \left( \frac { { x }^{ 3 } }{ 3 } -\frac { { x }^{ 2 } }{ 2 } -2x \right) }_{ -3 }^{ -1 }+{ \left( 2x+\frac { { x }^{ 2 } }{ 2 } -\frac { { x }^{ 3 } }{ 3 } \right) }_{ -1 }^{ 2 }+{ \left( \frac { { x }^{ 3 } }{ 3 } -\frac { { x }^{ 2 } }{ 2 } -2x \right) }_{ 2 }^{ 3 }\)
\(=\left( -\frac { 1 }{ 3 } -\frac { 1 }{ 2 } +2 \right) -\left( -9-\frac { 9 }{ 2 } +6 \right) +\left( 4+2-\frac { 8 }{ 3 } \right) -\left( -2+\frac { 1 }{ 2 } +\frac { 1 }{ 3 } \right) +\left( 9-\frac { 9 }{ 2 } -6 \right) -\left( \frac { 8 }{ 3 } -2-4 \right) \)
\(\\ =\left( \frac { -2-3+6 }{ 6 } \right) -\left( \frac { -6-9 }{ 2 } \right) +\left( \frac { 18-8 }{ 3 } \right) -\left( \frac { -12+3+2 }{ 6 } \right) +\left( \frac { 6-9 }{ 2 } \right) -\left( \frac { 8-24 }{ 3 } \right) \)
\(=\frac { 1 }{ 6 } +\frac { 15 }{ 2 } +\frac { 10 }{ 3 } +\frac { 7 }{ 6 } -\frac { 3 }{ 2 } +\frac { 16 }{ 3 } \)
\(=\frac { 1+45+20+7-9+32 }{ 6 } =\frac { 90 }{ 6 } =15 \)

Given
\(f(x)=\begin{cases} \begin{matrix} { Ke }^{ \frac { -x }{ 3 } } & \begin{matrix} for & x>0 \end{matrix} \end{matrix} \\ \begin{matrix} 0 & \begin{matrix} for & x\le 0 \end{matrix} \end{matrix} \end{cases}\)
(i) Since f(x) is a probability density function
\(\int _{ -\infty }^{ \infty }{ f(x) } dx=1\) 
\(\Rightarrow \int _{ 0 }^{ \infty }{ K.{ e }^{ \frac { -x }{ 3 } } } dx=1\Rightarrow k\frac { \left[ { e }^{ \frac { -x }{ 3 } } \right] ^{ \infty } }{ -\frac { 1 }{ 3 } } \) 
\(\Rightarrow -3k\left[ { e }^{ -\infty }-{ e }^{ 0 } \right] =1\) [∵ e^∞ = 0, e⁰ = 1]
\(\Rightarrow 3k=1\Rightarrow k=\frac { 1 }{ 3 } \) 
\(\therefore k=\cfrac { 1 }{ 3 } \) 
(ii) The distribution function F(x) = \(\int _{ -\infty }^{ x }{ f(u)du } \)
Case 1: x < 0, 
\(F(x)=\int _{ -\infty }^{ x }{ f(x)dx=0 } \) 
Case 2: x > 0, 
\(f(x)=\int _{ -\infty }^{ x }{ f(x)dx } \) 
= \(\int _{ -\infty }^{ 0 }{ f(x)dx+\int _{ 0 }^{ x }{ f(x) } dx } \) 
= \(0+k\int _{ 0 }^{ x }{ { e }^{ \frac { -x }{ 3 } } } dx\) 
= \(\frac { 1 }{ 3 } \left[ \cfrac { { e }^{ \frac { -x }{ 3 } } }{ -\frac { 1 }{ 3 } } \right] =-\left[ { e }^{ -\frac { x }{ 3 } }-{ e }^{ o } \right] \) 
= \(-[{ e }^{ -\frac { x }{ 3 } }-1]\) 
= \(1-{ e }^{ -\frac { x }{ 3 } }\) 
\(\therefore F(x)=\begin{cases} \begin{matrix} 0 & x\le 0 \end{matrix} \\ \begin{matrix} 1-{ e }^{ \frac { -x }{ 3 } } & x>0 \end{matrix} \end{cases}\) 
(iii) p(X < 3)
= \(\int _{ 0 }^{ 3 }{ ke^{ -\frac { x }{ 3 } } } dx=\cfrac { 1 }{ 3 } \int _{ 0 }^{ 3 }{ { e }^{ -\frac { x }{ 3 } }dx } \)
= \(\cfrac { 1 }{ 3 } \left[ \frac { { e }^{ -\frac { x }{ 3 } } }{ \frac { -1 }{ 3 } } \right] \) 
= -[e^-1-e⁰] = -[e^-1-1]
(iv) \(p(5\le X)=p(X\ge 5)=\int _{ 5 }^{ \infty }{ f(x)dx } \) 
= \(\int _{ 5 }^{ \infty }{ ke^{ -\frac { x }{ 3 } } } dx=\cfrac { 1 }{ 3 } \cfrac { \left[ { e }^{ -\frac { x }{ 3 } } \right] _{ 5 }^{ \infty } }{ \frac { -1 }{ 3 } } \) 
= \(-\left[ { e }^{ -\infty }-e^{ \frac { -3 }{ 5 } } \right] =\left[ 0-{ e }^{ \frac { -5 }{ 3 } } \right] \) 
= \({ e }^{ \frac { -5 }{ 3 } }\) 
(v) \(p(X\le 4)=\int _{ -\infty }^{ 4 }{ f(x)dx } \) 
= \(\int _{ -\infty }^{ 0 }{ f(x)dx+\int _{ 0 }^{ 4 }{ f(x)dx } } \) 
= \(0+\int _{ 0 }^{ 4 }{ { ke }^{ -\frac { x }{ 3 } }dx } =k\left[ \frac { { e }^{ -\frac { x }{ 3 } } }{ -\frac { 1 }{ 3 } } \right] _{ 0 }^{ 4 }\) 
= \(\cfrac { 1 }{ 3 } \cfrac { \left[ { e }^{ -\frac { x }{ 3 } } \right] _{ 0 }^{ 4 } }{ -\frac { 1 }{ 3 } } =-\left[ { e }^{ \frac { -4 }{ 3 } }-{ e }^{ o } \right] \) 
= \(-\left[ { e }^{ \frac { -4 }{ 3 } }-1 \right] =1-{ e }^{ \frac { -4 }{ 3 } }\)

Let there be N radioactive nuclei in a sample at any time t and let N_o be the initial number of radioactive nuclei.
Then \(\frac{dN}{dt}\infty N\)
\(\Rightarrow \frac { dN }{ dt } =-\lambda N\)
Where \(\lambda>0\) is a constant
\(\Rightarrow \frac { dN }{ N } =-\lambda dt\)
\(\int { \frac { dN }{ N } } =-\lambda dt\)
\(\int { \frac { dN }{ N } =-\lambda \int { dt } } \)
\(\Rightarrow log\ N=-\lambda t+C\ ...(1)\)
\(T\quad t=0,\ we\ have\ N={ N }_{ 0 }\)
\(\therefore log{ N }_{ 0 }=0+C\)
\(\Rightarrow C=log{ N }_{ 0 }\)
\(\therefore\)(1) becomes, log N = -\(\lambda t+log{ N }_{ 0 }\)
\(\Rightarrow log\frac { N }{ { N }_{ 0 } } =-\lambda t\quad ...(2)\)
It is given that 10% of the original number of nuclei have undergone disintegration in a period of 100 years.
Whent= 100 \(N={ N }_{ 0 }-\frac { 10 }{ 100 } \times { N }_{ 0 }=\frac { { 9N }_{ 0 } }{ 10 } \)
Substituting in (2) we get
\(log\quad \frac { 9 }{ 10 } =-100\lambda \)
\(\Rightarrow \lambda =-\frac { 1 }{ 100 } log\frac { 9 }{ 10 } \)
Substituting in (2) we get,
\(log\frac { N }{ { N }_{ 0 } } =\left( \frac { 1 }{ 100 } log\frac { 9 }{ 10 } \right) t\)
when t = 1000,
\(log\frac { N }{ { N }_{ 0 } } =\frac { 1 }{ 100 } log\left( \frac { 9 }{ 10 } \right) \times 1000\)
\(=10log\left( \frac { 9 }{ 10 } \right) \)
\(\Rightarrow log\frac { N }{ { N }_{ 0 } } ={ \left( \frac { 9 }{ 10 } \right) }^{ 10 }\)
\(\Rightarrow \frac { N }{ { N }_{ 0 } } ={ \left( \frac { 9 }{ 10 } \right) }^{ 10 }\)
\(\Rightarrow \frac { N }{ { N }_{ 0 } } \times 100={ \left( \frac { 9 }{ 10 } \right) }^{ 10 }\times 100=\frac { { 9 }^{ 10 } }{ { 10 }^{ 8 } } \)
Hence, \(\frac { { 9 }^{ 10 } }{ { 10 }^{ 8 } } \%\) of radioactive nuclei will remain after 1000 years,

Given \(v\frac { dv }{ dx } =g\left( 1-\frac { { v }^{ 2 } }{ { k }^{ 2 } } \right) =g\left( \frac { { k }^{ 2 }-{ v }^{ 2 } }{ { k }^{ 2 } } \right) \)
On separating the variables we get,
\(\frac { vdv }{ { k }^{ 2 }-{ v }^{ 2 } } =\frac { g }{ { k }^{ 2 } } .dx\)
Multiplying by -2 both sides we get,
\(v\frac { dv }{ dx } =g\left( 1-\frac { { v }^{ 2 } }{ { k }^{ 2 } } \right) =g\left( \frac { { k }^{ 2 }-{ v }^{ 2 } }{ { k }^{ 2 } } \right) \)
\(\frac { -2vdv }{ { k }^{ 2 }-{ v }^{ 2 } } =\frac { -2g }{ { k }^{ 2 } } dx\)
Taking integrating on both sides, we get
\(\int { \frac { -2v }{ { k }^{ 2 }-{ v }^{ 2 } } } =\frac { -2g }{ { k }^{ 2 } } .x+log\quad c\)
\(\Rightarrow log({ k }^{ 2 }-{ v }^{ 2 })=\frac { -2g }{ { k }^{ 2 } } .x+log\quad c\)
\(\Rightarrow log({ k }^{ 2 }-{ v }^{ 2 })-log\quad c=\frac { -2g }{ { k }^{ 2 } } .x\)
\(\Rightarrow log\left( \frac { { k }^{ 2 }-{ v }^{ 2 } }{ c } \right) -\frac { -2gx }{ { k }^{ 2 } } \)
\(\Rightarrow \frac { { k }^{ 2 }-{ v }^{ 2 } }{ e } ={ e }^{ -\frac { -2gx }{ { k }^{ 2 } } }\)
\(\Rightarrow { k }^{ 2 }-{ v }^{ 2 }{ ce }^{ \frac { -2gx }{ { k }^{ 2 } } }...(1)\)
Initial condition:
when v = 0, x = 0 we get
\( k^2-(0)^2 =C e^{\frac{-2g(0)}{k^2}} \)
\(k^2 =C e^0 \Rightarrow k^2=C \)
\((1) \Rightarrow k^2-v^2 =k^2 e^{\frac{-2 x^2}{k^2}} \)
\(k^2-k^2 e^{\frac{-2 s x}{k^1}} =\mathrm{v}^2 \)
\(k^2\left[1-e^{\frac{-2 s x}{k^2}}\right] =\mathrm{v}^2\)

\({ cot }^{ -1 }x-{ xot }^{ -1 }\left( x+2 \right) =\frac { \pi }{ 12 } ,x>0\)
\({ tan }^{ -1 }\left( \frac { 1 }{ x } \right) -{ tan }^{ -1 }\left( \frac { 1 }{ x+2 } \right) =\frac { \pi }{ 2 } \)
\(\left[ \because { cot }^{ 1 }\left( x \right) ={ tan }^{ -1 }\left( \frac { 1 }{ x } \right) ifx>0 \right] \)
\(\Rightarrow { tan }^{ -1 }\left( \frac { \frac { 1 }{ x } +\frac { 1 }{ x+2 } }{ 1+\frac { 1 }{ x } .\frac { 1 }{ x+2 } } \right) =\frac { \pi }{ 2 } \)
\(\Rightarrow \left( \frac { \frac { x+2-x }{ x(x+2) } }{ \frac { x(x+2)+1 }{ x(x+2) } } \right) ={ tan15 }^{ 0 }\)
\(\left[ \because { tan15 }^{ 0 }=tan\left( { 45 }^{ 0 }-30^{ 0 } \right) \right] \)
\(\frac { { tan45 }^{ 0 }-{ tan30 }^{ 0 } }{ 1+tan{ 45 }^{ 0 }tan{ 30 }^{ 0 } } \)
\(\frac { 1-\frac { 1 }{ \sqrt { 3 } } }{ 1+\frac { 1 }{ \sqrt { 3 } } } =\frac { \sqrt { 3 } -1 }{ \sqrt { 3 } +1 } \times \frac { \sqrt { 3 } -1 }{ \sqrt { 3 } -1 } \)
\(\frac { \left( \sqrt { 3 } -1 \right) ^{ 2 } }{ 3-1 } =\frac { \left( \sqrt { 3 } -1 \right) ^{ 2 } }{ 2 } \)

\(\Rightarrow \frac { 2 }{ { x }^{ 2 }+2x+1 } =\frac { \left( \sqrt { 3 } -1 \right) ^{ 2 } }{ 2 } \)
\(\Rightarrow \frac { 7 }{ (x+1)^{ 2 } } =\frac { \left( \sqrt { 3 } -1 \right) ^{ 2 } }{ 2 } \)
\(\Rightarrow 4=(x+1)^{ 2 }\left( \sqrt { 3 } -1 \right) ^{ 2 }\)
Taking square root both sides
\(2=(x+1)(\sqrt { 3 } -1)\)
\(\Rightarrow x+1=\frac { 2 }{ \sqrt { 3- } 1 } \times \frac { \sqrt { 3 } +1 }{ \sqrt { 3 } +1 } =\frac { 2\left( \sqrt { 3 } +1 \right) }{ 3-1 } =\sqrt { 3 } +1\)
\(x+1=\sqrt { 3 }+ 1\)
\(x=\sqrt { 3 } \)

Here the number of equations is 4 and the number of unknowns is 3. Reducing the augmented matrix to echelon-form, we get
[A | O] = \(\left[ \begin{matrix} 1 \\ \begin{matrix} 2 \\ \begin{matrix} 3 \\ 6 \end{matrix} \end{matrix} \end{matrix}\begin{matrix} 1 \\ \begin{matrix} -3 \\ \begin{matrix} -7 \\ -9 \end{matrix} \end{matrix} \end{matrix}\begin{matrix} -2 \\ \begin{matrix} 1 \\ \begin{matrix} 10 \\ 10 \end{matrix} \end{matrix} \end{matrix}|\begin{matrix} 0 \\ \begin{matrix} 0 \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{matrix} \end{matrix} \right] \overset { \begin{matrix} { R }_{ 2 }\rightarrow { R }_{ 2 }-2{ R }_{ 1 } \\ { R }_{ 3 }\rightarrow { R }_{ 3 }-3{ R }_{ 1 }, \\ { R }_{ 4 }\rightarrow { R }_{ 4 }-6{ R }_{ 1 } \end{matrix} }{ \longrightarrow } \left[ \begin{matrix} 1 \\ \begin{matrix} 0 \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{matrix} \end{matrix}\begin{matrix} 1 \\ \begin{matrix} -5 \\ \begin{matrix} -10 \\ -15 \end{matrix} \end{matrix} \end{matrix}\begin{matrix} -2 \\ \begin{matrix} 5 \\ \begin{matrix} 16 \\ 22 \end{matrix} \end{matrix} \end{matrix}|\begin{matrix} 0 \\ \begin{matrix} 0 \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{matrix} \end{matrix} \right] \overset { \begin{matrix} { R }_{ 2 }\rightarrow { R }_{ 2 }\div \left( -5 \right) \\ { R }_{ 3 }\rightarrow { R }_{ 3 }\div \left( -2 \right) \end{matrix} }{ \longrightarrow } \left[ \begin{matrix} 1 \\ \begin{matrix} 0 \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{matrix} \end{matrix}\begin{matrix} 1 \\ \begin{matrix} 1 \\ \begin{matrix} 5 \\ -15 \end{matrix} \end{matrix} \end{matrix}\begin{matrix} -2 \\ \begin{matrix} -1 \\ \begin{matrix} -8 \\ 22 \end{matrix} \end{matrix} \end{matrix}|\begin{matrix} 0 \\ \begin{matrix} 0 \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{matrix} \end{matrix} \right] \)
\(\overset { \begin{matrix} { R }_{ 3 }\rightarrow { R }_{ 3 }-5{ R }_{ 2 }, \\ { R }_{ 4 }\rightarrow { R }_{ 4 }+15{ R }_{ 2 } \end{matrix} }{ \longrightarrow } \left[ \begin{matrix} 1 \\ \begin{matrix} 0 \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{matrix} \end{matrix}\begin{matrix} 1 \\ \begin{matrix} 1 \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{matrix} \end{matrix}\begin{matrix} -2 \\ \begin{matrix} -1 \\ \begin{matrix} -3 \\ 7 \end{matrix} \end{matrix} \end{matrix}|\begin{matrix} 0 \\ \begin{matrix} 0 \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{matrix} \end{matrix} \right] \overset { \begin{matrix} { R }_{ 3 }\rightarrow { R }_{ 3 }\div \left( -3 \right) \\ { R }_{ 4 }\rightarrow { R }_{ 4 }\div 7 \end{matrix} }{ \longrightarrow } \left[ \begin{matrix} 1 \\ \begin{matrix} 0 \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{matrix} \end{matrix}\begin{matrix} 1 \\ \begin{matrix} 1 \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{matrix} \end{matrix}\begin{matrix} -2 \\ \begin{matrix} -1 \\ \begin{matrix} 1 \\ 1 \end{matrix} \end{matrix} \end{matrix}|\begin{matrix} 0 \\ \begin{matrix} 0 \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{matrix} \end{matrix} \right] \overset { { R }_{ 4 }\rightarrow { R }_{ 4 }-{ R }_{ 3 } }{ \longrightarrow } \left[ \begin{matrix} 1 \\ \begin{matrix} 0 \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{matrix} \end{matrix}\begin{matrix} 1 \\ \begin{matrix} 1 \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{matrix} \end{matrix}\begin{matrix} -2 \\ \begin{matrix} -1 \\ \begin{matrix} 1 \\ 0 \end{matrix} \end{matrix} \end{matrix}|\begin{matrix} 0 \\ \begin{matrix} 0 \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{matrix} \end{matrix} \right] \)
So, ρ(A) = ρ([A | O]) = 3 =  number of unknowns
Hence the system has trivial solution only.

Given plane is passing through the points
\(\vec { a } =2\hat { i } +2\hat { j } +2\hat { k }, \vec { b } =9\hat { i } +3\hat { j } +6\hat { k } \)
Equation of the given plane is 2x + 6y + 6z = 9. It can be written as \(\vec { r } .(2\hat { i } +6\hat { j } +6\hat { k } )=9\)
Since the given plane is perpendicular to \(2\hat { i } +6\hat { j } +6\hat { k } \), the required plane is parallel to \(\vec { c } =2\hat { i } +6\hat { j } +6\hat { k } \). Hence, parametric form of vector equation of plane passing through two points and parallel to a vector is
\(\vec { r } =\vec { a } +s(\vec { b } -\vec { a } )+t\vec { c } ,s,t\in R\)
\(\vec { r } =2\hat { i } +2\hat { j } +\hat { k } +s(7\hat { i } +\hat { j } +5\hat { k } )+t(2\hat { i } +6\hat { j } +6\hat { k } ),s,t\in R\)
Cartesian equation of the plane is
\(\left| \begin{matrix} x-{ x }_{ 1 } & y-{ y }_{ 1 } & z-{ z }_{ 1 } \\ { x }_{ 2 }-{ x }_{ 1 } & { y }_{ 2 }-{ y }_{ 1 } & { z }_{ 2 }-{ z }_{ 1 } \\ { c }_{ 1 } & { c }_{ 2 } & { c }_{ 3 } \end{matrix} \right| =0\)
\(\Rightarrow \left| \begin{matrix} x-2 & y-2 & z-1 \\ 7 & 1 & 5 \\ 2 & 6 & 6 \end{matrix} \right| =0\)
⇒ (x-2)(6-30) - (y-2)(42-10) + (z-1)(42-2) = 0
⇒ (x - 2)(-24) - (y - 2)(32) + (z - 1)(40) = 0
⇒ 24x + 48 - 32y + 64 + 40z - 40 = 0
⇒ -24x - 32y + 40z + 72 = 0
\(\div\) - 8 we get
3x+ 4y - 5z - 9 = 0 is the Cartesian form.
∴ The parametric form of vector equation is
\(\vec { r } =\vec { r } (3\vec { i } +4\vec { j } -5\vec { k } )=9\)

By taking the vertex; at the origin, the parabola is open downward.

Its equation is x² = -4ay
It passes through (6, -4)
∴ 36 = -4a(-4) ⇒ 4a = - \(\frac { 36 }{ 4 } \) = 9
∴ (1) becomes, x² = -9y
To find the slope at (-6, -4)
Differentiating (1) with respect to 'x' we get,
2x = -9\(\frac { dy }{ dx } \)
⇒ \(\frac { dy }{ dx } =\frac { -2x }{ 9 } \)
At (-6, -4), \(\frac { dy }{ dx } =-2\frac { (-6) }{ 9 } =\frac { 12 }{ 9 } =\frac { 4 }{ 3 } \)
∴ \(tan\theta =\frac { 4 }{ 3 } \Rightarrow \theta ={ tan }^{ -1 }\left( \frac { 4 }{ 3 } \right) \)
∴ The angle of projection is tan^-1 \(\left( \frac { 4 }{ 3 } \right) \)

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.

(i) 
Rearrange the terms as,
(x-5) (x+4) (x-7) (x+6) = 504
{-2, 3, -7, 8}
⇒ (x² -x - 20) (x² -x - 42) = 504
Put x²- x = y
⇒ (y-20)(y-42) = 504
⇒ y²-62y+840-504 = 0
⇒ y²-62y+336 = 0
⇒ (y - 56) (y - 6) = 0
⇒ y = 56, 6
Case (i)
When y = 56, x² - x = 56

⇒ x² - x - 56 = 0
⇒ (x - 8)(x + 7) = 0
⇒ x = 8, -7
Case (ii) 
When y = 6,
x²- x = 6
x²- x - 6 = 0
⇒ (x - 3)(x + 2) = 0
⇒ x = 3, - 2

Hence the roots are -2, 3, 8, -7

\({ I }_{ n }=\int _{ 0 }^{ \pi /2 }{ { cos }^{ 7 }xdx=\frac { n-1 }{ n } { I }_{ n-2 },n\ge 2 } \)
\(\therefore { I }_{ 7 }=\int _{ 0 }^{ \pi /2 }{ { cos }^{ 7 }xdx=\frac { 6 }{ 7 } \times \frac { 4 }{ 5 } \times \frac { 2 }{ 3 } \times 1 } =\frac { 16 }{ 35 } \)