~(p V q) V (~p ∧ q) and ~p

p	q	p V q	~(p ∧ q)	~p	~p ∧ q	~(p V q) V (~p ∧ q)
T	T	T	F	F	F	F
T	F	T	F	F	F	F
F	T	T	F	T	T	T
F	F	F	T	T	F	T

The entries in column (5) and column (7) are identical.
∴ ~(p V q) V (~p ∧ q) and ~p are logically equivalent.

It is known that Z₅ = {[0], [1], [2], [3], [4]}. The table corresponding to addition modulo 5 is as follows: We take reminders {0,1,2,3,4} to represent the classes {[0], [1], [2], [3], [4]}.

+5	0	1	2	3	4
0	0	1	2	3	4
1	1	2	3	4	0
2	2	3	4	0	1
3	3	4	0	1	2
4	4	0	1	2	3

(i) Since each box in the table is filled by exactly one element of Z₅, the output a +₅ b is unique and hence +₅ is a binary operation.
(ii) The entries are symmetrically placed with respect to the main diagonal. So +₅ has commutative property
(iii) The table cannot be used directly for the verification of the associative property. So it is to be verified as usual
For instance, (2+₅3)+₅ 4 = 0+₅ 4 = 4(mod 5)
and 2+₅(3+₅4) = 2 +₅ 2 = 4(mod5)
Hence (2+₅3)+₅4 = 2+₅(3+₅4)
Proceeding like this one can verify this for all possible triples and ultimately it can be shown that +₅ is associative
(iv) The row headed by 0 and the column headed by 0 are identical. Hence the identity element is 0.
(v) The existence of inverse is guaranteed provided the identity 0 exists in each row and each column. From Table, it is clear that this property is true in this case. The method of finding the inverse of any one of the elements of Z₅, say 2 is outlined below.
First find the position of the identity element 0 in the III row headed by 2. Move horizontally along the III row and after reaching 0, move vertically above 0 in the IV column, because 0 is in the III row and IV column. The element reached at the topmost position of IV column is 3. This element 3 is nothing but the inverse of 2, because, 2+₅ 5+ = 0 (mod5). In this way, the inverse of each and every element of Z₅ can be obtained. Note that the inverse of 0 is 0, that of 1 is 4, that of 2 is 3, that of 3 is 2, and, that of 4 is 1.

Equation of the given circle is x² + y² = 16 ...(1)
and the parabola is y² = 6x. ...(2)
Substituting (2) in (1) we get,
x² + 6x - 16 = 0 \(\Rightarrow\) (x + 8) (x - 2) = 0
\(\Rightarrow\) (x-4) (x-2) = 0 \(\Rightarrow\) x = 2, 4
\(\Rightarrow\) x = -8, 2

\(\therefore\) Required area = 2
\(\\ \\ \\ \\ \\ \\ \\ \int _{ 0 }^{ 2 }{ Area\ below\ the\ parabola } +\int _{ 2 }^{ 4 }{ Area\ below\ the\ circle } \)
\(=2\left[ \int _{ 0 }^{ 2 }{ \sqrt { 6x } } dx+\int _{ 2 }^{ 4 }{ \sqrt { 16-{ x }^{ 2 } } dx } \right] \)
\(=2\left[ { \left( \frac { \sqrt { 6 } .{ x }^{ \frac { 3 }{ 2 } } }{ \frac { 3 }{ 2 } } \right) }_{ 0 }^{ 2 }+{ \left( \frac { x }{ 2 } \sqrt { 16-{ x }^{ 2 } } +\frac { 16 }{ 2 } { sin }^{ -1 }\left( \frac { x }{ 4 } \right) \right) }_{ 0 }^{ 4 } \right] \)
\(=2\left[ \frac { 2 }{ 3 } \sqrt { 6. } 2\sqrt { 2 } +8{ sin }^{ -1 }\left( \frac { 1 }{ 2 } \right) \right] \)
\(=2\left[ \frac { 4\sqrt { 12 } }{ 3 } +8\left( \frac { \pi }{ 2 } \right) -\sqrt { 12 } -8\left( \frac { \pi }{ 6 } \right) \right] \)
\(=2\left[ \frac { 4\sqrt { 12 } }{ 3 } +4\pi -2\sqrt { 3 } -\frac { 4\pi }{ 3 } \right] \)
\(=2\left[ \frac { 8\sqrt { 3 } -6\sqrt { 3 } }{ 3 } +\frac { 12\pi -4\pi }{ 3 } \right] \)
\(=\left[ \frac { 2\sqrt { 3 } }{ 3 } +\frac { 8\pi }{ 3 } \right] =2\times \frac { 2 }{ 3 } \left[ \sqrt { 3 } +4\pi \right] \)
\(=\frac { 4 }{ 3 } \left[ 4\pi +\sqrt { 3 } \right] \)sq.units

\(\frac { \partial z }{ \partial x } =x.\frac { 1 }{ 1+{ x }^{ 2 }{ y }^{ 2 } } (y)+{ tan }^{ -1 }(xy)\)
\(\frac { \partial z }{ \partial y } =x.\frac { 1 }{ 1+{ x }^{ 2 }{ y }^{ 2 } } \)(x)
\(\frac { \partial z }{ \partial z } =\frac { xy }{ 1+{ x }^{ 2 }{ y }^{ 2 } } \) + tan^-1 (xy)
\(\frac { \partial z }{ \partial y } =\frac { x^2 }{ 1+{ x }^{ 2 }{ y }^{ 2 } } \)
\(\frac { \partial z }{ \partial y } =\frac { x^2 }{ 1+{ x }^{ 2 }{ y }^{ 2 } } \)
\(\frac { \partial z }{ \partial x } =\frac { { t }^{ 2 }{ se }^{ t } }{ 1+{ t }^{ 4 }{ s }^{ 2 }{ e }^{ 2t } } +{ tan }^{ -1 }({ t }^{ 2 }{ se }^{ 2 })\)
\(\frac { \partial z }{ \partial y } =\frac { { t }^{ 4 } }{ 1+{ t }^{ 4 }{ s }^{ 2 }{ e }^{ 2t } } \)
\(\frac { dx }{ dt } =2t;\frac { dy }{ dt } ={ s.e }^{ t }\)
Also, \(\frac { dx }{ ds } =0;\frac { dy }{ ds } ={ e }^{ t }\)
By chain rule
\(\frac { dw }{ ds } =\frac { \partial z }{ \partial x } .\frac { dx }{ ds } +\frac { \partial z }{ \partial y } .\frac { dy }{ ds } \)
= \(\frac { { t }^{ 2 }{ se }^{ t } }{ 1+{ t }^{ 4 }{ s }^{ 2 }{ e }^{ 2t } } (0)+\frac { { t }^{ 4 } }{ 1+{ t }^{ 4 }{ s }^{ 2 }{ e }^{ 2t } } { (e }^{ t })=\frac { { e }^{ t }{ t }^{ 4 } }{ 1+{ t }^{ 4 }{ s }^{ 2 }{ e }^{ 2t } } \)
\(\therefore { \left( \frac { \partial z }{ \partial s } \right) }_{ s=t=1 }=\frac { e(1) }{ 1+{ e }^{ 2 } } =\frac { e }{ 1+{ e }^{ 2 } } \)
By chain rule
= \(\frac { dz }{ dt } =\frac { \partial z }{ \partial x } .\frac { dx }{ dt } +\frac { \partial z }{ \partial y } .\frac { dy }{ dt } \)
\(\therefore { \left( \frac { \partial z }{ \partial t } \right) }_{ (s=t=1) }=\frac { e }{ 1+{ e }^{ 2 } } =\frac { e }{ 1+{ e }^{ 2 } } \)
= \(\frac { 2e+e }{ 1+{ e }^{ 2 } } =\frac { 3e }{ 1+{ e }^{ 2 } } \)+ 2 tan^-1(e)

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 = \(\left( \begin{matrix} 8 \\ 2 \end{matrix} \right) =28\)

Values of Random Variable X	40	10	-20	Total
Number of elements in inverse images	6	32	28	66

Probability mass function is

X	40	10	-20	Total
f (x)	\(\cfrac { 6 }{ 66 } \)	\(\cfrac { 32 }{ 66 } \)	\(\cfrac { 28 }{ 66 } \)	1

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 \(V(x)=\frac { 4000 }{ 11 } \)

We recall that the equation of the tangent to the circle x² + y² = a² at (x₁, y₁) is xx₁+yy₁ = a². So, the equation of the tangent to the circle x² + y² + = 4 at (1, \(\sqrt 3\)) is x+y\(\sqrt 3\) = 4; that is y = -\(\frac{1}{\sqrt 3}\)( x - 4). The tangent meets the x-axis at the point (4,0). The slope of the tangent is-\(\frac{1}{\sqrt 3}\). So the slope of the normal is \(\sqrt 3\) and hence equation of the normal is y-\(\sqrt 3\) =\(\sqrt 3\)(x-1); that is y = \(\sqrt 3\) x and it passes through the origin. The area to be found is shaded in the adjoining figure. It can be found by two methods.

(i) By definition \(F(x)=\le x)=\int _{ -\infty }^{ x }{ f(u)dx } \) 
When x < 1 \(F(x)=P\left( x\le x \right) =\int _{ 1 }^{ x }{ odu+\int _{ 1 }^{ 0 }{ (u-1) } du } \) 
When 1 ≤ x < 2 \(F(x)=P(X\le x)=\int _{ -\infty }^{ x }{ odu } =0\) 
When 1 ≤ x < 2 \(F(x)=P\left( x\le x \right) =\int _{ 1 }^{ x }{ odu+\int _{ 1 }^{ 0 }{ (u-1) } du } \) 
= \(0+\left[ \frac { \left( u-1 \right) ^{ 2 } }{ 2 } \right] =\frac { \left( x-1 \right) ^{ 2 } }{ 2 } \) 
When 2 ≤ x <3 \(F(x)=P(X\le x)=\int _{ -\infty }^{ 1 }{ du } +\int _{ 1 }^{ 2 }{ \left( u-1 \right) du } +\int _{ 2 }^{ x }{ \left( 3-u \right) du } \) 
= \(0+\left[ \frac { \left( u-1 \right) ^{ 2 } }{ 2 } \right] +\left[ \frac { (3-u)^{ 2 } }{ 2 } \right] \) 
= \(\frac { { 1 }^{ 2 }-0 }{ 2 } +\frac { 1-(3-x)^{ 2 } }{ 2 } =1\frac { \left( 3-x \right) ^{ 2 } }{ 2 } \) 
When x ≥ 3, \(F(x)=P\left( X\le x \right) =\int _{ -\infty }^{ 1 }{ odu } +\int _{ 1 }^{ 3 }{ (u-1) } +\int _{ 2 }^{ 1 }{ (3-u) } +\int _{ 3 }^{ x }{ odu } \) 
= \(\int _{ -\infty }^{ 1 }{ 0du } +\int _{ 1 }^{ 2 }{ (u-1)du } +\int _{ 2 }^{ 3 }{ (3-u) } du+\int _{ 3 }^{ x }{ 0du } \) 
= \(0+\left[ \frac { \left( u-1 \right) ^{ 2 } }{ 2 } \right] +\left[ \frac { \left( 3-u \right) ^{ 2 } }{ 2 } \right] _{ 2 }^{ 3 }+0\) 
= \(\cfrac { 1 }{ 2 } +\cfrac { 1 }{ 2 } =1\) 
These give 


(ii) P(1.5 ≤ X ≤ 2.5) = F(2.5) − F(1.5)
= \(\left( 1-\frac { \left( 3-2.5 \right) ^{ 2 } }{ 2 } \right) -\left( \frac { \left( 1.5-1 \right) ^{ 2 } }{ 2 } \right) \) 
= \(\cfrac { 1.75-0.25 }{ 2 } =0.75\) 
\(P\left( 1.5\le X\le \right) =\int _{ 1.5 }^{ 2.5 }{ f(x)dx } =\int _{ 1.5 }^{ 2 }{ (x-1) } dx+\int _{ 2 }^{ 2.5 }{ (-x+3) } dx=0.75\) 

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)]
= \(1-\left[ ^{18}{ C }_{ 0 }\left( \frac { 1 }{ 3 } \right) ^{ 0 }\left( \frac { 2 }{ 3 } \right) ^{ 18 }+^{ 18}{C }_{ 1 }\left( \frac { 1 }{ 3 } \right) ^{ 1 }\left( \frac { 2 }{ 3 } \right) ^{ 17 } \right] \)
=  \(1-\left[ \left( \frac { 2 }{ 3 } \right) ^{ 18 }+6\left( \frac { 2 }{ 3 } \right) ^{ 17 } \right] \) 
= \(1-\left( \frac { 2 }{ 3 } \right) ^{ 17 }\left[ \frac { 2 }{ 3 } +6 \right] \) 
= \(1-\left( \frac { 2 }{ 3 } \right) ^{ 17 }\left( \frac { 20 }{ 3 } \right) \)
= \(1-\frac { 20 }{ 3 } \left( \frac { 2 }{ 3 } \right) ^{ 17 }\)

Let x = tan \(\theta \Rightarrow\)dx = sec²\(\theta d \theta\)

x	0	1
\(\theta\)	0	\(\frac{\pi}{4}\)

\(\therefore I=\int _{ 0 }^{ \frac { \pi }{ 4 } }{ \frac { log(1+tan\theta ) }{ { sec }^{ 2 }\theta } } { sec }^{ 2 }\theta d\theta \)
\(=\int _{ 0 }^{ \frac { \pi }{ 4 } }{ log(1+tan\theta )d\theta } \quad ...(1)\)
Using property,
\(I=\int _{ 0 }^{ \frac { \pi }{ 4 } }{ log(1+tan(\frac { \pi }{ 4 } -\theta ))d\theta } \)
\(I=\int _{ 0 }^{ \frac { \pi }{ 4 } }{ log\left( \frac { 1+tan\theta +1-tan\theta }{ 1+tan\theta } \right) } d\theta \)
\(I=\int _{ 0 }^{ \frac { \pi }{ 4 } }{ log\left( \frac { 2 }{ 1+tan\theta } \right) } d\theta \quad ..(2)\)
(1)+(2)
\(2I=\int _{ 0 }^{ \frac { \pi }{ 4 } }{ log(1+tan\theta )d\theta } +\int _{ 0 }^{ \frac { \pi }{ 4 } }{ log\left( \frac { 2 }{ 1+tan\theta } \right) } d\theta \)

\(=\int _{ 0 }^{ \frac { \pi }{ 4 } }{ log2d\theta =log2{ [\theta ] }_{ 0 }^{ \frac { \pi }{ 4 } } } \)
\(2I=log2(\frac { \pi }{ 4 } -0)=\frac { \pi }{ 4 } log2\)
\(\therefore I=\frac { \pi }{ 8 } log2\)

First we calculate  \(\frac { { \partial }w }{ { \partial x } } (x,y)=\frac { { \partial }(xy) }{ { \partial x } } +\frac { \partial \left( \frac { { e }^{ y } }{ { y }^{ 2 }+1 } \right) }{ \partial x } \)
This gives  \(\frac { { \partial }^{ }w }{ { \partial x } } \) (x, y) = y + 0 and hence \(\frac { { \partial }^{ 2 }w }{ \partial y\partial x } \) (x, y) = 1 On the other hand,
\(\frac { { \partial }w }{ { \partial y } } (x,y)=\frac { { \partial }(xy) }{ { \partial y } } +\frac { \partial \left( \frac { { e }^{ y } }{ { y }^{ 2 }+1 } \right) }{ \partial y } \)
\(=x+\frac { \left( { y }^{ 2 }+1 \right) { e }^{ y }-{ e }^{ y }2y }{ \left( { y }^{ 2 }+1 \right) } \)
Hence, \(\frac { { \partial }^{ 2 }w }{ { \partial x\partial y } } \) (x, y) = 1

\(f(x)=\begin{cases} \begin{matrix} x+1 & -1\le x<0 \end{matrix} \\ \begin{matrix} -x+1 & 0\le x<1 \end{matrix} \\ \begin{matrix} 0 & otherwise \end{matrix} \end{cases}\)
(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

Put I = \(\int _{ 0 }^{ \frac { \pi }{ 4 } } \frac { dx }{ a^{ 2 }sin^{ 2 }x+b^{ 2 }cos^{ 2 }x } =\int _{ 0 }^{ \frac { \pi }{ 4 } } \frac { sec^{ 2 }xdx }{ a^{ 2 }tan^{ 2 }x+b^{ 2 } } \)
Put u = tan x 
Then du = sec² x dx
When x = 0 , we have u = tan 0 = 0 
When x = \(\frac{\pi}{4}\), we have u = tan\(\frac{\pi}{4}\) = 1
\(\therefore I=\int _{ 0 }^{ 1 } \frac { du }{ a^{ 2 }u^{ 2 }+b^{ 2 } } =\frac { 1 }{ a^{ 2 } } \int _{ 0 }^{ 1 } \frac { du }{ u^{ 2 }+\left( \frac { b }{ a } \right) ^{ 2 } } =\frac { 1 }{ a^{ 2 } } { \left[ \frac { a }{ b } tan^{ -1 }(\frac { au }{ b } ) \right] }_{ 0 }^{ 1 }=\frac { 1 }{ ab } tan^{ -1 }\left( \frac { a }{ b } \right) \)
We derive some more properties of definite integrals

Here a = 0, b = 1,f (x) = 5x + 4
\(\therefore f(a+(b-a)\frac { r }{ n } )=f\left( 0+1(\frac { r }{ n } ) \right) =f(\frac { r }{ n } )=5(\frac { r }{ n } )+4\)
\(\int _{ a }^{ b }{ f(x)dx } =\underset { n\rightarrow \infty }{ lim } \frac { b-a }{ n } \sum _{ r=1 }^{ n }{ f(a+(b-a)\frac { r }{ n } } \)
\(\therefore \int _{ 0 }^{ 1 }{ (5x+4)dx } =\underset { n\rightarrow \infty }{ lim } \frac { 1 }{ n } \sum _{ r=1 }^{ n }{ \left( \frac { 5r }{ n } +4 \right) } \)
\(=\underset { n\rightarrow \infty }{ lim } \left[ \frac { 1 }{ n } \sum _{ r=1 }^{ n }{ \frac { 5r }{ n } +\frac { 1 }{ n } \sum _{ r=1 }^{ n }{ 4 } } \right] \)
\(=\underset { n\rightarrow \infty }{ lim } \left[ \frac { 1 }{ n } .\frac { 5 }{ n } .(1+2+3+...+n)+\frac { 1 }{ n } .4n \right] \)
\(=\underset { n\rightarrow \infty }{ lim } \left[ \frac { 5 }{ { n }^{ 2 } } \frac { n(n+1) }{ 2 } +4 \right] \)
\([\because \sum { r=\frac { n(n+1) }{ 2 } ;\sum { { r }^{ 2 }=\frac { n(n+1)(2n+1) }{ 6 } } } ]\)
\(=\underset { n\rightarrow \infty }{ lim } \frac { 5 }{ { n }^{ 2 } } { n }^{ 2 }\frac { (1+\frac { 1 }{ n } ) }{ 2 } +4\)
\(=\frac { 5 }{ 2 } (1+0)+4=\frac { 5 }{ 2 } +4\)
\([wehen\quad n\rightarrow \infty ,1/n\quad \rightarrow 0]\)
\(=\frac { 5+8 }{ 2 } =\frac { 13 }{ 2 } \)
\(\therefore \int _{ 0 }^{ 1 }{ (5x+4)dx } =\frac { 13 }{ 2 } \)

Let the given function be written as f (x) = x² ( 10− x)² . Now
f(x) = x²(100 - 20x + x²) = x⁴-20x³+100x²
Therefore, f'(x) = 4x³- 60x³ + 200x = 4x(x²-15x+50)
f'(x) = 4x(x²-15x + 50) = 0 ⇒ x = 0, 5, 10
and f"(x) = 12x²-120x + 200
The stationary points of f(x) are x = 0, 5, 10 at these points the values of f′′(x) are respectively 200, −100 and 200. At x = 0, it has local minimum and its value is f(0) = 0. At x = 5, it has local maximum and its value is f(5) = 625. At x = 10, it has local minimum and its value is f(10) = 0.

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
\(\left( -\infty ,-1 \right) \left( -1,\frac { 1 }{ 2 } \right) \left( \frac { 1 }{ 2 } ,\infty \right) \)

Interval	(∞,-1)	\(\left( -1,\frac { 1 }{ 2 } \right) \)	\(\left( \frac { 1 }{ 2 } ,\infty \right) \)
Sign of f'(x)	Say x = -2 12(-2)2 + 6 (-2)-6 = +ve	Say x = 0 = -6 -ve	Say x = 1 I2(1)2 + 6(1) - 6 = +ve
Monoto nicity	strictly increasmg	Strictly decreasing	Strictly increasing

∴ 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 \(\left( -\infty ,\frac { -1 }{ 4 } \right) \left( \frac { -1 }{ 4 } ,\infty \right) \) 

Interval	\(\left( -\infty ,\frac { -1 }{ 4 } \right) \)	\(\left( \frac { -1 }{ 4 } ,\infty \right) \)
Sign of f'(x)	Say x = -1 24(-1) + 6 = -ve	Say x = 0  +ve
Concavity	Concave down	Concave up

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-\(\\ \\ \\ \\ \\ \\ \\ { e }^{ -\frac { 3t }{ 50 } }\))

Let f(x) = \(y=\frac { 1 }{ 1+{ e }^{ -x } } \)
1. The domain of y is R
2. It is not symmetric
3. Putting y = 0, we get x does not exist 
∴ It has no X intercept
4. Putting x = 0, we get \(y=\frac { 1 }{ 1+{ e }^{ o } } =\frac { 1 }{ 1+1 } =\frac { 1 }{ 2 } \)
∴ Y-intercept is \(\left( 0,\frac { 1 }{ 2 } \right) \)
5. \(f'(x)=\frac { d }{ dx } \left( 1+e^{ -x } \right) ^{ -1 }=-1\left( 1+{ e }^{ -x } \right) \left( -e^{ -x } \right) \)
= \(\frac { { e }^{ -x } }{ \left( 1+e^{ -x } \right) ^{ 2 } } ={ e }^{ -x }\left( 1+{ e }^{ -x } \right) ^{ -2 }\)
\(f'(x)=0\Rightarrow { e }^{ -x }0\Rightarrow -x=log0\Rightarrow x=\infty \)
Also, f(x) > 0 ∀ X ∈ R
6. Since there is no sign change inf'(x), it has no local extreme.
7. f"(x) = e^-x(-2)(1+e^-x)^-3(^_e^-x)+(1+e^-x)^-2(^_e^-x)
= 2e^-2x (I+e^-x )^-3 -e^-x ( 1+e-x )^-2
= e^-2x (1+e-x r³ [2 - e^-x (1+e^-x)]
= e^-2X(I+e-xr³[2-e^x -1]
= \(\frac { { e }^{ -2x }\left( 1-{ e }^{ x } \right) }{ \left( 1+{ { e }^{ x } } \right) ^{ 3 } } \)
f"(x) = 0 ⇒ x = 0
The intervals of concavity is tabulated as follows.

Interval	(-∞,0)	(0,∞)
Sign of f"(x)	+	-
Concavity	Concave up	Concave down

8.\(\underset { x\rightarrow \infty }{ lim } \frac { 1 }{ 1+{ e }^{ -x } } =1\)

⇒ y = 1 is the horizontal asymptote and
\(\underset { x\rightarrow \infty }{ lim } \frac { 1 }{ 1+{ e }^{ -x } } =0\)
⇒ y = 0 is the horizontal asymptote
ஃ The rough sketch is

Let A be the number of bacteria at any time t
Given \(\frac { dA }{ A } \alpha A\)
\(\Rightarrow \frac { dA }{ dt } =kA\)
\(\Rightarrow \frac { dA }{ dt } =kdt\)
\(\Rightarrow \int { \frac { dA }{ dt } } =k\int { \frac { dt }{ t } } \)
\(\Rightarrow log\left( \frac { A }{ c } \right) =kt\)
\(\Rightarrow log\left( \frac { A }{ c } \right) =kt\)
\(\Rightarrow \frac { A }{ c } ={ e }^{ kt }\)
\(\Rightarrow A=c.{ e }^{ kt }\) ...(1)
Initially when t = 0, assume that A = A₀
\(\therefore\) becomes, = ce₀ \(\Rightarrow\) c = A₀ 
\(\therefore\) A = A0ekt  ...(2)
Given when t = 5, A = 3A₀
3A₀ = A₀e^5k \(\Rightarrow\) 3 = 35k 
when t = 10, 2 becomes,
A = A₀e^10k
= A₀(e^5k)2 = A₀(3)²
\(\Rightarrow\) A = 9A₀
Hence, the amount of bacteria will be present after 10 hours is 9 times the initial amount.

\(\mathrm{y} \mathrm{d} x=-\left(2 x-10 \mathrm{y}^3\right) \mathrm{dy}  \)
\(y \frac{d x}{d y}=-2 x+10 \mathrm{y}^3 \)
\(\div y, \frac{y}{y} \frac{d x}{d y}+\frac{2}{y} x=\frac{10 y^3}{y}  \)
\(\frac{d x}{d y}+\left(\frac{2}{y}\right) x=10 \mathrm{y}^2\)
This is of the form \( \frac{d x}{d y}+P x=Q \)
where \(\mathrm{P}=\frac{2}{y} \quad \mathrm{Q}=10 \mathrm{y}^2 \)
Thus, the given equation is linear. \( I.F  =e^{\int p d y}=e^{\int \frac{2}{y} d y}=e^{2 \log y}=\mathrm{y}^2\)
So, the required solution is
\(x \times \text { I.F }   =\int(Q \times I . F) d y+c \)
\(x \mathrm{y}^2   =\int 10 y^2 \times y^2 d y+c  \)
\(=\int 10 y^4 d y+c  \)
\(=\frac{10 y^5}{5}+c=2 \mathrm{y}^5+\mathrm{c}\)
\(x y^2=2 y^5+c\) is a required solution

Let us take the semi circle with center (0, 0) an radius r

x = r cos θ,
y = r sin θ
ஃ Length of the rectangle = 2x = 2r cos θ
Breadth of the rectangle = y = r sin θ
ஃ Area = 2xy = 2r² cos θ sin θ
= r² sin 2θ
Letf'(θ) = r² 2 cos 2θ
f(θ) = r²cosθ sinθ
⇒ 2r² cos 2θ = 0
\(\Rightarrow cos2\theta =0=cos\frac { \pi }{ 2 } \)
\(\Rightarrow 2\theta =\frac { \pi }{ 2 } \)
\(\Rightarrow \theta =\frac { \pi }{ 4 } \) 
\(f''={ 2r }^{ 2 }(2)-(-sin2\theta )\) 
= \(-4{ r }^{ 2 }sin2\theta \) 
\(\therefore f''\left( \frac { 4\pi }{ 4 } \right) =-4{ r }^{ 2 }sin2\left( \frac { \pi }{ 4 } \right) \) 
= \(-4{ r }^{ 2 }sin\frac { \pi }{ 2 } ={ 4r }^{ 2 }<0\) 
ஃ f(θ) is maximum when \(\theta =\frac { \pi }{ 4 } \)
ஃ Length of the rectangle \((x)=2ros\frac { \pi }{ 4 } \)
= \(2r\times \frac { 1 }{ \sqrt { 2 } } =\sqrt { 2 } r\) 
ஃ Breadth of the rectangle = y = r sin θ
= \(rsin\frac { \pi }{ 4 } =\frac { r }{ \sqrt { 2 } } \)

Given x + y = 10
⇒ y = 10 -x ...(1)
Let f(x) = x²+y
= x²+ (10 -x)²
= x + 100 +x²- 20x
f(x) = 2x² - 20x+ 100
f'(x) = 4x - 20
f'(x) = 0
4x-20 = 0
4x = 20
⇒ x = 5
∴The critical number is 5
f"(x) = 4
ஃ f"(5) = 4 > 0
ஃf(x) is minimum when x = 5
When x = 5, y = 10- 5 = 5
[From (1)]
ஃ Smallest possible value of x²+y²
= 5² + 5² = 25 + 25 = 50

\(\Rightarrow \frac { dy }{ dx } =\frac { { y }^{ 2 }-2xy }{ { x }^{ 2 }-2xy } \)  ...(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,
\( \int \frac{(2 v-1) d v}{\left(v^2-v\right)} =\int-3 \frac{d x}{x} \)
\(\log \left(v^2-v\right) =-3 \log x+\log \mathrm{c} \)
\(\log \left(v^2-v\right)+\log \left(x^3\right) =\log \mathrm{c} \)
\(\left(v^2-v\right)\left(x^3\right) =\mathrm{c} \)
\(\left(\frac{y^2}{x^2}-\frac{y}{x}\right) x^3 =c \)
\(x y^2-x^2 \mathrm{y} =\mathrm{c}\)

f(x) is defined and differentiable in (-∞, ∞)
\(\therefore f'(x)=\frac { (1-{ e }^{ x }){ e }^{ x }-{ e }^{ x }\left( -{ e }^{ x } \right) }{ (1-{ e }^{ x })^{ 2 } } \) 
= \(\frac { 1-{ e }^{ 2 }+{ e }^{ 2 }x }{ (1-{ e }^{ x })^{ 2 } } =\frac { 1 }{ \left( 1-{ e }^{ x } \right) ^{ 2 } } \) 
\(f'\left( x \right) =0\Rightarrow \frac { 1 }{ (1-{ e }^{ x })^{ 2 } } \neq 0\)
Thus there is no stationary point 
Since \(f'\left( x \right) =\frac { 1 }{ \left( 1-{ e }^{ x } \right) ^{ 2 } } >0\)  for-all x ∈ (-∞, ∞)
Since there is no stationary print, the curve does not change its position.
Hence there is no local extremum.

Let the point of intersection of the two curves be (a,b) . Hence,
 \(a^{2}+4b^{2}=8\) and  \(a^{2}-2b^{2}=4\) ...(4)
It is enough if we show that the product of the slopes of the two curves evaluated at (a, b) is −1.
Differentiation of \(x^{2}+4y^{2}=8\)  with respect x, gives
\(2x+8y=\frac{dy}{dx}=0\).
Therefore \(\frac{dy}{dx}= -\frac{-x}{4y}\),
\((\frac{dy}{dx})_{(a,b)}=m_{1}= -\frac{a}{4b}\)
Differentiation of x²-2y² = 4 with respect to x, gives
\(2x-4y\frac{dy}{dx}=0\)
Therefore, \(\frac{dy}{dx}=\frac{x}{2y}\), 
at \((a,b)(\frac{dy}{dx})=m_{2}= \frac{a}{4b}\).
Therefore, \(m_{1}\times m_{2}=(-\frac{a}{4b})\times (\frac{a}{2b})= -\frac{a^{2}}{8b^{2}}\)  ...(5)
Applying the ratio of proportions in (4), we get
\(\frac{a^{2}}{-16-16}=\frac{b^{2}}{-8+4}=\frac{1}{-2-4}\)
Therefore, \(\frac{a^{2}}{b^{2}}=\frac{32}{4}=8\) Substituting in (5), we get \(m_{1}\times m_{2}=-1\) Hence, the curves cut orthogonally.

Given s (t) = 2t³ − 9t² + 12t ≥ 0

Let a(t) be the distance of car A north of P at time t, and b (t) the distance of car B east of P at time t, and let c(t) be the distance from car A to car B at time t. By the Pythagorean Theorem, c(t)² = a(t)² + b(t)²

Taking derivatives, we get 2c(t)c'(t) = 2a(t)a'(t) + 2b(t)b'(t).
So, c′ = \(\frac { { aa }^{ ' }+{ bb }^{ ' } }{ c } =\frac { { aa }^{ ' }+{ bb }^{ ' } }{ \sqrt { { a }^{ 2 }+{ b }^{ 2 } } } \)
Substituting known values, we get
\(c' =\frac { (10\times 80)+(15\times 100) }{ \sqrt { { 10 }^{ 2 }+{ 15 }^{ 2 } } } =\frac { 460 }{ \sqrt { 13 } } \) ≈ 127.6 km/hr at the time of intersect

(x-4)(x-7)(x-2)(x+1) = 16
Given (x - 4)(x -7)(x - 2)(x + 1) = 16
Rearrange the terms as
(x - 4)(x - 2)(x -7)(x + 1) = 16
⇒ (x² - 6x + 8)(x2 - 6x -7) = 16
Put x²-6x = y
⇒ (y + 8) (y - 7) = 16
⇒ y² +y - 56 - 16 = 0
⇒ (y+9)(y-8) = 0
⇒ y = -9, 8

case (i)
 When y = -9
⇒ x²-6x = -9
⇒ x²-6x+9 = 0
⇒ (x-3)² = 0
⇒ x = 3, 3

case ii
When y = 8,
\(\Rightarrow x=\frac { 6\pm \sqrt { 36-4(1)(-8) } }{ 2 } \)
\(\Rightarrow x=\frac { 6\pm \sqrt { 36-32 } }{ 2 } \Rightarrow x=\frac { 6\pm \sqrt { 68 } }{ 2 } \)
\(\Rightarrow x=\frac { 6\pm 2\sqrt { 17 } }{ 2 } \Rightarrow x=\frac { 2(3\pm \sqrt { 17 } ) }{ 2 } \)
\(\Rightarrow x=3\pm \sqrt { 17 } \)
∴ The roots are 3, 3, 3 +\(\sqrt{17}\), and 3 -\(\sqrt{17}\)

 \(\left( -\sqrt { 3 } +3i \right) ^{ 31 }\)
Let \(-\sqrt { 3 } +3i=r\left( cos\theta +isin\theta \right) \). Then, we get
\(r=\sqrt { \left( -\sqrt { 3 } \right) ^{ 2 }+{ 3 }^{ 2 } } =\sqrt { 12 } =2\sqrt { 3 } \)
\(\alpha ={ tan }^{ -1 }\left| \frac { 3 }{ -\sqrt { 3 } } \right| ={ tan }^{ -1 }\sqrt { 3 } =\frac { \pi }{ 3 } \)
\(\theta =\pi -\alpha =\pi -\frac { \pi }{ 3 } =\frac { 2\pi }{ 3 } \) (\(\because\)  \(\sqrt { 3 } +3i\) lies in II Quadrant)
Therefore,\(-\sqrt { 3 } +3i=2\sqrt { 3 } \left( cos\frac { 2\pi }{ 3 } +isin\frac { 2\pi }{ 3 } \right) \)
Raising power 31 on both sides,
\(\left( -\sqrt { 3 } +3i \right) ^{ 31 }=\left( 2\sqrt { 3 } \right) ^{ 31 }\left( cos\frac { 2\pi }{ 3 } +isin\frac { 2\pi }{ 3 } \right) ^{ 31 }\)
= \(\left( 2\sqrt { 3 } \right) ^{ 31 }\left( cos\left( 20\pi +\frac { 2\pi }{ 3 } \right) +isin\left( 20\pi +\frac { 2\pi }{ 3 } \right) \right) \)
= \(\left( 2\sqrt { 3 } \right) ^{ 31 }\left( cos\frac { 2\pi }{ 3 } +isin\frac { 2\pi }{ 3 } \right) \)
= \(\left( 2\sqrt { 3 } \right) ^{ 31 }\left( cos\left( \pi -\frac { \pi }{ 3 } \right) +isin\left( \pi -\frac { \pi }{ 3 } \right) \right) \)
= \(\left( 2\sqrt { 3 } \right) ^{ 31 }\left( -cos\frac { \pi }{ 3 } +isin\frac { \pi }{ 3 } \right) =\left( 2\sqrt { 3 } \right) ^{ 31 }\left( -\frac { 1 }{ 2 } +i\frac { \sqrt { 3 } }{ 2 } \right) \).

\(\left[ \begin{matrix} 1 & 2 & 3 \\ 2 & 5 & 3 \\ 1 & 0 & 8 \end{matrix} \right] \)
Let A =\(\left[ \begin{matrix} 1 & 2 & 3 \\ 2 & 5 & 3 \\ 1 & 0 & 8 \end{matrix} \right] \)
Applying Gauss - Jordan method, we get
[A|I₃] =\(\left[ \begin{matrix} 1 & 2 & 3 \\ 2 & 5 & 3 \\ 1 & 0 & 8 \end{matrix}|\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} \right] \)
\(\overset { { R }_{ 2 }\rightarrow { R }_{ 2 }-2{ R }_{ 1 }\\ { R }_{ 3 }\rightarrow { R }_{ 3 }-{ R }_{ 1 } }{ \longrightarrow } \left[ \begin{matrix} 1 & 2 & 3 \\ 0 & 1 & -3 \\ 0 & -2 & 5 \end{matrix}|\begin{matrix} 1 & 0 & 0 \\ -2 & 1 & 0 \\ -1 & 0 & 1 \end{matrix} \right] \)
\(\overset { { R }_{ 1 }\rightarrow { R }_{ 1 }+{ R }_{ 3 } }{ \longrightarrow } \left[ \begin{matrix} 1 & 0 & 8 \\ 0 & 1 & -3 \\ 0 & -2 & 5 \end{matrix}|\begin{matrix} 0 & 0 & 1 \\ -2 & 1 & 0 \\ -1 & 0 & 1 \end{matrix} \right] \)
\(\\ \overset { { R }_{ 3 }\rightarrow { R }_{ 3 }+2{ R }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} 1 & 0 & 8 \\ 0 & 1 & -3 \\ 0 & 0 & -1 \end{matrix}|\begin{matrix} 0 & 0 & 1 \\ -2 & 1 & 0 \\ -5 & 2 & 1 \end{matrix} \right] \)
\(\overset { { R }_{ 2 }\rightarrow { R }_{ 2 }-3{ R }_{ 3 } }{ \longrightarrow } \left[ \begin{matrix} 1 & 0 & 8 \\ 0 & 1 & 0 \\ 0 & 0 & -1 \end{matrix}|\begin{matrix} 0 & 0 & 1 \\ 13 & -5 & -3 \\ -5 & 2 & 1 \end{matrix} \right] \)
\(\overset { R_{ 1 }\rightarrow { R }_{ 1 }+8R_{ 3 } }{ \longrightarrow } \left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1 \end{matrix}|\begin{matrix} -40 & 16 & 9 \\ 13 & -5 & -3 \\ -5 & 2 & 1 \end{matrix} \right] \)
\(\overset { { R }_{ 3 }\rightarrow { R }_{ 3 }\times (-1) }{ \longrightarrow } \left[ \begin{matrix} 11 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}|\begin{matrix} -40 & 16 & 9 \\ 13 & -5 & -3 \\ 5 & -2 & -1 \end{matrix} \right] \)
So we get A^-1 =\(\left[ \begin{matrix} -40 & 16 & 9 \\ 13 & -5 & -3 \\ 5 & -2 & -1 \end{matrix} \right] \).

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.

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 \(\left( \frac { 4 }{ 3 } \right) \)

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.

2x+3y = 9, 7x+3y-5z = 8, 2x+3y+⋋z = μ
The matrix form of the system is AX = B where
A =\(\left[ \begin{matrix} 2 & 3 & 5 \\ 7 & 3 & -5 \\ 2 & 3 & \lambda \end{matrix} \right] ,X=\left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] ,B=\left[ \begin{matrix} 9 \\ 8 \\ \mu \end{matrix} \right] \)
Applying elementary row operations augmented matrix [A|B] we get
[A|B] =\(\left[ \begin{matrix} 2 & 3 & 5 \\ 7 & 3 & -5 \\ 2 & 3 & \lambda \end{matrix}|\begin{matrix} 9 \\ 8 \\ \mu \end{matrix} \right] \overset { { R }_{ 1 }\leftrightarrow { R }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} 7 & 3 & -5 \\ 2 & 3 & 5 \\ 2 & 3 & \lambda \end{matrix}|\begin{matrix} 8 \\ 9 \\ \mu \end{matrix} \right] \)
\(\overset { { R }_{ 2 }\rightarrow { R }_{ 2 }-\frac { 2 }{ 7 } { R }_{ 1 }\\ { R }_{ 3 }\rightarrow { R }_{ 3 }-{ R }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} 7 & 3 & -5 \\ 0 & \frac { 15 }{ 7 } & \frac { 45 }{ 7 } \\ 0 & 0 & \lambda -5 \end{matrix}|\begin{matrix} -8 \\ \frac { 45 }{ 7 } \\ 4-9 \end{matrix} \right] \)
\(\overset { { R }_{ 2 }\rightarrow { R }_{ 2 }\times 7 }{ \longrightarrow } \left[ \begin{matrix} 7 & 3 & -5 \\ 0 & 15 & 45 \\ 0 & 0 & \lambda -5 \end{matrix}|\begin{matrix} -8 \\ 47 \\ \mu -9 \end{matrix} \right] \)
Case (i): when λ = 5
[A|B] =\(\left[ \begin{matrix} 7 & 3 & -5 \\ 0 & 15 & 45 \\ 0 & 0 & 0 \end{matrix}|\begin{matrix} -8 \\ 47 \\ -4 \end{matrix} \right] \)
Here \(\rho \)(A) = 2 and \(\rho \)[A|B] = 3
So, \(\rho \)(A) ≠ \(\rho \)[A|B]
Hence the system is inconsistent and has no solution
Case (ii) : When λ ≠ 5, μ ≠ 9
[A|B] =\(\\ \left[ \begin{matrix} 7 & 3 & -5 \\ 0 & 15 & 45 \\ 0 & 0 & not\quad zero \end{matrix}|\begin{matrix} -8 \\ 47 \\ not\quad zero \end{matrix} \right] \)
Here \(\rho \)(A) = 3 and \(\rho \)[A|B] = 3
∴ \(\rho \)(A) = \(\rho \)[A|B] = 3 = number of unknowns
Hence, the system is consistent with solution
Case (iii) : When λ = 5, μ = 9
[A|B] =\(\left[ \begin{matrix} 7 & 3 & -5 \\ 0 & 15 & 45 \\ 0 & 0 & 0 \end{matrix}|\begin{matrix} -8 \\ 47 \\ 0 \end{matrix} \right] \)
Here \(\rho \)(A) = 2 and \(\rho \)[A|B] = 2
∴ \(\rho \)(A) = \(\rho \)[A|B] = 2
∴ The system is consistent and has infinite number of solutions.

Here the number of unknowns is 3.
The matrix form of the system is AX = B, where
A = \(\left[ \begin{matrix} 4 & -2 & 6 \\ 1 & 1 & -3 \\ 15 & -3 & 9 \end{matrix} \right] \), X = \(\left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] \), B = \(\left[ \begin{matrix} 8 \\ -1 \\ 21 \end{matrix} \right] \)
Applying elementary row operations on the augmented matrix [A | B], we get
[A | B] = \(\left[ \begin{matrix} 4 & -2 & 6 \\ 1 & 1 & -3 \\ 15 & -3 & 9 \end{matrix}|\begin{matrix} 8 \\ -1 \\ 21 \end{matrix} \right] \) \(\overset { { R }_{ 1 }\leftrightarrow { R }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} 1 & 1 & -3 \\ 4 & -2 & 6 \\ 15 & -3 & 9 \end{matrix}|\begin{matrix} -1 \\ 8 \\ 21 \end{matrix} \right] \overset { \begin{matrix} { R }_{ 2 }\longrightarrow { R }_{ 2 }-4{ R }_{ 1 }, \\ { R }_{ 3 }\longrightarrow { R }_{ 3 }-15{ R }_{ 1 } \end{matrix} }{ \longrightarrow } \left[ \begin{matrix} 1 & 1 & -3 \\ 0 & -6 & 18 \\ 0 & -18 & 54 \end{matrix}|\begin{matrix} -1 \\ 12 \\ 36 \end{matrix} \right] \)\(\overset { \begin{matrix} { R }_{ 2 }\longrightarrow { R }_{ 2 }\div \left( -6 \right) , \\ { R }_{ 3 }\longrightarrow { R }_{ 3 }\div \left( -18 \right) \end{matrix} }{ \longrightarrow } \left[ \begin{matrix} 1 & 1 & -3 \\ 0 & 1 & -3 \\ 0 & 1 & -3 \end{matrix}|\begin{matrix} -1 \\ -2 \\ -2 \end{matrix} \right] \overset { { R }_{ 3 }\longrightarrow { R }_{ 3 }-{ R }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} 1 & 1 & -3 \\ 0 & 1 & -3 \\ 0 & 0 & 0 \end{matrix}|\begin{matrix} -1 \\ -2 \\ 0 \end{matrix} \right] \).
So, ρ(A)= ρ([A | B]) = 2 < 3. From the echelon form, we get the equivalent equations
x + y − 3z = −1, y − 3z = −2 , 0 = 0.
The equivalent system has two non-trivial equations and three unknowns. So, one of the unknowns should be fixed at our choice in order to get two equations for the other two unknowns. We fix z arbitrarily as a real number t , and we get y = 3t − 2, x = −1−(3t − 2) + 3t = 1. So, the solution is (x = 1, y = 3t − 2, z = t), where t is real. The above solution set is a one-parameter family of solutions. Here, the given system is consistent and has infinitely many solutions which form a one parameter family 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 = \(\pm \frac { 1 }{ \sqrt { 2 } } \)

The path y = ax² + bx + c passes through the points (10, 8), (20, 16), (40, 22). So, we get the system of equations 100a + 10b + c = 8, 400a + 20b + c = 16, 1600a + 40b + c = 22. To apply Cramer’s rule, we find
Δ = \(\left| \begin{matrix} 100 & 10 & 1 \\ 400 & 20 & 1 \\ 1600 & 40 & 1 \end{matrix} \right| =1000\left| \begin{matrix} 1 & 1 & 1 \\ 4 & 2 & 1 \\ 16 & 4 & 1 \end{matrix} \right| \) = 1000 [-2 + 12 - 6] = -6000,
Δ₁ = \(\left| \begin{matrix} 8 & 10 & 1 \\ 16 & 20 & 1 \\ 22 & 40 & 1 \end{matrix} \right| =20\left| \begin{matrix} 4 & 1 & 1 \\ 8 & 2 & 1 \\ 11 & 4 & 1 \end{matrix} \right| \) = 20[-8 + 3 + 10] = 100,
Δ₂ = \(\left| \begin{matrix} 100 & 8 & 1 \\ 400 & 16 & 1 \\ 1600 & 22 & 1 \end{matrix} \right| =200\left| \begin{matrix} 1 & 4 & 1 \\ 4 & 8 & 1 \\ 16 & 11 & 1 \end{matrix} \right| \) = 200[-3 + 48 - 84] = -7800,
Δ₃ = \(\left| \begin{matrix} 100 & 10 & 8 \\ 400 & 20 & 16 \\ 1600 & 40 & 22 \end{matrix} \right| =2000\left| \begin{matrix} 1 & 1 & 4 \\ 4 & 2 & 8 \\ 16 & 4 & 11 \end{matrix} \right| \) = 2000[-10 + 84 - 64] = 20000.
By Cramer’s rule, we get a = \(\frac { { \Delta }_{ 1 } }{ \Delta } =-\frac { 1 }{ 60 } \), b = \(\frac { { \Delta }_{ 2 } }{ \Delta } =\frac { 7800 }{ 6000 } =\frac { 78 }{ 60 } =\frac { 13 }{ 10 } \), c = \(\frac { { \Delta }_{ 3 } }{ \Delta } =\frac { 20000 }{ 6000 } =-\frac { 20 }{ 6 } =-\frac { 10 }{ 3 } \).
So, the equation of the path is y = \(\frac { 1 }{ 60 } { x }^{ 2 }+\frac { 13 }{ 10 } x-\frac { 10 }{ 3 } \).
When x = 70, we get y = 6. So, the ball went by 6 metres high over the boundary line and it is impossible for a fielder standing even just before the boundary line to jump and catch the ball. Hence the ball went for a super six and the team won the match.

For the parabola,
x²- 4x - 5y -1 = 0
x²- 4x = 5y +1
x²−4x +4 = 5y +1+ 4
(x − 2)² = 5(y +1) which is in standard form.
Therefore 4a = 5 and the vertex is (2, -1) , and focus is \(\left( 2,\frac { 1 }{ 4 } \right) \)
Equation of directrix is 
y-k+ a = 0
y+1+\(\frac { 5 }{ 4 } \)
4y +9 = 0 
Length of latus rectum is 5 units.

We can solve this fourth degree equation by rewriting it suitably and adopting a technique of substitution. Rewriting the equation as
(x−2)(x−3)(x−7)(x+2)+19 = 0
the given equation becomes
(x²−5x+6)(x²−5x−14)+19 = 0 .
If we take x² − 5x as y, then the equation becomes (y+6)(y−14)+19 = 0;
that is, y²-8y-65 = 0
Solving this we get solutions y = 13 and y = −5. Substituting this we get two quadratic equations
x²-5x-13 = 0 and x²-5x+5 = 0
which can be solved by usual techniques. The solutions obtained for these two equations together give solutions as \(\frac { 5\pm \sqrt { 77 } }{ 2 } ,\frac { 5\pm \sqrt { 5 } }{ 2 } \).

If A is orthogonal, then AA^T = A^TA = I₃. So, we have
AA^T = I₃ ⇒ \(\frac { 1 }{ 7 } \left[ \begin{matrix} 6 & -3 & a \\ b & -2 & 6 \\ 2 & c & 3 \end{matrix} \right] \frac { 1 }{ 7 } \left[ \begin{matrix} 6 & -3 & a \\ b & -2 & 6 \\ 2 & c & 3 \end{matrix} \right] =\left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} \right] \)
⇒ \(\left[ \begin{matrix} 45+{ a }^{ 2 } & 6b+6+6a & 12-3c+3a \\ 6b+6+6a & { b }^{ 2 }+40 & 2b-2c+18 \\ 12-3c+3a & 2b-2c+18 & { c }^{ 2 }+13 \end{matrix} \right] =49\left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} \right] \)
⇒ \(\left\{ \begin{matrix} 45+{ a }^{ 2 }=49 \\ \begin{matrix} { b }^{ 2 }+40=49 \\ \begin{matrix} { c }^{ 2 }+13=49 \\ \begin{matrix} 6b+6+6a=0 \\ \begin{matrix} 12-3c+3a=0 \\ 2b-2c+18=0 \end{matrix} \end{matrix} \end{matrix} \end{matrix} \end{matrix} \right\} \) ⇒ \(\left\{ \begin{matrix} { a }^{ 2 }=4,{ b }^{ 2 }=9,{ c }^{ 2 }=36 \\ a+b=-1,a-c=-4,b-c=-9 \end{matrix} \right\} \) ⇒ a = 2, b = −3, c = 6
So we get A = \(\frac { 1 }{ 7 } \left[ \begin{matrix} 6 & -3 & 2 \\ -3 & -2 & 6 \\ 2 & 6 & 3 \end{matrix} \right] \) and hence, we get A^-1 = A^T = \(\frac { 1 }{ 7 } \left[ \begin{matrix} 6 & -3 & 2 \\ -3 & -2 & 6 \\ 2 & 6 & 3 \end{matrix} \right] \).

\(Let\ u={ x }^{ 2 }\ v=cos 2x\ dx\)
\(u'=2x\quad { v }_{ 1 }=\frac { sin2x }{ 2 } \)
\(u"=2\quad { v }_{ 2 }=-\frac { cos2x }{ 4 } \)
\(\\ { v }_{ 3 }=-\frac { sinx }{ 8 } \)
Bernoulli's' formula:
\(\int { uvdx } =u{ v }_{ 1 }-u'{ v }_{ 2 }+u''{ v }_{ 3 }\)
\(\therefore \int _{ 0 }^{ \pi /2 }{ { x }^{ 2 }cos2x\quad dx={ \left[ \frac { { x }^{ 2 }sin2x }{ 2 } +\frac { 2xcos2x }{ 4 } -\frac { 2sin2x }{ 8 } \right] }_{ 0 }^{ \frac { \pi }{ 2 } } } \)
\(=\left[ \frac { { \pi }^{ 2 } }{ 8 } sin\pi +\frac { 2\pi }{ 8 } cos\pi -\frac { sin\pi }{ 8 } \right] -[0+0-0]\)
\(=\frac {- 2\pi }{ 8 } (-1)=\frac { \pi }{ 4 }\quad [\because sin\pi =0,cos\pi =-1]\)

Given that \(\frac { dy }{ dx } =\frac { x-y+5 }{ 2(x-y)+7 } .\)
Put z = x-y
\(\frac { dz }{ dx } =1-\frac { dy }{ dx } \)
\(\frac { dy }{ dx } =1-\frac { dz }{ dx } \)
Thus, the given equation reduces to
\(1-\frac { dz }{ dx } =\frac { z+5 }{ 2z+7 } \) 
\(\frac { dz }{ dx } =1+\frac { z+5 }{ 2z+7 } \)
\(\frac { dz }{ dx } =\frac { z+2 }{ 2z+7 } \)
Separating the variables, we get
\(\frac { 2z+7 }{ z+2 } dz=dx\)
\(\frac { 2(z+2)+3 }{ z+2 } =dx\)
\(\left( 2+\frac { 3 }{ z+2 } \right) dz=dx\)
Integrating both sides, we get
2z + 3log |z+ 2| = x + C
That is, 2(x − y) + 3log |x −y+2| = x + C

Every point on the line \(\frac { x-1 }{ 2 } =\frac { y-2 }{ 3 } =\frac { z-3 }{ 4 } \) = s (say) is of the form (2s + 1, 3s + 2, 4s + 3) and every point on the line \(\frac { x-4 }{ 5 } =\frac { y-1 }{ 2 } =z=t\) (say) is of the form (5t + 4, 2t + 1, t).
So, at the point of intersection, for some values of s and t, we have
(2s + 1, 3s + 2, 4s + 3) = (5t + 4, 2t + 1, t)
Therefore, 2s - 5t = 3, 3s - 2t = -1 and 4s - t = -3. Solving the first two equations we get t = -1, s = -1. 
These values of s and t satisfy the third equation. Therefore, the given lines intersect. 
Substituting, these values of t or s in the respective points, the point of intersection is (-1,- 1, -1)

AS = 94.5 × 10⁶ km,
SA' = 152 × 10⁶ km
a+c = 152 × 10⁶
a-c = 94.5 × 10⁶
Subtracting 2c = 57.5 × 10⁶ = 575 × 10⁵ km
Distance of the Sun from the other focus is SS' = 575 × 105 km.

Let \(\frac { \sqrt { 3 } }{ 2 } +\frac { i }{ 2 } \) = r(cos θ + i sin θ)
r = \(\sqrt { \left( \frac { \sqrt { 3 } }{ 2 } \right) ^{ 2 }+\left( \frac { 1 }{ 2 } \right) ^{ 2 } } =\sqrt { \frac { 3 }{ 4 } +\frac { 1 }{ 4 } } =\sqrt { \frac { 4 }{ 4 } } \)=1
α = \(tan^{ -1 }\left| \frac { y }{ x } \right| =tan^{ -1 }\left| \frac { \frac { 1 }{ 2 } }{ \frac { \sqrt { 3 } }{ 2 } } \right| =tan^{ -1 }\left( \frac { 1 }{ \sqrt { 3 } } \right) =\frac { \pi }{ 6 } \)
Since \(\frac { \sqrt { 3 } }{ 2 } +\frac { i }{ 2 } \) lies is the I quadrant, θ = α
∴ \(\frac { \sqrt { 3 } }{ 2 } +\frac { i }{ 2 } =1\left( cos\frac { \pi }{ 6 } +isin\frac { \pi }{ 6 } \right) \)
∴ \(\left( \frac { \sqrt { 3 } }{ 2 } +\frac { i }{ 2 } \right) ^{ 5 }=1^{ 5 }\left( cos\frac { \pi }{ 6 } +isin\frac { \pi }{ 6 } \right) ^{ 5 }\)
= \(cos\frac { 5\pi }{ 6 } +isin\frac { 5\pi }{ 6 } \)  ....(1) [De moivres theorem]
Similarly \(\frac { \sqrt { 3 } }{ 2 } +\frac { i }{ 2 } =1\left( cos\frac { \pi }{ 6 } -isin\frac { \pi }{ 6 } \right) \)
∴ \(\left( \frac { \sqrt { 3 } }{ 2 } -\frac { i }{ 2 } \right) ^{ 5 }=1^{ 5 }\left[ cos\frac { \pi }{ 6 } +isin\frac { \pi }{ 6 } \right] ^{ 5 }\)
= \(cos\frac { 5\pi }{ 6 } -isin\frac { 5\pi }{ 6 } \) ....(2)
Adding (1) and (2) we get,
\(\left( \frac { \sqrt { 3 } }{ 2 } +\frac { i }{ 2 } \right) ^{ 5 }+\left( \frac { \sqrt { 3 } }{ 2 } -\frac { i }{ 2 } \right) ^{ 5 }\)

= \(2cos\frac { 5\pi }{ 6 } =2cos\left( \pi -\frac { \pi }{ 6 } \right) \)
= \(-2cos\ \frac { \pi }{ 6 } \) [∵ \(\frac { 5\pi }{ 6 } \) lies in the II quard]
= \(-2\left( \frac { \sqrt { 3 } }{ 2 } \right) =-\sqrt { 3 } \).

With usual notations in triangle, ABC let \(\vec { BC } =\vec { a } \), \(\vec { CA } =\vec { b } \) and \(\vec { AB } =\vec { c} \). Then \(|\vec { BC }| =\vec { a } \), \(|\vec { CA } |=\vec { b } \) and \(|\vec { AB }| =\vec { c } \)
Since in ΔABC, \(\vec { BC }+\vec { CA }+\vec { AB}=0\) we have \(\vec { BC }\times(\vec { BC }+\vec { CA }+\vec { AB })=\vec { 0 }\)
Simplifying, we get,
\(\vec { BC }\times\vec { CA }=\vec { AB }\times\vec { BC }\) ....(1)

Similarly, since \(\vec { BC }+\vec { CA }+\vec { AB}=\vec 0\), we have
\(\vec {CA} \times (\vec { BC }+\vec { CA }+\vec { AB})=\vec 0\) .... (2)
On Simplification, we obtain \(\vec { BC }\times\vec { CA }=\vec {CA }\times\vec {AB }\)
From equations (1) and (2), we get 
\(\vec { AB }\times\vec {BC }\) = \(\vec { CA }\times\vec {AB }\)=\(\vec { BC}\times\vec {CA}\)
So, \(\left| \overrightarrow { AB} \times \overrightarrow { BC } \right| =\left| \overrightarrow { CA } \times \overrightarrow { AB } \right| =\left| \overrightarrow { BC } \times \overrightarrow { CA } \right| \). Then, we get
ca sin(π − B) = bc sin(π - A) = ab sin (π - C)
That is, ca sin B = bc sin A = absinC . Dividing by abc, we get
\(\frac { sinA }{ A } =\frac { sinB }{ b } =\frac { sinC }{ c } \) or \(\frac { a }{ sinA }= \frac { b }{ sinB } =\frac { c }{ sinC } \)

Given \(f(x)=\frac { { x }^{ 2 }-6x-1 }{ x+3 } \)
\(\underset { x\rightarrow -3^{ + } }{ lim } \frac { { x }^{ 2 }-6x-1 }{ x+3 } =\underset { h\rightarrow { 0 }^{ + } }{ lim } \frac { \left( -3+h \right) ^{ 2 }-6(-3+h)-1 }{ -3+h+3 } \) 
= \(\underset { h\rightarrow 0^{ + } }{ lim } \frac { 9+{ h }^{ 2 }-6h+18-6h-1 }{ h } \)
= \(\underset { h\rightarrow 0^{ + } }{ lim } \frac { { h }^{ 2 }-12h+26 }{ h } =\infty \)
and \(\underset { x\rightarrow 3^{ - } }{ lim } \frac { { x }^{ 2 }-6x-1 }{ x+3 } =\underset { h\rightarrow { o }^{ + } }{ lim } \frac { \left( -3-h \right) ^{ 2 }-6(-3-h)-1 }{ -3-h+3 } \)
= -∞
ஃx = -3 is the vertical asymptote,
Also

ஃ y = x - 9 is the slanting asymptote,