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 M = \(\left\{ \left( \begin{matrix} x & x \\ x & x \end{matrix} \right) :x\in R-\{ 0\} \right\} \) and * be the matrix multipilication.
Let A \(=\left( \begin{matrix} x & x \\ x & x \end{matrix} \right) \) and B = \(\left( \begin{matrix} y & y \\ y & y \end{matrix} \right) \in M\)
Where x, y ∈R-{0}.
\(A*B=\left( \begin{matrix} x & x \\ x & x \end{matrix} \right) \left( \begin{matrix} y & y \\ y & y \end{matrix} \right) \)
\(\\ =\left( \begin{matrix} 2xy & 2xy \\ 2xy & 2xy \end{matrix} \right) \in M\)
[∵ 2xy ∈R-{0}]
∴ M is closed under *.
Commutative property:
we know A*B =\(\left( \begin{matrix} 2xy & 2xy \\ 2xy & 2xy \end{matrix} \right) ..(1)\)
Let x,y∈R-{0}
Now B + A \(=\left( \begin{matrix} y & y \\ y & y \end{matrix} \right) \left( \begin{matrix} x & x \\ x & x \end{matrix} \right) \)
\(=\left( \begin{matrix} xy+xy & xy+xy \\ xy+xy & xy+xy \end{matrix} \right) \)
\(=\left( \begin{matrix} 2xy & 2xy \\ 2xy & 2xy \end{matrix} \right) \\ \)
From (1) &(2), A*B = B*A
∴ *has commutative property on M
Associative property:
Let A = \(\left( \begin{matrix} x & x \\ x & x \end{matrix} \right) \)
B =\(\left( \begin{matrix} y & y \\ y & y \end{matrix} \right) \) and
C = \(\left( \begin{matrix} z & z \\ z & z \end{matrix} \right) \)
for x, y, z ∈R-{0}
\((A*B)*C=\left( \begin{matrix} 2xy & 2xy \\ 2xy & 2xy \end{matrix} \right) *\left( \begin{matrix} z & z \\ z & z \end{matrix} \right) \)
\(=\left( \begin{matrix} 2xyz+2xyz & 2xyz+2xyz \\ 2xyz+2xyz & 2xyz+2xyz \end{matrix} \right) \)
\(=\left( \begin{matrix} 4xyz & 4xyz \\ 4xyz & 4xyz \end{matrix} \right) ...(1)\)
Now\(A*(B*C)=A*\left( \begin{matrix} 2yz & 2yz \\ 2yz & 2yz \end{matrix} \right) \)
\(=\left( \begin{matrix} x & x \\ x & x \end{matrix} \right) *\left( \begin{matrix} 2yz & 2yz \\ 2yz & 2yz \end{matrix} \right) \)
\(=\left( \begin{matrix} 4xyz & 4xyz \\ 4xyz & 4xyz \end{matrix} \right) ...(2)\\ \)
From (1)&(2), (a*B)*C = A*B*C)
Since matrix multiplication is associative, this axiom holds good for M.

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]}.

(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.

(i) The output m+ n - mn is clearly an integer and hence∗ is a binary operation on Z.
(ii) m*n = m+ n − mn = n + m − nm = n*m, ∀m,n∈Z. So ∗ has commutative property.
(iii) Consider (m*n)*p = (m+ n −m n)* p= (m+ n −mn) + p − (m+ n −m n) p
= m+ n + p −mn −m p − n p + m n p ... (1)
Similarly m*(n*p) = m*(n + p − n p) = m+ (n + p − n p) −m (n + p − n p)
= m+ n + p − n p −m n −mp + m n p ... (2)
From (1) and (2), we see that m*(n*p) = (m*n)*p. Hence * has associative property.
(iv) An integer e is to be found such that 
m*e = e*m = m, ∀m∈Z ⇒ m + e - m e = m
⇒e(1-m) = 0 ⇒ e = 0 or m = 1. But m is an arbitrary integer and hence need not be equal to 1. So the only possibility is e = 0. Also m*0 = 0*m = m, ∀m∈Z. Hence 0 is the identity element and hence the existence of identity is assured.
(v) An element m'∈ Z is to be found such that m*m' = m' * m = e = 0, ∀m∈Z.
m*m' = 0 ⇒ m+m'-m m'= 0⇒ m m' = 0 ⇒ \(\frac{m}{m-1}\). when m = 1, m' is not defined.
When m = 2, m' is an integer. But except m = 2, m′ need not be an integer for all values of m. Hence inverse does not exist in Z.

Volume of the right circular conical frustum is obtained by revolving the line y = x between x = a and x = b around the x - axis

\(\therefore\) Height of the frustum h = b - a
\(\therefore\)Volume \(=\pi \int _{ a }^{ b }{ { x }^{ 2 }dx } =\pi { \left[ \frac { { x }^{ 3 } }{ 3 } \right] }_{ a }^{ b }\)
\(=\frac { \pi }{ 3 } [{ b }^{ 3 }-{ a }^{ 3 }]\)
\(=\frac { \pi }{ 3 } (b-a)({ b }^{ 2 }+ab+{ a }^{ 2 })\)
Now, substitute h = b - a, r = a and R = b we get Volume of the conical frustum
\(\frac { \pi }{ 3 } [h({ R }^{ 2 }+rR+{ r }^{ 2 })]\)
Given h = 2 m, r = 1 m, R = 2 m we get
Required volume \(=\frac { \pi }{ 3 } [2(4+2+1)]\)
\(=\frac { \pi }{ 3 } (14)\)
\(=\frac { 14\pi }{ 3 } \)

Equation of the given curves are y² = 4x and x² = 4y

\(\therefore \) Required area \(=\int _{ 0 }^{ 4 }{ \left( \sqrt { 4x } -\frac { { x }^{ 2 } }{ 4 } \right) dx } \)
\(=\int _{ 0 }^{ 4 }{ \left( 2\sqrt { x } -\frac { { x }^{ 2 } }{ 4 } \right) dx } \)
\(={ \left[ \frac { { 2x }^{ \frac { 3 }{ 2 } } }{ \frac { 3 }{ 2 } } \right] }_{ 0 }^{ 4 }{ \left[ \frac { 4 }{ 3 } x\sqrt { x } -\frac { { x }^{ 3 } }{ 12 } \right] }_{ 0 }^{ 4 }\)
\(=\frac { 4 }{ 3 } (4)(2)-\frac { 64 }{ 12 } \)
\(=\frac { 32 }{ 3 } -\frac { 32 }{ 6 } =\frac { 64-32 }{ 6 } =\frac { 32 }{ 6 } \)
\(=\frac { 16 }{ 3 } \) sq.units
Yes the area can be divided into 3 equal parts and the area to the divided among his, wife daughter and son is  \(=\frac { 16 }{ 3 } \)sq.units

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

Probability mass function is

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

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

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

The given function is a polynomial of degree 4. Now,

f′(x) = (x −1)³ + 3(x −1)² . (x − 5) 
= 4(x-1)².(x-4)
f"(x) = 4(x-1)²+2(x-1).(x-4)
= 12(x −1)  (x − 3)
Now,
f''(x) = 0 ⇒ x = 1, x = 3
The intervals of concavity are tabulated in the table 7.7.

Interval	(-∞, 1)	(1, 3)	(3, ∞)
Sign of f'(x)	+	-	+
Concavity	concave up	concave down	concave up

The curve is concave upwards on (∞, 1) and (3, ∞) .
The curve is concave downwards on (1, 3) .
As f′′(x) changes its sign when it passes through x = 1 and x = 3, (1, f(1)) = (1, 0) and (3, f(3)) = (3, −16) are points of inflection for the graph y = f(x). This may be observed from the adjoining figure of the curve f′′(x) .

Let T represent the temperature of the boiling water and Tm represents the temperature of the kitchen.
By Newton's law of cooling
\(\Rightarrow \int { \frac { dT }{ T-{ T }_{ m } } =K\int { dt } } \)
\(\Rightarrow log(T-{ T }_{ m })=Kt+logC\)
\(\Rightarrow log(T-{ T }_{ m })-logC=Kt\)
\(\Rightarrow log\left( \frac { T-{ T }_{ m } }{ C } \right) =Kt\)
\(\Rightarrow T-{ T }_{ m }={ Ce }^{ Kt } ...(1)\)
when t=0,T=100
\(\therefore 100-{ T }_{ m }={ Ce }^{ 0 }\)
\(\Rightarrow C=100-{ T }_{ m }\)
\(\Rightarrow becomes,\ T-{ T }_{ m }=(100-{ T }_{ m }){ e }^{ Kt }\)
Also when t = 5, T = 80
\(\therefore 80-{ T }_{ m }=(100-{ T }_{ m }){ e }^{ 5K }\)
\(\Rightarrow { e }^{ 5K }=\frac { 80-{ T }_{ m } }{ 100-{ T }_{ m } } ..(2)\)
When t = 10, T = 65
(2) \(\Rightarrow\) 65 - T = (100-T_m)e^10K
= (100-T_m)(e^5K)²
\(=(100-{ T }_{ m }){ \left( \frac { 80-{ T }_{ m } }{ 100-{ T }_{ m } } \right) }^{ 2 }\)
[using(2)]
\(\Rightarrow 65-{ T }_{ m }=\frac { { (80-{ T }_{ m } })^{ 2 } }{ 100-{ T }_{ m } } \)
\(\Rightarrow\) 6500-65T_m-100T_m+T_m² = 6400+T_m²-160T_m
\(\Rightarrow\) 6500-6400 = 165Tm-160Tm
\(\Rightarrow\) 100 = 5Tm
\(\\ \Rightarrow { T }_{ m }=\frac { 100 }{ 5 } ={ 20 }^{ o }C\)
Hence the temperature of the kitchen is 20^oC

\(\frac { dy }{ dx } +\frac { 3y }{ x } =\frac { 1 }{ { x }^{ 2 } } \), given that y = 2 when x = 1 
This is a linear differential equation.
\(\therefore P=\frac { 3 }{ x } ;Q=\frac { 1 }{ { x }^{ 2 } } \)
\(\int { pdx } =3\int { \frac { 1 }{ x } dx=3logx=log{ x }^{ 3 } } \)
\(\therefore I.F.={ e }^{ \int { pdx } }={ e }^{ log{ x }^{ 3 } }={ x }^{ 3 }\)
\(\therefore\) The solution is \({ ye }^{ \int { pdx } }=\int { Q{ e }^{ \int { pdx } }dx+c } \)
\(\Rightarrow { yx }^{ 3 }=\int { \frac { 1 }{ { x }^{ 2 } } .{ x }^{ 3 } } dx+c\)
\(\Rightarrow { yx }^{ 3 }=\int { xdx+c } \)
\(\Rightarrow { yx }^{ 3 }=\frac { { x }^{ 2 } }{ 2 } +c...(1)\)
When x = 1, y = 2
\(\Rightarrow 2{ (1) }^{ 3 }=\frac { 1 }{ 2 } +c\Rightarrow 2-\frac { 1 }{ 2 } =\frac { 3 }{ 2 } \)
\({ yx }^{ 3 }=\frac { { x }^{ 2 } }{ 2 } +\frac { 3 }{ 2 } \)
\({ 2x }^{ 3 }y={ x }^{ 2 }+3\)

Let x be the length of the garden and y be the breadth of the garden.

Given 2 (x +y) = 40
[ஃ length of the wire = 40 Perimeter = 40 m]
⇒ x + y = 20
⇒ y = 20 - x ...(1)
Let f(x) = xy = x (20 - x²) = 20x- xl²
f'(x) = 20 - 2x
f'(x) = 0
⇒ 20-2x = 0
⇒ 20 = 2x
x = 10
ஃ The critical number is 10
f"(x) = -2
Now f"(10) = -2 < 0
ஃ f(x) is minimum at x = 10
When x = 10, y = 20 -10 = 10
ஃ Area = f(x) = xy
= 10(10) = 100m²
ஃ Largest possible area of the garden = 100 m²

(x²+y²)dy = xy dx
\(\frac { dy }{ dx } =\frac { xy }{ { x }^{ 2 }+{ y }^{ 2 } } ...(1)\)
\(\therefore put=vx\Rightarrow \frac { dy }{ dx } =v+x\frac { dv }{ dx } \)
\(\therefore\)(1) becomes,
\(v+x\frac { dv }{ dx } =\frac { xvx }{ { x }^{ 2 }+{ v }^{ 2 }{ x }^{ 2 } } \)
\(=\frac { { x }^{ 2 }v }{ { x }^{ 2 }(1+{ v }^{ 2 }) } =\frac { v }{ 1+{ v }^{ 2 } } \)
\(x\frac { dv }{ dx } =\frac { v }{ 1+{ v }^{ 2 } } v=\frac { v-v-{ v }^{ 3 } }{ 1+{ v }^{ 2 } } =\frac { -{ v }^{ 3 } }{ 1+{ v }^{ 2 } } \)
Separating the variables we get,
\(\frac { 1+{ v }^{ 2 } }{ { v }^{ 3 } } dv=\frac { -dx }{ x } \)
\(\Rightarrow \frac { 1 }{ { v }^{ 3 } } +\frac { { v }^{ 2 } }{ { v }^{ 3 } } dv=\frac { -dx }{ x } \)
\(\Rightarrow \int { { v }^{ -3 }dv } +\int { \frac { dv }{ v } =-\int { \frac { dx }{ x } } } \)
\(\Rightarrow \frac { { v }^{ -2 } }{ -2 } +log\ v=-logx+logc\)
\(\Rightarrow -\frac { 1 }{ 2{ v }^{ 2 } } +log\ v=-log\ x+log\quad c\)
\(\Rightarrow \frac { 1 }{ 2{ v }^{ 2 } } -log\ v=logx-logc\)
\(\Rightarrow \frac { 1 }{ 2{ v }^{ 2 } } =logv+logx-logc\)
\(\Rightarrow \frac { 1 }{ 2{ v }^{ 2 } } =logv+logx-logc\)
\(\Rightarrow \frac { 1 }{ 2{ v }^{ 2 } } =log\left( \frac { vx }{ c } \right) \)
\(\Rightarrow \frac { { x }^{ 2 } }{ 2{ y }^{ 2 } } =log\left( \frac { y }{ c } \right) \Rightarrow { e }^{ \frac { { x }^{ 2 } }{ { e }^{ 2{ y }^{ 2 } } } }=\frac { y }{ c } \)
\(\Rightarrow y={ ce }^{ \frac { { x }^{ 2 } }{ 2{ y }^{ 2 } } } ...(2)\)
Given y(1) = 1
\(1={ ce }^{ \frac { 1 }{ 2 } }\Rightarrow 1=c\sqrt { e } \)
\(\Rightarrow c=\frac { 1 }{ \sqrt { e } } \)
\(\therefore\)(2) becomes,
\(y=\frac { 1 }{ \sqrt { e } } { e }^{ \frac { { x }^{ 2 } }{ 2{ y }^{ 2 } } }\)
\(Also\ y({ x }_{ 0 })=e\Rightarrow e=\frac { 1 }{ \sqrt { e } } { e }^{ \frac { { x }_{ 0 }^{ 2 } }{ 2{ e }^{ 2 } } }\)
\(\Rightarrow e\sqrt { e } ={ e }^{ \frac { { x }_{ 0 }^{ 2 } }{ 2{ e }^{ 2 } } }\)
\(\Rightarrow \frac { { x }_{ 0 }^{ 2 } }{ 2{ e }^{ 2 } } =log\quad e\sqrt { e } =log{ e }^{ \frac { 3 }{ 2 } }\)
\(\Rightarrow \frac { { x }_{ 0 }^{ 2 } }{ 2{ e }^{ 2 } } =\frac { 3 }{ 2 } { log }_{ e }^{ e }=\frac { 3 }{ 2 } (1)\)
\(\left[ \because { log }_{ e }^{ e }=1 \right] \)
\(\Rightarrow { x }_{ 0 }^{ 2 }=\frac { 3 }{ 2 } (2{ e }^{ 2 })={ 3e }^{ 2 }\)
\(\Rightarrow { x }_{ 0 }=\pm \sqrt { 3{ e }^{ 2 } } =\pm \sqrt { 3 } .e\)
\(\therefore { x }_{ 0 }=\pm \sqrt { 3 } .e\)

Given slope curve
\(\Rightarrow \frac { dy }{ dx } =\frac { y-1 }{ { x }^{ 2 }+x } \) ..... (1)
\(\Rightarrow \frac { dy }{ y-1 } =\frac { dx }{ { x }^{ 2 }+x } \)
\( \Rightarrow \frac{d y}{y-1}=\frac{d x}{x^2+x}=\frac{d x}{x^2+2\left(\frac{x}{2}\right)+\left(\frac{1}{2}\right)^2-\left(\frac{1}{2}\right)^2} \)
\(ie) \frac{d y}{y-1}=\frac{d x}{\left(x+\frac{1}{2}\right)^2-\left(\frac{1}{2}\right)^2} \)
Integrating on both sides, we get
 \( \int \frac{d y}{\log (y-1)}=\int \frac{d x}{\left(x+\frac{1}{2}\right)^2-\left(\frac{1}{2}\right)^2} \\  \text { ie) } \log (y-1)=\left(\frac{1}{2\left(\frac{1}{2}\right)}\right) \log \left[\frac{\left(x+\frac{1}{2}\right)-\left(\frac{1}{2}\right)}{\left(x+\frac{1}{2}\right)+\left(\frac{1}{2}\right)}\right]+\log C \\  \log (y-1)=\log \left(\frac{x}{x+1}\right)+\log C\\   ie)  (y-1)=\frac{C x}{x+1}  \) 
\(\Rightarrow log(y-1)=log\left( \frac { cx }{ x+1 } \right) \)
\(\Rightarrow y-1=\frac { cx }{ x+1 } \)
Since the curve passes through (1, 0) we get,
\(0-1=\frac { c }{ 2 } \Rightarrow c=-2\)
\(\Rightarrow y-1=\frac { -2x }{ x+1 } \)
\(\Rightarrow y=1-\frac { -2x }{ x+1 } \)
\(\Rightarrow y=\frac { x+1-2x }{ x+1 } =\frac { 1-x }{ x+1 } \)
\(\therefore y=\frac { 1-x }{ 1+x } \)

Given s (t) = 2t³ − 9t² + 12t ≥ 0
On differentiating we get
V(t) = 6t²-18t+ 12 ... (1)
= 6 (t² - 3t+ 2)
= 6 (t - 1) (t - 2)
Now V(t) = 0
⇒ 6 (t-1)(t- 2) = 0
⇒ t = 1, 2
The particle changes direction when V(t) changes its sign.
If 0 ≤ t < 1 then both (t - 1) and (t - 2) < 0
⇒ V(t) > 0
If 1 < t < 2 then (t -1) > 0 and (t - 2) < 0
⇒ V(t) < 0
If t > 2 then both (t - 1) and (t - 2) > 0
⇒ V(t) > 0
∴ The particle changes direction when t = 1 and t = 2 sec.
(ii) Total distance travelled by the particle in the first 4 seconds is |s(0)- s (1)| + |s (1) - s (2)| + |s (2) -s (4)|
s(0) = -4
s(1) = 2(1)3 - 9(1)2 + 12 (1) - 4
= 2 - 9 + 12 - 4 = 1
s (2) = 2 \(\times\) 2³ - 9 \(\times\) 2² + 12 \(\times\) 2 - 4
= 16 - 36 + 24 - 4
= 0
s (4) = 2(4)³ - 9(4)² + 12 (4) - 4
= 128 - 144 + 48 - 4 = 28
∴ Is (0) -s (1)|+ Is (1) -s (2)|+ Is (2) -s(4)|
= |-4 - 1| + |1 - 0| + |0 - 28|
= |-5| + |1| + |0 - 28|
= 5 + 1 + 28 = 34 m
(iii) Given s (t) = 2t³ − 9t² + 12t ≥ 0
[acceleration = \(\frac { dv }{ dt } \)]
When t = 1,
Acceleration = 12 (1) - 18 = -6 m/sec²
When t = 2
Acceleration = 12 (2) - 18 = 6 m/sec²

\(\left[ \begin{matrix} 1 & 2 & 3 \\ 2 & 5 & 3 \\ 1 & 0 & 8 \end{matrix} \right] \)
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 =\(\left[ \begin{matrix} -40 & 16 & 9 \\ 13 & -5 & -3 \\ 5 & -2 & -1 \end{matrix} \right] \).

Let the of the parbola be x² = 4ay (1)

Since (30, 16) is a point on (1),
we get 30² = 4 \(\times\) a \(\times\) 16
⇒ a = \(\frac { 30\times 30 }{ 4\times 16 } =\frac { 225 }{ 16 } \)
∴ becomes, x² = \({ x }^{ 2 }=\frac { 4\times 225 }{ 16 } y=\frac { 225 }{ 4 } y\)
Let AC = h m and BD = lm
∴ A(6, h) is a point on the parabola [∵ OD = 6]
∴ \({ 6 }^{ 2 }=\frac { 225 }{ 4 } \times h\)
⇒ \(h=\frac { 36\times 4 }{ 225 } \Rightarrow h=0.52\)
∴ AD = 3 + h = 3 + 0.52 = 3.52 m
Also (12, 1) is a point on the parabola
[∵ ON = 6 + 6 = 12]
∴ \({ 12 }^{ 2 }=\frac { 225 }{ 4 } \times l\)
⇒ l = \(\frac { 12\times 12\times 4 }{ 225 } =\frac { 576 }{ 225 } =2.08\) = 5.08 m
Hence the length of first two vertical cables are 3.52 m and 5.08 m.

Let \(\hat { a } =\vec { OA } \) and \(\hat { b } =\vec { OB } \) be the unit vectors and which make angles α,β respectively with positive x-axis
Draw AL and BM 丄 to x-axis
Then \(|\vec { OL } |=|\vec { OA } |cos\alpha \Rightarrow \vec { OL } =\vec { |OL| } \hat { i } =cos\alpha \hat { i } \)
\(|\vec { LA } |=|\vec { OB } |\) sin α
⇒ \(\vec { LA } =|\vec { OB } |\hat { j } =sin\alpha (-\hat { j } )=-sin\alpha \hat { j } \)
[\(\vec { LA } \) is in the opp direction of y axis]
\(\hat { a } =\vec { OA } =\vec { OL } +\vec { LA } =cos\alpha \hat { i } -sin\alpha \check { j } \) ..(1)
Similarly \(\hat { b } =\vec { OB } =\vec { OM } +\vec { MB } =cos\beta \hat { i } +sin\beta \hat { j } \) ...(2)
Now \(\hat { a } \times \hat { b } =|\hat { a } ||\hat { b } |sin(\alpha +\beta )\hat { k } =sin(\alpha +\beta )\hat { k } \) ....(3)
[\(|\hat { a } |=|\hat { b } |\) = 1]
Also \(\hat { a } \times \hat { b } =\left| \begin{matrix} \hat { i } & \hat { j } & \hat { k } \\ cos\alpha & -sin\alpha & 0 \\ cos\beta & cos\beta & 0 \end{matrix} \right| \)
= \(\hat { i } (0)-\hat { j } (0)+\hat { k } \)(cosα sinβ + sinα cosβ)
= (sin α cos β + cos α sin β)\(\hat { k } \) .(4)
using (3) and (4), sin(α+β) = sin α cos β + cos α sin β

Consider a triangle ABC in which the two altitudes AD and BE intersect at O. Let CO be produced to meet AB at F. We take O as the origin and let \(\vec { OA } =\vec { a } \), \(\vec { OB } =\vec { b} \) and  \(\vec { OC } =\vec { c } \)

Since \(\vec { AD } \) is perpendicular to \(\vec { BC } \), we have \(\vec { OA } \) is perpendicular to \(\vec { BC } \), and
hence we get \(\vec { OA } \) . \(\vec { BC } \) = 0. That is, \(\vec { a } .(\vec { c } -\vec { b } )=0\), which means
\(\vec { a } .\hat{c}-\hat{a}.\hat{b}=0\)....(1)
Similarly, since \(\vec { BE } \) is perpendicular to \(\vec { CA } \), we have \(\vec { OB } \) is perpendicular to \(\vec { CA } \), and hence we get \(\vec { OB } .\vec { CA } \) = 0.
That is, \(\vec {b } .(\vec {a } -\vec { c } )=0\)
\(\vec { a } .\hat{c}-\hat{b}.\hat{c}=0\).......(2)
Adding equations (1) and (2), gives  \(\vec { a } .\hat{c}-\hat{b}.\hat{c}=0\). That is, \(\hat{c}(\hat{a}-\hat{b})=0\)
That is \(\vec { OC } \) . \(\vec { BA } \) = 0.
Therefore, \(\vec { BA } \) is perpendicular to \(\vec { OC} \). 
Which implies that \(\vec { CF} \) is perpendicular to \(\vec { AB } \). 
Hence, the perpendicular drawn from C to the side AB passes through O. Therefore, the altitudes are concurrent.

Given cubic equation is 2x³-6x²+3x+k = 0
Here, a = 2, b = -6, c = 3, d = k
Let ∝, β, ૪ be the roots
Given ∝ = 2(β+૪) ⇒ \(\frac{\alpha}{2}\) = β+૪ ...(1)
Now, \(\alpha +\beta +\gamma =\frac { -b }{ a } =-\frac { (-6) }{ 2 } =3\)
\(\frac { \alpha }{ 2 } +\alpha =3\Rightarrow \frac { \alpha +2\alpha }{ 2 } =3\Rightarrow \frac { 3\alpha }{ 2 } =3\)
\(\Rightarrow \alpha =2\)
\(\alpha \beta \gamma =\frac { -d }{ a } =\frac { -k }{ 2 } \Rightarrow 2.\beta \gamma =\frac { -k }{ 2 } \)
\(\beta \gamma =\frac { -k }{ 4 } ...(2)\)
Also, \(\alpha \beta +\beta \gamma +\gamma \alpha =\frac { c }{ a } \) 
\(2\beta +\beta \gamma +2\gamma =\frac { 3 }{ 2 } \)
\(2(\beta +\gamma )+\beta \gamma =\frac { 3 }{ 2 } \)
\(\alpha \frac { -k }{ 4 } =\frac { 3 }{ 2 } \quad [from(1)\& (2)]\)
Also, \(2-\frac { k }{ 4 } =\frac { 3 }{ 2 } [\because \alpha =2]\)
\(2-\frac { 3 }{ 2 } =\frac { k }{ 4 } \Rightarrow \frac { 1 }{ 2 } =\frac { k }{ 4 } \)
\(\\ k=\frac { 4 }{ 2 } \Rightarrow k=2\)
From(2), \(\beta \gamma =\frac { -k }{ 4 } =\frac { -2 }{ 4 } =\frac { -1 }{ 2 } \)\(\Rightarrow \gamma =\frac { -1 }{ 2\beta } \)
From \((1),\beta +\gamma =\frac { \alpha }{ 2 } =\frac { 2 }{ 2 } =1\)
Substituting \(\gamma =\frac { -1 }{ 2\beta } \) We get
\(\beta -\frac { 1 }{ 2\beta } =1\Rightarrow 2{ \beta }^{ 2 }-1=2\beta \Rightarrow 2\beta -2\beta -1=0\)
\(\beta =\frac { 2\pm \sqrt { 4-4(2)(-1) } }{ 4 } =\frac { 2\pm \sqrt { 4+8 } }{ 4 } \)
\(=\frac { 2\pm \sqrt { 12 } }{ 4 } =\frac { 2\pm 2\sqrt { 3 } }{ 4 } \)
\(\beta =\frac { 1\pm \sqrt { 3 } }{ 2 } \)
Hence the roots are \(2,\frac { 1+\sqrt { 3 } }{ 2 } ,\frac { 1-\sqrt { 3 } }{ 2 } \) and k = 2

Rearranging the terms, the equation of ellipse is 4x² + 24x + y²− 2y + 21 = 0

That is, 4(x² + 6x + 9 − 9) + (y² − 2y + 1 − 1) + 21 = 0,
 4(x + 3)² − 36 + (y−1)² −1 + 21 = 0,
 4(x + 3)² + (y − 1)² = 16,
\(\frac { { \left( x+3 \right) }^{ 2 } }{ 4 } +\frac { { \left( y+1 \right) }^{ 2 } }{ 16 } =1\)
Centre is (-3, 1) a = 4, b = 2, and the major axis is parallel to y-axis c² = 16−4 = 12
c = ±2\(\sqrt { 3 } \)
Therefore, the foci are (−3, 2\(\sqrt { 3 } \) +1) and (−3, −2\(\sqrt { 3 } \) +1).
Vertices are (3, ±4 +1).
That is the vertices are (3, 5) and (3, -3) and the length of Latus rectum = \(\frac { { 2b }^{ 2 } }{ a } \) = 2 units.

\(put\ t={ sin }^{ -1 }x\Rightarrow dt=\frac { dx }{ \sqrt { 1-{ x }^{ 2 } } } \)
\(\therefore \int _{ 0 }^{ \frac { \pi }{ 4 } }{ { e }^{ at }t\quad dt=\int _{ 0 }^{ \frac { \pi }{ 4 } }{ t } { e }^{ at }dt } \)
\( u=t\ \ v={ e }^{ at }dt\)
\(u'=1\quad { v }_{ 1 }= e ^t\)
\(u'' = 0 \ \ { v }_{ 2 }=e^t\)
\(\int { uvdx } ={ uv }_{ 1 }-u'{ v }_{ 2 }\)
\(\int _{ 0 }^{ \frac { \pi }{ 4 } }{ t{ e }^{ at }dt } ={ \left[ t\frac { { e }^{ at } }{ a } -1\frac { { e }^{ at } }{ { a }^{ 2 } } \right] }_{ 0 }^{ \frac { \pi }{ 4 } }\)
\(dt = \frac{1}{\sqrt 1-x^2}dx\)
I = \(\int ^\frac{\pi}{4}_0 e^t t \ dt\)
\([t e^t-t]^{\frac{\pi}{4}}_0\)
\(=\frac { { e }^{ \frac { \pi }{ 4 } } }{ { a }^{ 2 } } \left( \frac { a\pi }{ 4 } -1 \right) -\frac { { e }^{ 0 } }{ { a }^{ 2 } } (-1)\)
\(=1+ e ^ \frac { \pi }{ 4 } (\frac { \pi }{ 4 }-1) \)

We know that the given equation is homogeneous
Now, we rewrite the given equation as \(\frac { dy }{ dx } =\frac { 3y }{ 2x } -\frac { x }{ 2y } \)
Taking y = vx , we have \(v+x\frac { dv }{ dx } =\frac { 3v }{ 2 } -\frac { 1 }{ 2v } orx\frac { dv }{ dx } =\frac { { v }^{ 2 }-1 }{ 2v }\)
Separating the variables, we obtain \(\frac { 2vdv }{ { v }^{ 2 }-1 } =\frac { dx }{ x } \)
On integration, we get log\(|{ v }^{ 2 }-1|=log|x|+log|C|,\)
Hence, |v²-1| = |Cx|, where C is an arbitrary constant
Now, replace v by\(\frac{y}{x}\) to get \(|\frac { { y }^{ 2 } }{ { x }^{ 2 } } -1|\) = |Cx|.
Thus, we have |y²-x²| = |Cx³|
Hence, y² − x² = ±Cx³ (or) y² − x2 = kx³ gives the general solution