First, we find the points of intersection of x + y = 3 and y² − 4x:
x + y = 3 ⇒ y = 3− x.
\(\therefore\) y² = 4x ⇒ (3 -x)² = 4x
⇒ x² −10x + 9 = 0
⇒ x = 1, x = 9 .
\(\therefore\) x = 1 in x + y = 3 \(\Rightarrow\) y = 2, and x = 9 in x + y = 3 ⇒ y = −6 .
\(\therefore\) (1, 2) and (9,−6) are the points of intersection.
The line x + y = 3 meets the y -axis at (0, 3).
The required area is sketched

Viewing in the direction of y -axis, on the right bounding curve is given by
\(x=\begin{cases} \frac { { y }^{ 2 } }{ 4 } ,0\le y\le 2 \\ 3-y,2\le y\le 3 \end{cases}\)
\(\therefore A=\int _{ 0 }^{ 2 }{ xdy+\int _{ 2 }^{ 3 }{ xdy } =\int _{ 0 }^{ 2 }{ \frac { { y }^{ 2 } }{ 4 } dy+\int _{ 2 }^{ 3 }{ (3-y) } dy } } \)
\(={ \left( \frac { { y }^{ 3 } }{ 12 } \right) }_{ 0 }^{ 2 }+{ \left( 3y-\frac { { y }^{ 3 } }{ 2 } \right) }_{ 2 }^{ 3 }=\left( \frac { 8 }{ 12 } -0 \right) +\left( 9-\frac { 9 }{ 2 } \right) -\left( 6-\frac { 4 }{ 2 } \right) =\frac { 7 }{ 6 } \)

Here a = 1, b = 2,f(x) = 4x²-1
\(\therefore f(a+(b-a)\frac { r }{ n } )=f(1+1(\frac { r }{ n } ))\)
\(=f\left( 1+\frac { r }{ n } \right) \)
\(=4{ \left( 1+\frac { r }{ n } \right) }^{ 2 }-1=4\left( 1+\frac { { r }^{ 2 } }{ { n }^{ 2 } } +\frac { 2r }{ n } \right) -1\)
\(=4+\frac { { 4r }^{ 2 } }{ { n }^{ 2 } } +\frac { 8r }{ n } -1=3+\frac { { 4r }^{ 2 } }{ { n }^{ 2 } } +\frac { 8r }{ n } \)
\(\int _{ a }^{ b }{ f(x)dx= } \underset { n\rightarrow \infty }{ lim } \frac { b-a }{ n } \sum _{ r=1 }^{ n }{ f } \left( a+(b-a)\frac { r }{ n } \right) \)
\(\therefore \int _{ 1 }^{ 2 }{ ({ 4x }^{ 2 }-1)dx } =\underset { n\rightarrow \infty }{ lim } \frac { 1 }{ n } \sum _{ r=1 }^{ n }{ \left( 3+\frac { { 4r }^{ 2 } }{ { n }^{ 2 } } +\frac { 8r }{ n } \right) } \)
\(=\underset { n\rightarrow \infty }{ lim } \left[ \frac { 1 }{ n } \sum _{ r=1 }^{ n }{ 3+\frac { 1 }{ n } } \sum _{ r=1 }^{ n }{ \frac { { 4r }^{ 2 } }{ { n }^{ 2 } } } +\frac { 1 }{ n } \sum _{ r=1 }^{ n }{ \frac { 8r }{ n } } \right] \)
\(=\left[ \underset { n\rightarrow \infty }{ lim } \frac { 1 }{ n } .3n+\frac { 1 }{ n } \frac { 4 }{ { n }^{ 2 } } ({ 1 }^{ 2 }+{ 2 }^{ 2 }+...+{ n }^{ 2 })+\frac { 1 }{ n } .\frac { 8 }{ n } (1+2+3....+n) \right] \)\(=\underset { n\rightarrow \infty }{ lim } \left[ 3+\frac { 4 }{ { n }^{ 3 } } \frac { n(n+1)(2+1) }{ 6 } +\frac { 8 }{ { n }^{ 2 } } \frac { n(n+1) }{ 2 } \right] \)
\(\left[ \because \sum { r } =\frac { n(n+1) }{ 2 } \sum { { r }^{ 2 }=\frac { n(n+1)(2n+1) }{ 6 } } \right] \)
\(=\underset { n\rightarrow \infty }{ lim } \left[ 3+\frac { 4 }{ { n }^{ 3 } } \frac { { n }^{ 3 }(1+\frac { 1 }{ n } )(2+\frac { 1 }{ n } ) }{ 6 } +\frac { 8 }{ { n }^{ 2 } } \frac { { n }^{ 2 }(1+\frac { 1 }{ n } ) }{ 2 } \right] \)
\(=[3+\frac { 2 }{ 3 } (1+0)(2+0)+4(1+0)]\)
\([\because when\quad n\rightarrow \infty ,1/n\rightarrow 0]\)
\(=3+\frac { 4 }{ 3 } +4=\frac { 9+4+12 }{ 3 } =\frac { 25 }{ 3 } \)
\(\therefore \int _{ 1 }^{ 2 }{ (4{ x }^{ 2 }-1)dx=\frac { 25 }{ 3 } } \)

The given function is defined and is differentiable at all  \(x\in (-\infty, \infty) \). As
\(f(x)=\frac{1}{1+x^{2}}\).
We have \(f'(x)=-\frac{2x}{(1+x^{2})^{2}}\)
The stationary points are given by \(-\frac{2x}{(1+x^{2})^{2}}=0\) that is x = 0
Hence the intervals of monotonicity are \((-\infty,0)\) and \((0,\infty)\)
On the interval \((-\infty,0)\) the function strictly increases because f'(x) > 0 in that interval.
The function f(x) strictly decreases in the interval \((0,\infty)\) because f'(x) < 0 in that interval.
Since f′(x) changes from positive to negative when passing through x = 0, the first derivative test tells us there is local maximum at x = 0 and the local maximum value is f (0) = 1.

We have,
\(f(x)=log(1+x)-\frac{x}{1+x}\)
Therefore, \(f'(x)=\frac{1}{1+x}-\frac{1}{(1+x)^{2}}\)
= \(\frac{x}{(1+x)^{2}}\).
Hence, f′(x) is \(\begin{cases} <0 \ when-10 \ whenx>0 \end{cases}\)
Therefore f (x) is strictly increasing for x > 0 and strictly decreasing for x < 0. Since f′(x) changes from negative to positive when passing through x = 0, the first derivative test tells us there is a local minimum at x = 0 which is f (0) = 0. Further, for x > 0, f(x) > f (0) = 0 gives
\(log(1+x)-\frac{x}{1+x}>0 \Rightarrow log(1+x)>\frac{x}{1+x}\).

Let (x) = cos2 x
f^I(x) = 2cos x (- sin x)
= - sin 2x ⇒ f^l(0) = 0
f^Il(x) = - 2 cos 2x ⇒ f^Il(0) = -2
f^III(x) = + 4 sin 2 x ⇒ f^III(0) = 0
f^IV(x) = 8 cos 2 x ⇒ f^IV(0) = 8
f^V(x) = -16 sin 2x ⇒ f^V(0) = 0
f^VI(x) = - 32 cos 2x ⇒ f^VI(0) = -32
∴ Maclaurin's series
\(f(x)=f(0)+\frac { { f }^{ 1 }(0) }{ 1! } x+\frac { { f }^{ II }(0) }{ 2! } { x }^{ 2 }+\frac { { f }^{ III }(0) }{ 3! } { x }^{ 3 }+\) .......
∴ cos x = 1 - \(\frac { 2{ x }^{ 2 } }{ 2! } +\frac { 8{ x }^{ 4 } }{ 4! } +\frac { { 32x }^{ 6 } }{ 6! } \) + ...
= 1 - \(\frac { 2{ x }^{ 2 } }{ 2! } +\frac { { 2 }^{ 3 }{ x }^{ 4 } }{ 4! } +\frac { { { 2 }^{ 5 }x }^{ 6 } }{ 6! } \)+ ...

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

Let h and r be the height and the base radius. Therefore h = 2r. Let V be the volume of the salt cone.

\(V=\frac{1}{3}\pi r^{2}h=\frac{1}{12}\pi h^{3}; \frac{dV}{dt}=30\) mtr³ / min.
Hence,  \(\frac{dV}{dt}=\frac{1}{4}\pi h^{2}\frac{dh}{dt}\)
Therefore,  \(\frac{dh}{dt}=4 \frac{dV}{dt}.\frac{1}{\pi h^{2}}\)
That is,  \(\frac{dh}{dt}=4\times30\times \frac{1}{100 \pi}\)
=\(\frac{6}{5\pi}\) mtr / min. 

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

By definition,
\(\vec { a } \times \vec { b } =\left| \begin{matrix} \hat { i } & \hat { j } & \hat { k } \\ 1 & -1 & 0 \\ 1 & -1 & -4 \end{matrix} \right| =4\hat { i } +4\hat { j } ,\vec { c } \times \vec { d } =\left| \begin{matrix} \hat { i } & \hat { j } & \hat { k } \\ 0 & 3 & -1 \\ 2 & 5 & 1 \end{matrix} \right| =8\hat { i } -2\hat { j } -6\hat { k } \)
\((\vec { a } \times \vec { b } )(\vec { c } \times \vec { d } )=\left| \begin{matrix} \hat { i } & \hat { j } & \hat { k } \\ 4 & 4 & 0 \\ 8 & -2 & -6 \end{matrix} \right| =-24\hat { i } +24\hat { j } -40\hat { k } \) ..............(1)
On the other hand, we have
\([\vec { a }, \vec { b }, \vec { d } ]\vec { c } -[\vec { a } ,\vec { b } ,\vec { c } ]\vec { d } =28(3\vec { j } -\vec { k } )-12(2\hat { i } +5\hat { j } +\hat { k } )=-24\hat { i } +24\hat { j } -40\hat { k } \) ..............(2)
Therefore, from equations (1) and (2), identity (i) is verified.
The verification of identity (ii) is left as an exercise to the reader

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 the equation of the parabola be
(x - h)² = -4a(y - k).
Here the vertex is (0.5, 4)
Equation of the parabola is (x - 0.5)²
= -4a(y-4) ...(1)
O(0, 0) is a point on the parabola
(0 - 0.5)² = -4a (0 - 4)
⇒ \({ \left( \frac { -1 }{ 2 } \right) }^{ 2 }=-4a(-4)\)
⇒ \(\frac { 1 }{ 4 } =16a\Rightarrow a=\frac { 1 }{ 64 } \)
∴ (1) becomes as (x - 0.5)² = \(-4\times \frac { 1 }{ 64 } (y-4)\)
Also D(0.75, y₁) is a point on the parabola
∴ (0.75 - 0.5)² = \(\frac { -1 }{ 16 } ({ y }_{ 1 }-4)\)
⇒ \({ \left( \frac { 1 }{ 4 } \right) }^{ 2 }=\frac { -1 }{ 6 } ({ y }_{ 1 }-4)\)
\(\Rightarrow \frac{1}{\not 16}=\frac{1}{\not16}\left(y_{1}-4\right)\)
⇒ 1 = -y₁ + 4
⇒  y₁ = -1 + 4 = 3m
Height of the water at a horizontal distance of 0.75m is 3m

kx-2y+z = 1, -2ky+z = -2, x-2y+k = 1 
The matrix form of the system is AX = B where
\(\left[ \begin{matrix} k & -2 & 1 \\ 1 & -2k & 1 \\ 1 & -2 & k \end{matrix} \right] ,X=\left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] ,B=\left[ \begin{matrix} 1 \\ -2 \\ 1 \end{matrix} \right] \)
Applying elementary row operation on the augment matrix [A|B] we get
[A|B] =\(\left[ \begin{matrix} k & -2 & 1 \\ 1 & -2k & 1 \\ 1 & -2 & k \end{matrix}|\begin{matrix} 1 \\ -2 \\ 1 \end{matrix} \right] \overset { { R }_{ 1 }\leftrightarrow { R }_{ 3 } }{ \longrightarrow } \left[ \begin{matrix} 1 & -2 & k \\ 1 & -2k & 1 \\ k & -2 & k \end{matrix}|\begin{matrix} 1 \\ -2 \\ 1 \end{matrix} \right] \)
\(\overset { { R }_{ 2 }\rightarrow { R }_{ 2 }+{ R }_{ 1 }\\ { R }_{ 3 }\rightarrow { R }_{ 3 }-k{ R }_{ 1 } }{ \longrightarrow } \left[ \begin{matrix} 1 & -2 & k \\ 1 & -2k+2 & k \\ 0 & -2+2k & 1-k^{ 2 } \end{matrix}|\begin{matrix} 1 \\ - \\ 1-k \end{matrix} \right] \)
\(\overset { { R }_{ 3 }\rightarrow { R }_{ 3 }+{ R }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} 1 & -2 & k \\ 0 & -2k+2 & k \\ 0 & 0 & 1-k^{ 2 } \end{matrix}|\begin{matrix} 1 \\ -3 \\ 1-k \end{matrix} \right] \)
\(\rightarrow \left[ \begin{matrix} 1 & -2 & k \\ 0 & -2k+2 & 1-k \\ 0 & 0 & { k }^{ 2 }-k+2 \end{matrix}\begin{matrix} 1 \\ -3 \\ -k-2 \end{matrix} \right] \)
\(\rightarrow \left[ \begin{matrix} 1 & -2 & k \\ 0 & -2k+2 & 1-k \\ 0 & 0 & (k+2)(1-k) \end{matrix}|\begin{matrix} 1 \\ -3 \\ -k-2 \end{matrix} \right] \).........(1)
Case (i): when k = 1
\([A|B]\rightarrow \left[ \begin{matrix} 1 & -2 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix}\begin{matrix} 1 \\ -3 \\ -3 \end{matrix} \right] \overset { { R }_{ 3 }\rightarrow { R }_{ 3 }-{ R }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} 1 & -2 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix}|\begin{matrix} 1 \\ -3 \\ 0 \end{matrix} \right] \)
Here \(\rho \)(A) = 3 and \(\rho \)[A|B] = 3
So, \(\rho \)(A) ≠ \(\rho \)[A|B] ⇒ The system has no solution
Case (ii): when k ≠ 2, k ≠ -2 
\(\left[ \begin{matrix} 1 & -2 & k \\ 0 & -2k+2 & 1-k \\ 0 & 0 & not\quad zero \end{matrix}|\begin{matrix} 1 \\ -3 \\ not\quad zero \end{matrix} \right] \)
⇒ \(\rho \)(A) = 3 and \(\rho \)[A|B] = 3
so, \(\rho \)(A) =\(\rho \)[A|B] = 3 = the number of unknowns Hence, the system has unique solution.
Case (iii): when k = -2
\(\rho [A|B]\rightarrow \left[ \begin{matrix} 1 \\ 1 \\ 0 \end{matrix}\begin{matrix} -2 \\ 6 \\ 0 \end{matrix}\begin{matrix} -2 \\ 3 \\ 0 \end{matrix}\begin{matrix} 1 \\ -3 \\ 0 \end{matrix} \right] \)
Here \(\rho \) (A) = 2 and \(\rho \)[A|B] = 2
∴ \(\rho \)(A) = \(\rho \)[A|B] = 2<3 the number of unknowns so the system is consistent with infinitely many solutions.

2x - y + z = 2, 6x - 3y + 3z = 6, 4x - 2y + 2z = 4
The matrix form of the given system is AX = B where
A =\(\left[ \begin{matrix} 2 & -1 & 1 \\ 6 & -3 & 3 \\ 4 & -2 & 2 \end{matrix} \right] \)
\(X=\left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] B=\left[ \begin{matrix} 2 \\ 6 \\ 4 \end{matrix} \right] \)
Applying elementary row operations on the augment matrix [A|B], we get,
[A|B] =\(\left[ \begin{matrix} 2 & -1 & 1 \\ 6 & -3 & 3 \\ 4 & -2 & 2 \end{matrix}|\begin{matrix} 2 \\ 6 \\ 4 \end{matrix} \right] \overset { { R }_{ 2 }\rightarrow { R }_{ 2 }-3{ R }_{ 1 }\\ { R }_{ 3 }\rightarrow { R }_{ 3 }-2{ R }_{ 1 } }{ \longrightarrow } \left[ \begin{matrix} 2 & -1 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix}|\begin{matrix} 2 \\ 0 \\ 0 \end{matrix} \right] \)
Here \(\rho \)(A) = 1 [∵ only one non zero row]
and \(\rho \)[A|B] = 1 [∵ only one-zero row]
∴ \(\rho \)(A) =\(\rho \)(A|B] = 1< 3 the given system is consistent and has two parameter family of solutions.
So, z = t and y = s where, t \(\in \)R
Writing the equivalent equations from the rowechelon matrix, we get
2x-y+z = 2 .............(1)
y = s
z = t
Substituting (2) and (3) In (1) we get
2x-s+t = 2
⇒ 2x-s+t = 2
⇒ x = \(\frac{1}{2}\)[s-t+2]
∴ Solution set is {\(\frac{1}{2}\)(s-t+2),s,t} here s, t \(\in \) R.

z³ = -27 = (-1 \(\times\) 3)³ = -1 \(\times\) 33
z = \((-1)^{ \frac { 1 }{ 3 } }\times 3^{ 3\times \frac { 1 }{ 3 } }=(-1)^{ \frac { 1 }{ 3 } }\)\(\times\) 3
∴ z = 3\(\left[ cos\pi +isin\pi \right] ^{ \frac { 1 }{ 3 } }\)
[∵ cos π = -1 and sin π = 0]
= 3\(\left[ cos\frac { 1 }{ 3 } (2k\pi +\pi )isin\frac { 1 }{ 3 } (2k\pi +\pi ) \right] \)
k = 0, 1, 2
When k = 0,
z = 3\(\left[ cos\frac { 1 }{ 3 } (\pi )isin\frac { 1 }{ 3 } (\pi ) \right] =3cos\frac { \pi }{ 3 } \)
When k = 1
z = 3\(\left[ cos\frac { 1 }{ 3 } (3\pi )isin\frac { 1 }{ 3 } (3\pi ) \right] \)
= 3[cos π + i sin π] = 3(-1+0) 
When k = 2
z = 3\(\left[ cos\frac { 1 }{ 3 } (5\pi )isin\frac { 1 }{ 3 } (5\pi ) \right] =3\left[ cos5\frac { \pi }{ 3 } \right] \)
Hence, the roots are 3 cis\(\frac { \pi }{ 3 } \), -3, 3 c is 5\(\frac { \pi }{ 3 } \)

Given equation of hyperbola is \(\frac { { x }^{ 2 } }{ 16 } -\frac { { y }^{ 2 } }{ 64 } =1\)
Let y = mx + c be the required tangent
⇒ a² = 16 and b² = 64
The tangents are parallel to 10x - 3y + 9 = 0.
∴ Slope of tangent (m)
= Slope of the line 10x - 3y + 9 = 0.
∴ m = \(\frac{-co - efficient of \ x}{co - efficient of \ y}\)= \(\frac { -10 }{ -3 } =\frac { 10 }{ 3 } \)
The condition for y = mx + c to be a tangent to the hyperbola is c² = a²m² - b²
⇒ \({ c }^{ 2 }=16\left( \frac { 100 }{ 9 } \right) -64=\frac { 1600-64\times 9 }{ 9 } \)
⇒ \(\frac { 1024 }{ 9 } \Rightarrow c=\pm \frac { 32 }{ 3 } \)
∴ The required tangents are
y = \(\frac { 10 }{ 3 } x+\frac { 32 }{ 3 } \) or \(y=\frac { 10x }{ 3 } -\frac { 32 }{ 3 } \)
⇒ 3y = 10x + 32 or 3y = 10x - 32
⇒ 10x−3y+32 = 0, or 10x+3y−32 = 0

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 = \(\pm \frac { 1 }{ \sqrt { 2 } } \)

We find AB = \(\left[ \begin{matrix} -4 & 4 & 4 \\ -7 & 1 & 3 \\ 5 & -3 & -1 \end{matrix} \right] \left[ \begin{matrix} 1 & -1 & 1 \\ 1 & -2 & -2 \\ 2 & 1 & 3 \end{matrix} \right] =\left[ \begin{matrix} -4+4+8 & 4-8+4 & -4-8+12 \\ -7+1+6 & 7-2+3 & -7-2+9 \\ 5-3-2 & -5+6-1 & 5+6-3 \end{matrix} \right] \)
= \(\left[ \begin{matrix} 8 & 0 & 0 \\ 0 & 8 & 0 \\ 0 & 0 & 8 \end{matrix} \right] \) = 8I₃
and BA = \(\left[ \begin{matrix} 1 & -1 & 1 \\ 1 & -2 & -2 \\ 2 & 1 & 3 \end{matrix} \right] \left[ \begin{matrix} -4 & 4 & 4 \\ -7 & 1 & 3 \\ 5 & -3 & -1 \end{matrix} \right] =\left[ \begin{matrix} -4+7+5 & 4-1-3 & 4-3-1 \\ -4+14-10 & 4-2+6 & 4-6+2 \\ -8-7+15 & 8+1-9 & 8+3-3 \end{matrix} \right] \)
= \(\left[ \begin{matrix} 8 & 0 & 0 \\ 0 & 8 & 0 \\ 0 & 0 & 8 \end{matrix} \right] \) = 8I₃
So we get AB = BA = 8I₃. That is, (\(\frac { 1 }{ 8 } A\))B = B(\(\frac { 1 }{ 8 } A\)) = I₃. Hence, B^-1 = \(\frac { 1 }{ 8 } A\).
Writing the given system of equations in matrix form, we get
\(\left[ \begin{matrix} 1 & -1 & 1 \\ 1 & -2 & -2 \\ 2 & 1 & 3 \end{matrix} \right] \left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] =\left[ \begin{matrix} 4 \\ 9 \\ 1 \end{matrix} \right] \).
That is B \(\left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] =\left[ \begin{matrix} 4 \\ 9 \\ 1 \end{matrix} \right] \).
So, \(\left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] ={ B }^{ -1 }\left[ \begin{matrix} 4 \\ 9 \\ 1 \end{matrix} \right] \)
= \(\left( \frac { 1 }{ 8 } A \right) \left[ \begin{matrix} 4 \\ 9 \\ 1 \end{matrix} \right] =\frac { 1 }{ 8 } \left[ \begin{matrix} -4 & 4 & 4 \\ -7 & 1 & 3 \\ 5 & -3 & -1 \end{matrix} \right] \left[ \begin{matrix} 4 \\ 9 \\ 1 \end{matrix} \right] =\frac { 1 }{ 8 } \left[ \begin{matrix} -16+36+4 \\ -28+9+3 \\ 20-27-1 \end{matrix} \right] =\frac { 1 }{ 8 } \left[ \begin{matrix} 24 \\ -16 \\ -8 \end{matrix} \right] =\left[ \begin{matrix} 3 \\ -2 \\ -1 \end{matrix} \right] \)
Hence, the solution is (x = 3, y = -2, z = -1).

The matrix form of the system is AX = B, where
A = \(\left[ \begin{matrix} 2 & 3 & 3 \\ 1 & -2 & 1 \\ 3 & -1 & -2 \end{matrix} \right] \),X = \(\left[ \begin{matrix} { x }_{ 1 } \\ { x }_{ 2 } \\ { x }_{ 3 } \end{matrix} \right] \),B = \(\left[ \begin{matrix} 5 \\ -4 \\ 3 \end{matrix} \right] \).
We find |A| = \(\left| \begin{matrix} 2 & 3 & 3 \\ 1 & -2 & 1 \\ 3 & -1 & -2 \end{matrix} \right| \) = 2(4 + 1) - 3(-2 - 3) + 3(-1 + 6) = 10 + 15 + 15 = 40 ≠ 0.
So, A⁻¹ exists and
A^-1 = \(\frac { 1 }{ \left| A \right| } \) (adj A) = \(\frac { 1 }{ 40 } { \left[ \begin{matrix} +\left( 4+1 \right) & -\left( -2-3 \right) & +\left( -1+6 \right) \\ -\left( -6+3 \right) & +\left( -4-9 \right) & -\left( -2-9 \right) \\ +\left( 3+6 \right) & -\left( 2-3 \right) & +\left( -4-3 \right) \end{matrix} \right] }^{ T }=\frac { 1 }{ 40 } \left[ \begin{matrix} 5 & 3 & 9 \\ 5 & -13 & 1 \\ 5 & 11 & -7 \end{matrix} \right] \)
Then, applying X = A⁻¹B, we get
\(\left[ \begin{matrix} { x }_{ 1 } \\ { x }_{ 2 } \\ { x }_{ 3 } \end{matrix} \right] =\frac { 1 }{ 40 } \left[ \begin{matrix} 5 & 3 & 9 \\ 5 & -13 & 1 \\ 5 & 11 & -7 \end{matrix} \right] \left[ \begin{matrix} 5 \\ -4 \\ 3 \end{matrix} \right] =\frac { 1 }{ 40 } \left[ \begin{matrix} 25-12+27 \\ 25+52+3 \\ 25-44-21 \end{matrix} \right] =\frac { 1 }{ 40 } \left[ \begin{matrix} 40 \\ 80 \\ -40 \end{matrix} \right] =\left[ \begin{matrix} 1 \\ 2 \\ -1 \end{matrix} \right] \)
So, the solution is (x₁ = 1, x₂ = 2, x₃ = -1).

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

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 \(\frac { 5\pm \sqrt { 77 } }{ 2 } ,\frac { 5\pm \sqrt { 5 } }{ 2 } \).

Let ∝, β, ૪ be the roots of the equation 
Given \(\frac { \alpha }{ \beta } =\frac { 3 }{ 2 } \Rightarrow 2\alpha =3\beta \Rightarrow \alpha =\frac { 3 }{ 2 } \beta \)
\(\therefore \frac { 3 }{ 2 } \beta ,\beta ,\gamma \) are the roots of the given equation
Then by Vieta's formula,
\(\frac { 3 }{ 2 } \beta +\beta +\gamma =\frac { -b }{ a } =\frac { -(-9) }{ 1 } =9\)
\(\frac { 5 }{ 2 } \beta +\gamma =9\Rightarrow \gamma =9-\frac { 5 }{ 2 } \beta \)
\(\Rightarrow \gamma =\frac { 18-5\beta }{ 2 } ...(2)\)
Also \(\frac { 3 }{ 2 } \beta (\beta )+\beta \gamma +\left( \frac { 3 }{ 2 } \beta \right) \gamma =\frac { c }{ a } =\frac { 14 }{ 1 } =14\)
\(\Rightarrow \frac { 3 }{ 2 } { \beta }^{ 2 }+\frac { 5 }{ 2 } \beta \left( \frac { 18-5\beta }{ 2 } \right) =14\ [using\ (2)]\)
\(\Rightarrow \frac { 3 }{ 2 } { \beta }^{ 2 }+\frac { 90\beta }{ 4 } -\frac { 25{ \beta }^{ 2 } }{ 4 } =14\)
Multiplying by \(4,6{ \beta }^{ 2 }+90{ \beta }-25{ \beta }^{ 2 }=56\)
\(19{ \beta }^{ 2 }-90{ \beta }+56=0\)
\(\Rightarrow ({ \beta }-4)(19{ \beta }-14)=0\)
\(\Rightarrow { \beta }=4\)
\({ \beta }=\frac { 14 }{ 19 } \)
When \(\beta\) = 4, the other roots are \(\frac { 3 }{ 2 } (4),4,\frac { 18-5 }{ 2 } (4)\)
\(\Rightarrow 6,4,-1\)

When \(\\ \beta =\frac { 14 }{ 19 } ,\) the other roots are \(\frac { 3 }{ 2 } \beta ,\beta \frac { 18-5\beta }{ 2 } [by(2)]\)
\(\Rightarrow \frac { 3 }{ 2 } \left( \frac { 14 }{ 19 } \right) ,\frac { 14 }{ 19 } ,\frac { 18-5\left( \frac { 14 }{ 19 } \right) }{ 2 } \Rightarrow \frac { 21 }{ 19 } ,\frac { 14 }{ 19 } ,\frac { 136 }{ 19 } \)

Given equation of the curve is y = 1 + x³ and the line is x + 12y = 12
Slope of the tangent to the curve
m₁ = \(\frac { dy }{ dx } \) = 3 x² and the
Slope of the line = m₂
= \(\frac{-1}{2}\) \(\left[ \because m=\frac { co-efficient\ of\ x }{ co-efficient\ of\ y } \right] \)
Since the slope of the tangent to the curve and the line are orthogonal, m₁ m₂ = - 1.
∴ 3x²\(\left( \frac { -1 }{ 2 } \right) \) = -1
⇒ \(\frac{x^2}{4}\) = 1
⇒ x² = 4
⇒ x = ±2
When x = 2, y = 1 + 2³ = 9
When x = -2, y = 1+ (-2)³
= 1-8 = -7 
∴ Equation of the tangent at (2, 9) is
y-9 = 12(x-2) [∵ m₁ = 3x² = 3(2)² = 12]
∴ y - 9 = 12x - 24
∴ 12x - y = 15
Equation of the tangent at (-2, -7) is
y+7= 12(x + 2)
⇒ y + 7 = 12x + 24
⇒ 12x-y+17 = 0

Given \(f(x)=\frac { { x }^{ 2 } }{ { x }^{ 2 }-1 } \)
\(\quad \underset { x\rightarrow 1^{ + } }{ lim } \frac { { x }^{ 2 } }{ { x }^{ 2 }-1 } =\underset { x\rightarrow 1^{ + } }{ lim } \frac { { x }^{ 2 } }{ \left( x+1 \right) \left( x-1 \right) } \) 
= \(\underset { h\rightarrow { 0 }^{ + } }{ lim } \frac { \left( 1+h \right) ^{ 2 } }{ \left( 1+h+1 \right) \left( 1+h-1 \right) } \)
= \(\underset { h\rightarrow 0^{ + } }{ lim } \frac { (1+h)^{ 2 } }{ (2+h)(h) } \infty \) 
Also  \(\underset { x\rightarrow 1^{ - } }{ lim } \frac { { x }^{ 2 } }{ (x+1)/(x-1) } =\underset { h\rightarrow 0^{ + } }{ lim } \frac { \left( 1-h \right) ^{ 2 } }{ \left( 1-h+1 \right) \left( 1-h-1 \right) } \)
= \(\underset { h\rightarrow { 0 }^{ + } }{ lim } \frac { \left( 1-h \right) ^{ 2 } }{ (2-h)(-h) } =-\infty \) 
ஃ x = -1 and x = 1are vertical asymptotes,
Also
\(\underset { x-\rightarrow \infty }{ lim } \frac { { x }^{ 2 } }{ { x }^{ 2 }-1 } =\underset { \frac { 1 }{ x } \rightarrow 0 }{ lim } \frac { 1 }{ 1-\frac { 1 }{ { x }^{ 2 } } } =1\)
[Divide numerator and denominator by x²]
ஃ y = 1 is a horizontal asymptote,