Since A² = \(\left[ \begin{matrix} 4 & 3 \\ 2 & 5 \end{matrix} \right] \left[ \begin{matrix} 4 & 3 \\ 2 & 5 \end{matrix} \right] =\left[ \begin{matrix} 22 & 27 \\ 18 & 31 \end{matrix} \right] \).
A² + xA + yI₂ = O₂ ⇒ \(\left[ \begin{matrix} 22 & 27 \\ 18 & 31 \end{matrix} \right] +x\left[ \begin{matrix} 4 & 3 \\ 2 & 5 \end{matrix} \right] +y\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix} \right] =\left[ \begin{matrix} 0 & 0 \\ 0 & 0 \end{matrix} \right] \)
⇒ \(\left[ \begin{matrix} 22+4x+y & 27+3x \\ 18+2x & 31+5x+y \end{matrix} \right] =\left[ \begin{matrix} 0 & 0 \\ 0 & 0 \end{matrix} \right] \).
So, we get 22 + 4x + y = 0, 31 + 5x + y = 0, 27 + 3x = 0 and 18 + 2x = 0.
Hence x = −9 and y = 14. Then, we get  A² - 9A + 14I₂ = O₂.
Postmultiplying this equation by A^-1, we get A - 9I₂ + 14A^-1 = O₂. Hence, we get
A^-1 = \(\frac { 1 }{ 14 } \) (9I₂ - A) = \(\frac { 1 }{ 14 } \left( 9\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix} \right] -\left[ \begin{matrix} 4 & 3 \\ 2 & 5 \end{matrix} \right] \right) =\frac { 1 }{ 14 } \left[ \begin{matrix} 5 & -3 \\ -2 & 4 \end{matrix} \right] \).

Let P_o be the initial population and the population after t year be P.
Given \(\frac { dp }{ dt } =\frac { 2p }{ 100 } \Rightarrow \frac { dp }{ dt } =\frac { p }{ 50 } \)
⇒ \(\frac { dp }{ p } =\frac { dt }{ 50 } \Rightarrow \int { \frac { dp }{ p } } =\int { \frac { dt }{ 50 } } \)
⇒ log p =\(\frac { t }{ 50 } \) + c ...(1)
when t = 0, p = 0
⇒ log p₀ = 0+c ⇒ log p₀  ....(1)
∴ (1) becomes, log p =\(\frac { t }{ 50 } \)+log P₀.
⇒ log\(\left( \frac { P }{ { p }_{ 0 } } \right) =\frac { t }{ 50 } \)
⇒ t = 50 log\(\left( \frac { P }{ { p }_{ 0 } } \right) \)
when p = 2p₀, t = 50 log\(\left( \frac { 2P_{ 0 } }{ { p }_{ 0 } } \right) \) = 50 log 2
= 50(0.3) = 15 years.
Hence the population doubles in 15 years.

Since the entries in the last column are F, (p ∧ ~p) ∧ (~q ∧ P) is a contradiction

Let \(I=\int _{ 0 }^{ \frac { \pi }{ 2 } }{ sin2x \ log(tan \ x)dx } \) ...(1)
Using the property \(\int _{ 0 }^{ a }{ f(x) } dx=\int _{ 0 }^{ a }{ f(a-x) } dx\) we get
\(I=\int _{ 0 }^{ \frac { \pi }{ 2 } }{ sin \ 2(\frac { \pi }{ 2 } -x)log \ tan(\frac { \pi }{ 2 } -x)dx } \)
\(=\int _{ 0 }^{ \frac { \pi }{ 2 } }{ sin(\pi -2x) } log\ cot \ x \ dx\)
\(=\int _{ 0 }^{ \frac { \pi }{ 2 } }{ sin2 \ x \ log\left( \frac { 1 }{ tanx } \right) } dx\)
\(=\int _{ 0 }^{ \frac { \pi }{ 2 } }{ sin2x( } log1-log \ tan \ x)dx\)
[\(\because \) log 1 = 0]
\(=-\int _{ 0 }^{ \frac { \pi }{ 2 } }{ sin \ 2 \ x } log \ (tan \ x)d\) ...(2)
\(\therefore (1)+(2)\rightarrow \)
\(2I=\int _{ 0 }^{ \frac { \pi }{ 2 } }{ sin2x } log \ (tanx)dx-\int _{ 0 }^{ \frac { \pi }{ 2 } }{ sin \ 2 \ x } \ log(tan \ x)dx=0\)
\(\therefore I=0\)

Let f(x) = cos x, x = 60° dx = 1°
f(x_o) = cos 60° = \(\frac12\) = 0.5
f'(x) = - sinx dx
f'(x_o) = - sin x_o dx
f'(60) = - sin 60° (1°)
= - (0.86603) (0.01745)
= - 0.0154
∴ f(x) = f(x_o) +f(x_o) dx 
f(61) = 0.5 - 0.0154
∴ tan 46° = f(x_o) +f(x_o) dx 
cos 61° = 0.4849

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

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

Let us take the axis AX as the and the tangent AY at A as y-axis.
Equation of the parabola is y² = 4ax
CA 15, FG = 120
CF = CG= 60
F is (15, 60)
Since F lies on (1), 602 = Aa(15) ∴ ⇒ a = 60
y² = 240x
When y = 24, 24² = 240(x)
⇒ x = \(\frac { 24\times 24 }{ 240 } \)
x = \(\frac{24}{10}\) = \(\frac{12}{5}\) = 2.4
From the diagram
BD = BE - ED = 15-2.4 = 12.6m.
Hence the required height is 12.6 m.

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

Given \(\overset { \rightarrow }{ OA } =2\overset { \wedge }{ i } -\overset { \wedge }{ j } +\overset { \wedge }{ k } ,\overset { \rightarrow }{ OB } =\overset { \wedge }{ i } -3\overset { \wedge }{ j } -5\overset { \wedge }{ k } ,\) and \(\overset { \rightarrow }{ OC } =3\overset { \wedge }{ i } -4\overset { \wedge }{ j } +4\overset { \wedge }{ k } \)

\(\therefore \overset { \rightarrow }{ AB } =\overset { \rightarrow }{ OB } -\overset { \rightarrow }{ OA } =-\overset { \wedge }{ i } -2\overset { \wedge }{ j } -6\overset { \wedge }{ k } \)
\(\overset { \rightarrow }{ BC } =\overset { \rightarrow }{ OC } -\overset { \rightarrow }{ OB } =2\overset { \wedge }{ i } -\overset { \wedge }{ j } +\overset { \wedge }{ k } \)
\(\overset { \rightarrow }{ CA } =\overset { \rightarrow }{ OA } -\overset { \rightarrow }{ OC } =\overset { \wedge }{ -i } +3\overset { \wedge }{ j } +5\overset { \wedge }{ k } \)
\(\overset { \rightarrow }{ AB } +\overset { \rightarrow }{ BC } +\overset { \rightarrow }{ CA } =\overset { \rightarrow }{ 0 } \)
Also, \(\overset { \rightarrow }{ BC } \). \(\overset { \rightarrow }{ CA } \)=\(\left( 2\overset { \wedge }{ i } -\overset { \wedge }{ j } +\overset { \wedge }{ k } \right) .\left( \overset { \wedge }{ -i } +3\overset { \wedge }{ j } +5\overset { \wedge }{ k } \right) \)
= -2 -3 + 5 = 0
\(\Rightarrow \overset { \rightarrow }{ BC } .\overset { \rightarrow }{ CA } \Rightarrow \angle BCA=\frac { \pi }{ 2 } \)
Hence, ABC is a right angled triangle.
\(\cos { A } =\frac { \overset { \rightarrow }{ AB } .\overset { \rightarrow }{ AC } }{ \left| \overset { \rightarrow }{ AB } \right| \left| \overset { \rightarrow }{ AC } \right| } \)
\(=\frac { \left( -\overset { \wedge }{ i } -2\overset { \wedge }{ j } -6\overset { \wedge }{ k } \right) .\left( \overset { \wedge }{ -i } -3\overset { \wedge }{ j } -5\overset { \wedge }{ k } \right) }{ \sqrt { 1+4+36 } .\sqrt { 1+9+25 } } \)
\(=\frac { 35 }{ \sqrt { 41 } .\sqrt { 35 } } =\sqrt { \frac { 35 }{ 41 } } =A={ Cos }^{ -1 }\left( \sqrt { \frac { 35 }{ 41 } } \right) \)
\(\cos { B } =\frac { \overset { \rightarrow }{ BA. } \overset { \rightarrow }{ BC } }{ \left| \overset { \rightarrow }{ BA } \right| \left| \overset { \rightarrow }{ BC } \right| } =\frac { \left( \overset { \wedge }{ i } +2\overset { \wedge }{ j } +6\overset { \wedge }{ k } \right) .\left( 2\overset { \wedge }{ i } -\overset { \wedge }{ j } +\overset { \wedge }{ k } \right) }{ \sqrt { 1+4+36 } +\sqrt { 4+1+1 } } \)
\(=\sqrt { \frac { 6 }{ 41 } } \Rightarrow B={ Cos }^{ -1 }\left( \sqrt { \frac { 6 }{ 41 } } \right) \)

Augmented matrix = [A|B] = \(\left[ \begin{matrix} 1 & 1 & 1 \\ 1 & 2 & 4 \\ 1 & 4 & 10 \end{matrix}|\begin{matrix} 1 \\ \lambda \\ { \lambda }^{ 2 } \end{matrix} \right] \)
[A|B] =  \(\overset { { R }_{ 2 }\rightarrow { R }_{ 2 }-{ R }_{ 1 }\\ { R }_{ 3 }\rightarrow { R }_{ 3 }-{ R }_{ 1 } }{ \longrightarrow } \left[ \begin{matrix} 1 & 1 & 1 \\ 0 & 1 & 3 \\ 0 & 3 & 9 \end{matrix}|\begin{matrix} 1 \\ \lambda -1 \\ { \lambda }^{ 2 }-1 \end{matrix} \right] \)
\(\overset { { R }_{ 3 }\rightarrow { R }_{ 3 }-3{ R }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} 1 & 1 & 1 \\ 0 & 1 & 3 \\ 0 & 0 & 0 \end{matrix}|\begin{matrix} 1 \\ \lambda -1 \\ { \lambda }^{ 2 }-1-3\lambda +3 \end{matrix} \right] \)
⟶ \(\left[ \begin{matrix} 1 & 1 & 1 \\ 0 & 1 & 3 \\ 0 & 0 & 0 \end{matrix}|\begin{matrix} 1 \\ \lambda -1 \\ { \lambda }^{ 2 }-3\lambda +2 \end{matrix} \right] \)
Here \(\rho\)(A) = 2
The given system of equations is consistent only when \(\rho\)([AIB]) = 2
\(\rho\)([AIB]) = 2 only when λ²-3λ + 2 = 0
⇒ (λ-1) (λ-2) = 0 ⇒ λ = 1 or λ = 2
∴ The given system is consistent when the values of A are 1 and 2.

Given \(\frac { p }{ 2x } =\frac { a }{ x+x } +\frac { b }{ x-c } \)
\(\frac { p }{ 2x } =\frac { (a+b)x+c(-a) }{ { x }^{ 2 }-{ c }^{ 2 } } \)
⇒ P (x² - c²) = 2 (a + b) x² - 2c(a-b)x
⇒ (2a + 2b - p)x2 - 2c (a - b)x +pc² = 0
This equation has equal roots
if b²-4ac = 0
⇒ c²(a-b)² - pc² (2a + 2b - b) = 0
⇒ (a - b)² - 2p (a + b) +p² = 0 [∵ c² ≠ 0]
⇒ [p - (a + b)]² = (a + b)² - (a - b)² = 4ab
⇒ p-(a+b) = 土2\(\sqrt{ab}\)
⇒ p-(a+b)土2\(\sqrt{ab}\) = (\(\sqrt{a}\) 土\(\sqrt{b}\))²

 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.

3x + 3y − z = 11, 2x − y + 2z = 9, 4x + 3y + 2z = 25.
Δ = \(\left| \begin{matrix} 3 & 3 & -1 \\ 2 & -1 & 2 \\ 4 & 3 & 2 \end{matrix} \right| \)
= \(3\left| \begin{matrix} -1 & 2 \\ 3 & 2 \end{matrix} \right| -3\left| \begin{matrix} 2 & 2 \\ 4 & 2 \end{matrix} \right| -1\left| \begin{matrix} 2 & -1 \\ 4 & 3 \end{matrix} \right| \)
= 3(-2-6)-3(4-8)-1(6+4)·
= 3(- 8) - 3(- 4) - 1(10)
= - 24 + 12 - 10 = - 22
Δ₂ = \(\left| \begin{matrix} 11 & 3 & -1 \\ 9 & -1 & 2 \\ 25 & 3 & 2 \end{matrix} \right| \)
= \(11\left| \begin{matrix} -1 & 2 \\ 3 & 2 \end{matrix} \right| -3\left| \begin{matrix} 9 & 2 \\ 25 & 2 \end{matrix} \right| -1\left| \begin{matrix} 9 & -1 \\ 25 & 3 \end{matrix} \right| \)
= 11(-2-6)-3(18-50)-1(27+25)
= 11(-8)-3(-32)-1(52)
= -88+96-52 = -44
Δ₂ = \(\left| \begin{matrix} 3 & 11 & -1 \\ 2 & 9 & 2 \\ 4 & 25 & 2 \end{matrix} \right| \)
= \(3\left| \begin{matrix} 9 & 2 \\ 25 & 2 \end{matrix} \right| -11\left| \begin{matrix} 2 & 2 \\ 4 & 2 \end{matrix} \right| -1\left| 2\begin{matrix} 2 & 9 \\ 4 & 25 \end{matrix} \right| \)
= 3(18 - 50) -11(4 - 8) - 1(50- 36)
= 3(- 32) - 11(- 4) - 1(14)
= -96+44-14 = - 66
Δ₃ = \(\left| \begin{matrix} 3 & 3 & 11 \\ 2 & -1 & 9 \\ 4 & 3 & 25 \end{matrix} \right| \)
\(3\left| \begin{matrix} -1 & 9 \\ 3 & 25 \end{matrix} \right| -3\left| \begin{matrix} 2 & 9 \\ 4 & 25 \end{matrix} \right| -11\left| \begin{matrix} 2 & -1 \\ 4 & 3 \end{matrix} \right| \)
= 3(-25-27)-3(50-36)+ 11(6+4)
= 3(- 52) - 3(14) + 11(10)
= -156 - 42 + 110= - 88
∴ x = \(\frac { { \triangle }_{ 1 } }{ \triangle } =\frac { -44 }{ -22 } \) = 2
y = \(\frac { { \triangle }_{ 2 } }{ \triangle } =\frac { -66 }{ -22 } \) = 3
z = \(\frac { { \triangle }_{ 3 } }{ \triangle } =\frac { -88 }{ -22 } \) = 4
∴ Solution set is {2, 3, 4}

Here the number of unknowns is 3. So, if the system is consistent and has a non-trivial solution, then the rank of the coefficient matrix is equal to the rank of the augmented matrix and is less than 3.
So the determinant of the coefficient matrix should be 0.
Hence we get
\(\left| \begin{matrix} 3\lambda -8 & 3 & 3 \\ 3 & 3\lambda -8 & 3 \\ 3 & 3 & 3\lambda -8 \end{matrix} \right| \) = 0 or \(\left| \begin{matrix} 3\lambda -2 & 3\lambda -2 & 3\lambda -2 \\ 3 & 3\lambda -8 & 3 \\ 3 & 3 & 3\lambda -8 \end{matrix} \right| \) = 0 (by applying R₁ ➝ R₁ + R₂ + R₃)
or (3λ - 2) \(\left| \begin{matrix} 1 & 1 & 1 \\ 3 & 3\lambda -8 & 3 \\ 3 & 3 & 3\lambda -8 \end{matrix} \right| \) = 0 (by taking out (3λ − 2) from R₁)
or (3λ - 2) \(\left| \begin{matrix} 1 & 1 & 1 \\ 3 & 3\lambda -11 & 3 \\ 3 & 3 & 3\lambda -11 \end{matrix} \right| \) = 0 (by applying R₂ ➝ R₂ - 3R₁, R₃ ➝ R₃ - 3R₁)
or (3λ - 2)(3λ - 11)² 0. So λ = \(\frac { 2 }{ 3 } \) and λ = \(\frac { 11 }{ 3 } \).
We now give an application of system of linear homogeneous equations to chemistry. You are already aware of balancing chemical reaction equations by inspecting the number of atoms present on both sides.

Equation of the plane passing through the point
\(=\vec { a } =\hat { i } -2\hat { j } +4 \) .......(1)
Equation of the given plane is x + 2y - 3z = 11
\(\Rightarrow \vec { r } .(\hat { i } +2\hat { j } -3\hat { k } )=11\)
The given plane is perpendicular to the vector \(\hat { i } +2\hat { j } -3\hat { k } \)
∴ The required plane is parallel to the vector 
\(\vec { b } =\hat { i } +2\hat { j } -3\hat { k } \).......(2)
The given plane is parallel to the line 
\(\frac { x+7 }{ 3 } =\frac { y+3 }{ -1 } =\frac { z }{ 1 } \) Whose direction ratios are  3,-1,1
The required plane is parallel to the vector
 \(\vec { c } =3\hat { i } -\hat { j } +\hat { k } \)     (3)
∴  The non-parametric vector equation of the plane passing through a point (\(\vec { a } \)) and parallel to two vectors \(\vec { b } \) and \(\vec { c } \) is
\((\vec { r } -\vec { a } ).(\vec { b } \times \vec { c } )=0\)
\(\vec { b } \times \vec { c } =\left| \begin{matrix} \hat { i } & \hat { j } & \hat { k } \\ 1 & 2 & -3 \\ 3 & -1 & 1 \end{matrix} \right| \)
\(=\hat { i } (2-3)-\hat { j } (1+9)+\hat { k } (-1-6)\)
\(=-\hat { i } -10\hat { j } -7\hat { k } \)
\(\therefore (\vec { r } -(\hat { i } -2\hat { j } +4\hat { k } )).(-\hat { i } -10\hat { j } -7\hat { k } )=0\)
\([\vec { r } -(-\hat { i } -10\hat { j } -7\hat { k } )]-[(\hat { i } -2\hat { j } +4\hat { k } ).(-\hat { i } -10\hat { j } -7\hat { k } )]=0\)
\(\Rightarrow \vec { r } .(-\hat { i } -10\hat { j } -7\hat { k } )-[-1+20-28]=0\)
\(\Rightarrow \vec { r } .(-\hat { i } -10\hat { j } -7\hat { k } )\times 9=0\)
\(\Rightarrow \vec { r } .(\hat { i } +10\hat { j } +7\hat { k } )-9=0\)
\(\Rightarrow \vec { r } .(\hat { i } +10\hat { j } +7\hat { k } )=9\)
Let \(\Rightarrow \vec { r } =x\hat { i } +y\hat { j } +z\hat { k } \)
\(\therefore (x\hat { i } +y\hat { j } +z\hat { k } ).(\hat { i } +10\hat { j } +7\hat { k } )=9\)
⇒ x+10y+7z = 9 which is the required Cartesian equation of the plane.

Let \({ z }^{ 3 }+8i=0\)
\(\Rightarrow\) z³ = -8i
= \(8(-i)=8\left( cos\left( -\frac { \pi }{ 2 } +2k\pi \right) isin\left( -\frac { \pi }{ 2 } +2k\pi \right) \right) \),k\(\in Z\)
\(z=\sqrt [ 3 ]{ 8 } \left( cos\left( \frac { -\pi +4k\pi }{ 6 } \right) +isin\left( \frac { -\pi +4k\pi }{ 6 } \right) \right) \)
Taking k = 0, 1, 2 we get,
k = 0, \(z=2\left( cos\left( -\frac { \pi }{ 6 } \right) +isin\left( -\frac { \pi }{ 6 } \right) \right) =2\left( -\frac { 1 }{ 2 } -i\frac { \sqrt { 3 } }{ 2 } \right) =2\left( \frac { \sqrt { 3 } }{ 2 } -i\frac { 1 }{ 2 } \right) \)
k = 1, \(z=2\left( cos\left( \frac { \pi }{ 2 } \right) +isin\left( \frac { \pi }{ 2 } \right) \right) =2=\left( 0+i \right) =0+2i=2i\)
k = 2,\(z=2\left( xcos\left( \frac { 7\pi }{ 6 } \right) +isim\left( \frac { 7\pi }{ 6 } \right) \right) =2\left( cos\left( \pi +\frac { \pi }{ 6 } \right) \right) +isin\left( \pi +\frac { \pi }{ 6 } \right) \)
= \(2\left( -cos\left( \frac { \pi }{ 6 } \right) -isin\left( \frac { \pi }{ 6 } \right) \right) =2\left( -\frac { \sqrt { 3 } }{ 2 } -i\frac { 1 }{ 2 } \right) =-\sqrt { 3 } -i\)
The values of z are \(\sqrt { 3 } -i,2i\) and \(-\sqrt { 3 } -i\)

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.

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.

Let f(x) x⁶-3x⁵-5x⁴+22x³-39x²-39x+135
Given (1+2i) is a root \(\Rightarrow\)(-2i) is also a root
Also \(\sqrt3\) is a root \(\Rightarrow\)-\(\sqrt3\) is also a root.
Hence, the factors of f(x) are [x - (1 + 2i)]
[x-(1-2i)] [x\(\sqrt3\)] [x+\(\sqrt3\)]
[(x-1)-2i] [(x-1)+2i] [x-\(\sqrt3\)][x+\(\sqrt3\)]
((x-1)²+2²)(x²-3) = (x²-2x+1+4)(x²-3)
\(\Rightarrow\) factor of f(x) is (x²-2x+5)(x²-3)
\(\Rightarrow\)x⁴-3x²-2x³+6x+5x²-15
\(\Rightarrow\)(x⁴-3x²-2x³+6x-15) is a factor of f(x) 
To find the other factor, let us divide f(x) by
x⁴ - 2x³ + 2x² + 6x - 15

The other factor is x² - x - 9
\(\Rightarrow x=\frac { 1\pm \sqrt { { (-1) }^{ 2 }-4(1)(-9) } }{ 2 } \left[ \because x=\frac { -b\pm \sqrt { { b }^{ 2 }-4ac } }{ 2a } \right] \)
\(\Rightarrow x=\frac { 1\pm \sqrt { 37 } }{ 2 } \)
Hence the roots are
1 - 2i, 1 + 2i, \(\sqrt { 3 }, -\sqrt { 3 }, \frac { 1+\sqrt { 37 } }{ 2 } ,\frac { 1-\sqrt { 37 } }{ 2 } \).

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.