Equation of tangent is y – 2 = 0
Equation of normal is \(x=\frac { \pi }{ 6 } \)

(−∞,−2)concave downward
(−2,1)concave upward
(1,∞)concave downward
Points of inflections are (−2,48),(1,9)

Local min value =−4
Local max.value = 0

(−∞,−2)decreasing
(−2,−1)increasing
(−1,∞)decreasing
Local maximum value = 6
Local minimum value =5

\(a=2,b=-\frac { 1 }{ 2 } \)

\(\frac { 35 }{ 38 } m/h\)

\(\left( -\frac { 2 }{ 3 } ,-\frac { 14 }{ 7 } \right) \) and (2,−2)

Given 2y² = x³
⇒ y² = \(\frac { { x }^{ 3 } }{ 2 } \) ....(1)
⇒ y² = 32x  ...(2)
From (1) & (2),
\(\frac { { x }^{ 3 } }{ 2 } \) = 32x
⇒ x³ = 64x
⇒ x(x²-64) = 0
⇒ x = 0, 8, -8
when x = 0, y = 0
when x = 8, y²= 32(8)
⇒ y = ±16
when x = -8, y² = 32(-8)
which is not possible
∴ The point are (0, 0) (8, 16) (8, -16)
Differentiating 2y² = x³, with respect to 'x'
⇒ 4y\(\frac { dy }{ dx } \) = 3x²
⇒ \(\frac { dy }{ dx } =\frac { 3x^{ 2 } }{ 4y } \)
∴ m₁ = \(\left( \frac { dy }{ dx } \right) _{ (8,16) }=\frac { 3\times 8\times 8 }{ 4\times 16 } \)= 3
Differentiating y² = 32x with respect to 'x',
2y\(\frac { dy }{ dx } \) = 32
⇒ \(\frac { dy }{ dx } =\frac { 32 }{ 2y } =\frac { 16 }{ y } \)
∴ m₂ =\(\left( \frac { dy }{ dx } \right) _{ (8,16) }=\frac { 16 }{ 16 } \) = 1
Let θ be the angle between the given curves at (8, 16)
∴ tan θ = \(\left| \frac { { m }_{ 1 }-{ m }_{ 2 } }{ 1+{ m }_{ 1 }{ m }_{ 2 } } \right| =\frac { 3-1 }{ 1+3(1) } \)
=\(\frac { 2 }{ 4 } =\frac { 1 }{ 2 } \)
∴ θ = \({ tan }^{ -1 }\left( \frac { 1 }{ 2 } \right) \).

The normal vector to the planes
x + 2y + 3z - 7 = 0, 2x - 3y + 4z = 0 are 
\(\overset { \rightarrow }{ b } =\overset { \wedge }{ i } +2\overset { \wedge }{ j } +3\overset { \wedge }{ k } \) and \(\overset { \rightarrow }{ c } =2\overset { \wedge }{ i } -3\overset { \wedge }{ j } +4\overset { \wedge }{ k } \)
∴ The required planes passes through the point \(\overset { \rightarrow }{ a } =\overset { \wedge }{ i } +\overset { \wedge }{ j } +\overset { \wedge }{ k } \) and parallel to two vector 5 namely \(\overset { \rightarrow }{ b } \) and \(\overset { \rightarrow }{ c } \) 
∴ The Parametric form of vectors equation of the plans is \(\overset { \rightarrow }{ r } =\overset { \rightarrow }{ a } +s\overset { \rightarrow }{ b } +t\overset { \rightarrow }{ c } \) s, t ∈ R
\(\overset { \rightarrow }{ r } =\left( \overset { \rightarrow }{ i } +\overset { \rightarrow }{ j } -\overset { \rightarrow }{ k } \right) +s\left( \overset { \rightarrow }{ i } +2\overset { \rightarrow }{ j } +3\overset { \rightarrow }{ k } \right) +t\left( 2\overset { \rightarrow }{ i } -3\overset { \rightarrow }{ j } +4\overset { \rightarrow }{ k } \right) ,\)
Cartesian equation is \(\left| \begin{matrix} x-{ x }_{ 1 } \\ { b }_{ 1 } \\ { c }_{ 1 } \end{matrix}\begin{matrix} y-{ { y }_{ 1 } } \\ { b }_{ 2 } \\ { c }_{ 2 } \end{matrix}\begin{matrix} z-{ { z }_{ 1 } } \\ { b }_{ 3 } \\ { c }_{ 3 } \end{matrix} \right| =0\)
\(\Rightarrow \left| \begin{matrix} x-1 \\ 1 \\ 2 \end{matrix}\begin{matrix} y-1 \\ 2 \\ -3 \end{matrix}\begin{matrix} z+1 \\ 3 \\ 4 \end{matrix} \right| =0\)
⇒ (x - 1) (8 + 9) - (y - 1)(4 - 6) + (z + 1)(-3 -4) = 0
⇒ 17 (x - 1) +2 (y - 1) -7 (z + 1) = 0
⇒ 17x - 17 + 2y - 2 - 7z - 7 = 0
⇒ 17x + 2y - 7z - 26 = 0

Let \(\overset { \rightarrow }{ B } =x\overset { \wedge }{ i } +y\overset { \wedge }{ j } +z\overset { \wedge }{ k } \)
Given \(\overset { \rightarrow }{ A } \times \overset { \rightarrow }{ B } =\overset { \rightarrow }{ C } \Rightarrow \left| \begin{matrix} \overset { \wedge }{ i } \\ 1 \\ x \end{matrix}\begin{matrix} \overset { \wedge }{ j } \\ 1 \\ y \end{matrix}\begin{matrix} \overset { \wedge }{ k } \\ 1 \\ z \end{matrix} \right| =\overset { \wedge }{ j } -\overset { \wedge }{ k } \)
\(\Rightarrow \overset { \wedge }{ i } (z-y)-\overset { \wedge }{ j } (z-x)+\overset { \wedge }{ k } (y-x)\quad \overset { \wedge }{ j } -\overset { \wedge }{ k } \)
Equating the like components on both sides, we get
z - y = 0   .....(1)
x - y = 1   .....(2)
y - x = -1  .....(3)
Also, \(\overset { \rightarrow }{ A } .\overset { \rightarrow }{ B } =3\Rightarrow \left( \overset { \wedge }{ i } +\overset { \wedge }{ j } +\overset { \wedge }{ k } \right) .\left( x\overset { \wedge }{ i } +y\overset { \wedge }{ j } +z\overset { \wedge }{ k } \right) =3\)
⇒ x + y + z = 3   ....(4)
Solving (1), (2), (3) and (4), we get \(x=\frac { 5 }{ 3 } ,y=\frac { 2 }{ 3 } \)and \(z=\frac { 2 }{ 3 } \)
\(\therefore \overset { \rightarrow }{ B } =\frac { 5 }{ 3 } \overset { \wedge }{ i } +\frac { 2 }{ 3 } \overset { \wedge }{ j } +\frac { 2 }{ 3 } \overset { \wedge }{ k } \)

arg(1+i) + arg(1-i) = arg[(1+i) (1-i)]
LHS = arg (1+i) + arg(1-i)
1+i = \(\sqrt { 2 } \left( \frac { 1 }{ \sqrt { 2 } } +\frac { i }{ \sqrt { 2 } } \right) \)
= \(\sqrt { 2 } \left( cos\frac { \pi }{ 4 } +isin\frac { \pi }{ 4 } \right) \)
∴ arg (1+i) = π/4
-1+i =\(\sqrt { 2 } \left( \frac { -1 }{ \sqrt { 2 } } +\frac { i }{ \sqrt { 2 } } \right) \)
= \(\sqrt { 2 } \left( cos3\frac { \pi }{ 4 } +isin3\frac { \pi }{ 4 } \right) \)
∴ (-1+i) = 3\(\frac { \pi }{ 4 } \)
∴ LHS = \(\frac { \pi }{ 4 } +\frac { 3\pi }{ 4 } =\frac { 4\pi }{ 4 } =\pi \)
RHS = arg[(1+i) (-1+i)]
= arg[-1-i + i + i²]
= (-1-i + i-1) = arg(-2)
= arg(2) - (1) = 2 arg(-1)
= 2 (cos π + isin π) = π
∴ LHS = RHS

LHS = \(\left( \frac { i+\sqrt { 3 } }{ -i+\sqrt { 3 } } \right) ^{ 2\omega }+\left( \frac { i-\sqrt { 3 } }{ i+\sqrt { 3 } } \right) ^{ 2\omega }\)
= \(\left( \frac { \sqrt { 3 } +i }{ \sqrt { 3 } -i } \times \frac { \sqrt { 3 } +i }{ \sqrt { 3 } +i } \right) ^{ 2\omega }+\left( \frac { -\sqrt { 3 } +i }{ \sqrt { 3 } +i } \times \frac { \sqrt { 3 } -i }{ \sqrt { 3 } -1 } \right) ^{ 2\omega }\)
= \(\left( \frac { 3-1+2\sqrt { 3 } i }{ 3+1 } \right) ^{ 2\omega }+\left( \frac { -3+1+2\sqrt { 3 } i }{ 3+1 } \right) ^{ 2\omega }\)
= \(\left( \frac { 2+2\sqrt { 3 } i }{ 4 } \right) ^{ 2\omega }+\left( \frac { -2+2\sqrt { 3 } i }{ 4 } \right) ^{ 2\omega }\)
= \(\left( \frac { 1+\sqrt { 3 } i }{ 2 } \right) ^{ 2\omega }+\left( \frac { -2+2\sqrt { 3 } i }{ 2 } \right) ^{ 2\omega }\)
=\(\left[ -\left( \frac { -1-\sqrt { 3 } i }{ 2 } \right) \right] ^{ 2\omega }+\left[ \frac { -1+\sqrt { 3 } i }{ 2 } \right] ^{ 2\omega }\)
= (-ω²)^2ω+(ω)^2ω
[∴ ω = \(\frac { -1+i\sqrt { 3 } }{ 2 } \), ω² = \(\frac { -1-i\sqrt { 3 } }{ 2 } \)]
= ω^4ω+ω^2ω
= (ω³)¹³³. ω¹ + (ω³)⁶⁶.ω²
= 1.ω+1.ω²  [∴ 1+ω+ω² = 0 & ω³ = 1]
= ω + ω²
= -1 = RHS

Equation of the ellipse is \(\frac { { x }^{ 2 } }{ 25 } +\frac { y^{ 2 } }{ 9 } =1\)
∴ a² = 25, b² = 9
∴ e = \(\sqrt { 1-\frac { { b }^{ 2 } }{ { a }^{ 2 } } } =\sqrt { 1-\frac { 9 }{ 25 } } =\sqrt { \frac { 16 }{ 25 } } =\frac { 4 }{ 5 } \)
Focus is (ae, 0) = \(\left( 5\times \frac { 4 }{ 5 } \right) \) = (4, 0)
Since the focus of the hyperbola coincides with the focus of the ellipse, foci of the hyperbola are (±4,0).
Let A be the length of the semi-transverse axis
∴ Ae - 4 ⇒ 2A = \(\frac { 4 }{ e } =\frac { 4 }{ 2 } =2\) [∵ e = 2]
Let B b th length of the semi conjugate axis
B² = A²(e² - 1) = 4(4 - 1) = 12
Equation of the hyperbola is \(\frac { { x }^{ 2 } }{ { A }^{ 2 } } -\frac { { y }^{ 2 } }{ { B }^{ 2 } } =1\)
⇒ \(\frac { { x }^{ 2 } }{ 4 } -\frac { { y }^{ 2 } }{ 12 } =1\)

Let the equation of the parabola be y² = 4ax
Let AB = I.

Since ABC is an equilateral triangle,
∠BAM = ∠MAC = 30^o
In MBM, cos 30°= \(\frac{AM}{l}\)
⇒ AM = l cos 30° = l\(\left( \frac { \sqrt { 3 } }{ 2 } \right) \)
sin 30^o = \(\frac { BM }{ AM } \)
⇒ sin 30^o = \(\frac { BM }{ l } \)
⇒ BM = l sin 30^o = l\(\left( \frac { 1 }{ 2 } \right) \)
ஃ Co-ordinates of B are (AM, BM) ⇒ \(B\left( \frac { \sqrt { 3 } l }{ 2 } ,\frac { 1 }{ 2 } \right) \)
Also B lies on (1)
\({ \left( \frac { l }{ 2 } \right) }^{ 2 }=4a\left( \frac { \sqrt { 3 } l }{ 2 } \right) \)
⇒ \(\frac { { l }^{ 2 } }{ 4 } =4a\left( \frac { \sqrt { 3 } l }{ 2 } \right) \)
\(\frac { { l }^{ 2 } }{ 4 } =2\sqrt { 3 } al\)
\(\frac { l }{ 4 } =2\sqrt { 3 } \) a ⇒ 1 = 8 \(a\sqrt { 3 } \)
Length of the side of the equilateral triangle is 8 \(a\sqrt { 3 } \) units.

LHS =\({ tan }^{ -1 }\left( \frac { 1-x }{ 1+x } \right) -{ tan }^{ -1 }\left( \frac { 1-y }{ 1+y } \right) \)
= tan^-1(1) - tan^-1 (x) - (tan^-1(1) - tan^-1(y)
\(\left[ \because { tan }^{ -1 }(\frac { x-y }{ 1+xy } )={ tan }^{ -1 }x-{ tan }^{ -1 }y \right] \)
= tan^-1(1) - tan^-1 (x) - tan^-1(1) + tan^-1(y)
= tan^-1(y) - tan^-1(x)
= \({ tan }^{ -1 }\left( \frac { y-x }{ 1+xy } \right) \)
= \({ sin }^{ -1 }\left( \frac { y-x }{ \sqrt { 1+\left( yx \right) ^{ 2 }+\left( y-x \right) ^{ 2 } } } \right) \)
= \({ sin }^{ -1 }\left( \frac { y-x }{ \sqrt { (1+{ x }^{ 2 })(1+{ x }^{ 2 }) } } \right) \)
RHS

\({ sin }^{ -1 }\left( \frac { sinx+cosx }{ \sqrt { 2 } } \right) \)
\(\left[\because \frac{-\pi}{4}
= \({ sin }^{ -1 }\left( \frac { 1 }{ \sqrt { 2 } } sinx+\frac { 1 }{ \sqrt { 2 } } cosx \right) \)
= \({ sin }^{ -1 }\left( sin x\ cos\frac { \pi }{ 4 } +cosx\ sin\frac { \pi }{ 4 } \right) \)
= \({ sin }^{ -1 }\left( sin\left( x+\frac { \pi }{ 4 } \right) \right) =\pi +\frac { \pi }{ 4 } \)

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.
\(\cos { A } =\frac { \overset { \rightarrow }{ AB } .\overset { \rightarrow }{ AC } }{ \left| \overset { \rightarrow }{ AB } \right| \left| \overset { \rightarrow }{ AC } \right| } \)
\(=\frac { \left( -\overset { \wedge }{ i } -2\overset { \wedge }{ j } -6\overset { \wedge }{ k } \right) .\left( \overset { \wedge }{ -i } -3\overset { \wedge }{ j } -5\overset { \wedge }{ k } \right) }{ \sqrt { 1+4+36 } .\sqrt { 1+9+25 } } \)
\(=\frac { 35 }{ \sqrt { 41 } .\sqrt { 35 } } =\sqrt { \frac { 35 }{ 41 } } =A={ Cos }^{ -1 }\left( \sqrt { \frac { 35 }{ 41 } } \right) \)
\(\cos { B } =\frac { \overset { \rightarrow }{ BA. } \overset { \rightarrow }{ BC } }{ \left| \overset { \rightarrow }{ BA } \right| \left| \overset { \rightarrow }{ BC } \right| } =\frac { \left( \overset { \wedge }{ i } +2\overset { \wedge }{ j } +6\overset { \wedge }{ k } \right) .\left( 2\overset { \wedge }{ i } -\overset { \wedge }{ j } +\overset { \wedge }{ k } \right) }{ \sqrt { 1+4+36 } +\sqrt { 4+1+1 } } \)
\(=\sqrt { \frac { 6 }{ 41 } } \Rightarrow B={ Cos }^{ -1 }\left( \sqrt { \frac { 6 }{ 41 } } \right) \)

The augmented matrix [A|B] is \(\left[ \begin{matrix} 2 & -3 & 5 \\ 3 & 1 & \lambda \\ 1 & -7 & 8 \end{matrix}|\begin{matrix} 12 \\ \mu \\ 17 \end{matrix} \right] \)
\(\overset { { R }_{ 1 }\leftrightarrow { R }_{ 3 } }{ \longrightarrow } \left[ \begin{matrix} 1 & -7 & 8 \\ 3 & 1 & \lambda \\ 2 & -3 & 5 \end{matrix}|\begin{matrix} 17 \\ \mu \\ 12 \end{matrix} \right] \)
\(\overset { { R }_{ 2 }\rightarrow { R }_{ 2 }-{ R }_{ 1 }\\ { R }_{ 3 }\rightarrow { R }_{ 3 }-2R_{ 1 } }{ \longrightarrow } \left[ \begin{matrix} 1 & -7 & 8 \\ 0 & 22 & \lambda -51 \\ 0 & 11 & -11 \end{matrix}|\begin{matrix} 17 \\ \mu -51 \\ -22 \end{matrix} \right] \)
\(\overset { { R }_{ 2 }\rightarrow { R }_{ 2 }-2{ R }_{ 3 }\\ { R }_{ 3 }\rightarrow { R }_{ 3 }\div 11 }{ \longrightarrow } \left[ \begin{matrix} 1 & -7 & 8 \\ 0 & 0 & \lambda -2 \\ 0 & 1 & -1 \end{matrix}|\begin{matrix} 17 \\ \mu -7 \\ -2 \end{matrix} \right] \)
\(\overset { { R }_{ 2 }\leftrightarrow { R }_{ 3 } }{ \longrightarrow } \left[ \begin{matrix} 1 & -7 & 8 \\ 0 & 1 & -1 \\ 0 & 0 & \lambda -2 \end{matrix}|\begin{matrix} 17 \\ -2 \\ \mu -7 \end{matrix} \right] \)
Case (i) : when λ ≠ 2,
\(\rho\) ([A|B]) = 3 and \(\rho\)(A) = 3
∴ \(\rho\)([AIB])= \(\rho\)(A) = 3 = the number of unknowns
∴ The system has unique solution
Case (ii) : when λ = 2, μ =7
\(\left[ \begin{matrix} 1 & -7 & 8 \\ 0 & 1 & -1 \\ 0 & 0 & 0 \end{matrix}|\begin{matrix} 17 \\ -2 \\ 0 \end{matrix} \right] \)
Here \(\rho\)(A) = 2 and \(\rho\)([A|B]) = 2
∴ \(\rho\)(A) = \(\rho\)([A|B]) = 2 < number of unknowns
Thus the system is consistent with infinitely many solutions.
Case (iii) : When λ= 2 and μ ≠ 7
\(\rho\) (A) = 2 and \(\rho\) ([A|B]) = 3
∴ \(\rho\) (A) ≠ \(\rho\) ([A|B])
Thus, the given system of equations is inconsistent.

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.

Put \(\frac { 1 }{ x } \) = a, \(\frac { 1 }{ y } \) = b, \(\frac { 1 }{ z } \) = c
∴ 2a + 3b + 10c = 4  ....(1)
4a- 6b - 5c = 1 .....(2)
6a + 9b -20c = 2 ...(3)
Δ = \(\left| \begin{matrix} 2 & 3 & 10 \\ 4 & -6 & 5 \\ 6 & 9 & -20 \end{matrix} \right| \)
= \(2\left| \begin{matrix} -6 & 5 \\ 9 & -20 \end{matrix} \right| -3\left| \begin{matrix} 4 & 5 \\ 6 & -20 \end{matrix} \right| +10\left| \begin{matrix} 4 & -6 \\ 6 & 9 \end{matrix} \right| \)
= 2 (120 - 45) -3 (-80 - 30) + 10 (36 + 36)
= 150 + 330 + 720 = 1200
Δ₁ = \(\left| \begin{matrix} 4 & 3 & 10 \\ 1 & -6 & 5 \\ 2 & 9 & -20 \end{matrix} \right| \)
= \(4\left| \begin{matrix} -6 & 5 \\ 9 & -20 \end{matrix} \right| -3\left| \begin{matrix} 1 & 5 \\ 2 & -20 \end{matrix} \right| +10\left| \begin{matrix} 1 & -6 \\ 2 & 9 \end{matrix} \right| \)
= 4 (120 - 45) -3 (-20 - 10) + 10 (9 + 12)
= 300 + 90 + 120 = 600
Δ₂ = \(\left| \begin{matrix} 2 & 4 & 10 \\ 4 & 1 & 5 \\ 6 & 2 & -20 \end{matrix} \right| =2\left| \begin{matrix} 1 & 5 \\ 2 & -20 \end{matrix} \right| -4\left| \begin{matrix} 4 & 5 \\ 6 & -20 \end{matrix} \right| +10\left| \begin{matrix} 4 & 1 \\ 6 & 2 \end{matrix} \right| \)
= 2 (-2 - 10) - 4 (-80 - 30) + 10 (8 - 6)
= -60 + 440 + 20 = 400
Δ₃ = \(\left| \begin{matrix} 2 & 3 & 4 \\ 4 & -6 & 1 \\ 6 & 9 & 2 \end{matrix} \right| 2\left| \begin{matrix} -6 & 1 \\ 9 & 2 \end{matrix} \right| -3\left| \begin{matrix} 4 & 1 \\ 6 & 2 \end{matrix} \right| +10\left| \begin{matrix} 4 & -6 \\ 6 & 9 \end{matrix} \right| \)
= 2 (-12 - 9) -3 (8 - 6) + 4 (36 + 36)
= - 42 - 6 + 288 = 240
∴ a = \(\frac { { \triangle }_{ 1 } }{ \triangle } =\frac { 600 }{ 1200 } =\frac { 1 }{ 2 } \Rightarrow \frac { 1 }{ x } =\frac { 1 }{ 2 } \) ⇒ x = 2
∴ b = \(\frac { { \triangle }_{ 2 } }{ \triangle } =\frac { 400 }{ 1200 } =\frac { 1 }{ 3 } \Rightarrow \frac { 1 }{ y } =\frac { 1 }{ 3 } \) ⇒ y = 1 
∴ c = \(\frac { { \triangle }_{ 3 } }{ \triangle } =\frac { 240 }{ 1200 } =\frac { 1 }{ 5 } \Rightarrow \frac { 1 }{ z } =\frac { 1 }{ 5 } \) ⇒ z = 5
∴ Solution set is {2, 3, 5}

Let ∝, β be the roots x²+ bx + ca = 0
∝ +β = -b, ∝β = ca
and a, ૪ be the roots of x² + cx + ab = 0
∝+૪ = -c, ∝૪ = ab
Then, a² + b∝ + ca = 0 and
∝²+ c∝ + ab = 0
⇒ (b-c)∝+a(c-b) = 0 ⇒ ∝ = a
Also ∝β = ca and ∝૪ = ab
∴ β = c and ૪ = c
since ∝ = a is a root of x² + bx + ca = 0,
we get a² + ba + ca = 0 ⇒ a + b + c = 0
Thus, β+૪ = b + c = -a and β૪ = bc
Hence β, ૪ are the roots of the equation
x² + ax + bc = 0

Given equation is (a²+b²+c²) p²-2
(ab+bc+cd) p+(b²+c²+d²) ≤ 0...(1)
(1) can be rewritten as
(a²p² - 2abp + b²) + (b²p² - 2bcp + c²) + (c²p2 - 2cdp + d2) ≤ 0
Since a, b, c, d, p∈ R
(ap-b)² ≥ 0, (bp-c)² ≥ 0 and (cp-d)² ≥ 0
∴ (2) will be satisfied only if
ap - b = 0, bp - c = 0, cp - d = 0
\(\Rightarrow \frac { b }{ a } =\frac { c }{ b } =\frac { d }{ c } =p\)
⇒ a, b, c, d are in G.P and ad = bc