\(\frac { 5 }{ 18 } \)

(i) Local min value = \(\frac { 2 }{ 3\sqrt { 3 } } \) and   local max value = \(\frac { 2 }{ 3\sqrt { 3 } } \) 
(ii) local min value = −3 and local max.value = 1

(i) f(x) is strictly increasing on\(\left( -\infty ,-\frac { 1 }{ 2 } \right) \) and f(x) is strictly decreasing on \(\left[ -\frac { 1 }{ 2 } ,\infty \right] \) Local maximum value = \(\frac { 81 }{ 4 } \)
(ii) f(x) is strictly decreasing on \(\left( 0,\frac { 1 }{ \sqrt { 2 } } \right) \) and strictly increasing on \(\left( \frac { 1 }{ \sqrt { 2 } } ,2 \right) \)
Local minimum value = \(-\frac { 2 }{ 2\sqrt { 2 } } \)

\(C=\frac { 3\pi }{ 4 } \)

Let OA = x m, OB = y m
Then x² + y² = 25

Differentiating, \(2x\frac { dx }{ dt } +2y\frac { dy }{ dt } \) = 0
⇒ \(\frac { dy }{ dx } =-\frac { x }{ y } \frac { dx }{ dt } \)
When \(\frac { dx }{ dt } =\frac { 1 }{ 2 } ,\frac { dy }{ dt } =\frac { -x }{ 2y } \)
When y = 4, x² = 25-y²
⇒ x =\(\sqrt { 25-16 } \) = 3
Thus \(\frac { dy }{ dt } =-\frac { 3 }{ 2\times 4 } =\frac { -3 }{ 8 } \).

y = sin²x
\(\frac { dy }{ dx } \) = 2 sin x cos x = sin 2x
∴ m = \(\left( \frac { dy }{ dx } \right) _{ \left( \frac { \pi }{ 3 } ,\frac { 3 }{ 4 } \right) }=sin\frac { 2\pi }{ 3 } =\frac { \sqrt { 3 } }{ 2 } \)
∴ Slope of the normal = \(-\frac { 1 }{ m } =-\frac { 2 }{ \sqrt { 3 } } \)
∴ Equation of normal is y-y₁ = \(-\frac { 1 }{ m } \)(x-x₁)
⇒ \(y-\frac { 3 }{ 4 } =-\frac { 2 }{ \sqrt { 3 } } \left( x-\frac { \pi }{ 3 } \right) \)
⇒ 12\(\sqrt { 3 } \)y-9\(\sqrt { 3 } \) = -24x + 8π [ multiply 12\(\sqrt { 3 } \)]
∴ 24x + 12\(\sqrt { 3 } \)y = 8π + 9\(\sqrt { 3 } \).

AB = |(3+2i) - (0+5i)| = |3-3i|
=\(\sqrt { 9+9 } =\sqrt { 18 } =3\sqrt { 2 } \)
BC = |(0+5i) - (-3+2i)| = |3+3i|
= \(\sqrt { 9+9 } =\sqrt { 18 } =3\sqrt { 2 } \)
CD = |(-3+2i) - (0-i) = |-3+i|
= \(\sqrt { 9+9 } =\sqrt { 18 } =3\sqrt { 2 } \)
DA = |(0-i) - (3+2i)| = |-3-3i|
=\(\sqrt { 9+9 } =\sqrt { 18 } =3\sqrt { 2 } \)
∴ AB = BC = CD = DA
Also AC = |(3+2i) - (-3+2i)|
= |6| = \(\sqrt { 36 } \) = 6
∴ AC = BD
Hence ABCD is a square

Given line is x + y + 1 = 0
⇒ y = -x-1
m = -1, c = -1
Equation of the hyperbola is \(\frac { { x }^{ 2 } }{ 16 } -\frac { { y }^{ 2 } }{ 15 } \) = 1
a² = 16, b² = 15
The condition for the line y = mx + c to be a tangent to the hyperbola is c² = a²m² - b²
∴ (-1)² = 16(-1)² - 15
1 = 16 -15
1 = 1
Since the condition is satisfied, x + y + 1 = 0 touches the hyperbola \(\frac { { x }^{ 2 } }{ 16 } -\frac { { y }^{ 2 } }{ 15 } \) = 1
The point of contact is \(\left( \frac { -{ a }^{ 2 }m }{ c } ,\frac { -{ b }^{ 2 } }{ c } \right) \) = \(\left( \frac { -16(-1) }{ -1 } ,\frac { -15 }{ -1 } \right) \) = (-16, 15)
Hence, the point of contact is (-16, 15)

Given equation of the hyperbola is
3x² - 6y² = -18; \(\div \)by (-18) we get \(\frac{y^2}{3}- \frac{x^2}{6}\) = 1
The transverse axis is long y-axis.
Here a² = 3, b² = 6
Length of transverse axis is 2a = 2\(\sqrt { 3 } \)
Length of conjugate axis is 2b = 2\(\sqrt { 3 } \)
\(e=\sqrt { 1+\frac { { b }^{ 2 } }{ { a }^{ 2 } } } =\sqrt { 1+\frac { 6 }{ 3 } } =\sqrt { 1+2 } =\sqrt { 3 } \)

Let the equation of the ellipse be \(\frac { { x }^{ 2 } }{ { a }^{ 2 } } +\frac { { y }^{ 2 } }{ { b }^{ 2 } } =1\)
Given e = \(\frac { 2 }{ 3 } \) and \(\frac { 2{ b }^{ 2 } }{ a } \) = 5 ⇒ 2b² = 5a  ...(1)
∴ b² = a²(1 - e²) = a² \({ a }^{ 2 }\left( 1-\frac { 4 }{ 9 } \right) ={ a }^{ 2 }\left( \frac { 5 }{ 9 } \right) \)
∴ \({ 2b }^{ 2 }=\frac { 10{ a }^{ 2 } }{ 9 } \)  ...(2)
From (1) and (2),
\(\frac { 10{ a }^{ 2 } }{ 9 } \) = 5a ⇒ 10a² = 45a
10a² - 45a = 0 ⇒ 5a(2 - 9a) = 0
⇒ a = 0 or a = \(\frac { 9 }{ 2 } \) [∵ a = 0 is not possible]
∴ \({ a }^{ 2 }=\frac { 81 }{ 4 } \)
∴ \(2{ b }^{ 2 }=\frac { 5\times 9 }{ 2 } =\frac { 45 }{ 2 } \Rightarrow { b }^{ 2 }=\frac { 45 }{ 4 } \)
∴ Equation of the ellipse is \(\frac { { x }^{ 2 } }{ \frac { 81 }{ 4 } } +\frac { { y }^{ 2 } }{ \frac { 45 }{ 4 } } =1\)
⇒ \(\frac { 4{ x }^{ 2 } }{ 81 } +\frac { 4{ y }^{ 2 } }{ 45 } =1\)

Equation of circle is x² + y² = 64
∴ a² = 64 ⇒ a = 8
Given line is 3x+ 4y = P
4y = -3x + p
y = \(y=\frac { -3 }{ 4 } x+\frac { p }{ 4 } \)
m = \(\frac { -3 }{ 4 } \) and c = \(\frac { p }{ 4 } \)
The condition for y = mx + c to be a tangent to the circle in c² = a²(1 + m²).
∴ \({ \left( \frac { p }{ 4 } \right) }^{ 2 }=64\left( 1+\frac { 9 }{ 16 } \right) \)
\(\Rightarrow \frac{p^{2}}{\not 16}=64\left(\frac{16+9}{\not 16}\right) \Rightarrow p^{2}=64(25)\)
\(p=\pm \sqrt { 64(25) } =\pm 8(5)\)
∴ p = ±40

cos (tan^-1x) = \(sin\left( { cot }^{ -1 }\frac { 3 }{ 4 } \right) \)
\(\Rightarrow sin\left( { tan }^{ -1 }\frac { 4 }{ 3 } \right) =sin\left( { sin }^{ -1 }\frac { 4 }{ \sqrt { { 3 }^{ 2 }+{ 4 }^{ 2 } } } \right) \)
\(\left[ \because { tan }^{ -1 }x={ sin }^{ -1 }\left( \frac { x }{ \sqrt { 1+{ x }^{ 2 } } } \right) \right] \)
\(\Rightarrow sin\left( { sin }^{ -1 }\left( \frac { 4 }{ 5 } \right) \right) =\frac { 4 }{ 5 } \)
\(\Rightarrow cos\left( { tan }^{ -1 }x \right) =cos\left(- { tan }^{ -1 }x \right) =\frac { 4 }{ 5 } \)
\(\left[ \because cosx=cos(-x) \right] \)
\(\Rightarrow { tan }^{ -1 }x={- tan }^{ -1 }x={ cos }^{ -1 }\left( \frac { 4 }{ 5 } \right) \)
\(\Rightarrow \tan ^{-1} x=-\tan ^{-1} x=\tan ^{-1} \frac{3}{4}\)
\(\left[ \because { cos }^{ -1 }x={ tan }^{ -1 }\sqrt { \frac { 1-{ x }^{ 2 } }{ x } } ;{ cos }^{ -1 }\left( \frac { 4 }{ 5 } \right) ={ tan }^{ -1 }\sqrt { \frac { 1-\frac { 16 }{ 25 } }{ \frac { 4 }{ 5 } } } \right] \)
\(\Rightarrow x=\frac { 3 }{ 4 } ,\frac { 3 }{ 4 } \) 

\(cos\left[ { cos }^{ -1 }\left( \frac { -\sqrt { 3 } }{ 2 } +\frac { \pi }{ 6 } \right) \right] \)
= \(cos\left[ \pi -{ cos }^{ -1 }\left( \frac { -\sqrt { 3 } }{ 2 } +\frac { \pi }{ 6 } \right) \right] \)
\(\left[ \because { cos }^{ -1 }\left( -x \right) =\pi -{ cos }^{ -1 }x \right] \)
= \(cos\left[ \pi -\frac { \pi }{ 6 } +\frac { \pi }{ 6 } \right] \)
\(\left[ \because { cos }^{ -1 }\frac { \sqrt { 3 } }{ 2 } =x\Rightarrow \frac { \sqrt { 3 } }{ 2 } =cosx\Rightarrow x=\frac { \pi }{ 6 } \right] \)
= \(cos\pi -1\)

Let \({ sin }^{ -1 }\left( \frac { 3 }{ 5 } \right) =A\Rightarrow \frac { 3 }{ 5 } =sinA\)

\(cosA=\frac { adj }{ hyp } =\frac { 4 }{ 5 } \)
Let \({ sin }^{ -1 }\left( \frac { 5 }{ 13 } \right) =B\Rightarrow sinB=\frac { 5 }{ 13 } \)

\(\Rightarrow cosB=\frac { 12 }{ 13 } \)
\(\therefore cos\left[ { sin }^{ -1 }\frac { 3 }{ 5 } +{ sin }^{ -1 }\frac { 5 }{ 13 } \right] =cos(A+B)\)
= cos A cos B-sin A sin B
= \(\frac { 4 }{ 5 } .\frac { 12 }{ 13 } -\frac { 3 }{ 5 } .\frac { 5 }{ 13 } =\frac { 48 }{ 65 } -\frac { 15 }{ 65 } \)
= \(\frac { 33 }{ 65 } \)

Given \(\overset { \rightarrow }{ a } +\overset { \rightarrow }{ b } +\overset { \rightarrow }{ c } =0\)
\(\overset { \rightarrow }{ a } +\overset { \rightarrow }{ b } \)= -\(\overset { \rightarrow }{ c } \) ... (1)
Taking cross product with \(\overset { \rightarrow }{ a } \) both sides, we get 
\(\overset { \rightarrow }{ a } \left( \overset { \rightarrow }{ a } +\overset { \rightarrow }{ b } \right) =-\left( \overset { \rightarrow }{ a } \times \overset { \rightarrow }{ c } \right) \)
\(\overset { \rightarrow }{ a } \times \overset { \rightarrow }{ a } +\overset { \rightarrow }{ a } +\overset { \rightarrow }{ b } =\overset { \rightarrow }{ c } \times \overset { \rightarrow }{ a } \)
\(\left[ \because -\left( \overset { \rightarrow }{ a } \times \overset { \rightarrow }{ c } \right) =\overset { \rightarrow }{ c } \times \overset { \rightarrow }{ a } \right] \)
\(\Rightarrow \overset { \rightarrow }{ a } \times \overset { \rightarrow }{ b } =\overset { \rightarrow }{ c } \times \overset { \rightarrow }{ a } \left( \because \overset { \rightarrow }{ a } \times \overset { \rightarrow }{ a } =\overset { \rightarrow }{ 0 } \right) \)
Taking cross product with \(\overset { \rightarrow }{ b } \) both sides, we get
\(\overset { \rightarrow }{ b } \times \left( \overset { \rightarrow }{ a } +\overset { \rightarrow }{ b } \right) =\overset { \rightarrow }{ b } \times \left( -\overset { \rightarrow }{ c } \right) \)
\(\overset { \rightarrow }{ b } \times \overset { \rightarrow }{ a } +\overset { \rightarrow }{ b } \times \overset { \rightarrow }{ b } =-\overset { \rightarrow }{ b } \times \overset { \rightarrow }{ c } \)
\(-\overset { \rightarrow }{ a } \times \overset { \rightarrow }{ b } =-\overset { \rightarrow }{ b } \times \overset { \rightarrow }{ c } \)
\(\overset { \rightarrow }{ a } \times \overset { \rightarrow }{ b } =\overset { \rightarrow }{ b } \times \overset { \rightarrow }{ c } \)
From (2) and (3) we get
\(\overset { \rightarrow }{ a } \times \overset { \rightarrow }{ b } =\overset { \rightarrow }{ b } \times \overset { \rightarrow }{ c } =\overset { \rightarrow }{ c } \times \overset { \rightarrow }{ a } \)

The given line is parallel to the vector \(\overset { \rightarrow }{ b } =3\overset { \wedge }{ i } -\overset { \wedge }{ j } +2\overset { \wedge }{ k } \) and the given plane is normal to the vector \(\overset { \rightarrow }{ n } =3\overset { \wedge }{ i } +4\overset { \wedge }{ j } +\overset { \wedge }{ k } \) 
Let θ be the angle between the line and the plane, then
\(\sin { \theta } =\frac { \overset { \rightarrow }{ b } .\overset { \rightarrow }{ n } }{ \left| \overset { \rightarrow }{ b } \right| \left| \overset { \rightarrow }{ n } \right| } =\frac { 9-4+2 }{ \sqrt { 9+1+4 } .\sqrt { 9+16+1 } } \)
\(=\frac { 7 }{ \sqrt { 14 } .\sqrt { 26 } } =\frac { 7 }{ \sqrt { 2 } \times \sqrt { 7 } \times \sqrt { 26 } } =\frac { \sqrt { 7 } }{ \sqrt { 52 } } \)
\(\therefore \theta ={ sin }^{ -1 }\left( \frac { \sqrt { 7 } }{ \sqrt { 52 } } \right) \)

Let \(\overset { \rightarrow }{ r } =x\overset { \wedge }{ i } +y\overset { \wedge }{ j } +z\overset { \wedge }{ k } \)
\(\therefore x\overset { \wedge }{ i } +y\overset { \wedge }{ j } +z\overset { \wedge }{ k } =\left( s-2t \right) \overset { \wedge }{ i } +\left( 3-t \right) \overset { \wedge }{ j } +\left( 2s+t \right) \overset { \wedge }{ k } \)
Equating the co-efficients of like components both sides,
We get, x = s - 2t
y = 3 - t
z = 2s + t
Eliminating x and t using determinates we get
\(\left| \begin{matrix} x \\ y-3 \\ z \end{matrix}\begin{matrix} 1 \\ 0 \\ 2 \end{matrix}\begin{matrix} -2 \\ -1 \\ 1 \end{matrix} \right| =0\)
⇒ x (0+2) -1(y - 3 + z) -2 (2y - 6 - 0) = 0
⇒ 2x - y + 3 - z- 4y + 12 = 0
⇒ 2x - 5y - z + 15 = 0

Augmented matrix [A|B] =\(\left[ \begin{matrix} 1 & 1 & 3 \\ 2 & 2 & 6 \\ 2 & 1 & 1 \end{matrix}|\begin{matrix} 4 \\ 7 \\ 10 \end{matrix} \right] \)
[A|B] \(\overset { { R }_{ 2 }\rightarrow { R }_{ 2 }-2{ R }_{ 1 }\\ { R }_{ 3 }\rightarrow { R }_{ 3 }-2{ R }_{ 1 } }{ \longrightarrow } \left[ \begin{matrix} 1 & 1 & 3 \\ 0 & 0 & 0 \\ 0 & -1 & -5 \end{matrix}|\begin{matrix} 4 \\ -1 \\ 2 \end{matrix} \right] \)
\(\overset { { R }_{ 2 }\leftrightarrow { R }_{ 3 } }{ \longrightarrow } \left[ \begin{matrix} 1 & 1 & 3 \\ 0 & -1 & -5 \\ 0 & 0 & 0 \end{matrix}|\begin{matrix} 4 \\ 2 \\ -1 \end{matrix} \right] \)
Here \(\rho\) (A) = 2 [only 2 two-non zero rows]
And \(\rho\) ([A|B]) = 3 [There are 3 non-zero rows]
∴ \(\rho\) (A) ≠ \(\rho\) ([A|B])
∴ The system is inconsistent.

Let A = \(\left[ \begin{matrix} 1 & 0 & 0 \\ 0 & h-2 & 2 \\ \begin{matrix} 0 \\ 0 \end{matrix} & \begin{matrix} 0 \\ 0 \end{matrix} & \begin{matrix} h+2 \\ 3 \end{matrix} \end{matrix} \right] \)
The rank of A will be less than 3 if every minor of order 3 vanishes
∴ \(\left| \begin{matrix} 1 & 0 & 0 \\ 0 & h-2 & 0 \\ 0 & 0 & 3 \end{matrix} \right| \) = 0
⇒ 1\(\left| \begin{matrix} h-2 & 0 \\ 0 & 3 \end{matrix} \right| \) + 0 + 0 =0 ⇒ 3(h-2) = 0
⇒ h-2 = 0 ⇒ h = 2

Δ = \(\left| \begin{matrix} 3 & a \\ 2 & a \end{matrix} \right| \) = 3a - 2a = a
Δ₁ =\(\left| \begin{matrix} 4 & a \\ 2 & a \end{matrix} \right| \) = 4a - 2a = 2a
Δ₂ = \(\left| \begin{matrix} 3 & 4 \\ 2 & 2 \end{matrix} \right| \)= 6 - 8 = -2

\(\therefore x=\frac{\Delta_{1}}{\Delta}=\frac{2 \not a}{\not a}=2=y y=\frac{\Delta_{2}}{\Delta}=\frac{-2}{a}\)
∴ Solution set is {2,\(\frac { -2 }{ a } \)}

The given equation can be written as
\({ 2 }^{ z }\left( 1+\frac { 1 }{ 2 } +\frac { 1 }{ 4 } \right) ={ 7 }^{ x }\left( 1+\frac { 1 }{ 7 } +\frac { 1 }{ 49 } \right) \)
\(\Rightarrow { 2 }^{ x }\left( \frac { 8+4+2 }{ 8 } \right) ={ 7 }^{ x }\left( \frac { 49+7+1 }{ 49 } \right) \)
\(\Rightarrow { 2 }^{ x }\left( \frac { 7 }{ 4 } \right) ={ 7 }^{ x }\left( \frac { 57 }{ 49 } \right) \Rightarrow \frac { 7 }{ 4 } \times \frac { 49 }{ 57 } =\frac { { 7 }^{ x } }{ { 2 }^{ x } } \)
\(\Rightarrow \frac { 7 }{ 4 } \times \frac { 49 }{ 57 } ={ \left( \frac { 7 }{ 4 } \right) }^{ x }\Rightarrow \frac { { 7 }^{ 3 } }{ 4\times 57 } ={ \left( \frac { 7 }{ 4 } \right) }^{ x }\)
\(\Rightarrow xlog\left( \frac { 7 }{ 4 } \right) =3log\ 7-log4-log57\)
\(\Rightarrow x=\frac { 3log7-log4-log57 }{ log\left( \frac { 7 }{ 2 } \right) } \)

\({ tan }^{ -1 }\left( { \frac { x-1 }{ x-2 } } \right) +{ tan }^{ -1 }\left( \frac { x+1 }{ x+2 } \right) =\frac { \pi }{ 4 } \)
\(\Rightarrow { tan }^{ -1 }\left( \cfrac { \frac { x-1 }{ x-2 } +\frac { x+1 }{ x+2 } }{ 1-\left( \frac { x-1 }{ x-2 } \right) \left( \frac { x+1 }{ x+2 } \right) } \right) =\frac { \pi }{ 4 } \)

\(\Rightarrow \frac { 2{ x }^{ 2 }-4 }{ { x }^{ 2 }-4-{ x }^{ 2 }+1 } =1\)
\(\Rightarrow\) 2x²- 4 = -3
\(\Rightarrow \) 2x²- 4 = -3
\(\Rightarrow\) 2x² = -3 + 4 = 1
\(\Rightarrow\) \({ x }^{ 2 }=\frac { 1 }{ 2 } \)
\(\Rightarrow\) \(x=\frac { 1 }{ \sqrt { 2 } } \)