\(x\frac { dy }{ dx } =2y\)

\(\frac { { 208a }^{ 2 } }{ 3 } \)

0

\(\frac { { \vartheta }^{ 2 }u }{ \vartheta x\vartheta y } ={ e }^{ xy }\left[ 3xy+1+{ x }^{ 2 }{ y }^{ 2 } \right] \)

f(x) is strictly increasing \(\left[ -\frac { 1 }{ 2 } ,\frac { 1 }{ 2 } \right] \)

c = 3

Given  x \(\frac{dy}{dx}=x+y\)
\(\frac{dy}{dx}=\frac{x+y}{x}\)    ...(1)
This is a homogeneous differential equation
put y = vx
\(\Rightarrow \frac { dy }{ dx } =v+x\frac { dv }{ dx } \)
∴ (1) becomes, 
\(v+x\frac { dv }{ dx } =\frac { x+vx }{ x } =1+v\)
\(\Rightarrow x\frac { dv }{ dx } =1+v-v=1\)
\(\Rightarrow dv=\frac { dx }{ x } \)
\(\Rightarrow \int { dv } =\int { \frac { dx }{ x } } \)
\(\Rightarrow v=log\quad x+c\)
\(\\ \Rightarrow \frac { y }{ x } =log\ x+c[\because v=\frac { y }{ x } ]\)

Let a be any element and e be the identity element.
The a * e = e * a = a
a * e = a ⇒ a + e -1 = a ⇒ e-1 = 0 ⇒ e = 1
∴ The identity element is 1

Let y = f(x) = \(\sqrt x\)
Let x_o = 25, dx = 25.2 - 25 = 0.2
y = \(\sqrt x\)
dy = \(\frac{1}{2\sqrt{x}}\) dx
dy = \(\frac{1}{2\sqrt{x}}\) (0.2) = 0.02
∴\(\sqrt{25.2}\) = f(x₀) + f'(x₀) dx
= \(\sqrt{25}\) + 0.02
= 5 + 0.02 = 5.02

Let AB be the lamp post. Let the man CD be at distance x m from lamp post and y m be the length of his shadow at any time t.
Given \(\frac { dx }{ dt } \) = 5 km / hr = 5000 m/hr
ΔABE and CDE are similar

∴ \(\frac { DE }{ CD } =\frac { BE }{ AB } \)
⇒ \(\frac { y }{ 2 } =\frac { x+y }{ 6 } \)
⇒ 6y = 2x+2y
⇒ 4y = 2x
⇒ y = \(\frac { x }{ 2 } \)
⇒ \(\frac { dy }{ dt } =\frac { 1 }{ 2 } \frac { dx }{ dt } =\frac { 1 }{ 2 } \)(5000)
= 2500 m/hr = 2.5 km/hr.

|(1+3i)³| = |1+3i|³ = \(\left[ \sqrt { { 1 }^{ 2 }+{ 3 }^{ 2 } } \right] ^{ 3 }\) =\(\left( \sqrt { 10 } \right) ^{ 3 }\)
=\((\sqrt { 10 } )^{ 3 }=\sqrt { 10 } \times \sqrt { 10 } \times \sqrt { 10 } \times \sqrt { 10 } =10\sqrt { 10 } \).

(cos θ + i sin θ )² = cos 2θ + isin 2θ
[By De moivre's theorem]
⇒ cos 2θ + isin 2θ = x + iy
Equating the real and imaginary parts we get,
x = cos 2θ, y = sin 2θ
∴ x² + y² = cos²2θ + sin²2θ = 1
Hence proved

Since the foci are one the x-axis, the equation of the hyperbola is \(\frac { { x }^{ 2 } }{ { a }^{ 2 } } +\frac { { y }^{ 2 } }{ { b }^{ 2 } } =1\)
Given conjugate axis = \(\frac34\) (transverse axis)
⇒ 2b = \(\frac34\) (2a) ⇒ b = \(\frac{3a}4\) ⇒ b² = \(\frac { { 9a }^{ 2 } }{ 16 } \)
∴ e = \(\sqrt { 1-\frac { { b }^{ 2 } }{ { a }^{ 2 } } } =\sqrt { 1-\frac { 9{ a }^{ 2 } }{ { 16a }^{ 2 } } } =\sqrt { 1+\frac { 9 }{ 16 } } =\sqrt { \frac { 25 }{ 16 } } =\frac { 5 }{ 4 } \)
∴ e = \(\frac { 5 }{ 4 } \)

Since the parabola is symmetric about x-axis, it is either open upward or downward.
Let the equation be x² = 4ay ...(1)
Since (2, -3) lies on the parabola,
2² = 4a(-3) ⇒ a = \(\frac { -1 }{ 3 } \)
Substituting a = \(\frac { -1 }{ 3 } \) in (1) we get,
x² = 4 \(\left( \frac { -1 }{ 3 } \right) \) y ⇒ 3x² = -4y. Which is the required equation of the parabola.

LHS = \(2{ tan }^{ -1 }\left( \frac { 2 }{ 3 } \right) \)
= \({ tan }^{ -1 }\left( \frac { 2 }{ 3 } \right) +{ tan }^{ -1 }\left( \frac { 2 }{ 3 } \right) \)
= \({ tan }^{ -1 }\left( \frac { \frac { 2 }{ 3 } +\cfrac { 2 }{ 3 } }{ 1-\left( \frac { 2 }{ 3 } \right) \left( \frac { 2 }{ 3 } \right) } \right) ={ tan }^{ -1 }\left( \frac { \frac { 4 }{ 3 } }{ \frac { 9-2 }{ 9 } } \right) \)
= \({ tan }^{ -1 }\left( \frac { 4 }{ 3 } \times \frac { 9 }{ 7 } \right) ={ tan }^{ -1 }\left( \frac { 12 }{ 7 } \right) \)
= RHS
Hence proved

Let \({ cos }^{ -1 }\left( \frac { -1 }{ 2 } \right) =y\) where \(0\le y\le \pi \)
Then \({ cos }^{ -1 }\left( \frac { -1 }{ 2 } \right) =y\Rightarrow cosy=\frac { -1 }{ 2 } \)
\(\Rightarrow cos\ y=-cos\frac { \pi }{ 3 } =cos\left( \pi -\frac { \pi }{ 3 } \right) =cos\left( \frac { 2\pi }{ 3 } \right) \)
\(\Rightarrow y=\frac { 2\pi }{ 3 } \left[ \because \frac { 2\pi }{ 3 } \in \left[ 0,\pi \right] \right] \)
\(\therefore \) The principal value of \({ cos }^{ -1 }\left( \frac { -1 }{ 2 } \right) \frac { 2\pi }{ 3 } \)

We have \(x=\sqrt { 2+x } \)
\(\Rightarrow { x }^{ 2 }=2+x \Rightarrow { x }^{ 2 }-x-2=0\)
\(\Rightarrow x=\frac { 1\pm \sqrt { 1+8 } }{ 2 } \Rightarrow x=\frac { 1\pm 3 }{ 2 } \)
\(\Rightarrow x=\frac { 1+3 }{ 2 } ,\frac { 1-3 }{ 2 } \Rightarrow x=2,-1\)
Also x>0, we get x = 2

Agumented matrix
[A|B] \(\left[ \begin{matrix} 2 & 5 \\ 6 & 15 \end{matrix}|\begin{matrix} 7 \\ 13 \end{matrix} \right] \overset { { R }_{ 2 }\rightarrow { R }_{ 2 }-3{ R }_{ 1 } }{ \longrightarrow } \left[ \begin{matrix} 2 & 5 \\ 0 & 0 \end{matrix}|\begin{matrix} 7 \\ -8 \end{matrix} \right] \)
Here \(\rho\) (A) = 2 and \(\rho\)([A|B]) = 3
∴ \(\rho\) (a) ≠ \(\rho\) ([AIB])
Hence the system is inconsistent.

The equation of the required plane through the intersection of the given planes is
( x + y + z - 6 ) + λ (2x + 3y + 4z + 5) = 0 ......(1)
This passes through (1, 1, 1)
∴ ( 1+ 1 + z - 6) + λ (2 + 3+ 4 + 5) = 0
⇒ -3 +14λ = 0 \(\Rightarrow \lambda =\frac { 3 }{ 14 } \)
Substituting \(\lambda =\frac { 3 }{ 14 } \) in (1) we get
( x + y + z - 6 )+\(\frac { 3 }{ 14 } \) (2x + 3y + 4z + 5) = 0
⇒ 14( x + y + z - 6 ) +3 (2 + 3+ 4 + 5) = 0
⇒ 20x + 23y + 26z - 69 = 0

\(\overset { \rightarrow }{ F } =\frac { 6\left( \overset { \wedge }{ 2i } -2\overset { \wedge }{ j } +\overset { \wedge }{ k } \right) }{ \sqrt { 4+4+1 } } =\frac { 6 }{ 3 } \left( \overset { \wedge }{ 2i } -2\overset { \wedge }{ j } +\overset { \wedge }{ k } \right) =\overset { \wedge }{ 4i } -4\overset { \wedge }{ j } +2\overset { \wedge }{ k } \)
\(\overset { \rightarrow }{ d } \) = (5, 3, 7) - (1, 2, 3) = (4, 1, 4) =\(\overset { \wedge }{ 4i } +\overset { \wedge }{ j } +4\overset { \wedge }{ k } \)
∴ Work done (w)
= \(\overset { \rightarrow }{ F } .\overset { \rightarrow }{ d } =\left( \overset { \wedge }{ 4i } -4\overset { \wedge }{ j } +2\overset { \wedge }{ k } \right) .\left( \overset { \wedge }{ 4i } +\overset { \wedge }{ j } +4\overset { \wedge }{ k } \right) \)
= 16 - 4 + 8 = 20 units