y=tan^-1(x+y+1)

prove.

\(\frac { 9\sqrt { 3 } { a }^{ 2 } }{ 45 } \)

\( \frac { \partial w }{ \partial u } =\frac { 2u }{ \sqrt { 1-\left( { u }^{ 2 }-{ v }^{ 2 } \right) } } ;\frac { \partial w }{ \partial v } =\frac { -2v }{ \sqrt { 1-\left( { u }^{ 2 }-{ v }^{ 2 } \right) } } \)

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

100 m / s, t = 4 sec, 200 m / s, −100 m / s

Let s be the surface area of the balloon after t sec.
s = 4πr²
\(\frac { ds }{ dt } =8\pi r.\frac { dr }{ dt } \)
Given that \(\frac { ds }{ dt } \) = constant = k
∴ \(8\pi r\frac { dr }{ dt } \) = k
⇒ 8πr dr = k dt
⇒ \(\int { 8\pi r } dr=k\int { dt } \)
⇒ \(8\pi .\frac { { r }^{ 2 } }{ 2 } \) = kt+c
⇒ 4πr² = kt+c ...(1)
when t = 0, r = 3
⇒ 4π(32) = 36(0) + c
c = 36π
⇒ 4π(52) = 25(k) +36
⇒ 100π = 2k+36π
⇒ k = 32π
(1) becomes, 4πr² = 32πt+36π
⇒ r² = 8t + 9
⇒ r = \(\sqrt { 8t+9 } \).

Area under the curve y = sin x between x = 0 and x = \(\frac { \pi }{ 3 } \) is
\({ A }_{ 1 }=\int _{ 0 }^{ \frac { \pi }{ 3 } }{ ydx } =\int _{ 0 }^{ \frac { \pi }{ 3 } }{ sinxdx } =-{ \left[ cosx \right] }_{ 0 }^{ \frac { \pi }{ 3 } }\)
\(=-(cos\frac { \pi }{ 3 } -cos0)=-\left( \frac { 1 }{ 2 } -1 \right) \)
\(=-\left( -\frac { 1 }{ 2 } \right) =\frac { 1 }{ 2 } \)
Area under the curve y = sin 2x between x = 0 and \(\frac { \pi }{ 3 } \) is
\({ A }_{ 2 }=\int _{ 0 }^{ \frac { \pi }{ 3 } }{ { sin2 \ x \ dx=-\left[ \frac { cos2 \ x }{ 2 } \right] }_{ 0 }^{ \frac { \pi }{ 3 } } } \)
\(=-\frac { 1 }{ 2 } [cos2\frac { \pi }{ 3 } -cos0]-\frac { 1 }{ 2 } \left[ -\frac { 1 }{ 2 } -1 \right] =-\frac { 1 }{ 2 } \left( -\frac { 3 }{ 2 } \right) =\frac { 3 }{ 4 } \)
\(\therefore \frac { { A }_{ 1 } }{ { A }_{ 2 } } =\frac { \frac { 1 }{ 2 } }{ \frac { 3 }{ 4 } } =\frac { 1 }{ 2 } \times \frac { 4 }{ 3 } =\frac { 2 }{ 3 } \)
∴ A₁:A₂ =2 : 3 

Given f(x, y) = 2x² + 3y² - 2xy
\(\frac { \partial f }{ \partial x } \) = 4x - 8y
\({ \left( \frac { \partial f }{ \partial x } \right) }_{ (2,3) }\) = 4(2) - 8(3)
= 8 - 24 = -16
\(\frac { \partial f }{ \partial y } \) = 6y-8x
\({ \left( \frac { \partial f }{ \partial y } \right) }_{ (2,3) }\) = 6(3)- 8(2)
= 18-16 = 2
\(\frac { { \partial }^{ 2 }f }{ { \partial x }^{ 2 } } =\frac { \partial }{ \partial x } { \left( \frac { \partial f }{ \partial x } \right) }=4\)
\(\frac { { \partial }^{ 2 }f }{ { \partial y }^{ 2 } } =\frac { \partial }{ \partial y } { \left( \frac { \partial f }{ \partial y } \right) }=6\)

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 = \({ tan }^{ -1 }\left( \frac { 1 }{ 2 } \right) \).

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 \(z=\frac { 2 }{ 3 } \)
\(\therefore \overset { \rightarrow }{ B } =\frac { 5 }{ 3 } \overset { \wedge }{ i } +\frac { 2 }{ 3 } \overset { \wedge }{ j } +\frac { 2 }{ 3 } \overset { \wedge }{ k } \)

Let 2-2i = r(cosθ + isinθ)
r =\(\sqrt { { 2 }^{ 2 }+(-2)^{ 2 } } =\sqrt { 4+4 } =\sqrt { 8 } =2\sqrt { 2 } \)
The principal value α =tan^-1\(\left| \frac { y }{ x } \right| =tan^{ -1 }\left| \frac { -z }{ z } \right| |\)
= tan^-1(1) = \(\frac { \pi }{ 4 } \)
Since the complex number 2 - 2i lies in the quadrant
θ = -α = -\(\frac { \pi }{ 4 } \)
∴ 2-2i = \(2\sqrt { 2 } \left[ cos\left( -\frac { \pi }{ 4 } \right) +isin\left( \frac { \pi }{ 4 } \right) \right] ^{ \frac { 1 }{ 3 } }\)
∴ \((2\sqrt { 2 } )^{ \frac { 1 }{ 3 } }\left[ cos\left( -\frac { \pi }{ 4 } \right) +isin\left( \frac { \pi }{ 4 } \right) \right] ^{ \frac { 1 }{ 3 } }\)
= \(8^{ \frac { 1 }{ 6 } }\left[ cos\frac { 1 }{ 3 } \left( 2k\pi -\frac { \pi }{ 4 } \right) +isin\frac { 1 }{ 3 } \left( 2k\pi -\frac { \pi }{ 4 } \right) \right] \)
k = 0, 1, 2
The roots are
∴ When k = 0, \(8^{ \frac { 1 }{ 6 } }cis\left( -\frac { \pi }{ 12 } \right) \)
when k = 1, \(8^{ \frac { 1 }{ 6 } }cis\left( \frac { 7\pi }{ 12 } \right) \)
when k = 2, \(8^{ \frac { 1 }{ 6 } }cis\left( \frac { 15\pi }{ 12 } \right) \)
∴ The product of the root
= \(8^{ \frac { 1 }{ 6 } }cis\left( -\frac { \pi }{ 12 } +\frac { 7\pi }{ 12 } +\frac { 15\pi }{ 12 } \right) \)
= \(8^{ \frac { 1 }{ 6 } }cis\left( \frac { 21\pi }{ 12 } \right) =8^{ \frac { 1 }{ 6 } }cis\left( \frac { 7\pi }{ 12 } \right) \)
= \(8^{ \frac { 1 }{ 6 } }cis\left( 2\pi -\frac { \pi }{ 4 } \right) =8^{ \frac { 1 }{ 6 } }cis\left( -\frac { \pi }{ 4 } \right) \)
= \(8^{ \frac { 1 }{ 6 } }\left[ cos\left( -\frac { \pi }{ 4 } \right) +isin\left( -\frac { \pi }{ 4 } \right) \right] \)
= \(8^{ \frac { 1 }{ 6 } }\left[ cos\left( \frac { \pi }{ 4 } \right) +isin\left( \frac { \pi }{ 4 } \right) \right] \)
= \(8^{ \frac { 1 }{ 6 } }\left[ \frac { 1 }{ \sqrt { 2 } } -\frac { i }{ \sqrt { 2 } } \right] =2^{ 3\times \frac { 1 }{ 6 } }\left( \frac { 1-i }{ \sqrt { 2 } } \right) =2^{ 1/2 }\left( \frac { 1-i }{ \sqrt { 2 } } \right) \)
= 1-i

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 \(\frac { { x }^{ 2 } }{ { A }^{ 2 } } -\frac { { y }^{ 2 } }{ { B }^{ 2 } } =1\)
⇒ \(\frac { { x }^{ 2 } }{ 4 } -\frac { { y }^{ 2 } }{ 12 } =1\)

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

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