\(\frac { 2log5 }{ lof2 } hours\)

\({ tan }^{ -1 }y=\frac { \pi }{ 4 } +\frac { 1 }{ 2 } -\frac { 1 }{ 2 } \left( 1+logx \right) ^{ 2 }\)

prove.

\(\frac { 4 }{ 2 } \left( 4\pi +\sqrt { 3 } \right) \)

240

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

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.

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 r² = x² +y² + z²
log r² = log (x² + y² + z²)
⇒ 2 log r = log (x² + y² + z²)
∴ 2V = log (x² + y² + z²) [∵ V = log r ]
⇒ V = \(\frac12\) log (x² + y² + z²)
\(\frac { \partial V }{ \partial x } =\frac { 1 }{ 2 } \frac { 2x }{ { x }^{ 2 }+{ y }^{ 2 }+{ z }^{ 2 } } =\frac { x }{ { x }^{ 2 }+{ y }^{ 2 }+{ z }^{ 2 } } \)
\(\frac { { \partial }^{ 2 }V }{ \partial { x }^{ 2 } } =\frac { ({ x }^{ 2 }+{ y }^{ 2 }+{ z }^{ 2 })(1)-x(2x) }{ { { (x }^{ 2 }+{ y }^{ 2 }+{ z }^{ 2 }) }^{ 2 } } \)
\(=\frac { { x }^{ 2 }+{ y }^{ 2 }-{ x }^{ 2 } }{ { { (x }^{ 2 }+{ y }^{ 2 }+{ z }^{ 2 }) }^{ 2 } } \)
III^ty \(\frac { { \partial }^{ 2 }V }{ \partial { y }^{ 2 } } =\frac { { x }^{ 2 }+{ y }^{ 2 }-{ y }^{ 2 } }{ { { (x }^{ 2 }+{ y }^{ 2 }+{ z }^{ 2 }) }^{ 2 } } \)
∴ \(\frac { { \partial }^{ 2 }V }{ \partial { z }^{ 2 } } =\frac { { x }^{ 2 }+{ y }^{ 2 }-{ z }^{ 2 } }{ { { (x }^{ 2 }+{ y }^{ 2 }+{ z }^{ 2 }) }^{ 2 } } \)
\(\therefore \frac { { \partial }^{ 2 }V }{ \partial { x }^{ 2 } } +\frac { { \partial }^{ 2 }V }{ \partial { y }^{ 2 } } +\frac { { \partial }^{ 2 }V }{ \partial { z }^{ 2 } } =\frac { { y }^{ 2 }+{ z }^{ 2 }-{ x }^{ 2 }+{ z }^{ 2 }+{ x }^{ 2 }-{ y }^{ 2 }+{ x }^{ 2 }+{ y }^{ 2 }-{ z }^{ 2 } }{ { ({ x }^{ 2 }+{ y }^{ 2 }{ +z }^{ 2 }) }^{ 2 } } \)
= \(\frac { { x }^{ 2 }+{ y }^{ 2 }{ +z }^{ 2 } }{ { ({ x }^{ 2 }+{ y }^{ 2 }{ +z }^{ 2 }) }^{ 2 } } =\frac { 1 }{ { x }^{ 2 }+{ y }^{ 2 }{ +z }^{ 2 } } \)
= \(\frac { 1 }{ { r }^{ 2 } } \)
Hence proved.

Given curves are 4x = y² ....(1)
⇒ 4xy = k....(2)
Substituting (1) in (2) we get
y³ = k  ⇒ y = \(k^{ \frac { 1 }{ 3 } }\)
∴ (1) becomes, 4x = \(\left( { k }^{ \frac { 1 }{ 3 } } \right) ^{ 2 }=k^{ \frac { 2 }{ 3 } }\)
⇒ x = \(\frac { { k }^{ \frac { 2 }{ 3 } } }{ 4 } \)
∴ The point of intersection of the given curves is \(\left( \frac { k^{ \frac { 2 }{ 3 } } }{ 4 } ,{ k }^{ \frac { 1 }{ 3 } } \right) \)
Differentiating 4x = y²
4 = 2y\(\frac { dy }{ dx } \)
⇒ \(\frac { dy }{ dx } =\frac { 2 }{ y } \)
m₁ = \(\frac { 2 }{ { k }^{ \frac { 1 }{ 3 } } } \)
Differentiating 4xy = k,
4x\(\frac { dy }{ dx } \)+4y = 0
⇒  \(\frac { dy }{ dx } =\frac { -y }{ x } \)
∴ m₂ = \(\frac { { -k }^{ \frac { 1 }{ 3 } } }{ { k }^{ \frac { 2 }{ 3 } } } \)(4)
Since the given curves cut at right angles, m₁m₂ = -1.
∴ \(\left( \frac { 2 }{ { k }^{ \frac { 1 }{ 3 } } } \right) \left( \frac { -4.k^{ \frac { 1 }{ 3 } } }{ k^{ \frac { 2 }{ 3 } } } \right) \) = 1
⇒ \(\frac { 8 }{ { k }^{ \frac { 2 }{ 3 } } } \) = 1 ⇒ \({ k }^{ \frac { 2 }{ 3 } }\) = 8
⇒ \(\left( { k }^{ \frac { 2 }{ 3 } } \right) ^{ 3 }\) = 8³
⇒ k² = 512.

From the line  \(\frac { x-3 }{ 3 } =\frac { y-8 }{ -1 } =\frac { z-3 }{ 1 } \), we get
\(\overset { \rightarrow }{ a } =3\overset { \wedge }{ i } +8\overset { \wedge }{ j } +3\overset { \wedge }{ k } \), \(\overset { \rightarrow }{ b } =3\overset { \wedge }{ i } -\overset { \wedge }{ j } +\overset { \wedge }{ k } \)
From the line \(\frac { x+3 }{ -3 } =\frac { y+7 }{ 2 } =\frac { z-6 }{ 4 } \) we get
\(\overset { \rightarrow }{ c } =-3\overset { \wedge }{ i } -7\overset { \wedge }{ j } +6\overset { \wedge }{ k } \) and \(\overset { \rightarrow }{ d } =-3\overset { \wedge }{ i } +2\overset { \wedge }{ j } +4\overset { \wedge }{ k } \)
Since the given lines are not parallel, the shortest distance between the line is 
\(d=\left| \frac { \left( \overset { \rightarrow }{ c } -\overset { \rightarrow }{ a } \right) .\left( \overset { \rightarrow }{ b } \times \overset { \rightarrow }{ d } \right) }{ \left| \overset { \rightarrow }{ b } \times \overset { \rightarrow }{ d } \right| } \right| \)
\(\overset { \rightarrow }{ c } -\overset { \rightarrow }{ a } =-6\overset { \wedge }{ i } -15\overset { \wedge }{ j } +3\overset { \wedge }{ k } \)
\(\overset { \rightarrow }{ b } \times \overset { \rightarrow }{ d } =\left| \begin{matrix} \overset { \wedge }{ i } \\ 3 \\ -3 \end{matrix}\begin{matrix} \overset { \wedge }{ j } \\ -1 \\ 2 \end{matrix}\begin{matrix} \overset { \wedge }{ k } \\ 1 \\ 4 \end{matrix} \right| \)
\(=\overset { \wedge }{ i } (-4-2)-\overset { \wedge }{ j } (12+3)+\overset { \wedge }{ k } (6-3)\\ \)
\(=-6\overset { \wedge }{ i } -15\overset { \wedge }{ j } +3\overset { \wedge }{ k } \)
\(\therefore \left| \overset { \rightarrow }{ b } \times \overset { \rightarrow }{ d } \right| =\sqrt { 36+225+9 } \)
\(=\sqrt { 270 } \)
\(\therefore \left( \overset { \rightarrow }{ c } -\overset { \rightarrow }{ a } \right) .\left( \overset { \rightarrow }{ b } \times \overset { \rightarrow }{ d } \right) =\left( -6\overset { \wedge }{ i } -15\overset { \wedge }{ j } +3\overset { \wedge }{ k } \right) .\left( -6\overset { \wedge }{ i } -15\overset { \wedge }{ j } +3\overset { \wedge }{ k } \right) \)
= -6 (-6) + 15 _(15) + 3(3)
= 270 ≠ 0
Since the given lines are neither intersecting, nor parallel they are skew lines
\(\therefore d=\frac { 270 }{ \sqrt { 270 } } =\sqrt { 270 } units\)

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

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

\({ sin }^{ -1 }\left( \frac { sinx+cosx }{ \sqrt { 2 } } \right) \)
\(\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