2

\(\frac { dw }{ d\theta } =0\)

(x-2)+(x-2)^z

Equation of family of straight lines in my plane is y = mx - c where m and c are arbitrary constraints.
Differentiating, y' = m
Differentiating again, y" = 0, is the required differential equation.

∴p v (q Λ r) is a contingency

Let \(\int _{ 0 }^{ \frac { \pi }{ 2 } }{ log(tan \ x)dx } \)  ....(1)
Applying the property \(\int _{ 0 }^{ a }{ f(x) } dx=\int _{ 0 }^{ a }{ f(a-x)dx } \)
we get
I = \(\int _{ 0 }^{ \pi /2 }{ log(tan(\frac { \pi }{ 2 } -x))dx } \)
= \(\int _{ 0 }^{ \pi /2 }{ log(cot \ x) } dx\) ......(2)
\((1)+(2)\longrightarrow 2I\int _{ 0 }^{ \pi /2 }{ log(tan \ x)+log(cot \ x)dx } \)
\( =\int_{0}^{\pi / 2} \log \tan x \cdot \cot x d x \)
\(\int _{ 0 }^{ \pi /2 }{ log1dx } =0\)
⇒ I = 0 Hence proved

Given f(x, y) = 2x³ - 11x²y + 3y³
f(tx, ty) = 2t³ x3 - 11 t² x²ty + 3t³y³
= t³(2x³ - 11x²y + 3y³)
= t³. f(x,y)
∴ f (x, y) is a homogeneous function of degree 3.
∴ By Euler's theorem,
\(x\frac { \partial f }{ \partial x } +y\frac { \partial f }{ \partial y } =3f\)

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.

LHS = (1+ω²)³ - (1+ω)³
= (-ω)³ - (-ω²)³
[∴ 1 + ω + ω² = 0]
= -ω³ + ω⁶  [∴ ω³ = 1]
= -1 + 1 = 0

Since the vertices are (0, ±7), equation of the hyperbola is \(\frac { { x }^{ 2 } }{ { a }^{ 2 } } +\frac { { y }^{ 2 } }{ { b }^{ 2 } } =1\)
a = 7 and e = \(\frac { 4 }{ 3 } \)
b² = a²(e² - 1) = 49\(\left( \frac { 16 }{ 9 } -1 \right) =49\left( \frac { 16-9 }{ 9 } \right) \)
= \(49\left( \frac { 7 }{ 9 } \right) =\frac { 343 }{ 9 } \)
∴ Equation of the hyperbola is \(\frac { { y }^{ 2 } }{ 49 } -\frac { { x }^{ 2 } }{ \frac { 343 }{ 9 } } =1\) 
⇒ \(\frac { { y }^{ 2 } }{ 49 } -\frac { 9{ x }^{ 2 } }{ 343 } =1\)

Let AOB be the vertical section of the reflector and m is the mid-point of AB. Let the equation of the parabola be y² = 4ax A(6, 12) lies on (1)

∴ 12² = 4a(6) ⇒ a = 6
∴ Focus is (a, 0) = (b, 0)
Hence focus coincides with m, the mid-point of AB.

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 sin-1(-1) = y where \(\frac { -\pi }{ 2 } \le y\le \frac { \pi }{ 2 } \)
Then \(sin^{ -1 }(-1)=y\Rightarrow sin\quad y=-1\)
\(-1=sin\left( \frac { -\pi }{ 2 } \right) \Rightarrow y=\frac { -\pi }{ 2 } \) \(\left[ \because \frac { -\pi }{ 2 } \epsilon \left[ \frac { -\pi }{ 2 } ,\frac { \pi }{ 2 } \right] \right] \)
\( \therefore\)  The principal value of \({ sin }^{ -1 }(-1)\ is \ \frac { -\pi }{ 2 } \)

Let ∝, β are the roots of the equation
Sum of the roots \(\alpha +\beta =\frac { -b }{ a } \)
 \(=\frac { [-(a-2)] }{ 1 } =a-2\)
Product of the roots \(=\alpha \beta =\frac { c }{ a } \)
\(=\frac { -(a+1) }{ 1 } =-(a+1)\)
we have \({ \alpha }^{ 2 }{ \beta }^{ 2 }=({ \alpha +\beta ) }^{ 2 }-2\alpha \beta \)
\(={ (a-2) }^{ 2 }+2(a+1)\)
\(={ a }^{ 2 }-4a+4+2a+2\)
\(=(a-1{ ) }^{ 2 }+5\)
Thus \(\\ { \alpha }^{ 2 }+{ \beta }^{ 2 }\) is least if a = 1

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.

Given (x₁, y₁, z₁) is (2, -1, 3) (x₂, y₂, z₂) is (4, 2, 1)
Cartesian equation of a line passing through two points is \(\frac { x-{ x }_{ 1 } }{ { x }_{ 2 }-{ x }_{ 1 } } =\frac { y-{ y }_{ 1 } }{ { y }_{ 2 }-{ y }_{ 1 } } =\frac { z-{ z }_{ 1 } }{ { z }_{ 2 }-{ z }_{ 1 } } \)
\(\Rightarrow \frac { x-2 }{ 2 } =\frac { y+1 }{ 3 } =\frac { z-3 }{ -2 } \)