order 2 ; degree 1

\(\frac { { a }^{ 2 } }{ 6 } \)

0

\(sin\left( \frac { 22 }{ 14 } \right) =1\)

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

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

2x – y + 3 = 0

Let I = \(\int _{ 0 }^{ 1 }{ \frac { { e }^{ x } }{ 1+{ e }^{ 2x } } dx } \)

Put e^x = t ⇒ e^x dx = dt
∴ \(\int _{ 1 }^{ e }{ \frac { dt }{ 1+{ t }^{ 2 } } dx } { \left[ { tan }^{ -1 }(t) \right] }_{ 1 }^{ e }\)
= tan^-1(e) - tan^-1(1)
= tan^-1(e) -\(\frac { \pi }{ 4 } \)

Given f(x,y) = x² + y³ + 2xy²
f_x = 3x² + 2y²
f_xx = 6x
f_y = 0+ 3y² + 4xy
= 3y² + 4xy
f_yy = 6y + 4x
f_xy = 4y
f_yx = 4y

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

⇒ 3x + (2x - 3y)i = 6 + 3i⁹
⇒ 3x + (2x - 3y)i = 6 + 3 . i⁴ . i⁴ . i¹
⇒ 3x + (2x - 3y)i = 6 + 3i
Equating the real and imaginary parts we get,
3x = 6 ⇒ x = 2
2x - 3y = 3 ⇒ 2(2) - 3y = 3
⇒ 4 - 3y = 3
⇒ 4 - 3 = 3y
⇒ 3y = 1 ⇒ y = \(\frac{1}{3}\)
∴ x = 2, y = \(\frac{1}{3}\).

Let z = -2
z = 2(-1) = 2(cos π + i sin π)
∴ arg(z) = π

z₁z₂ = (1 - i)(-2 + 4i) = -2 + 4i + 2i - 4i²
= -2 + 6i + 4 = 2 + 6i
\(\bar { { z }_{ 1 } } \) = 1+i
∴ \(\frac { { z }_{ 1 }{ z }_{ 2 } }{ \bar { { z }_{ 1 } } } =\frac { 2+6i }{ 1+6i } \times \frac { 1-i }{ 1-i } =\frac { 2(1-i+3i-i^{ 2 }) }{ 1+1 } \)
= 1 + 2i + 3
= 4 + 2i
∴ Im\(\left( \frac { { z }_{ 1 }{ z }_{ 2 } }{ \bar { { z }_{ 1 } } } \right) \) = 2

Given z = 5i¹¹ + 7i³
= 5i⁴ . i⁴ . i² . i¹ + 7. i² . i¹
= 5(1)(1)(-1)(i) + 7(-1)(i)
= -5i - 7i = -12i
∴ Re(z) = 0 and In(z) = -12

Given equation is x² + 3y² = a²
\(\div \) a² we get, \(\frac { { x }^{ 2 } }{ { a }^{ 2 } } +\frac { { y }^{ 2 } }{ \frac { { a }^{ 2 } }{ 3 } } =1\)
Here a² and b2 = \(\frac{a^2}{3}\) ⇒ b = \(\frac{a}{\sqrt3}\)
Length of major axis is 2a and
Length of minor axis is 2b = \(\frac { 2a }{ \sqrt { 3 } } \times \frac { \sqrt { 3 } }{ \sqrt { 3 } } =\frac { 2a\sqrt { 3 } }{ 3 } \)

Given equation of tangent is y = 3x + 1
The condition for any line y = mx + c to be a tangent to y² = 4ax is c = \(\frac{a}{m}\)
1= \(\frac{a}{m}\) ⇒ 1 = \(\frac{a}{m}\) ⇒ a = 3
Length of the latus rectum is 4a = 4(3) = 12 units.

Given 
\({ tan }^{ -1 }x={ sin }^{ -1 }\left( \frac { 1 }{ 2 } \right) ={ sin }^{ -1 }\left( sin\frac { \pi }{ 6 } \right) \)
\(\Rightarrow { tan }^{ -1 }x=\frac { \pi }{ 6 } \)
\(\Rightarrow x=tan\frac { \pi }{ 6 } =\frac { 1 }{ \sqrt { 3 } } \)
\(\therefore x=\frac { 1 }{ \sqrt { 3 } } \)

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 \({ cos }^{ -1 }\left( \frac { -1 }{ 2 } \right) \frac { 2\pi }{ 3 } \)

Δ = \(\left| \begin{matrix} 6 & -7 \\ 9 & -5 \end{matrix} \right| \) = -30 + 63 = 33
Δ₁ = \(\left| \begin{matrix} 16 & -7 \\ 35 & -5 \end{matrix} \right| \) = -80 + 245 = 165
Δ₂ = \(\left| \begin{matrix} 6 & 16 \\ 9 & 35 \end{matrix} \right| \) = 210 - 144 = 66
∴ x = \(\frac { { \triangle }_{ 1 } }{ \triangle } =\frac { 165 }{ 33 } \) = 5
y = \(\frac { { \triangle }_{ 2 } }{ \triangle } =\frac { 66 }{ 33 } \) = 2
∴ Solution set is { 5, 2}

Matrix form of the given system of equations is
\(\left[ \begin{matrix} 1 & 2 & 2 \\ 1 & -3 & -3 \\ 2 & 1 & k \end{matrix} \right] \left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] =\left[ \begin{matrix} 0 \\ 0 \\ 0 \end{matrix} \right] \)
AX = B where A =\(\left[ \begin{matrix} 1 & 2 & 2 \\ 1 & -3 & -3 \\ 2 & 1 & k \end{matrix} \right] \)
Homogeneous system of equations has trivial solution only if |A| ≠ 0.
∴ \(\left[ \begin{matrix} 1 & 2 & 2 \\ 1 & -3 & -3 \\ 2 & 1 & k \end{matrix} \right] \) ≠ 0
Expanding along R₁,
\(1\left| \begin{matrix} -3 & -3 \\ 1 & k \end{matrix} \right| -2\left| \begin{matrix} 1 & -3 \\ 2 & k \end{matrix} \right| +2\left| \begin{matrix} 1 & -3 \\ 2 & 1 \end{matrix} \right| \neq 0\)
⇒ 1 (-3k + 3) - 2 (k + 6) + 2 (1 + 6) ≠ 0
⇒ -3k + 3 -2k - 12 + 14 ≠ 0
⇒ -5k + 5 ≠ 0
⇒ -5k ≠ -5 ⇒ k ≠ \(\frac { -5 }{ -5 } \) =1
⇒ k ≠ 1

A =\(\left[ \begin{matrix} 4 \\ 7 \end{matrix}\begin{matrix} 5 \\ -3 \end{matrix}\begin{matrix} -6 \\ 0 \end{matrix}\begin{matrix} 1 \\ 8 \end{matrix} \right] \)
\(\overset { { R }_{ 2 }\rightarrow { R }_{ 2 }-2{ R }_{ 1 } }{ \longrightarrow } \left[ \begin{matrix} 4 \\ -1 \end{matrix}\begin{matrix} 5 \\ -3 \end{matrix}\begin{matrix} -6 \\ 12 \end{matrix}\begin{matrix} 1 \\ 6 \end{matrix} \right] \)
\(\overset { { R }_{ 1 }\leftrightarrow { R }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} -1 \\ 4 \end{matrix}\begin{matrix} -3 \\ 5 \end{matrix}\begin{matrix} 12 \\ -6 \end{matrix}\begin{matrix} 6 \\ 1 \end{matrix} \right] \)
\(\overset { { R }_{ 1 }+(-1){ R }_{ 1 } }{ \longrightarrow } \left[ \begin{matrix} 1 \\ 4 \end{matrix}\begin{matrix} 13 \\ 5 \end{matrix}\begin{matrix} -12 \\ -6 \end{matrix}\begin{matrix} 6 \\ 1 \end{matrix} \right] \)
\(\overset { { R }_{ 2 }\rightarrow { R }_{ 2 }-4{ R }_{ 1 } }{ \longrightarrow } \left[ \begin{matrix} 1 \\ 0 \end{matrix}\begin{matrix} 13 \\ -47 \end{matrix}\begin{matrix} -12 \\ 42 \end{matrix}\begin{matrix} -6 \\ 25 \end{matrix} \right] \)
The equivalent row-echelon matrix hats two non zero rows.
∴ \(\rho\) (A) = 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.

Since the plane passing through the point \(\overset { \rightarrow }{ a } =\overset { \wedge }{ i } -\overset { \wedge }{ j } +2\overset { \wedge }{ k } \) and is normal to the vector \(\overset { \rightarrow }{ n } =2\overset { \wedge }{ i } +3\overset { \wedge }{ j } +2\overset { \wedge }{ k } \)
the vector equation of the plane is \(\overset { \rightarrow }{ r } .\overset { \rightarrow }{ n } =\overset { \rightarrow }{ a } .\overset { \rightarrow }{ n } \)
\(\Rightarrow { \overset { \rightarrow }{ r } }.\left( 2\overset { \wedge }{ i } +3\overset { \wedge }{ j } +2\overset { \wedge }{ k } \right) =\left( \overset { \wedge }{ i } -\overset { \wedge }{ j } +2\overset { \wedge }{ k } \right) .\left( 2\overset { \wedge }{ i } +3\overset { \wedge }{ j } +2\overset { \wedge }{ k } \right) \)
\(\Rightarrow { \overset { \rightarrow }{ r } }.\left( 2\overset { \wedge }{ i } +3\overset { \wedge }{ j } +2\overset { \wedge }{ k } \right) \)= 2 - 3 + 4 = 3
\(\therefore { \overset { \rightarrow }{ r } }.\left( 2\overset { \wedge }{ i } +3\overset { \wedge }{ j } +2\overset { \wedge }{ k } \right) \)= 3