y = x² + cx

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

\(\frac { 28 }{ 3 } \)

\(\frac { 20\sqrt { 2 } }{ 3 } \)

0.677

\(\left( -\infty ,-\frac { 5 }{ 6 } \right) \) concave downward
\(\left( -\frac { 5 }{ 6 } ,\infty \right) \) concave upward
Points of inflection is \(\left( -\frac { 5 }{ 6 } ,\frac { 305 }{ 54 } \right) \)

4x + 3y = 7a

Let I = \(\int _{ 0 }^{ 1 }{ \sqrt { 9-4{ x }^{ 2 } } dx } =\int _{ 0 }^{ 1 }{ 2\sqrt { \left( \frac { 3 }{ 2 } \right) ^{ 2 }-{ x }^{ 2 } } } dx\)
\(={ 2\left[ \frac { x }{ 2 } \sqrt { \left( \frac { 3 }{ 2 } \right) ^{ 2 }-{ x }^{ 2 } } +\frac { 9 }{ 8 } { sin }^{ -1 }\left( \frac { x }{ \frac { 3 }{ 2 } } \right) \right] }_{ 0 }^{ 1 }\)
\(\left[ \because \sqrt { { a }^{ 2 }-{ x }^{ 2 } } =\frac { x }{ 2 } \sqrt { { a }^{ 2 }-{ x }^{ 2 } } +\frac { { a }^{ 2 } }{ 2 } \left( \frac { x }{ a } \right) \right] \)
\(=2\left[ \frac { 1 }{ 2 } \sqrt { \frac { 9 }{ 4 } -1 } +\frac { 9 }{ 8 } { sin }^{ -1 }\left( \frac { 2 }{ 3 } \right) -0 \right] \)
\(=2\left[ \frac { 1 }{ 2 } \sqrt { \frac { 5 }{ 4 } } +\frac { 9 }{ 8 } { sin }^{ -1 }\left( \frac { 2 }{ 3 } \right) \right] \)
\(=\frac { \sqrt { 5 } }{ 2 } +\frac { 9 }{ 4 } { sin }^{ -1 }\left( \frac { 2 }{ 3 } \right) \)

w = xy + z
\(\frac { \partial w }{ \partial x } =y;\frac { \partial w }{ \partial y } =x;\frac { \partial w }{ \partial z } =1\)
⇒ \(\frac { \partial w }{ \partial x } \) = sin t; \(\frac { \partial w }{ \partial y } \) = cos t; \(\frac { \partial w }{ \partial z } \) = 1
\(\frac { dx }{ dt } \) = - sin t; \(\frac { dy}{ dt } \) = cos t; \(\frac { dz }{ dt } \) = 1
∴  \(\frac { dw }{ dt } \) = \(\frac { \partial w }{ \partial x } .\frac { dx }{ dt } +\frac { \partial w }{ \partial y } .​​\frac { dy }{ dt } +\frac { \partial w }{ \partial z } .\frac { dz }{ dt } \)
= sin t(-sin t) + cos t(cost) + 1 (1) 
= - sin² t + cos² t + 1
= cos² t + 1 - sin² t
= cos² t + cos² t  [∵ 1- sin² t = cos² t]
\(\frac { dw }{ dt } \) = 2 cos² t

Let z = x + iy be a complex number
∴ \(\left| \frac { x+iy-3 }{ x+iy+3 } \right| \)  = 2
⇒ \(\left| \frac { x+iy-3 }{ x+iy+3 } \right| \) = 2
⇒ |(x-3) + iy| = 2|(x+3)+iy|
⇒ \(\sqrt { (x-3)^{ 2 }+{ y }^{ 2 } } =2\sqrt { (x+3)^{ 2 }+{ y }^{ 2 } } \)
Squaring both sides we get,
(x - 3)² + y² = 4[(x + 3)² + y²]
⇒ x² + 9 - 6x + y² = 4 [x²+ 9 + 6x + y²]
x² + 9 - 6x + 1 = 4x² + 36 + 24x + 4y²
⇒ 3x² + 3y² + 30x + 27 = which represent a circle.

Given line is y = x + c
m = 1, c = c
Equation of the hyperbola is 9x² - 16y² = 144
\(\div \)144 we get \(\frac { { x }^{ 2 } }{ 16 } +\frac { { y }^{ 2 } }{ 9 } \) = 1
a² = 16, b² = 9
The condition for the line y = mx + c to be a tangent to the hyperbola is c² = a²m² - b²
∴ c² = 16(1)² - 9 = 16 - 9 = 7
∴ c = ±\(\sqrt7\)

Equation of circle is x² + y² = 64
∴ a² = 64 ⇒ a = 8
Given line is 3x+ 4y = P
4y = -3x + p
y = \(y=\frac { -3 }{ 4 } x+\frac { p }{ 4 } \)
m = \(\frac { -3 }{ 4 } \) and c = \(\frac { p }{ 4 } \)
The condition for y = mx + c to be a tangent to the circle in c² = a²(1 + m²).
∴ \({ \left( \frac { p }{ 4 } \right) }^{ 2 }=64\left( 1+\frac { 9 }{ 16 } \right) \)
\(\Rightarrow \frac{p^{2}}{\not 16}=64\left(\frac{16+9}{\not 16}\right) \Rightarrow p^{2}=64(25)\)
\(p=\pm \sqrt { 64(25) } =\pm 8(5)\)
∴ p = ±40

\(cos\left[ { cos }^{ -1 }\left( \frac { -\sqrt { 3 } }{ 2 } +\frac { \pi }{ 6 } \right) \right] \)
= \(cos\left[ \pi -{ cos }^{ -1 }\left( \frac { -\sqrt { 3 } }{ 2 } +\frac { \pi }{ 6 } \right) \right] \)
\(\left[ \because { cos }^{ -1 }\left( -x \right) =\pi -{ cos }^{ -1 }x \right] \)
= \(cos\left[ \pi -\frac { \pi }{ 6 } +\frac { \pi }{ 6 } \right] \)
\(\left[ \because { cos }^{ -1 }\frac { \sqrt { 3 } }{ 2 } =x\Rightarrow \frac { \sqrt { 3 } }{ 2 } =cosx\Rightarrow x=\frac { \pi }{ 6 } \right] \)
= \(cos\pi -1\)

Let AOB be right angled triangle, right angled at O.
Take O as origin.
Then \(\overset { \rightarrow }{ OA } =\overset { \rightarrow }{ a } ,\overset { \rightarrow }{ OB } =\overset { \rightarrow }{ b } \)
\(\therefore \overset { \rightarrow }{ AB } =\overset { \rightarrow }{ OB } -\overset { \rightarrow }{ OA } =\overset { \rightarrow }{ b } -\overset { \rightarrow }{ a } \)
\(\therefore \overset { \rightarrow }{ OA } \bot \overset { \rightarrow }{ OA } =\overset { \rightarrow }{ a } \bot \overset { \rightarrow }{ b } \Rightarrow \overset { \rightarrow }{ a } .\overset { \rightarrow }{ b } =0\)
\(\therefore { AB }^{ 2 }={ \left| \overset { \rightarrow }{ AB } \right| }^{ 2 }=\overset { \rightarrow }{ AB } .\overset { \rightarrow }{ AB } \)
\(=0=\left( \overset { \rightarrow }{ b } -\overset { \rightarrow }{ a } \right) .\left( \overset { \rightarrow }{ b } -\overset { \rightarrow }{ a } \right) \)
\(=\overset { \rightarrow }{ b } .\overset { \rightarrow }{ b } -\overset { \rightarrow }{ b } .\overset { \rightarrow }{ a } -\overset { \rightarrow }{ a } .\overset { \rightarrow }{ b } +\overset { \rightarrow }{ a } .\overset { \rightarrow }{ a } \)
\(={ \left| \overset { \rightarrow }{ b } \right| }^{ 2 }-2\overset { \rightarrow }{ a } .\overset { \rightarrow }{ b } +{ \left| \overset { \rightarrow }{ a } \right| }^{ 2 }\)
= OB² - 0 + OA²    \(\left[ \because \overset { \rightarrow }{ a } .\overset { \rightarrow }{ b } =0 \right] \)
⇒ AB² = OA² + OB²
Hence the Pythagoras theorem

|A| =\(\left[ \begin{matrix} -2 & -3 \\ 5 & -6 \end{matrix} \right] \) = 12+15 = 27
∴ A^-1 = \(\frac { 1 }{ |A| } adjA=\frac { 1 }{ 27 } \left[ \begin{matrix} -6 & 3 \\ -5 & 2 \end{matrix} \right] \)
(A^-1)^T = \(\frac { 1 }{ 27 } \left[ \begin{matrix} -6 & -5 \\ 3 & 2 \end{matrix} \right] \)...(1)
A^T =\(\left[ \begin{matrix} -2 & 5 \\ -3 & -6 \end{matrix} \right] \)
|A^T| =\(\left[ \begin{matrix} -2 & 5 \\ -3 & -6 \end{matrix} \right] \) = 12+15 = 27
∴ (A^T)^-1 = \(\frac { 1 }{ |A^{ T }| } adj(A^{ T })=\frac { 1 }{ 27 } \left[ \begin{matrix} -6 & -5 \\ 3 & -2 \end{matrix} \right] \)...(2)
From (1) and (2), (A^-1)^T = (A^T)^-1

Put y = \(\frac { x-1+x-5 }{ 2 } =-3\)
⇒ x = y + 3
∴ (x-1)⁴+(x-5)⁴ = 82
⇒ (y+3-1)⁴+(y+3-5)⁴ = 82
⇒ (y+2)⁴+(y-2)⁴ = 82
⇒ 2(y⁴+24y²+16) = 82
⇒ y⁴+24y²+16 = 41
⇒ y⁴+24y²-25 = 0
⇒ (y²+25)(y²-1) = 0
⇒ y = 土5i, y = 士1
∴ x = 3土5i, 4, 2.