(1+ sin x)(1 + cos y) = c

\(\left[ 1+\left( \frac { dy }{ dx } \right) ^{ 2 } \right] ^{ 2 }={ y }^{ 2 }\left( \frac { { d }^{ 2 }y }{ { dx }^{ 2 } } \right) ^{ 2 }\)

1

0

\(L(z)={ 2 }^{ 1/4 }+\frac { 1 }{ 4 } \left( { 2 }^{ -3/4 } \right) \left( z-2 \right) \)

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

(i) f(x) is strictly increasing on\(\left( -\infty ,-\frac { 1 }{ 2 } \right) \) and f(x) is strictly decreasing on \(\left[ -\frac { 1 }{ 2 } ,\infty \right] \) Local maximum value = \(\frac { 81 }{ 4 } \)
(ii) f(x) is strictly decreasing on \(\left( 0,\frac { 1 }{ \sqrt { 2 } } \right) \) and strictly increasing on \(\left( \frac { 1 }{ \sqrt { 2 } } ,2 \right) \)
Local minimum value = \(-\frac { 2 }{ 2\sqrt { 2 } } \)

\(\frac { 8 }{ 3 } { cm }^{ 2 }/sec\)

2:π

Given \(\frac { dy }{ dx } +y=cosx\)
This is a linear differential equation
Here p = 1, Q = cos x
\(\therefore \int { p\ dx } =\int { dx } =x\)
\(I.F={ e }^{ \int { p\ dx } }={ e }^{ x }\)
The solution is
\({ y }^{ \int { p\ dx } }=\int { Q{ e }^{ \int { p\ dx } }dx+c } \)
\(\Rightarrow { ye }^{ x }=\int { cosx.{ e }^{ x }dx+c } \)
\(\Rightarrow { ye }^{ x }=\frac { { e }^{ x } }{ 2 } \left( cosx+sinx \right) +c\)
\(\Rightarrow y=\frac { 1 }{ 2 } \left( cosx+sinx \right) +{ ce }^{ x }\)
\(\therefore \int { { e }^{ ax }cos\ bx\ dx=\frac { { e }^{ ax } }{ { a }^{ 2 }+{ b }^{ 2 } } \left[ acos\ bx+sin\ ax \right] } \)

Given f(x, y) = \(\frac { x }{ { x }^{ 2 }+{ y }^{ 2 } } \)
f(tx, ty) = \(\frac { tx }{ { t }^{ 2 }{ x }^{ 2 }+{ t }^{ 2 }{ y }^{ 2 } } =\frac { tx }{ { t }^{ 2 }(x^{ 2 }+{ y }^{ 2 }) } \)
= \(\frac { x }{ t({ x }^{ 2 }+{ y }^{ 2 }) } ={ t }^{ -1 }.f(x,y)\)
∴ f(x,y) is a homogeneous function of degree - 1
∴ By Euler's theorem,
\(x\frac { \partial f }{ \partial x } +y\frac { \partial f }{ \partial y } \) = nf = -1.f
∴ \(x\frac { \partial f }{ \partial x } +y\frac { \partial f }{ \partial y } \) = -f
Hence proved

Let x_o = 3 and dx = 0.02
f(x_o) = f(3) = 3 (3²) + 5 (3) + 3
= 27 + 15 + 3 = 45
f'(x) = 6x + 5
f'(x) = f'(3) = 6 (3) + 5 = 23
∴ f(3. 02) = f(x_o) +f'(x_o) dx
= 45 + 23 (.02)
= 45 + 0.46 = 45.46

Let z = x+iy ⇒ \(\bar { z } \) = x+iy
∴ \(\\ \frac { \bar { z } +1 }{ z-1 } =\frac { z-iy+1 }{ x-iy-i } =\frac { (x+1)iy }{ x-i(y+1) }\)
= \(\frac { (x+1)-iy }{ x-i(y+1) } \times \frac { x+i(y+1) }{ x+i(y+1) } \)
Choosing the real part alone we get,
\(\frac { x(x+1)+y(y+1) }{ { x }^{ 2 }+(y+1)^{ 2 } } \) = 0
⇒ x(x+1) + y(y+1) = 0
⇒ x²+x+y²+y = 0 which is the locus of z.

AB = |(3+2i) - (0+5i)| = |3-3i|
=\(\sqrt { 9+9 } =\sqrt { 18 } =3\sqrt { 2 } \)
BC = |(0+5i) - (-3+2i)| = |3+3i|
= \(\sqrt { 9+9 } =\sqrt { 18 } =3\sqrt { 2 } \)
CD = |(-3+2i) - (0-i) = |-3+i|
= \(\sqrt { 9+9 } =\sqrt { 18 } =3\sqrt { 2 } \)
DA = |(0-i) - (3+2i)| = |-3-3i|
=\(\sqrt { 9+9 } =\sqrt { 18 } =3\sqrt { 2 } \)
∴ AB = BC = CD = DA
Also AC = |(3+2i) - (-3+2i)|
= |6| = \(\sqrt { 36 } \) = 6
∴ AC = BD
Hence ABCD is a square

\({ tan }^{ -1 }\sqrt { x(x+1) } +{ sin }^{ -1 }\sqrt { { x }^{ 2 }+x+1 } =\frac { \pi }{ 2 } \)
This equation holds if
\({ x }^{ 2 }+x\ge 0\) and \({ x }^{ 2 }+x+1\le 1\)
Now, \({ x }^{ 2 }+x\le 0\) and \(0\le { x }^{ 2 }+x+1\le 1\)
\(\Rightarrow { x }^{ 2 }+x\ge 0\quad { x }^{ 2 }+x+1\le 1\) [\( \because { x }^{ 2 }+x+1\ge 0 \) for all x]
\( \Rightarrow x^{2}+x \geq 0 \text { and } x^{2}+x \leq 0 \)
\( \Rightarrow x^{2}+x=0 \Rightarrow x=0,-1 \)
Hence, x = 0, -1 are the solutions of the given equation.

Given \(\overset { \rightarrow }{ a } +\overset { \rightarrow }{ b } +\overset { \rightarrow }{ c } =0\)
\(\overset { \rightarrow }{ a } +\overset { \rightarrow }{ b } \)= -\(\overset { \rightarrow }{ c } \) ... (1)
Taking cross product with \(\overset { \rightarrow }{ a } \) both sides, we get 
\(\overset { \rightarrow }{ a } \left( \overset { \rightarrow }{ a } +\overset { \rightarrow }{ b } \right) =-\left( \overset { \rightarrow }{ a } \times \overset { \rightarrow }{ c } \right) \)
\(\overset { \rightarrow }{ a } \times \overset { \rightarrow }{ a } +\overset { \rightarrow }{ a } +\overset { \rightarrow }{ b } =\overset { \rightarrow }{ c } \times \overset { \rightarrow }{ a } \)
\(\left[ \because -\left( \overset { \rightarrow }{ a } \times \overset { \rightarrow }{ c } \right) =\overset { \rightarrow }{ c } \times \overset { \rightarrow }{ a } \right] \)
\(\Rightarrow \overset { \rightarrow }{ a } \times \overset { \rightarrow }{ b } =\overset { \rightarrow }{ c } \times \overset { \rightarrow }{ a } \left( \because \overset { \rightarrow }{ a } \times \overset { \rightarrow }{ a } =\overset { \rightarrow }{ 0 } \right) \)
Taking cross product with \(\overset { \rightarrow }{ b } \) both sides, we get
\(\overset { \rightarrow }{ b } \times \left( \overset { \rightarrow }{ a } +\overset { \rightarrow }{ b } \right) =\overset { \rightarrow }{ b } \times \left( -\overset { \rightarrow }{ c } \right) \)
\(\overset { \rightarrow }{ b } \times \overset { \rightarrow }{ a } +\overset { \rightarrow }{ b } \times \overset { \rightarrow }{ b } =-\overset { \rightarrow }{ b } \times \overset { \rightarrow }{ c } \)
\(-\overset { \rightarrow }{ a } \times \overset { \rightarrow }{ b } =-\overset { \rightarrow }{ b } \times \overset { \rightarrow }{ c } \)
\(\overset { \rightarrow }{ a } \times \overset { \rightarrow }{ b } =\overset { \rightarrow }{ b } \times \overset { \rightarrow }{ c } \)
From (2) and (3) we get
\(\overset { \rightarrow }{ a } \times \overset { \rightarrow }{ b } =\overset { \rightarrow }{ b } \times \overset { \rightarrow }{ c } =\overset { \rightarrow }{ c } \times \overset { \rightarrow }{ a } \)

Let \(\overset { \rightarrow }{ r } =x\overset { \wedge }{ i } +y\overset { \wedge }{ j } +z\overset { \wedge }{ k } \)
\(\therefore x\overset { \wedge }{ i } +y\overset { \wedge }{ j } +z\overset { \wedge }{ k } =\left( s-2t \right) \overset { \wedge }{ i } +\left( 3-t \right) \overset { \wedge }{ j } +\left( 2s+t \right) \overset { \wedge }{ k } \)
Equating the co-efficients of like components both sides,
We get, x = s - 2t
y = 3 - t
z = 2s + t
Eliminating x and t using determinates we get
\(\left| \begin{matrix} x \\ y-3 \\ z \end{matrix}\begin{matrix} 1 \\ 0 \\ 2 \end{matrix}\begin{matrix} -2 \\ -1 \\ 1 \end{matrix} \right| =0\)
⇒ x (0+2) -1(y - 3 + z) -2 (2y - 6 - 0) = 0
⇒ 2x - y + 3 - z- 4y + 12 = 0
⇒ 2x - 5y - z + 15 = 0

Given rank of \(\left[ \begin{matrix} \lambda & -1 & 0 \\ 0 & \lambda & -1 \\ -1 & 0 & \lambda \end{matrix} \right] \) is 2
⇒ The value of the third order determinant is zero
⇒ \(\left[ \begin{matrix} \lambda & -1 & 0 \\ 0 & \lambda & -1 \\ -1 & 0 & \lambda \end{matrix} \right] \)=0
⇒ λ\(\left| \begin{matrix} \lambda & -1 \\ 0 & \lambda \end{matrix} \right| +1\left| \begin{matrix} 0 & -1 \\ -1 & \lambda \end{matrix} \right| \)+0 = 0
⇒ λ(λ² - 0) + 1(0 - 1) = 0
⇒ λ³ - 1 = 0 ⇒ λ³ = 1 ⇒ λ = 1
∴ λ = 1

|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

Δ = \(\left| \begin{matrix} 2 & 3 \\ 1 & 6 \end{matrix} \right| \) = 12 - 3 = 9 ≠ 0
Δ₁ = \(\left| \begin{matrix} 10 & 3 \\ 4 & 6 \end{matrix} \right| \) = 60 - 12 = 48
Δ₂ = \(\left| \begin{matrix} 2 & 10 \\ 1 & 4 \end{matrix} \right| \) = 8 - 10 = -2
∴ x = \(\frac { { \triangle }_{ 1 } }{ \triangle } =\frac { 48 }{ 9 } =\frac { 16 }{ 3 } \)
y = \(\frac { { \triangle }_{ 2 } }{ \triangle } =\frac { -2 }{ 9 } \)
∴ Solution set is \(\left\{ \frac { 16 }{ 3 } ,\frac { -2 }{ 9 } \right\} \).