y = log(x²+1)+c

Given R (x, y) = 80 x + 90 y + 0.04xy - 0.05 x² + 0.05 y² and
(x,y) = 8x + 6y + 2000
Profit function P (x,y) = Revenue - cost
P (x,y) = R (x,y) - C (x,y)
= -80 x + 90 y + 0.04xy - 0.05 x² - 0.05y² - 8x - 6y - 2000
P (x, y) = 72x + 84y + 0.04 xy - 0.05 x² - 0.05y² - 2000
(ii) \(\frac { { \partial P } }{ \partial { x } } \) = 72 + 0 + 0.04y - 0.05(2x) - 0 - 0
= 72 + 0.04y- 0.1x
∴ \(\frac { { \partial P } }{ \partial { x } } \) (1200, (1800)
= 72+ 0.04 (1800) - 0.1(1200)
= 72 + 72 - 120 = 24 .......(1)
\(\frac { \partial v }{ \partial y} \) = 0 + 84+ 0.4x-0-0.5(2y) - 0
= 84 + 0.04x - 0.1y
= 84 + 0.04 (1200) - 0.1(1800)
∴ \(\frac { \partial v }{ \partial y} \)_(1200,1800) = 84 + 48 - 180 = - 48 .......(2)
From (1) and (2), keeping y constant and 4 increasing x then increases profit.

i²⁵ = (i⁴)⁶ \(\times\) i¹ = i⁶ \(\times\) i = i
∴ |i²⁵| = |i| = 1

Since the foci are one the x-axis, the equation of the hyperbola is \(\frac { { x }^{ 2 } }{ { a }^{ 2 } } +\frac { { y }^{ 2 } }{ { b }^{ 2 } } =1\)
Given conjugate axis = \(\frac34\) (transverse axis)
⇒ 2b = \(\frac34\) (2a) ⇒ b = \(\frac{3a}4\) ⇒ b² = \(\frac { { 9a }^{ 2 } }{ 16 } \)
∴ e = \(\sqrt { 1-\frac { { b }^{ 2 } }{ { a }^{ 2 } } } =\sqrt { 1-\frac { 9{ a }^{ 2 } }{ { 16a }^{ 2 } } } =\sqrt { 1+\frac { 9 }{ 16 } } =\sqrt { \frac { 25 }{ 16 } } =\frac { 5 }{ 4 } \)
∴ e = \(\frac { 5 }{ 4 } \)

\(-\sqrt { 3 } +i\)

Modulus = 2 and
\(a={ tan }^{ -1 }\left| \frac { y }{ x } \right| ={ tan }^{ -1 }\frac { 1 }{ \sqrt { 3 } } =\frac { \pi }{ 6 } \)
Since the complex number  \(-\sqrt { 3 } +i\) lies in the second quadrant has the principal value
\(\theta =\pi -\alpha =\pi -\frac { \pi }{ 6 } =\frac { 5\pi }{ 6 } \)
Therefore the modulus and principal argument of \(-\sqrt { 3 } +i\) are 2 and \(\frac { 5\pi }{ 6 } \) respectively.

(1-i)¹⁰
Let z = (1- i)¹⁰
|z| = |1- i|¹⁰ = \(\left[ \sqrt { { 1 }^{ 2 }+(-1)^{ 2 } } \right] ^{ 10 }\)
= \(\left[ \sqrt { 2 } \right] ^{ 10 }\) = 2^{1/2 x 10} = 2⁵ = 32

\(\frac { 2i }{ 3+4i } \)
Let z = \(\frac { 2i }{ 3+4i } \)
|z| = \(\left| \frac { 2i }{ 3+4i } \right| =\frac { |2i| }{ |3+4i| } =\frac { \sqrt { 2^{ 2 } } }{ \sqrt { { 3 }^{ 2 }+{ 4 }^{ 2 } } } =\frac { 2 }{ \sqrt { 9+16 } } \)
= \(\frac { 2 }{ \sqrt { 25 } } =\frac { 2 }{ 5 } \).

Represent z, iz and z + iz in the Argand diagram.
z = 2 + 3i can be represented as (2,3)
iz = i(2 + 3i)
= 2i + 3i²
= 2i-3
= -3 + 2i can be represented as (-3, 2)
z + iz = 2 + 3i - 3 + 2i = -1 + 5i can be represented as (-1, 5) in the argand diagram.

Given h (x, y, z) = x sin (xy) + z²x
\(\frac { \partial h }{ \partial x } =x.cos(x,y).\frac { \partial }{ \partial x } (xy)+sin(xy)(1)+{ z }^{ 2 }(1)\)
= x cos (xy) (y)(1) + sin (xy) + z²
= xy cos (xy) + sin (xy) + z²
\(\therefore { \left( \frac { \partial h }{ \partial x } \right) }_{ \left( 2,\frac { \pi }{ 4 } ,1 \right) }=2\left( \frac { \pi }{ 4 } \right) cos\left( 2\frac { \pi }{ 4 } \right) +sin\left( 2\frac { \pi }{ 4 } (1) \right) +{ 1 }^{ 2 }\)
\(=\frac { \pi }{ 2 } cos\left( \frac { \pi }{ 2 } \right) +sin\left( \frac { \pi }{ 2 } \right) +1\)
\(=\frac { \pi }{ 2 } (0)+1+1=2\)
\(\frac { \partial h }{ \partial y } =x.cos(xy).\frac { \partial h }{ \partial y } (xy)+0\)
= x cos (xy) x(1)
= x² cos (xy)
\(\therefore { \left( \frac { \partial h }{ \partial x } \right) }_{ \left( 2,\frac { \pi }{ 4 } ,1 \right) }={ 2 }^{ 2 }cos\left( 2\frac { \pi }{ 4 } \right) \)
= \(4cos\left( \frac { \pi }{ 2 } \right) \)
= 4(0) = (0)
\(\left( \frac { \partial h }{ \partial z } \right) =0+x(2z)=2xz\)
\(\therefore { \left( \frac { \partial h }{ \partial x } \right) }_{ \left( 2,\frac { \pi }{ 4 } ,1 \right) }=2(2)(1)=4\)

I \(=\int _{ 0 }^{ 1 }{ \sqrt { \frac { (1-x)(1-x) }{ (1+x)(1-x) } } } dx\)
\(=\int _{ 0 }^{ 1 }{ \sqrt { \frac { (1-x) 2}{ (1-x)^2 } } } dx\)
\(=\int _{ 0 }^{ 1 }{ \frac { 1-x }{ \sqrt { 1-{ x }^{ 2 } } } } dx\)
Put x = sin u, dx = cosu.du, 0u = sin^-1(x)
For x = 0, u = 0 and x = 1, u = \(\frac{\pi}{2}\)
\(=\int _{ 0 }^{ 1 }{ \frac { 1-sin\ u }{ \sqrt { 1-{ sin }^{ 2 }u } } } \times cos u.du\)
\(=\int _{ 0 }^{ 1 } \frac { 1-sin\ u }{cos u} \times cos u.du\)
= \( [u+cos u]^\frac{\pi }{2}_0\)= (\(\frac{\pi}{2}\)+0)-(0+1) = \(\frac{\pi}{2}\) -1
\(=\frac { \pi }{ 2 } -0-1=\frac { \pi }{ 2 } -1\)

log₁₀e = 0.4343 to find log₁₀g 1003
f(1000) = log₁₀1000 = log₁₀10³ = 3log₁₀10³ = 3 log₁₀10
= 3(1) = 3
f'(x) = \(\frac1x\). log10^e
f'(1000) = \(\frac{1}{1000}\)(0.4343)
∴ L(x) = f(x₀) f'(x₀) (x - x₀)
= 3 + \(\frac{1}{1000}\) (0.4343) (3)
= 3 + \(\frac{1.3029}{1000}\)
= 3 + 0.0013029
log₁₀1003 = 3.0013029

f(x) = x² -12x + 10; [1, 2]
Given f(x) = x² -12x + 10 ; [1, 2]
f'(x) = 2x - 12
f'(x) = 0
\(\Rightarrow\) 2x-12 = 0
\(\Rightarrow\) 2x = 12
\(\Rightarrow\) x = 6
\(\therefore\) The critical number is 6
Evaluating f (x) at the end points x = 1,
x = 2 and at the critical number x = 6 we get
f(1) = 1²-12(1)+10 = -1
f(2) = 2²-12(2)+10 = -10
Absolute maximum f(1) = -1
Absolute minimum f(2) = -10

Given equation of the hyperbola is
3x² - 6y² = -18; \(\div \)by (-18) we get \(\frac{y^2}{3}- \frac{x^2}{6}\) = 1
The transverse axis is long y-axis.
Here a² = 3, b² = 6
Length of transverse axis is 2a = 2\(\sqrt { 3 } \)
Length of conjugate axis is 2b = 2\(\sqrt { 3 } \)
\(e=\sqrt { 1+\frac { { b }^{ 2 } }{ { a }^{ 2 } } } =\sqrt { 1+\frac { 6 }{ 3 } } =\sqrt { 1+2 } =\sqrt { 3 } \)

z = 2+3t
-iz = -i(2+3i)
= 2i -3i² = -2i+3
= 3 - 2i
z - iz = 2+ 3i-3+2i
= -1+5i

Equation of the given plane is 2x - 3y + 5z +7 = 0
Equation of the plane parallel to the given plane
is 2x - 3y + 5z + k = 0  ..(1)
Since this plane passes through the point (3, 4, -1). we get
2(3) - 3(4) + 5(-1) + k = 0
\(\Rightarrow\) 6 - 12 - 5 + k = 0
\(\Rightarrow\) 11 +k = 0  

k = 11
\(\therefore\) (1) becomes, 2x - 3y + 5z + 11 = 0 which is the equation of the required plane. Distance between two parallel planes.
= \(\frac { \left| { d }_{ 1 }-{ d }_{ 2 } \right| }{ \sqrt { { a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 } } } \)
\(d=\frac { \left| 7-11 \right| }{ \sqrt { { 2 }^{ 2 }+\left( -3 \right) ^{ 2 }+{ 5 }^{ 2 } } } \)
= \(\cfrac { \left| -4 \right| }{ \sqrt { 4+9+25 } } \)
= \(\cfrac { 4 }{ \sqrt { 38 } } \) units.

\(\vec { r } .(2\hat { i } -7\hat { j } +4\hat { k } )=3\) and 3x - 5y + 11 = 0,
The vector equation of a plane passing through the line of intersection of the planes
\(\vec { r } .\vec { { n }_{ 1 } } ={ d }_{ 1 }\) and \(\vec { r } .\vec { n_{ 2 } } ={ d }_{ 2 }\) is given by
\(\left( \vec { r } .\vec { { n }_{ 1 } } -{ d }_{ 1 } \right) +\lambda \left( \vec { r } .{ \vec { n } }_{ 2 }-{ d }_{ 2 } \right) =0\) ...(1)
put \(\vec { r } =x\hat { i } +y\hat { j } +z\hat { k } \)
\({ \vec { n } }_{ 1 }=2\hat { i } -7\hat { j } +4\hat { k } ,{ \vec { n } }_{ 2 }=3\hat { i } -5\hat { j } +4\hat { k } \)
\({ d }_{ 1 }=+3,{ d }_{ 2 }=-11\) in (1) we get
\(\left[ (x\hat { i } +y\hat { j } +z\hat { k } ).\left( 2\hat { i } -7\hat { j } +4\hat { k } -3 \right) \right] +\lambda \left[ \left( x\hat { i } +y\hat { j } +z\hat { k } \right) .\left( 3\hat { i } -5y+4z+11 \right) \right] =0\)
\(\Rightarrow \left( 2x-7y+4z-3 \right) +\lambda \left( 3x-5y+4z+11 \right) =0\)....2
Since the plane passes through the point (-2, 1, 3) we get,
\(\left[ 2(-2)-7(1)+4(3)-3 \right] +\lambda \left[ -6-5+12+11 \right] =0\)
\(\Rightarrow (-4-7+12-3)+\lambda (12)=0\)
\(\Rightarrow -2+12\lambda =0\Rightarrow 12\lambda =2\Rightarrow \lambda =\frac { 1 }{ 6 } \)
Substituting \(\lambda =\frac { 1 }{ 6 } \) in (1) we get,
\(\left( 2x-7y+4z-3 \right) +\frac { 1 }{ 6 } \left( 3x-5y+4z+11 \right) =0\)
Multiplying by 6, we get,
\(\Rightarrow 12x-42y+24z-18+3x-5y+4z+11=0\)
\(\Rightarrow 15x-47y+28z-7=0\) which is the required equation of the plane.

Given \((\sqrt { 5 } -\sqrt { 3 } )\) is a root
Another root \(\Rightarrow \sqrt { 5 } +\sqrt { 3 } \) 
∴ Sum of the roots \(=\sqrt { 5 } -\sqrt { 3 } +\sqrt { 5 } +\sqrt { 3 } =2\sqrt { 5 } \)
Product of the roots 
\(=(\sqrt { 5 } -\sqrt { 3 } )(\sqrt { 5 } +\sqrt { 3 } )\)
\(=(\sqrt { 5 } )^{ 2 }-{ (\sqrt { 3 } ) }^{ 2 }=5-3=2\)
∴ One of the factor is x² -x (sum of the roots) + product of the roots 
\(\Rightarrow { x }^{ 2 }-2x\sqrt { 5 } +2\)
The other factor also will be \({ x }^{ 2 }-2x\sqrt { 5 } +2\)
\(({ x }^{ 2 }-2x\sqrt { 5 } +2)({ x }^{ 2 }+2x\sqrt { 5 } +2)=0\)
\(\Rightarrow ({ x }^{ 2 }+2-2\sqrt { 5 } x)({ x }^{ 2 }+2+2\sqrt { 5 } x)=0\)
\(\Rightarrow \left( { x }^{ 2 }+2 \right) ^{ 2 }-{ (2\sqrt { 5 } x) }^{ 2 }=0\)
\([\because (a+b)(a-b)={ a }^{ 2 }-{ b }^{ 2 }]\)
\(\Rightarrow { x }^{ 4 }+{ 4x }^{ 2 }+4{ (5)x }^{ 2 }=0\)
\(\Rightarrow { x }^{ 4 }+{ 4x }^{ 2 }+4-{ 20x }^{ 2 }=0\)
\(\Rightarrow { x }^{ 4 }-{ 16x }^{ 2 }+4=0\) is a rational co-efficient polynomial equation.

Given 2i + 3 is a root
∴ Its conjugate 3 - 2i is also a root of the polynomial equation.
∴ Sum of the roots 3 + 2i + 3 - 2i = 6
Product of the roots = (3 + 2i) (3 - 2i)
= 3² + 2² = 9+ 4 = 13
∴ The polynomial equation of minimum degree with rational co-efficients is
x² - x (sum of the roots) + product of the roots = 0
⇒x² -x (6) + 13 = 0
⇒ x²- 6x + 13 = 0

Since 2-\(\sqrt{3}\)i is a root and the coefficients are rational numbers, 2+\(\sqrt{3}\)i  is also a root. A required polynomial equation is given by
x² −(Sum of the roots) x + Product of the roots = 0
and hence
x²- 4x +1 = 0 is a required equation.

Given polynomial equation is 
2x⁴+ 5x³−7x² + 8 = 0
Here a = 2, b = 5, c = -7, d = 0, e = 8
By Vieta's formula, 
\(\alpha +\beta +\gamma +\delta =\frac { -b }{ a } =\frac { -5 }{ 2 } \)
\(\alpha \beta +\alpha \gamma +\alpha \delta +\beta \gamma +\beta \delta +\gamma \delta =\frac { c }{ a } =\frac { -7 }{ 2 } \)
\(\alpha \beta \gamma +\alpha \beta \delta +\alpha \gamma \delta +\beta \gamma \delta =\frac { -d }{ a } =0\)
\(\alpha \beta \gamma \delta =\frac { e }{ a } =\frac { 8 }{ 2 } =4\)
Given roots of quadratic equation are 
∝ + β + ૪ + \(\delta \) and ∝β૪\(\delta \)
∴ sum of the roots = (∝+β+૪+\(\delta \)) (∝β૪\(\delta \))
\(=\left( \frac { -5 }{ 2 } +4 \right) =\frac { -5+8 }{ 2 } =\frac { 3 }{ 2 } \)
\(=\left( \alpha +\beta +\gamma +\delta \right) (\alpha \beta \gamma \delta )\)
\(=\left( \frac { -5 }{ 2 } \right) (4)=\frac { -20 }{ 2 } =-10\)
∴ The required quadratic equation is x²-x 
(sum of the roots) + product of the roots = 0
\(\Rightarrow { x }^{ 2 }-x\left( \frac { 3 }{ 2 } \right) -10=0\)
\(\Rightarrow { 2x }^{ 2 }-3x-20=0\)

\(log|x|=cos\left( \frac { y }{ x } \right) ,x\neq 0\)

\(y=\frac { 2x }{ { 2x }^{ 2-1 } } ,x\neq \pm \frac { 1 }{ \sqrt { 2 } } \)

y=log(x+1)+log(x+1)-x+5

3πa²

2 : 1

Given u (x, y, z) = xyz; x = e^-t y = e^-t cas t; z = sin t
\(\frac { \partial u }{ \partial x } \) = yz; \(\frac { \partial u }{ \partial y } \) = xz; \(\frac { \partial u }{ \partial z } \) = xy
⇒ \(\frac { \partial u }{ \partial x } \) = e^t cas t sin t
\(\frac { \partial u }{ \partial y } \) = e^t sin t
\(\frac { \partial u }{ \partial z } \) = e^-2t cas t and
\(\frac{dx}{dt}=-e^{-t}\)
⇒ \(\frac{dy}{dt}\) = e^-t (- sin t) - cas t e^-t
⇒ \(\frac{dz}{dt}\) = cos t
∴ By chain rule;
\(\frac { dw }{ dt } =\frac { \partial w }{ \partial x } .\frac { dx }{ dt } +\frac { \partial w }{ \partial y } .\frac { dy }{ dt } +\frac { \partial w }{ \partial z } .\frac { dz }{ dt } \)
∴ \(\frac { dw }{ dt } =\frac { \partial u }{ \partial x } .\frac { dx }{ dt } +\frac { \partial u }{ \partial y } .\frac { dy }{ dt } +\frac { \partial u }{ \partial z } .\frac { dz }{ dt } \)
= e^-1 cos t sin t (-e^-t) + e^-t sin t (-e^-t) + e^-t sin t (-e^-t sin t - e^-t cos t)+ (-e-^2t) cos t(cos t)
= -e^-2t [(sin t cos t + sin² t + sin t cos t - cos² t]
= -e^-2t [2 sin t cos t - (cos² t - sin² t)]
\(\frac { du }{ dt } \) = - e^-2t, [sin 2t - cos2t]
[∵ cas 2 t = cos² t - sin² t and sin 2t = 2 sin t cos t]

Recall that volume of a right circular cylinder is given by V = \(\pi \)r²h where r is the radius and h is the height. So we have V (r) = \(\pi \)r²h = 20\(\pi \)r²
V (10.1) −V (10)≈ \(\frac { dV }{ dr } { { | }_{ r=10 } }\) (10.1 10) = 20\(\pi \)²(10(0.1))
Thus the estimate for the change in the volume is 40 \(\pi \) cm³
Exact calculation of the volume change gives
V (10.1) −V (10) = 2040.2\(\pi \) -2000\(\pi \) = 40.2\(\pi \) cm³.
So relative error = \(\frac { 40.2\pi -40\pi }{ 40.2\pi } \) = \(\frac { 1 }{ 201 } \) = 0.00497 and hence
the percentage error = relative error x 100 = \(\frac { 1 }{ 201 } \)x100 = 0.497% 

The domain of sin^-1x is [-1, 1]
\(\therefore\) f(x) = sin^-1(2x - 3) is defined for all x, satisfying
\(-1\le 2x-3\le 1\)
\(\Rightarrow 3-1\le 2x\le 1+3\)
\(\Rightarrow 2\le 2x\le 4\Rightarrow 1\le x\le 2\Rightarrow x\epsilon \left[ 1,2 \right] \)
\(\therefore\) Domain of f(x) = sin^-1(2x - 3) is [1, 2].
(ii) The domain of f(x) is [-1, 1] and that of cosx is R
\(\therefore\) Domain of f(x) = sin^-1x + cos x is
\(\left[ -1,1 \right] \cap R=\left[ -1,1 \right] \)

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.

x⁸- 3x + 1 = 0
Here a_n = 1, a_o = 1
If \(\frac{p}{q}\) is a root of the polynomial, then as 
(p, q) = 1p is a factor of a_o = 1 and q is a factor of a_n = 1
Since 1 has no factors, the given equation has no rational roots.

We know that the vector equation of a plane passing through the line of intersection of the planes
\(\vec { r } .\vec { { n }_{ 1 } } ={ d }_{ 1 }\) and \(\vec { r } .\vec { { n }_{ 2 } } ={ d }_{ 2 }\) is given by \((\vec { r } .\vec { { n }_{ 1 } } -{ d }_{ 1 })+\lambda (\vec { r } .\vec { { n }_{ 2 } } -{ d }_{ 2 })=0\)
Substituting \(\vec { r } =x\hat { i } +y\hat { j } +z\hat { k } ,\vec { { n }_{ 1 } } =\hat { i } +\hat { j } +\hat { k } ,\vec { { n }_{ 2 } } =2\hat { i } -3\hat { j } +5\hat { k } \), \({ d }_{ 1 }=1,{ d }_{ 2 }=-2\) in the above equation, we get
(x + y + z + 1) + \(\lambda \) (2x - 3y + 5z - 2) = 0
Since this plane passes through the point (−1, 2,1) , we get λ = \(\frac{3}{5}\), and hence the required equation
of the plane is 11x−4y+20z=1 .