Given, marginal cost, C' (x) \(\frac{x}{3000}+2.50\)
\(\int { C'(x) } =\int { \left( \frac { x }{ 300 } +2.50 \right) dx } \)
\(C(x)=\frac { { x }^{ 2 } }{ 6000 } +2.50x+k\)
When x = 0, c = 0 ⇒ k = 0
∴ c(x) = \(\frac{x^2}{6000}+2.50x\)
When x = 3000
Cost of production
\(=\frac { { (3000) }^{ 2 } }{ 6000 } +2.50(3000)\)

= 1500 + 7500
= Rs. 9000

Given non-homogeneous equations are
2x + 3y = 5  6X + 5y = 11
\(\Delta =\left| \begin{matrix} 2 & 3 \\ 6 & 5 \end{matrix} \right| =10-18=-8\)
Since \(\Delta \neq 0\) Cramer's rule can be applied and the system is consistent with unique solution
\(\Delta x=\left| \begin{matrix} 5 & 3 \\ 11 & 5 \end{matrix} \right| =25-33=-8\)
\(\Delta y=\left| \begin{matrix} 2 & 5 \\ 6 & 11 \end{matrix} \right| =22-30=-8\)
\(\therefore x=\cfrac { \Delta x }{ \Delta } =\cfrac { -8 }{ -8 } =1\)
\(y=\cfrac { \Delta y }{ \Delta } =\cfrac { -8 }{ -8 } =1\)
\(\therefore \) Solution set is {1, 1}

Given
\(MR=10+3x-{ x }^{ 2 }\)
\(\frac { dR }{ dx } =10+3x-{ x }^{ 2 }\)
\(\Rightarrow dR=(10+3x-{ x }^{ 2 })dx\)
\(\Rightarrow \int { dR } =\int { (10+3x-{ x }^{ 2 }) } dx\)
\(\Rightarrow R=10x+\frac { { 3x }^{ 2 } }{ 2 } -\frac { { x }^{ 3 } }{ 3 } +k\)
When x = 0, R = 0 ⇒ k = 0
\(\Rightarrow R=10x+\frac { { 3x }^{ 2 } }{ 2 } -\frac { { x }^{ 2 } }{ 3 } \)
Demand function\(P=\frac { R }{ x } =\frac { 10x+\frac { { 3x }^{ 2 } }{ 2 } -\frac { { x }^{ 3 } }{ 3 } }{ x } \)
P = 10 + \(\frac { { 3x }^{ 2 } }{ 2 } -\frac { { x }^{ 3 } }{ 3 } \)

\(\int { \frac { { x }^{ 3 } }{ x+2 } } dx\)
\(=\int { \left( { x }^{ 2 }-2x+4-\frac { 8 }{ x+2 } \right) } dx\)

\(=\frac { { x }^{ 3 } }{ 3 } -{ x }^{ 2 }+4x-8\log\left| x+2 \right| +c\)

Let A =\(\left( \begin{matrix} 0 & -1 & 5 \\ 2 & 4 & -6 \\ 1 & 1 & 5 \end{matrix} \right) \)
Order of A is 3 \(\times\) 3  
∴ \(\rho \)(A)\(\le \)3
Consider the third order minor \(\left| \begin{matrix} 0 & -1 & 5 \\ 2 & 4 & -6 \\ 1 & 1 & 5 \end{matrix} \right| =6\neq 0\)
There is a minor of order 3, which is not zero 
\(\therefore \rho (A)=3\)

2x-y = 4

x	0	2
y	-4	0

y² = 4x and 2x = 4 + y ⇒ 4x = 8 + 2y
y² = 8+2y ⇒ y²-2y-8 = 0
(y-4)(y+2) = 0
y = -2, 4.
Required area \(=\int _{ -2 }^{ 4 }{ ({ x }_{ 1 }-{ x }_{ 2 })dy } \)
\(=\int _{ -2 }^{ 4 }{ \left( \frac { y+4 }{ 2 } -\frac { { y }^{ 2 } }{ 4 } \right) dy } \)
Where x₁ is the line  x₂ is the parabola
\(=\int _{ -2 }^{ 4 }{ \frac { y+4 }{ 2 } dy } -\frac { 1 }{ 4 } \int _{ -2 }^{ 4 }{ { y }^{ 2 }dy } \)
\(=\frac { 1 }{ 2 } { \left[ \frac { { y }^{ 2 } }{ 2 } +4y \right] }_{ -2 }^{ 4 }-\frac { 1 }{ 4 } { \left[ \frac { { y }^{ 3 } }{ 3 } \right] }_{ -2 }^{ 4 }\\ \)
\(=\frac { 1 }{ 2 } \left[ \left( \frac { 16 }{ 2 } +16 \right) -\left( \frac { 4 }{ 2 } -8 \right) \right] -\frac { 1 }{ 4 } \left[ \frac { 64 }{ 3 } +\frac { 8 }{ 3 } \right] \)
\(=\frac { 1 }{ 2 } \left[ 24+6 \right] -\frac { 1 }{ 4 } \left[ \frac { 64+8 }{ 3 } \right] \)
\(=\frac { 1 }{ 2 } (30)-\frac { 1 }{ 4 } \left( \frac { 72 }{ 3 } \right) \)
\(=15-\frac { 1 }{ 4 } (24)=15-6\)
= 9 sq.units.

Given, Marginal revenue R'(x) = 10 + e^−0.05x
Total revenue from sale of 100 units is
\(R=\int _{ 0 }^{ 100 }{ (10+{ e }^{ -0.05x }) } dx\)
= \({ \left[ 10x+\frac { { e }^{ -0.05x } }{ -0.05 } \right] }_{ 0 }^{ 100 }\)
= \(\left( 1000-\frac { { e }^{ 5 } }{ 0.05 } \right) -\left( 0-\frac { 100 }{ 5 } \right) \)
= 1000 + 20 − (20 × 0.0067)
= 1019.87
Total revenue = 1019.87 × 1000
= Rs. 10,19,870

The equation y² = 16x represents a parabola (Open rightward)
Required Area = 2\(\int _{ a }^{ b }{ y } dx\)
\(=2\int _{ 0 }^{ 4 }{ \sqrt { 16x } } \ dx\)
\(=8\int _{ 0 }^{ 4 }{ { x }^{ \frac { 1 }{ 2 } } } dx=8{ \left[ { \frac { { x }^{ { \frac { 3 }{ 2 } } } }{ \frac { 3 }{ 2 } } } \right] }_{ 0 }^{ 4 }=\frac { 16 }{ 3 } \left( { \left( 4 \right) }^{ \frac { 3 }{ 2 } } \right) =\frac { 128 }{ 3 } \) sq.units

Here \(\triangle =\left| \begin{matrix} 1 & 1 & 1 \\ 2 & -1 & 3 \\ 3 & 2 & -1 \end{matrix} \right| =13\neq 0\)
\(\therefore \) We can apply Cramer’s Rule and the system is consistent and it has unique solution.
\({ \triangle }_{ x }=\left| \begin{matrix} 1 & 1 & 1 \\ 2 & -1 & 3 \\ 3 & 2 & -1 \end{matrix} \right| =-13\)
\( { \triangle }_{ y }=\left| \begin{matrix} 1 & 4 & 1 \\ 2 & 1 & 3 \\ 3 & 1 & -1 \end{matrix} \right| =39\)
\( { \triangle }_{ z }=\left| \begin{matrix} 1 & 1 & 4 \\ 2 & -1 & 1 \\ 3 & 2 & 1 \end{matrix} \right| =26\)
\(\therefore \) By Cramer’s rule
\(x=\frac { { \triangle }x }{ { \triangle } } =\frac { -13 }{ 13 } =-1\)
\( y=\frac { { \triangle }y }{ { \triangle } } =\frac { 39 }{ 13 } =3\)
\( z=\frac { { \triangle }z }{ { \triangle } } =\frac { 26 }{ 13 } =2\)
\(\therefore \) The solution is (x, y, z) = (−1, 3, 2)

Given non-homogeneous equations are
x + yz = 9
2x + 5y + 7z = 52
2x - y - z = -6
The matrix equation corresponding to the given system is
\(\left( \begin{matrix} 1 & 1 & 1 \\ 2 & 5 & 7 \\ 2 & -1 & -1 \end{matrix} \right) \left( \begin{matrix} x \\ y \\ z \end{matrix} \right) =\left( \begin{matrix} 9 \\ 52 \\ -6 \end{matrix} \right) \)
Augmented matrix Elementary Transformation  \(\left( \begin{matrix} 1 & 1 & 1 \\ 2 & 5 & 7 \\ 2 & -1 & -1 \end{matrix}\begin{matrix} 9 \\ 52 \\ -6 \end{matrix} \right) \)   \(\sim \left( \begin{matrix} 1 & 1 \\ 0 & 3 \\ 2 & -1 \end{matrix}\begin{matrix} 1 & 9 \\ 5 & 34 \\ -1 & -6 \end{matrix} \right) \) \({ R }_{ 2 }\rightarrow { R }_{ 2 }-2{ R }_{ 1 }\) \(\sim \left( \begin{matrix} 1 & 1 \\ 0 & 3 \\ 0 & -3 \end{matrix}\begin{matrix} 1 & 9 \\ 5 & 34 \\ -3 & -24 \end{matrix} \right) \) \({ R }_{ 3 }\rightarrow { R }_{ 3 }-{ 2R }_{ 1 }\) \(\sim \left( \begin{matrix} 1 & 1 \\ 0 & 3 \\ 0 & 0 \end{matrix}\begin{matrix} 1 & 9 \\ 5 & 34 \\ 2 & 10 \end{matrix} \right) \) \({ R }_{ 3 }\rightarrow { R }_{ 3 }+{ R }_{ 2 }\) The last equivalent matrix is in echelon form andit has three non-zero rows.
\(\therefore \rho (A)=3\)
\(\rho \left( \left[ A,B \right] \right) =3\) Number of unknowns
\(\therefore\) The given system is consistent and has unique solution.
To find the solution, let us rewrite the above echelon from into the matrix form
\(\left( \begin{matrix} 1 & 1 & 1 \\ 0 & 3 & 5 \\ 0 & 0 & 2 \end{matrix} \right) \left( \begin{matrix} x \\ y \\ z \end{matrix} \right) =\left( \begin{matrix} 9 \\ 34 \\ 16 \end{matrix} \right) \)
\(\Rightarrow \) x + y + z = 9 .. (1)
3y + 5z = 34  ..(2)
2z = 16   ..(3)
From (3), \(2z=16\Rightarrow z=\cfrac { 16 }{ 2 } =8\)
Substituting Z = 8 in (2) we get,
3y + 5 (8) = 34
\(\Rightarrow\) 3y + 40 = 34
\(\Rightarrow\) 3y = 34 - 40
\(\Rightarrow\) 3y = -6
y = -2
Substituting y = -2 and z = 8 in (1) we
\(\Rightarrow\) x -2 + 8 = 9
\(\Rightarrow\) x + 6 = 9
\(\Rightarrow\)x = 9-6
\(\Rightarrow\)x = 3
\(\therefore\) Solution set is {3, -2, 8}

 \(\int { \frac { 2+3cosx }{ { sin }^{ 2 }x } } dx\) = \(\int { \frac { 2 }{ { sin }^{ 2 }x } } dx\) + \(\int { \frac { 3cosx }{ { sin }^{ 2 }x } } dx\)
= 2 \(\int { \frac { 1 }{ { sin }^{ 2 }x } } dx\) + 3 \(\int { \frac { cosx }{ { sin }x } .\frac { 1 }{ sinx } } dx\)
= 2 ∫ cosec² x dx + 3 ∫ cot x cosec x dx
= -2 cot x - 3 cosec x + c

ഽ\(\frac { dx }{ { 4x }^{ 2 }-1 } \) = ഽ\(\frac { dx }{ 4\left( { x }^{ 2 }-\frac { 1 }{ 4 } \right) } \)
= \(\frac { 1 }{ 4 } \)ഽ\(\frac { dx }{ { x }^{ 2 }-{ \left( \frac { 1 }{ 2 } \right) }^{ 2 } } \)
= \(\frac { 1 }{ 4 } \left[ \frac { 1 }{ 2\left( \frac { 1 }{ 2 } \right) } \log\left| \frac { x-\frac { 1 }{ 2 } }{ x-\frac { 1 }{ 2 } } \right| \right] +c\)
= \(\frac { 1 }{ 4 } \log\left| \frac { 2x-1 }{ 2x+1 } \right| \) + c

Given A =\(\left( \begin{matrix} 1 & -3 \\ 9 & 1 \end{matrix}\begin{matrix} 4 & 7 \\ 2 & 0 \end{matrix} \right) \)
\(\left( \begin{matrix} 1 & -3 \\ 0 & 28 \end{matrix}\begin{matrix} 4 & 0 \\ -34 & -63 \end{matrix} \right) { R }_{ 2 }\rightarrow { R }_{ 2 }-9{ R }_{ 1 }\)
\(-\left( \begin{matrix} 1 & -3 \\ 0 & 0 \end{matrix}\begin{matrix} 4 & 0 \\ \frac { 10 }{ 3 } & -63 \end{matrix} \right) { R }_{ 2 }\rightarrow { R }_{ 2 }+\frac { 28 }{ 3 } .{ R }_{ 1 }\)
The last equivalent matrix is in echelon form and there are 2 non-zero rows
\(\therefore \rho (A)=2\)