\( \eta_{d} =\frac{p}{x^{2}} \)
\(\frac{-p}{x} \frac{d x}{d p} =\frac{p}{x^{2}} \Rightarrow \frac{-d x}{d p}=\frac{1}{x} \)
\(-\int x d x =\int d p \)
\(-\frac{x^{2}}{2} =\mathrm{p}+\mathrm{k} \)
\(\text {If } x=2, \mathrm{p} =3 \)
\(-2 =3+\mathrm{k} \Rightarrow \mathrm{k}=-5 \)
\( p=5-\frac{x^{2}}{2} \\ R=n x=5 x-\frac{x^{3}}{2} \)

\( \eta_{d} =\frac{3-x}{x} \)
\(\frac{-p}{x} \frac{d x}{d p} =\frac{3-x}{x} \)
\(\frac{d x}{3-x} =\frac{-d p}{p} \)
\(\int \frac{d x}{x-3} =\int \frac{d p}{p} \)
log (x-3) = log p + log k
log(x-3) =  log pk
x-3 = pk
If p = 2, x = 1
-2 = 2k
k = -1
p = 3-x
R = px = 3x - 3x²

\(\text {(i) } C^{\prime}(x) =10.6 x \)
\( \mathrm{C}(x) =10.6 \int x d x+k_{1} \)
\( =10.6 \frac{x^{2}}{2}+k \)
\( =5.3 x^{2}+\mathrm{k} \)
\( \text {If } x =0, \mathrm{C}=50, \)
\(\mathrm{k}=50 \)
\(\mathrm{C} =5.3 x^{2}+50 \)
\((ii) \ \mathrm{R}(x) =5 x \)
\( \mathrm{P} =\mathrm{R}-\mathrm{C} \)
\( =5 x-\left(5.3 x^{2}+50\right) \)
\( =5 x-5.3 x^{2}-50 \)

\( C^{\prime}(x) =20+\frac{x}{20} \)
\(\mathrm{C}(x) =\int\left(20+\frac{x}{20}\right) d x=20 x+\frac{x^{2}}{40}+k \)
\( \text {If } x=0, \mathrm{C}=200, \mathrm{k}_{1}=200 \)
\(\mathrm{C} =20 x+\frac{x^{2}}{40}+200 \)
\(R^{\prime}(x) =30 \)
\(\mathrm{R}(x) =\int 30 d x+k_{2}=30 x+\mathrm{k}_{2} \)
\(\text {If } x=0, \mathrm{R}=0 \Rightarrow \mathrm{k}_{2}=0, \mathrm{R}(x)=30 x \)
\(\mathrm{P} =\mathrm{R}-\mathrm{C} \)
\( =30 x-20 x-\frac{x^{2}}{40}-200 \)
\( =10 x-\frac{x^{2}}{40}-200 \)
\(\frac{d P}{d x} =10-\frac{x}{20} \)
\(\frac{d^{2} P}{d x^{2}} =\frac{-1}{20}<0 \)
For maximum profit
\( \frac{d P}{d x} =0 \\ 10-\frac{x}{20} =0 \\ 10 =\frac{x}{20} \Rightarrow x=200 \)
Profit is marimum when x = 200
Maximum profit
\(P=2000-\frac{40000}{40}-200=Rs.800\)

R(x)  =\( \left(4+e^{-0 . a_{x}}\right) d x+k  \)
 =\(4 x-\frac{e^{-0.0 r_{x}}}{0.03 x}+k  \)
\(\text {If } x =0, R=0  \)
\(0  =-\frac{1}{0.03}+k \Rightarrow \mathrm{k}=\frac{1}{0.03}   \)
\(\mathrm{R}(x)  =4 x-\frac{c^{-2.3 x}}{0.03}+\frac{1}{0.03}   \)
\(\text {If } x=100  \)
\(\mathrm{R} =400-\frac{e^{-3}}{0.03}+\frac{1}{0.03}  \)
\( =400-\frac{0.05}{0.03}+\frac{1}{0.03}  \)
\( =431.667 \text { thousands. }   \)
\(\mathrm{R} =4.31,667 \text { thousands }\)

\( R^{\prime}(x) =\frac{d R}{d x}=7+\mathrm{e}^{-0.05 x} \)
\(\mathrm{R} =\int_{0}^{100}\left(7+\mathrm{e}^{-0.05 \mathrm{x}}\right) d x \)
\( =\left[7 x+\frac{\mathrm{e}^{-0.05 x}}{-0.05}\right]_{0}^{100}=700-\frac{100}{5}\left(\mathrm{e}^{-5}-1\right) \)
= 700 - 20(0.0067-1)  
 = 700 + 20 - 0.134 
 = 719.866\( \times \)1000
= Rs. 7,19,866

\( \eta_{d} =2 \\ \frac{-p}{x} \frac{d x}{d p} =2 \\ \frac{d x}{x} =-2 \frac{d p}{p} \\ \int \frac{d x}{x} =-2 \int \frac{d p}{p} \\ \log x =-2 \log \mathrm{p}+\log \mathrm{k} \\ \log x+2 \log \mathrm{p} =\log \mathrm{k} \\ \log x+\log \mathrm{p}^{2} =\log \mathrm{k} \\ \log \mathrm{p}^{2} x =\log \mathrm{k} \\ \mathrm{p}^{2} x =\mathrm{k} \)
\(If \ x=4, \mathrm{p}=1 \Rightarrow \mathrm{k}=4 \)
\(\mathrm{p}^{2} x =4 \text { (or) } \mathrm{p}^{2}=\frac{4}{x} \)
\(\mathrm{p} =\frac{2}{\sqrt{x}} ; \mathrm{R}=\mathrm{p} x=2 \sqrt{x}\)

\( \eta_{d} =\frac{x-5}{x} \)
\(\frac{-p}{x} \frac{d x}{d p} =\frac{x-5}{x} \)
\(\int \frac{d x}{x-5} =-\int \frac{d p}{p} \)
\(\log (x-5) =-\log \mathrm{p}+\log \mathrm{k} \)
\(\log \mathrm{p}+\log (x-5) =\log \mathrm{k} \)
\(\log \mathrm{p}(x-5) =\log \mathrm{k} \)
\(\mathrm{p}(x-5) =\mathrm{k} \)
\(\text {If } \mathrm{p}=2, x=7, \mathrm{k} =4 \)
\(\mathrm{p} =\frac{4}{x-5} \)
\(\mathrm{R} =\mathrm{p} x=\frac{4 x}{x-5} x>5\)

\( \mathrm{MR} =\frac{e^{x}}{100}+x+x^{2} \)
\(\mathrm{R} =\int\left(\frac{e^{x}}{100}+x+x^{2}\right) d x \)
\( =\frac{e^{x}}{100}+\frac{x^{2}}{2}+\frac{x^{3}}{3}+k \)
\(\text {If } x =0, \mathrm{R}=0 \)
\(0 =\frac{e^{0}}{100}+k \Rightarrow \mathrm{k}=\frac{-1}{100} \)
\(\mathrm{R} =\frac{e^{x}}{100}+\frac{x^{2}}{2}+\frac{x^{3}}{3}-\frac{1}{100}\)

Given \(R'(x)=\frac { 3 }{ { x }^{ 2 } } -\frac { 2 }{ x } \)
\(\Rightarrow \int { R'(x) } =\int { \left( \frac { 3 }{ { x }^{ 2 } } -\frac { 2 }{ x } \right) } dx\)
\(\Rightarrow R(x)=\frac { -3 }{ x } -2log\quad x+k\)
Given R(1) = 6 ⇒ when x = 1, R = 6
\(\Rightarrow 6=\frac { -3 }{ 1 } -2log1+k\)
⇒ 6 + 3 = k [∵ log 1 = 0]
⇒ k = 9 
\(\therefore R(x)=-\frac { 3 }{ x } -2log\quad x+9\)
Demand function \(P=\frac { R }{ x } \)
\(\\ =\frac { 3 }{ { x }^{ 2 } } -\frac { 2log\quad x }{ x } +\frac { 9 }{ x } \)

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

Here, the area lies below the x-axis
\(\therefore \ A=\int _{ -2 }^{ 0 }{ (-y)dx } =\int _{ 0 }^{ -2 }{ ydx } \)
\(\left[ \because \int _{ a }^{ b }{ f()dx } =-\int _{ b }^{ a }{ f(x)dx } \right] \)
\(={ \int _{ 0 }^{ -2 }{ (x-1)dx=\left[ \frac { { x }^{ 2 } }{ 2 } -x \right] } }_{ 0 }^{ -2 }\)
\(\left( \frac { { (-2) }^{ 2 } }{ 2 } -(2) \right) -0\)
\(=\frac { 4 }{ 2 } +2=2+2\)
A = 4 sq.units

The limits for one arch of the curve y = sin ax When y = 0 ⇒Sin ax = 0
⇒ sin ax = sin 0, sin \(\pi\)
⇒ ax = 0 or ax = \(\pi\)
⇒ x = 0, x = \(\frac{\pi}{a}\)
∴ The limits are from x = 0 to x = \(\frac{\pi}{a}\)
∴ Area =\(\int _{ a }^{ b }{ ydx } \)
\(=\int _{ 0 }^{ a }{ sin\quad ax\quad dx } \)
\(={ \left[ -\frac { cos\quad ax }{ a } \right] }_{ 0 }^{ \frac { \pi }{ a } }\)
\(=-\frac { 1 }{ a } \left[ cos\quad a\times \frac { \pi }{ a } -cos(a)(0) \right] \)
\(=-\frac { 1 }{ a } \left[ cos\quad \pi -cos0 \right] \)
\(=-\frac { 1 }{ a } (-1-1)[\because cos0=1\ cos\pi =-1]\)
\(A=\frac { 2 }{ a } \) sq.units.

Area \(=\int _{ a }^{ b }{ ydx } \)
\(=\int _{ 3 }^{ 9 }{ \frac { 1 }{ x } dx } \)
\(={ [log\quad x] }_{ 3 }^{ 9 }\)
= log9-log3
\(=log\left( \frac { 9 }{ 3 } \right) \)
A = log 3 sq.units.

Area \(=\int _{ -\frac { \pi }{ 2 } }^{ \frac { \pi }{ 2 } }{ ydx } \)
\(=\int _{ -\frac { \pi }{ 2 } }^{ \frac { \pi }{ 2 } }{ cos\quad xdx } \)
\(=2\int _{ 0 }^{ \frac { \pi }{ 2 } }{ cos\quad xdx } \) [∵ cos x is an even function]
\(=2{ [sinx] }_{ 0 }^{ \frac { \pi }{ 2 } }\)
\(=2\left[ sin\frac { \pi }{ 2 } -sin0 \right] =2[1-0]\)
A = 2 sq.units.