Exercise 3.4 - Chapter 3 Integral Calculus II 12th Maths Guide Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated On May 15, 2024

By SaraNextGen

Ex $3.4$
Choose the correct answer form the given alternatives.
Question 1.
Area bounded by the curve $y=x(4-x)$ between the limits 0 and 4 with $x$-axis is
(a) $\frac{30}{3}$ sq.units
(b) $\frac{31}{2}$ sq.units
(c) $\frac{32}{3}$ sq.units
(d) $\frac{15}{2}$ sq.units
Answer:
(c) $\frac{32}{3}$ sq.units
Hint:
$
\begin{aligned}
\text { Area } & =\int_0^4 x(4-x) d x=\int_0^4\left(4 x-x^2\right) d x \\
& =\left(2 x^2-\frac{x^3}{3}\right)_0^4=32-\frac{64}{3}=\frac{32}{3}
\end{aligned}
$
Question 2.
Area bounded by the curve $y=e^{-2 x}$ between the limits $0 \leq x \leq \infty$ is
(a) 1 sq.units
(b) $\frac{1}{2}$ sq.unit
(c) 5 sq.units
(d) 2 sq.units
Answer:
(b) $\frac{1}{2}$ sq.unit
Hint:
$
\text { Area }=\int_0^{\infty} e^{-2 x} d x=\left(\frac{e^{-2 x}}{-2}\right)_0^{\infty}=0+\frac{1}{2}=\frac{1}{2}
$

Question 3 .
Area bounded by the curve $\mathrm{y}=\frac{1}{x}$ between the limits 1 and 2 is
(a) $\log 2$ sq.units
(b) $\log 5$ sq.units
(c) $\log 3$ sq.units
(d) $\log 4$ sq.units
Answer:
(a) $\log 2$ sq.units
Hint:
$
\begin{aligned}
& \text { Area }=\int_1^2 \frac{1}{x} d x \\
& =(\log x)_1^2 \\
& =\log 2-\log 1 \\
& =\log 2(\text { Since } \log 1=0)
\end{aligned}
$
Question 4.
If the marginal revenue function of a firm is $\mathrm{MR}=e^{\frac{-x}{10}}$, then revenue is
(a) $-10 e^{\frac{-x}{10}}$
(b) $1-e^{\frac{-z}{10}}$
(c) $10\left(1-e^{\frac{-x}{10}}\right)$
(d) $e^{\frac{-z}{10}}+10$
Answer:
(c) $10\left(1-e^{\frac{-x}{10}}\right)$
Hint:
$
\mathrm{R}=\int e^{\frac{-x}{10}}+k=\frac{e^{\frac{-x}{10}}}{\frac{-1}{10}}+k
$
When $\mathrm{R}=0, x=0$
So $0=\frac{1}{\frac{-1}{10}}+k \Rightarrow k=10$
$
\Rightarrow R=-10 e^{\frac{-x}{10}}+10=10\left(1-e^{\frac{-x}{10}}\right)
$

Question 5.
If $\mathrm{MR}$ and $\mathrm{MC}$ denotes the marginal revenue and marginal cost functions, then the profit functions is
(a) $P=\int(\overline{M R-M C}) \mathrm{d} x+k$
(b) $\mathrm{P}=\int(\mathrm{MR}+\mathrm{MC}) \mathrm{dx}+\mathrm{k}$
(c) $P=\int(\mathrm{MR})(\mathrm{MC}) \mathrm{dx}+\mathrm{k}$
(d) $P=\int(R-C) d x+k$
Answer:
(a) $P=\int(\mathrm{MR}-\mathrm{MC}) \mathrm{dx}+\mathrm{k}$
Question 6.
The demand and supply functions are given by $D(x)=16-x^2$ and $S(x)=2 x^2+4$ are under perfect competition, then the equilibrium price $\mathrm{x}$ is
(a) 2
(b) 3
(c) 4
(d) 5
Answer:
(a) 2
Hint:
$
\mathrm{D}(\mathrm{x})=16-\mathrm{x}^2, \mathrm{~S}(\mathrm{x})=2 \mathrm{x}^2+4
$
Under equilibriuim, $\mathrm{D}(\mathrm{x})=\mathrm{S}(\mathrm{x})$
$
\begin{aligned}
& \Rightarrow 16-\mathrm{x}^2=2 \mathrm{x}^2+4 \\
& \Rightarrow 3 \mathrm{x}^2=12 \\
& \Rightarrow \mathrm{x}=\pm 2 .
\end{aligned}
$
Since $\mathrm{x}$ cannot be negative $\mathrm{x}=2$.

Question 7.
The marginal revenue and marginal cost functions of a company are $\mathrm{MR}=30-6 \mathrm{x}$ and $\mathrm{MC}$ $=-24+3 \mathrm{x}$ where $\mathrm{x}$ is the product, then the profit function is
(a) $9 x^2+54 x$
(b) $9 x^2-54 x$
(c) $54 \mathrm{x}-\frac{9 x^2}{2}$
(d) $54 \mathrm{x}-\frac{9 x^2}{2}+\mathrm{k}$
Answer:
(d) $54 \mathrm{x}-\frac{9 x^2}{2}+\mathrm{k}$
Hint:
$
\begin{aligned}
& \text { Profit }=\int(M R-M C) d x+k \\
& =\int(30-60)-(-24+3 x) d x+k \\
& =\int(54-9 x) d x+k \\
& =54 x-\frac{9 x^2}{2}+k
\end{aligned}
$
Question 8.
The given demand and supply function are given by $\mathrm{D}(\mathrm{x})=20-5 \mathrm{x}$ and $\mathrm{S}(\mathrm{x})=4 \mathrm{x}+8$ if they are under perfect competition then the equilibrium demand is
(a) 40
(b) $\frac{41}{2}$
(c) $\frac{40}{3}$
(d) $\frac{41}{5}$
Answer:
(c) $\frac{40}{3}$
Hint:
$
\begin{aligned}
& \mathrm{D}(\mathrm{x})=\mathrm{S}(\mathrm{x}) \text { in equilibrium } \\
& 20-5 \mathrm{x}=4 \mathrm{x}+8 \\
& 9 \mathrm{x}=12 \\
& \mathrm{x}=\frac{4}{3}=\mathrm{x}_0 \\
& \mathrm{p}_0=20-5\left(\frac{4}{3}[/ \text { ltex }]\right)=20-[\text { latex }] \frac{20}{3} \\
& =\frac{40}{3}
\end{aligned}
$
Question 9.
If the marginal revenue $M R=35+7 x-3 x^2$, then the average revenue $A R$ is
(a) $35 \mathrm{x}+\frac{7 x^2}{2}-\mathrm{x}^3$
(b) $35 \mathrm{x}+\frac{7 x}{2}-\mathrm{x}^2$
(c) $35+\frac{7 x}{2}+\mathrm{x}^2$

(d) $35+7 x+x^2$
Answer:
(b) $35 x+\frac{7 x}{2}-x^2$
Hint:
$
\begin{aligned}
\mathrm{R} & =\int(\mathrm{MR}) d x+k \\
& =\int\left(35+7 x-3 x^2\right) d x+k \\
& =35 x+\frac{7 x^2}{2}-x^3+k
\end{aligned}
$
When $R=0, x=0 \Rightarrow k=0$
So revenue function $\mathrm{R}=35 x+\frac{7 x^2}{2}-x^3$
$
\mathrm{AR}=\frac{R}{x}=35+\frac{7}{2} x-x^2
$
Question 10.
The profit of a function $\mathrm{p}(\mathrm{x})$ is maximum when
(a) $\mathrm{MC}-\mathrm{MR}=0$
(b) $\mathrm{MC}=0$
(c) $\mathrm{MR}=0$
(d) $\mathrm{MC}+\mathrm{MR}=0$
Answer:
(a) $\mathrm{MC}-\mathrm{MR}=0$
Hint:
$\mathrm{P}=$ Revenue $-$ Cost
$\mathrm{P}$ is maximum when $\frac{d p}{d x}=0$
$
\frac{d p}{d x}=\mathrm{R}^{\prime}(\mathrm{x})-\mathrm{C}^{\prime}(\mathrm{x})=\mathrm{MR}-\mathrm{MC}=0
$

Question 11.
For the demand function $p(x)$, the elasticity of demand with respect to price is unity then
(a) revenue is constant
(b) cost function is constant
(c) profit is constant
(d) none of these
Answer:
(a) revenue is constant
Hint:
$
\begin{aligned}
\eta_d & =1 \\
\Rightarrow \frac{-\mathrm{p}}{x} \frac{d x}{d p} & =1 \\
\frac{d x}{x} & =\frac{-d p}{p} \\
\log x & =-\log p+\log k \\
\Rightarrow p x & =k, \text { constant } \\
\text { But } p x & =\mathrm{R} \text { (revenue), which is a constant. }
\end{aligned}
$
Question 12.
The demand function for the marginal function $M R=100-9 x^2$ is
(a) $100-3 x^2$
(b) $100 \mathrm{x}-3 \mathrm{x}^2$
(c) $100 x-9 x^2$
(d) $100+9 x^2$
Answer:
(a) $100-3 x^2$

Hint:
$
\begin{aligned}
& \mathrm{R}=\int(\mathrm{MR}) d x+c_1 \\
& R=\int\left(100-9 x^2\right) d x+c_1 \\
& R=100 x-3 x^3+c_1
\end{aligned}
$
When $R=0, x=0, c_1=0$
$
\mathrm{R}=100 \mathrm{x}-3 \mathrm{x}^3
$
Demand function is $\frac{R}{x}=100-3 \mathrm{x}^2$
Question 13.
When $\mathrm{x}_0=5$ and $\mathrm{p}_0=3$ the consumer's surplus for the demand function $\mathrm{p}_{\mathrm{d}}=28-\mathrm{x}^2$ is
(a) 250 units
(b) $\frac{250}{3}$ units
(c) $\frac{251}{2}$ units
(d) $\frac{251}{3}$ units
Answer:
(b) $\frac{250}{3}$ units
Hint:
$
\begin{aligned}
\mathrm{CS} & =\int_0^5\left(28-x^2\right) d x-(5)(3) \\
& =\left(28 x-\frac{x^3}{3}\right)_0^5-15 \\
& =140-\frac{125}{3}-15=\frac{250}{3} \text { units }
\end{aligned}
$

Question 14.
When $\mathrm{x}_0=2$ and $\mathrm{P}_0=12$ the producer's surplus for the supply function $\mathrm{P}_{\mathrm{S}}=2 \mathrm{x}^2+4$ is
(a) $\frac{31}{5}$ units
(b) $\frac{31}{2}$ units
(c) $\frac{32}{3}$ units
(d) $\frac{30}{7}$ units
Answer:
(c) $\frac{32}{3}$ units
Hint:
$
\begin{aligned}
\text { PS } & =2(12)-\int_0^2\left(2 x^2+4\right) d x \\
& =24-\left[\frac{2}{3} x^3+4 x\right]_0^2 \\
& =24-\frac{16}{3}-8=\frac{32}{3} \text { units }
\end{aligned}
$
Question 15.
Area bounded by $\mathrm{y}=\mathrm{x}$ between the lines $\mathrm{y}=1, \mathrm{y}=2$ with $\mathrm{y}$-axis is
(a) $\frac{1}{2}$ sq.units
(b) $\frac{5}{2}$ sq.units
(c) $\frac{3}{2}$ sq.units
(d) 1 sq.unit
Answer:
(c) $\frac{3}{2}$ sq.units

Hint:
$
\text { Area }=\int_1^2 y d y=\left(\frac{y^2}{2}\right)_1^2=2-\frac{1}{2}=\frac{3}{2}
$
Question 16.
The producer's surplus when the supply function for a commodity is $\mathrm{P}=3+\mathrm{x}$ and $\mathrm{x}_0=3$ is
(a) $\frac{5}{2}$
(b) $\frac{9}{2}$
(c) $\frac{3}{2}$
(d) $\frac{7}{2}$
Answer:
(b) $\frac{9}{2}$
Hint:
$
\begin{aligned}
& \mathrm{x}_0=3, \mathrm{p}=3+\mathrm{x} \Rightarrow \mathrm{p}_0=3+3=6 \\
& \mathrm{PS}=(3)(6)-\int_0^3(x+3) d x \\
&=18-\left[\frac{x^2}{2}+3 x\right]_0^3=18-\frac{9}{2}-9=\frac{9}{2}
\end{aligned}
$
Question 17.
The marginal cost function is $\mathrm{MC}=100 \sqrt{\mathrm{x}}$. find $\mathrm{AC}$ given that $\mathrm{TC}=0$ when the output is zero is
(a) $\frac{200}{3} x^{\frac{1}{2}}$
(b) $\frac{200}{3} x^{\frac{3}{2}}$
(c) $\frac{200}{3 x^{\frac{3}{2}}}$
(d) $\frac{300}{3 x^{\frac{1}{2}}}$
Answer:
(a) $\frac{200}{3} x^{\frac{1}{2}}$

Hint:
$
\begin{aligned}
\mathrm{TC} & =\int(\mathrm{MC}) d x+k \\
& =\int 100 \sqrt{x} d x+k \\
\mathrm{TC} & =100 \frac{x^{\frac{3}{2}}}{\frac{3}{2}}+k
\end{aligned}
$
when $\mathrm{TC}=0, x=0 \Rightarrow k=0$
$
\begin{aligned}
& \mathrm{TC}=\frac{200}{3} x^{\frac{3}{2}} \\
& \mathrm{AC}=\frac{\mathrm{TC}}{x}=\frac{200}{3} x^{\frac{1}{2}}
\end{aligned}
$
Question 18.
The demand and supply function of a commodity are $\mathrm{P}(\mathrm{x})=(\mathrm{x}-5)^2$ and $\mathrm{S}(\mathrm{x})=\mathrm{x}^2+\mathrm{x}+3$ then the equilibrium quantity $\mathrm{x}_0$ is
(a) 5
(b) 2
(c) 3
(d) 19
Answer:
(b) 2
Hint:
At equilibrium, $\mathrm{P}(\mathrm{x})=\mathrm{S}(\mathrm{x})$
$
\begin{aligned}
& \Rightarrow(\mathrm{x}-5)^2=\mathrm{x}^2+\mathrm{x}+3 \\
& \Rightarrow \mathrm{x}^2-10 \mathrm{x}+25=\mathrm{x}^2+\mathrm{x}+3 \\
& \Rightarrow 11 \mathrm{x}=22 \\
& \Rightarrow \mathrm{x}=2
\end{aligned}
$
Question 19.
The demand and supply function of a commodity are $\mathrm{D}(\mathrm{x})=25-2 \mathrm{x}$ and $\mathrm{S}(\mathrm{x})=\frac{10+x}{4}$ then the equilibrium price $\mathrm{p}_0$ is
(a) 5
(b) 2

(c) 3
(d) 10
Answer:
(a) 5
Hint:
At equilibrium, $\mathrm{D}(\mathrm{x})=\mathrm{S}(\mathrm{x})$
$
\begin{aligned}
& 25-2 \mathrm{x}=\frac{10+x}{4} \\
& \Rightarrow 100-8 \mathrm{x}=10+\mathrm{x} \\
& \Rightarrow \mathrm{x}=10 \\
& \text { That is } \mathrm{x}_0=10 \\
& \mathrm{P}_0=25-2\left(\mathrm{x}_0\right)=25-20=5
\end{aligned}
$
Question 20.
If $\mathrm{MR}$ and $\mathrm{MC}$ denote the marginal revenue and marginal cost and $\mathrm{MR}-\mathrm{MC}=36 \mathrm{x}-3 \mathrm{x}^2-$ 81 , then the maximum profit at $\mathrm{x}$ is equal to
(a) 3
(b) 6
(c) 9
(d) 5
Answer:
(c) 9
Hint:
The maximum profit occurs when $\mathrm{MR}-\mathrm{MC}=0$
$
\begin{aligned}
& \Rightarrow 36 \mathrm{x}-3 \mathrm{x}^2-81=0 \\
& \Rightarrow \mathrm{x}^2-12 \mathrm{x}+27=0 \\
& \Rightarrow(\mathrm{x}-9)(\mathrm{x}-3)=0 \\
& \Rightarrow \mathrm{x}=9,3 \\
& \text { Now } \frac{d p}{d x}=36 \mathrm{x}-3 \mathrm{x}^2-81 \Rightarrow \frac{d^2 p}{d x^2}=36-9 \mathrm{x} \\
& \text { At } \mathrm{x}=9, \frac{d^2 p}{d x^2}=36-81<0 \\
& \text { At } \mathrm{x}=3, \frac{d^2 p}{d x^2}=36-27>0
\end{aligned}
$

Therefore profit is maximum when $\mathrm{x}=9$.
Question 21.

If the marginal revenue of a firm is constant, then the demand function is
(a) MR
(b) $\mathrm{MC}$
(c) $\mathrm{C}(\mathrm{x})$
(d) $\mathrm{AC}$
Answer:
(a) MR
Hint:
$
\mathrm{MR}=\mathrm{k} \text { (constant) }
$
Revenue function $\mathrm{R}=\int(\mathrm{MR}) \mathrm{dx}+\mathrm{c}_1$
$=\int \mathrm{kdx}+\mathrm{c}_1$
$=\mathrm{kx}+\mathrm{c}_1$
When $\mathrm{R}=0, \mathrm{x}=0, \Rightarrow \mathrm{c}_1=0$
$\mathrm{R}=\mathrm{kx}$
Demand function $\mathrm{p}=\frac{R}{x}=\frac{k x}{x}=\mathrm{k}$ constant $\Rightarrow \mathrm{p}=\mathrm{MR}$
Question 22.
For a demand function $\mathrm{p}$, if $\int \frac{d p}{p}=k \int \frac{d x}{x}$ then $\mathrm{k}$ is equal to
(a) $\eta_d$
(b) $-\eta_d$
(c) $\frac{-1}{\eta_d}$
(d) $\frac{1}{\eta_d}$
Answer:
(c) $\frac{-1}{\eta_d}$
Hint:

$
\int \frac{d p}{p}=k \int \frac{d x}{x}
$
Comparing with $\int \frac{d p}{p}=\frac{-1}{\eta_d} \int \frac{d x}{x} \Rightarrow k=\frac{-1}{\eta_d}$
Question 23.
Area bounded by $y=\mathrm{e}^{\mathrm{x}}$ between the limits 0 to 1 is
(a) (e $-1)$ sq.units
(b) $(\mathrm{e}+1)$ sq.units
(c) $\left(1-\frac{1}{e}\right)$ sq.units
(d) $\left(1+\frac{1}{e}\right)$ sq.units
Answer:
(a) (e - 1) sq.units
Hint:
$
\text { Area }=\int_0^1 e^x d x=\left(e^x\right)_0^1=e^1-e^0=e-1
$

Question 24.
The area bounded by the parabola $y^2=4 x$ bounded by its latus rectum is
(a) $\frac{16}{3}$ sq.units
(b) $\frac{8}{3}$ sq.units
(c) $\frac{72}{3}$ sq.units
(d) $\frac{1}{3}$ sq.units
Answer:
(b) $\frac{8}{3}$ sq.units
Hint:
$
y^2=4 x
$
Comparing with $\mathrm{y}^2=4 \mathrm{ax}$ gives $\mathrm{a}=1$
Since parabola is symmetric about $\mathrm{x}$ - axis,

$
\begin{aligned}
\text { Area } & =2 \int_0^a y d x=2 \int_0^1 2 \sqrt{x} d x \\
& =4\left[\frac{2}{3} x^{\frac{3}{2}}\right]_0^1=\frac{8}{3} \text { sq.units }
\end{aligned}
$
Question 25.
Area bounded by $y=|x|$ between the limits 0 and 2 is
(a) 1 sq.units
(b) 3 sq.units
(c) 2 sq.units
(d) 4 sq.units
Answer:

(c) 2 sq.units
Hint:
When $\mathrm{x}$ lies between 0 and 2
$
|\mathbf{x}|=\mathrm{x}
$
So area $=\int_0^2 x d x=\left(\frac{x^2}{2}\right)_0^2=\frac{4}{2}=2$
Area $=2$ sq.units