Case study (4 Marks)

Question 14 Marks

Read the following text and answer the following questions on the basis of the same:
Rakesh and Kartik are final year engineering students. Both are selected in their campus interviews and got placement in a reputed company. But both of them decide to start their own business. For this they are surveying the market. During their survey they find the demand curve for a certain item is $p=D(q)=\frac{20}{q+1}$ and the supply curve is p = S(q) = q + 2

Q.1. At equilibrium which of the following is true ?
(A) D(q) = S(q)
(B) D(q) > S(q)
(C) D(q) < S(q)
(D) none of these

Q.2. Find equilibrium quantity ?
(A) 5 (B) 3 (C) -6 (D)-3

Q.3. Find equilibrium price?
(A) 5 (B) 3 (C) -6 (D)-3

Q.4. Consumer's Profit (in ₹ ) is:
(A) 12 (C) 12.73 (B) 12.52 (D) 12.11

Answer

(1) (A) D(q) > S(q)
Explanation: At the equilibrium point, supply and demand curves meet.
So, D(q) = S(q)

(2) (B) 3
Explanation: To find the equilibrium quantity, we let D(q) = S(q) to obtain
$\frac{20}{q+1}=q+2$
Clearing the denominator gives 20 = (q + 1)(q + 2) which simplifies to $q^2+3 q-18=0$
The positive solution gives the equilibrium quantity $q_e=3$

(3) (A) 5
Explanation: Put q = 3 in D(q) or S(q) to get equilibrium price $\left(p_e\right)$
S(3) = 3 + 2 = 5

(4) (C) 12.73
Explanation ;
$\begin{aligned} C S & =\int_0^{q_e} D(q) d q-p_e q_e \\ & =\int_0^3 \frac{20}{q+1} d q-(5)(3) \\ & =\left.20 \ln (q+1)\right|_0 ^3-15 \\ & =20 \ln 4-15\end{aligned}$
=20 2log(2) - 15
$=202(0.3010)-15$
$\approx 12.73$.

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Question 24 Marks

Read the following text and answer the following questions on the basis of the same:
The demand function for a popular make of 12- speed bicycle is given by $p= D (x)=-0.001 x^2$= 250, where p is the unit price in Rupees and x is the quantity demanded in units of a thousands. The supply function for the same product is given by p $= S (x)=0.0006 x^2+0.02 x+100$,where p is the unit price in Rupees and x is the quantity supplied in units of a thousands.

Q.1. If at equilibrium, curves of demand function and supply function intersect at (x, y), then what does the point of intersection represents?
(A) x = equilibrium quantity, y = equilibrium price
(B) x = equilibrium price, y = equilibrium quantity
(C) x = equilibrium demand, equilibrium price
(D) x = equilibrium supply, y = equilibrium demand

Q.2. Find point of intersection.
(A) (300, 160)
(B) (160,300)
(C) (300,300)
(D) (160, 160)

Q.3. Formula for Consumer's Surplus is:
(A) $C S=\int_0^{Q_e} D(x) d x+Q_e \cdot P_e$
(B) CS $=\int_0^{Q_e} D(x) d x-Q_e \cdot P_e$
(C) $C S=-\int_0^{Q_e} D(x) d x-Q_e \cdot P_e$
(D) $CS =-\int_0^{Q_e} D(x) d x+Q_e \cdot P_e$

Q.4. Formula for Producer's Surplus is:
(A) $P S=Q_e \cdot P_e-\int_0^{Q_e} D(x) d x$
(B) $P S=Q_e \cdot P_e-\int_0^{Q_e} S(x) d x$
(C) $P S=Q_c \cdot P_e+\int_0^{Q_e} S(x) d x$
(D) $P S=-Q_e \cdot P_e+\int_0^{Q_e} D(x) d x$

Answer

(1) (A) x = equilibrium quantity, y = equilibrium price
Explanation: At equilibrium, the intersection point (x, y) of demand curve and supply curve represents equilibrium quantity $\left( Q _e\right)$ and equilibrium price $\left( P _e\right)$ respectively

(2) (A) (300, 160)
Explanation: At equilibrium, D(x) = S(x)
$\begin{array}{l}\Rightarrow \quad-0.001 x^2+250=0.0006 x^2+0.02 x+100 \\ \Rightarrow \quad(0.0006+0.001) x^2+0.02 x+(100-250)=0 \\ \Rightarrow \quad 2 x^2+25 x-187500=0 \\ \Rightarrow \quad 2 x^2+625 x-600 x-187500=0 \\ \Rightarrow(2 x+625)(x-300)=0 \\ \Rightarrow x=\frac{-625}{2} \text { or } x=300\end{array}$
The value of x cannot be negative, So x = 300
Now, substituting x = 300 in D(x) or S(x) we get p = 160
Thus, point of intersection is (300, 160).

(3) (B) $C S=\int_0^{Q_e} D(x) d x-Q_e \cdot P_e$
Explanation: Formula for Consumer's Surplus is
$C S=\int_0^{Q_e} D(x) d x-Q_e \cdot P_e$
Where, D(x) = Demand Function
$\begin{array}{l}P_e=\text { Equilibrium Price } \\ Q_e=\text { Equilibrium Quantity }\end{array}$

(4) (B) $P S=Q_e \cdot P_e-\int_0^{Q_e} S(x) d x$
Explanation: Formula for Producer's Surplus is $P S=Q_e \cdot P_e-\int_0^{Q_e} S(x) d x$
Where,
$\begin{aligned} S(x) & =\text { Supply Function } \\ P_e & =\text { Equilibrium Price } \\ Q_e & =\text { Equilibrium Quantity }\end{aligned}$

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Question 34 Marks

Read the following text and answer the following questions on the basis of the same:
Rishi and Nitish were discussing integration, in there discussion following points were discussed.
In question like $\int e^x\left[f(x)+f^{\prime}(x)\right] d x$ we should result $e^x f(x)+C$
If we have irrational terms in integration question, then we should try doing rationalising these terms

Q.1. $\int \frac{1}{\sqrt{3 x+4}-\sqrt{3 x+1}} d x$ is equal to:
(A) $\frac{2}{27}\left\{(x+4)^{3 / 2}+(3 x+1)^{3 / 2}\right\}+C$
(B) $\frac{2}{27}\left\{(3 x-4)^{1 / 2}+(3 x+1)^{3 / 2}\right\}+C$
(C) $\frac{2}{7}\left\{(3 x+4)^{3 / 2}+(3 x+1)^{3 / 2}\right\}+C$
(D) $\frac{2}{27}\left\{(3 x+4)^{3 / 2}+(3 x+1)^{3 / 2}\right\}+C$

Q.2. $\int e^x\left[\log x+\frac{1}{x}\right] d x$ equal to:
(A) $e^x \log x+C$
(B) $\frac{e^x}{x} \log x+C$
(C) $\frac{\log e}{e^x}+C$
(D) $\frac{e^x}{\log x}+C$

Q.3. $\int \frac{1}{\sqrt{1-2 x}+\sqrt{3-2 x}} d x$ equal to:
(A) $\frac{1}{6}(1-2 x)^{3 / 2}+\frac{1}{5}(3-2 x)^{3 / 2}+C$
(B) $\frac{1}{5}(1-2 x)^{3 / 2}+\frac{1}{6}(3-2 x)^{3 / 2}+C$
(C) $\frac{1}{6}(1-2 x)^{3 / 2}-\frac{1}{6}(3-2 x)^{3 / 2}+C$
(D) $\frac{1}{6}(1-x)^{3 / 2}+\frac{1}{6}(3-2 x)^{3 / 2}+C$

Q.4. $\int\left[\frac{1}{\log x}-\frac{1}{(\log x)^2}\right] d x$ is equal to:
(A) $\frac{\log x}{x}+C$
(B) $x \log x+C$
(C) $\frac{x}{\log x}+C$
(D) $e^x \log x+C$

Answer

(1) (D) $\frac{2}{27}\left\{(3 x+4)^{3 / 2}+(3 x+1)^{3 / 2}\right\}+C$
Explanation: Rationalising denominator we get:
$\begin{array}{l}=\int \frac{\sqrt{3 x+4}+\sqrt{3 x+1}}{3} d x \\ =\frac{1}{3} \int[\sqrt{3 x+4}+\sqrt{3 x+1}] d x \\ =\frac{1}{3}\left[\frac{(3 x+4)^{3 / 2}}{3 \times 3 / 2}+\frac{(3 x+1)^{3 / 2}}{3 \times 3 / 2}+C\right] \\ =\frac{2}{27}\left\{(3 x+4)^{3 / 2}+(3 x+1)^{3 / 2}\right\}+C\end{array}$

(2) (A) $e^x \log x+C$
Explanation: Let $I=\int e^x\left[\log x+\frac{1}{x}\right] d x$
Here $f(x)=\log x$ and $f^{\prime}(x)=\frac{1}{x}$
$\therefore$ By using the rule
$\int e^x\left[f(x)+f^{\prime}(x)\right] d x=e^x(f x)+C$
We get, $I=e^x \log x+C$

(3) (C) $\frac{1}{6}(1-2 x)^{3 / 2}-\frac{1}{6}(3-2 x)^{3 / 2}+C$
Explanation: Rationalising the denominator, we get:
$\int \frac{\sqrt{1-2 x}-\sqrt{3-2 x}}{-2} d x$
$\begin{array}{l}=-\frac{1}{2} \int \sqrt{1-2 x} d x+\frac{1}{2} \int \sqrt{3-2 x} d x \\ =\frac{1}{6}(1-2 x)^{3 / 2}-\frac{1}{6}(3-2 x)^{3 / 2}+C\end{array}$

(4) (C) $\frac{x}{\log x}+C$
Explanation: Let $I=\int\left[\frac{1}{\log x}-\frac{1}{(\log x)^2}\right] d x$
Let log(x) = t
$\Rightarrow \quad x=e^{\prime}$
$\Rightarrow \quad d x=e^{\prime} d t$
Thus, $I=\int e^t\left[\frac{1}{t}-\frac{1}{t^2}\right] d t$
Here $f(t)=\frac{1}{t}$ and $f(t)=-\frac{1}{t^2}$
$\begin{aligned} I & =e^t f(x) \\ & =e^{\log x} \times \frac{1}{\log x}+C \\ & =\frac{x}{\log x}+C\end{aligned}$

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Question 44 Marks

Read the following text and answer the following questions on the basis of the same:
The second new species named Puntius euspilurus is an edible freshwater fish found in the Mananthavady river in Wayanad. The epithet euspilurus is a Greek word referring to the distinct black spot on the caudal fin. The slender bodied fish prefers fast flowing, shallow and clear waters and occurs only in unpolluted areas. It appears in great numbers in paddy fields during the onset of the Southwest monsoon.

Suppose that the supply schedule of this Fish is given in the table below which follows a linear relationship between price and quantity supplied.

Price P per kg (in)	Quantity (X) of Fish supplied (in kg)
25	800
20	700
15	600
10	500
5	400

Suppose that this Fish can be sold only in the Kerala. The Kerala demand schedule for this Fish is as follows and there is a linear relationship between price and quantity demanded.

Price P per kg (in)	Quantity (X) of Fish supplied (in kg)
25	200
20	400
15	600
10	800
5	1000

Q.1. Which of the following represents the Price (p) - supply (x) relationship?
(A) $p=65-\frac{x}{20}$
(B) $p=65+\frac{x}{20}$
(C) $p=-15+\frac{x}{20}$
(D) $p=15-\frac{x}{20}$

Q.2. The equation of demand curve can be given by:
(A) $p=30-\frac{x}{40}$
(B) $p=30+\frac{x}{40}$
(C) $p=20-\frac{x}{40}$
(D) $p=20+\frac{x}{40}$

Q.3. The value of x at equilibrium is:
(A) 1400/3
(B) 600
(C) 15
(D) 200/3

Q.4. The equilibrium price is:
(A) 400
(B) 20
(C) 600
(D) 15

Answer

(1) (C) $p=-15+\frac{x}{20}$
Explanation: From the given table of price and supply we have the following price and supply coordinates i.e., (25, 800), (20, 700), (15, 600), (10, 500), (5, 400)
All these coordinates, satisfy the equation $p=-15+\frac{x}{20}$

(2) (A) $p=30-\frac{x}{40}$
Explanation: From the given table of price and demand, we have the following price and demand coordinates, i.e., (25, 200), (20, 400), (15, 600), (10, 800), (5, 1000)
All of these coordinates, lie on the equation $p=30-\frac{x}{40}$

(3) (B) 600
Explanation: At equilibrium Demand = Supply
Therefore, $30-\frac{x}{40}=-15+\frac{x}{20}$
$\Rightarrow \quad 30+15=\frac{x}{20}+\frac{x}{40}$
$\Rightarrow \quad 45=\frac{2 x+x}{40}$
$\Rightarrow \quad 3 x=45 \times 40$
$\Rightarrow \quad x=600$

(4) (D) 15
Explanation: At x = 600
$p=30-\frac{600}{40}$
=30-15
= 15

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Applied Maths Case study (4 Marks)

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