Samacheer Kalvi 12th Business Maths Solutions Chapter 3 Integral Calculus II Ex 3.4

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Tamilnadu Samacheer Kalvi 12th Business Maths Solutions Chapter 3 Integral Calculus II 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:

Question 2.
Area bounded by the curve y = e^-2x between the limits 0 ≤ x ≤ ∞ 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:

Question 3.
Area bounded by the curve 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:
Area = \(\int_{1}^{2} \frac{1}{x} d x\)
= \((\log x)_{1}^{2}\)
= log 2 – log 1
= log 2 (Since log 1 = 0)

Question 4.
If the marginal revenue function of a firm is MR = \(e^{\frac{-x}{10}}\), then revenue is ______
(a) \(-10 e^{\frac{-x}{10}}\)
(b) \(1-e^{\frac{-x}{10}}\)
(c) \(10\left(1-e^{\frac{-x}{10}}\right)\)
(d) \(e^{\frac{-x}{10}}+10\)
Answer:
(c) \(10\left(1-e^{\frac{-x}{10}}\right)\)
Hint:

Question 5.
If MR and MC denotes the marginal revenue and marginal cost functions, then the profit functions is ________
(a) P = ∫(MR – MC) dx + k
(b) P = ∫(MR + MC) dx + k
(c) P = ∫(MR) (MC) dx + k
(d) P = ∫(R – C) dx + k
Answer:
(a) P = ∫(MR – MC) dx + k

Question 6.
The demand and supply functions are given by D(x) = 16 – x² and S(x) = 2x² + 4 are under perfect competition, then the equilibrium price x is ________
(a) 2
(b) 3
(c)4
(d) 5
Answer:
(a) 2
Hint:
D(x) = 16 – x² , S(x) = 2x² + 4
Under equilibriuim, D(x) = S(x)
⇒ 16 – x² = 2x² + 4
⇒ 3x² = 12
⇒ x = ±2.
Since x cannot be negative x = 2.

Question 7.
The marginal revenue and marginal cost functions of a company are MR = 30 – 6x and MC = -24 + 3x where x is the product, then the profit function is _______
(a) 9x² + 54x
(b) 9x² – 54x
(c) 54x – \(\frac{9 x^{2}}{2}\)
(d) 54x – \(\frac{9 x^{2}}{2}\) + k
Answer:
(d) 54x – \(\frac{9 x^{2}}{2}\) + k
Hint:
Profit = ∫(MR – MC) dx + k
= ∫(30 – 60) – (-24 + 3x) dx + k
= ∫(54 – 9x) dx + k
= 54x – \(\frac{9 x^{2}}{2}\) + k

Question 8.
The given demand and supply function are given by D(x) = 20 – 5x and S(x) = 4x + 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:
D(x) = S(x) in equilibrium
20 – 5x = 4x + 8
9x = 12
x = \(\frac{4}{3}\) = x₀
p₀ = 20 – 5(\(\frac{4}{3}[/ltex])
= 20 – [latex]\frac{20}{3}\)
= \(\frac{40}{3}\)

Question 9.
If the marginal revenue MR = 35 + 7x – 3x², then the average revenue AR is _______
(a) 35x + \(\frac{7 x^{2}}{2}\) – x³
(b) 35x + \(\frac{7 x}{2}\) – x²
(c) 35 + \(\frac{7 x}{2}\) + x²
(d) 35 + 7x + x²
Answer:
(b) 35x + \(\frac{7 x}{2}\) – x²
Hint:

Question 10.
The profit of a function p(x) is maximum when _______
(a) MC – MR = 0
(b) MC = 0
(c) MR = 0
(d) MC + MR = 0
Answer:
(a) MC – MR = 0
Hint:
P = Revenue – Cost
P is maximum when \(\frac{d p}{d x}\) = 0
\(\frac{d p}{d x}\) = R'(x) – C'(x) = MR – 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:

Question 12.
The demand function for the marginal function MR = 100 – 9x² is _______
(a) 100 – 3x²
(b) 100x – 3x²
(c) 100x – 9x²
(d) 100 + 9x²
Answer:
(a) 100 – 3x²
Hint:
R = ∫(MR) dx + c₁
R = ∫(100 – 9x²) dx + c₁
R = 100x – 3x³ + c₁
When R = 0, x = 0, c₁ = 0
R = 100x – 3x³
Demand function is \(\frac{R}{x}\) = 100 – 3x²

Question 13.
When x₀ = 5 and p₀ = 3 the consumer’s surplus for the demand function p_d = 28 – x² 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:

Question 14.
When x₀ = 2 and P₀ = 12 the producer’s surplus for the supply function P_s = 2x² + 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:

Question 15.
Area bounded by y = x between the lines y = 1, y = 2 with 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:

Question 16.
The producer’s surplus when the supply function for a commodity is P = 3 + x and x₀ = 3 is _______
(a) \(\frac{5}{2}\)
(b) \(\frac{9}{2}\)
(c) \(\frac{3}{2}\)
(d) \(\frac{7}{2}\)
Ans:
(b) \(\frac{9}{2}\)
Hint:
x₀ = 3, p = 3 + x ⇒ p₀ = 3 + 3 = 6

Question 17.
The marginal cost function is MC = 100√x. find AC given that 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) \(\frac{200}{3 x^{\frac{1}{2}}}\)
Answer:
(a) \(\frac{200}{3} x^{\frac{1}{2}}\)
Hint:

Question 18.
The demand and supply function of a commodity are P(x) = (x – 5)² and S(x) = x² + x + 3 then the equilibrium quantity x₀ is ________
(a) 5
(b) 2
(c) 3
(d) 19
Answer:
(b) 2
Hint:
At equilibrium, P(x) = S(x)
⇒ (x – 5)² = x² + x + 3
⇒ x² – 10x + 25 = x² + x + 3
⇒ 11x = 22
⇒ x = 2

Question 19.
The demand and supply function of a commodity are D(x) = 25 – 2x and S(x) = \(\frac{10+x}{4}\) then the equilibrium price p₀ is _________
(a) 5
(b) 2
(c) 3
(d) 10
Answer:
(a) 5
Hint:
At equilibrium, D(x) = S(x)
25 – 2x = \(\frac{10+x}{4}\)
⇒ 100 – 8x = 10 + x
⇒ x = 10
That is x₀ = 10
P₀ = 25 – 2(x₀) = 25 – 20 = 5

Question 20.
If MR and MC denote the marginal revenue and marginal cost and MR – MC = 36x – 3x² – 81, then the maximum profit at x is equal to _______
(a) 3
(b) 6
(c) 9
(d) 5
Answer:
(c) 9
Hint:
The maximum profit occurs when MR – MC = 0
⇒ 36x – 3x² – 81 = 0
⇒ x² – 12x + 27 = 0
⇒ (x – 9)(x – 3) = 0
⇒ x = 9, 3
Now \(\frac{d p}{d x}\) = 36x – 3x² – 81 ⇒ \(\frac{d^{2} p}{d x^{2}}\) = 36 – 9x
At x = 9, \(\frac{d^{2} p}{d x^{2}}\) = 36 – 81 < 0
At x = 3, \(\frac{d^{2} p}{d x^{2}}\) = 36 – 27 > 0
Therefore profit is maximum when x = 9.

Question 21.
If the marginal revenue of a firm is constant, then the demand function is ________
(a) MR
(b) MC
(c) C(x)
(d) AC
Answer:
(a) MR
Hint:
MR = k (constant)
Revenue function R = ∫(MR) dx + c₁
= ∫kdx + c₁
= kx + c₁
When R = 0, x = 0, ⇒ c₁ = 0
R = kx
Demand function p = \(\frac{R}{x}=\frac{k x}{x}\) = k constant
⇒ p = MR

Question 22.
For a demand function p, if \(\int \frac{d p}{p}=k \int \frac{d x}{x}\) then k is equal to _______
(a) η_d
(b) -η_d
(c) \(\frac{-1}{\eta_{d}}\)
(d) \(\frac{1}{\eta_{d}}\)
Answer:
(c) \(\frac{-1}{\eta_{d}}\)
Hint:

Question 23.
Area bounded by y = e^x between the limits 0 to 1 is _________
(a) (e – 1) sq.units
(b) (e + 1) sq.units
(c) (1 – \(\frac{1}{e}\)) sq.units
(d) (1 + \(\frac{1}{e}\)) sq.units
Answer:
(a) (e – 1) sq.units
Hint:

Question 24.
The area bounded by the parabola y² = 4x 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:
y² = 4x
Comparing with y² =4ax gives a = 1
Since parabola is symmetric about x – axis,

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 x lies between 0 and 2
|x| = x

Area = 2 sq.units