Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Business Maths Guide Pdf Chapter 3 Integral Calculus II Ex 3.4 Text Book Back Questions and Answers, Notes.

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

Choose the most suitable answer from the given four 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.unit

(b) \(\frac { 31 }{2}\) sq.unit

(c) \(\frac { 32 }{3}\) sq.unit

(d) \(\frac { 15 }{3}\) sq.unit

Solution:

(c) \(\frac { 32 }{3}\) sq.unit

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.units

(c) 5 sq.units

(d) 2 sq.units

Solution:

(b) \(\frac { 1 }{2}\) sq.units

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

Solution:

(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) 1 – e^{-x/10}

(b) e^{-x/10} + 10

(c) 10(1 – e^{-x/10})

(d) -10e^{-x/10}

Solution:

(c) 10(1 – e^{-x/10})

Hint:

MR = e^{\(\frac { -x }{10}\)} then R = ∫MR dx

R = ∫e^{-x/10} dx = \(\frac { e^{-x/10} }{(-1/10)}\) + k

R = -10e^{-x/10} + k when x = 0, R = 0

⇒ 0 = -10e^{0} + k

0 = -10(1) + k

∴ k = 10

R = -10e^{-x/10} + 10 = 10(1 – e^{-x/10})

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 = ∫(R – C) dx + k

(c) P = ∫(MR + MC)dx + k

(d) P = ∫(MR) (MC) dx + k

Solution:

(a) P = ∫(MR – MC) dx + k

Hint:

Profit = Revenue – Cost

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

Solution:

(a) 2

Hint:

D(x) =16 – x² and S(x) = 2x² + 4

Under perfect competition D(x) = S(x)

16 – x² = 2x² + 4; 16 – 4 = 2x² + x²

3x² = 12 ⇒ x² = \(\frac { 12 }{3}\) = 4

∴ x = ± 2, x cannot be in negative

∴ x = 2

Question 7.

The marginal revenue and marginal coast functions of a company are MR = 30 – 6x and MC = -24 + 3x where x is the product, profit function is

(a) 9x² + 54x

(b) 9x² – 54x

(c) 54x – \(\frac { 9x^2 }{2}\)

(d) 54x – \(\frac { 9x^2 }{2}\) + k

Solution:

(d) 54x – \(\frac { 9x^2 }{2}\) + 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) \(\frac { 41 }{5}\)

Solution:

(c) \(\frac { 40 }{3}\)

Hint:

Under perfect competition D(x) = S(x)

20 – 5x = 4x + 8

20 – 8 = 4x + 5x ⇒ 9x = 12

x = \(\frac { 4 }{3}\)

when x = \(\frac { 4 }{3}\); D(x) = 20 – 5(\(\frac { 4 }{3}\)) = 20 – \(\frac { 20 }{3}\)

= \(\frac { 40 }{3}\)

Question 9.

If the marginal revenue MR = 35 +7x – 3x², then the average revenue AR is.

(a) 35x + \(\frac { 7x^2 }{2}\) – x³

(b) 35x + \(\frac { 7x }{2}\) – x²

(c) 35x + \(\frac { 7x }{2}\) + x²

(d) 35x + 7x + x²

Solution:

(c) \(\frac { 40 }{3}\)

Hint:

R = ∫MR dx = ∫(35 + 7x – 3x²) dx

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

Solution:

(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) a cost function is constant

(c) profit is constant

(d) none of these

Solution:

(a) Revenue is constant

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

Solution:

(a) 100 – 3x²

Hint:

R = ∫(MR) dx + c_{1}

R = ∫(100 – 9x^{2}) dx + c_{1}

R = 100x – 3x^{3} + c_{1}

When R = 0, x = 0, c_{1} = 0

R = 100x – 3x^{3}

Demand function is \(\frac{R}{x}\) = 100 – 3x^{2}

Question 13.

When x_{0} = 5 and p_{0} = 3 the consumer’s surplus for the demand function p_{d} = 28 – x²

(a) 250 units

(b) \(\frac { 250 }{3}\) units

(c) \(\frac { 251 }{2}\) units

(d) \(\frac { 251 }{3}\) units

Solution:

(b) \(\frac { 250 }{3}\) units

Hint:

Question 14.

When x_{0} = 2 and P_{0} = 12 the producer’s surplus for the supply function P_{0} = 2x² + 4 is

(a) \(\frac { 31 }{5}\) units

(b) \(\frac { 31 }{2}\) units

(c) \(\frac { 32 }{2}\) units

(d) \(\frac { 30 }{7}\) units

Solution:

(c) \(\frac { 32 }{2}\) units

Hint:

Producer’s Surplus

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 units

Solution:

(c) \(\frac { 3 }{2}\) sq units

Hint:

Question 16.

The producer’s surplus when supply the function for a commodity is p = 3 + x and x_{0} = 3 is

(a) \(\frac { 1 }{2}\)

(b) \(\frac { 9 }{2}\)

(c) \(\frac { 3 }{2}\)

(d) \(\frac { 7 }{2}\)

Solution:

(b) \(\frac { 9 }{2}\)

Hint:

p = 3 + x and x_{0} = 3

then p_{0} = 3 + 3 = 6

Question 17.

The marginal cost function is MC = 100√x find AC given that TC = 0 when the out put is zero is

(a) \(\frac { 200 }{3}\) x^{1/2}

(b) \(\frac { 200 }{3}\) x^{3/2}

(c) \(\frac { 200 }{3x^{3/2}}\)

(d) \(\frac { 200 }{3x^{1/2}}\)

Solution:

(a) \(\frac { 200 }{3}\) x^{1/2}

Hint:

TC = ∫MC dx = ∫100√x dx = 100 ∫(x)^{1/2} dx

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 x_{0} is

(a) 5

(b) 2

(c) 3

(d) 10

Solution:

(b) 2

Hint:

At equilibrium, P(x) = S(x)

⇒ (x – 5)^{2} = x^{2} + x + 3

⇒ x^{2} – 10x + 25 = x^{2} + x + 3

⇒ 11x = 22

⇒ x = 2

Question 19.

The demand and supply function of a commodity are D(x) = 25 – 2x and S(x) = \(\frac { 10+x }{2}\) then the equilibrium price P_{0} is

(a) 2

(b) 2

(c) 3

(d) 10

Solution:

(a) 2

Hint:

At equilibrium, D(x) = S(x)

25 – 2x = \(\frac{10+x}{4}\)

⇒ 100 – 8x = 10 + x

⇒ x = 10

That is x_{0} = 10

P_{0} = 25 – 2(x_{0}) = 25 – 20 = 5

Question 20.

If MR and MC denote the marginal revenue and marginal cost and MR – MC = 36x – 3x² – 81, then maximum profit at x equal to

(a) 3

(b) 6

(c) 9

(d) 10

Solution

(c) 9

Hint:

Profit P = ∫(MR – MC) dx = ∫(36x – 3x² – 81) dx

P = [\(\frac { 36x^2 }{2}\) – \(\frac { 3x^3 }{3}\) – 81x] = 18x² – x³ – 81x

when p = 0; 18x² – x³ – 81x = 0 ⇒ x² – 18x + 81 = 0

(x – 9)² = 0 ⇒ x – 9 = 0

∴ 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

Solution:

(a) MR

Hint:

MR = k (constant)

Revenue function R = ∫(MR) dx + c_{1}

= ∫kdx + c_{1}

= kx + c_{1}

When R = 0, x = 0, ⇒ c_{1} = 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 ∫\(\frac { dp }{p}\) = k ∫\(\frac { dx }{x}\) then k is equal to

(a) n_{d}

(b) -n_{d}

(c) \(\frac { -1 }{n_d}\)

(d) \(\frac { 1 }{n_d}\)

Solution:

(c) \(\frac { -1 }{n_d}\)

Question 23.

The 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

Solution:

(a) (e – 1) sq.units

Hint:

Area A = \(\int_{a}^{b}\)ydx = \(\int_{0}^{1}\)e^{x}dx = [e^{x}]\(_{0}^{1}\)

= [e^{x} – e°] = [e – 1]

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

Solution:

(b) \(\frac { 8 }{3}\) sq units

Hint:

y² = 4x ⇒ y = \(\sqrt { 4x}\) 2√x = 2(x)^{1/2}

In this parabola 4a = 4 ⇒ a = 1 and vertex V(0, 0)

Question 25.

The 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

Solution:

(c) 2 sq.units

Hint:

Area A = \(\int_{a}^{b}\)ydx = \(\int_{0}^{2}\)x dx = [ \(\frac { x^2 }{2}\) ]\(_{0}^{2}\)

= \(\frac { (2)^2 }{2}\) – (0) = \(\frac { 4 }{2}\) = 2

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