Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 9 Differentiation Ex 9.2 Questions and Answers.

Maharashtra State Board 11th Commerce Maths Solutions Chapter 9 Differentiation Ex 9.2

I. Differentiate the following functions w.r.t. x.

Question 1.
\(\frac{x}{x+1}\)
Solution:
Maharashtra Board 11th Commerce Maths Solutions Chapter 9 Differentiation Ex 9.2 I Q1
Maharashtra Board 11th Commerce Maths Solutions Chapter 9 Differentiation Ex 9.2 I Q1.1

Question 2.
\(\frac{x^{2}+1}{x}\)
Solution:
Maharashtra Board 11th Commerce Maths Solutions Chapter 9 Differentiation Ex 9.2 I Q2

Maharashtra Board 11th Commerce Maths Solutions Chapter 9 Differentiation Ex 9.2

Question 3.
\(\frac{1}{e^{x}+1}\)
Solution:
Maharashtra Board 11th Commerce Maths Solutions Chapter 9 Differentiation Ex 9.2 I Q3

Question 4.
\(\frac{e^{x}}{e^{x}+1}\)
Solution:
Maharashtra Board 11th Commerce Maths Solutions Chapter 9 Differentiation Ex 9.2 I Q4
Maharashtra Board 11th Commerce Maths Solutions Chapter 9 Differentiation Ex 9.2 I Q4.1

Question 5.
\(\frac{x}{\log x}\)
Solution:
Maharashtra Board 11th Commerce Maths Solutions Chapter 9 Differentiation Ex 9.2 I Q5

Question 6.
\(\frac{2^{x}}{\log x}\)
Solution:
Maharashtra Board 11th Commerce Maths Solutions Chapter 9 Differentiation Ex 9.2 I Q6

Maharashtra Board 11th Commerce Maths Solutions Chapter 9 Differentiation Ex 9.2

Question 7.
\(\frac{\left(2 e^{x}-1\right)}{\left(2 e^{x}+1\right)}\)
Solution:
Maharashtra Board 11th Commerce Maths Solutions Chapter 9 Differentiation Ex 9.2 I Q7
Maharashtra Board 11th Commerce Maths Solutions Chapter 9 Differentiation Ex 9.2 I Q7.1

Question 8.
\(\frac{(x+1)(x-1)}{\left(e^{x}+1\right)}\)
Solution:
Maharashtra Board 11th Commerce Maths Solutions Chapter 9 Differentiation Ex 9.2 I Q8

II. Solve the following examples:

Question 1.
The demand D for a price P is given as D = \(\frac{27}{P}\), find the rate of change of demand when the price is 3.
Solution:
Demand, D = \(\frac{27}{P}\)
Rate of change of demand = \(\frac{dD}{dP}\)
Maharashtra Board 11th Commerce Maths Solutions Chapter 9 Differentiation Ex 9.2 II Q1
When price P = 3,
Rate of change of demand,
\(\left(\frac{\mathrm{dD}}{\mathrm{dP}}\right)_{\mathrm{P}=3}=\frac{-27}{(3)^{2}}=-3\)
∴ When price is 3, Rate of change of demand is -3.

Question 2.
If for a commodity; the price-demand relation is given as D = \(\frac{P+5}{P-1}\). Find the marginal demand when the price is 2.
Solution:
Maharashtra Board 11th Commerce Maths Solutions Chapter 9 Differentiation Ex 9.2 II Q2

Question 3.
The demand function of a commodity is given as P = 20 + D – D2. Find the rate at which price is changing when demand is 3.
Solution:
Given, P = 20 + D – D2
Rate of change of price = \(\frac{dP}{dD}\)
= \(\frac{d}{dD}\)(20 + D – D2)
= 0 + 1 – 2D
= 1 – 2D
Rate of change of price at D = 3 is
\(\left(\frac{\mathrm{dP}}{\mathrm{dD}}\right)_{\mathrm{D}=3}\) = 1 – 2(3) = -5
∴ Price is changing at a rate of -5, when demand is 3.

Maharashtra Board 11th Commerce Maths Solutions Chapter 9 Differentiation Ex 9.2

Question 4.
If the total cost function is given by; C = 5x3 + 7x2 + 7; find the average cost and the marginal cost when x = 4.
Solution:
Total cost function, C = 5x3 + 7x2 + 7
Average cost = \(\frac{C}{x}\)
Maharashtra Board 11th Commerce Maths Solutions Chapter 9 Differentiation Ex 9.2 II Q4
When x = 4, Marginal cost = \(\left(\frac{\mathrm{dC}}{\mathrm{d} x}\right)_{x=4}\)
= 15(4)2 + 4(4)
= 240 + 16
= 256
∴ the average cost and marginal cost at x = 4 are \(\frac{359}{4}\) and 256 respectively.

Question 5.
The total cost function of producing n notebooks is given by
C = 1500 – 75n + 2n2 + \(\frac{n^{3}}{5}\)
Find the marginal cost at n = 10.
Solution:
The total cost function,
Maharashtra Board 11th Commerce Maths Solutions Chapter 9 Differentiation Ex 9.2 II Q5
Maharashtra Board 11th Commerce Maths Solutions Chapter 9 Differentiation Ex 9.2 II Q5.1
∴ Marginal cost at n = 10 is 25.

Question 6.
The total cost of ‘t’ toy cars is given by C = 5(2t) + 17. Find the marginal cost and average cost at t = 3.
Solution:
Total cost of ‘t’ toy cars, C = 5(2t) + 17
Maharashtra Board 11th Commerce Maths Solutions Chapter 9 Differentiation Ex 9.2 II Q6
∴ at t = 3, the Marginal cost is 40 log 2 and the Average cost is 19.

Maharashtra Board 11th Commerce Maths Solutions Chapter 9 Differentiation Ex 9.2

Question 7.
If for a commodity; the demand function is given by, D = \(\sqrt{75-3 P}\). Find the marginal demand function when P = 5.
Solution:
Demand function, D = \(\sqrt{75-3 P}\)
Now, Marginal demand = \(\frac{dD}{dP}\)
Maharashtra Board 11th Commerce Maths Solutions Chapter 9 Differentiation Ex 9.2 II Q7
Maharashtra Board 11th Commerce Maths Solutions Chapter 9 Differentiation Ex 9.2 II Q7.1

Question 8.
The total cost of producing x units is given by C = 10e2x, find its marginal cost and average cost when x = 2.
Solution:
Maharashtra Board 11th Commerce Maths Solutions Chapter 9 Differentiation Ex 9.2 II Q8

Question 9.
The demand function is given as P = 175 + 9D + 25D2. Find the revenue, average revenue, and marginal revenue when demand is 10.
Solution:
Given, P = 175 + 9D + 25D2
Total revenue, R = P.D
= (175 + 9D + 25D2)D
= 175D + 9D2 + 25D3
Average revenue = P = 175 + 9D + 25D2
Marginal revenue = \(\frac{dR}{dD}\)
= \(\frac{d}{dD}\) (175D + 9D2 + 25D3)
= 175 \(\frac{d}{dD}\) (D) + 9 \(\frac{d}{dD}\) (D2) + 25 \(\frac{d}{dD}\) (D3)
= 175(1) + 9(2D) + 25(3D2)
= 175 + 18D + 75D2
When D = 10,
Total revenue = 175(10) + 9(10)2 + 25(10)3
= 1750 + 900 + 25000
= 27650
Average revenue = 175 + 9(10) + 25(10)2
= 175 + 90 + 2500
= 2765
Marginal revenue = 175 + 18(10) + 75(10)2
= 175 + 180 + 7500
= 7855
∴ When Demand = 10,
Total revenue = 27650, Average revenue = 2765, Marginal revenue = 7855.

Maharashtra Board 11th Commerce Maths Solutions Chapter 9 Differentiation Ex 9.2

Question 10.
The supply S for a commodity at price P is given by S = P2 + 9P – 2. Find the marginal supply when the price is 7.
Solution:
Given, S = P2 + 9P – 2
Maharashtra Board 11th Commerce Maths Solutions Chapter 9 Differentiation Ex 9.2 II Q10
∴ The marginal supply is 23, at P = 7.

Question 11.
The cost of producing x articles is given by C = x2 + 15x + 81. Find the average cost and marginal cost functions. Find marginal cost when x = 10. Find x for which the marginal cost equals the average cost.
Solution:
Given, cost C = x2 + 15x + 81
Maharashtra Board 11th Commerce Maths Solutions Chapter 9 Differentiation Ex 9.2 II Q11
If marginal cost = average cost, then
2x + 15 = x + 15 + \(\frac{81}{x}\)
∴ x = \(\frac{81}{x}\)
∴ x2 = 81
∴ x = 9 …..[∵ x > 0]

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