HSC Commerce कक्षा १२ - Tamil Nadu Board of Secondary Education Question Bank Solutions for Business Mathematics and Statistics

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The demand and supply function of a commodity are D(x) = 25 – 2x and S(x) = `(10 + x)/4` then the equilibrium price p₀ is 

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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

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If the marginal revenue of a firm is constant, then the demand function is

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For a demand function p, if `int "dp"/"p" = "k" int ("d"x)/x` then k is equal to

A manufacture’s marginal revenue function is given by MR = 275 – x – 0.3x². Find the increase in the manufactures total revenue if the production is increased from 10 to 20 units

A company has determined that marginal cost function for x product of a particular commodity is given by MC = `125 + 10x - x^2/9`. Where C is the cost of producing x units of the commodity. If the fixed cost is ₹ 250 what is the cost of producing 15 units

The marginal revenue function for a firm given by MR = `2/(x + 3) - (2x)/(x + 3)^2 + 5`. Show that the demand function is P = `(2x)/(x + 3)^2 + 5`

For the marginal revenue function MR = 6 – 3x² – x³, Find the revenue function and demand function

The marginal cost of production of a firm is given by C'(x) = `20 + x/20` the marginal revenue is given by R’(x) = 30 and the fixed cost is ₹ 100. Find the profit function

The demand equation for a product is P_d = 20 – 5x and the supply equation is P_s = 4x + 8. Determine the consumers surplus and producer’s surplus under market equilibrium

A company requires f(x) number of hours to produce 500 units. It is represented by f(x) = 1800x^–0.4. Find out the number of hours required to produce additional 400 units. [(900)^0.6 = 59.22, (500)^0.6 = 41.63]

The price elasticity of demand for a commodity is `"p"/x^3`. Find the demand function if the quantity of demand is 3 when the price is ₹ 2.

Solve: `("d"y)/("d"x) = "ae"^y`

Solve: `(1 + x^2)/(1 + y) = xy ("d"y)/("d"x)`

Solve: `y(1 - x) - x ("d"y)/("d"x)` = 0

Solve: ydx – xdy = 0 dy

Solve: `("d"y)/("d"x) + "e"^x + y"e"^x = 0`

Solve : cos x(1 + cosy) dx – sin y(1 + sinx) dy = 0

Solve: (1 – x) dy – (1 + y) dx = 0

Solve: `log(("d"y)/("d"x))` = ax + by