Question 335 marksLong AnswerSection D

(a) Find the consumer's surplus for the demand function p=25−x−x2p = 25 - x - x^2, where the prevailing market price p0=19p_0 = 19.

OR

(b) Solve the following initial value differential equation : (x−1)dydx=2xy(x - 1) \frac{dy}{dx} = 2xy, when y(2)=1y(2) = 1.

Integration and its Applications / Differential Equations
Consumer's Surplus / Initial Value Differential Equation

Official answer

This is an internal-choice long-answer question; either part is worth the full 5 marks.


Part (a) — Consumer's Surplus


The demand function is p=25−x−x2p = 25 - x - x^2, and the market price is p0=19p_0 = 19.


  • Find the quantity x₀ at price 19: 19=25−x−x2  ⟹  x2+x−6=0  ⟹  (x+3)(x−2)=0  ⟹  x0=219 = 25 - x - x^2 \implies x^2 + x - 6 = 0 \implies (x + 3)(x - 2) = 0 \implies x_0 = 2 (reject x=−3x = -3, quantity cannot be negative).
  • Apply the formula: CS=∫0x0f(x)dx−p0⋅x0CS = \int_0^{x_0} f(x) dx - p_0 \cdot x_0.
  • ∫02(25−x−x2)dx=[25x−x2/2−x3/3]02=50−2−8/3=136/3\int_0^2 (25 - x - x^2) dx = [25x - x^2/2 - x^3/3]_0^2 = 50 - 2 - 8/3 = 136/3.
  • CS=136/3−(19×2)=136/3−38=22/3CS = 136/3 - (19 \times 2) = 136/3 - 38 = 22/3.

Consumer's Surplus = 22/3≈7.3322/3 \approx 7.33 (monetary units).


Part (b) — Initial Value Problem


(x−1)dy/dx=2xy(x - 1) dy/dx = 2xy is variable-separable.


  • Separate: dy/y=[2x/(x−1)]dxdy/y = [2x/(x - 1)] dx.
  • Split the RHS: 2x/(x−1)=2+2/(x−1)2x/(x - 1) = 2 + 2/(x - 1).
  • Integrate: ln⁡y=2x+2ln⁡∣x−1∣+C\ln y = 2x + 2 \ln|x - 1| + C.
  • Apply y(2)=1y(2) = 1: ln⁡1=4+2ln⁡1+C  ⟹  0=4+C  ⟹  C=−4\ln 1 = 4 + 2 \ln 1 + C \implies 0 = 4 + C \implies C = -4.

Particular solution: ln⁡y=2x+2ln⁡∣x−1∣−4\ln y = 2x + 2 \ln|x - 1| - 4, i.e. y=(x−1)2⋅e2x−4y = (x - 1)^2 \cdot e^{2x - 4}.

Answer keywords:consumer's surplusdemand functiondefinite integralmarket pricevariable separableinitial value problemparticular solutionintegrating factor of rational

Marking scheme

  1. 1.Part (a): 1 mark for setting up and solving 19=25−x−x219 = 25 - x - x^2 to get x0=2x_0 = 2 (rejecting x=−3x = -3).
  2. 2.Part (a): 1 mark for writing the correct surplus formula CS=∫0x0f(x)dx−p0⋅x0CS = \int_0^{x_0} f(x) dx - p_0 \cdot x_0.
  3. 3.Part (a): 2 marks for correct integration [25x−x2/2−x3/3]02=136/3[25x - x^2/2 - x^3/3]_0^2 = 136/3.
  4. 4.Part (a): 1 mark for the final answer CS=22/3CS = 22/3 ≈ Rs 7.33.
  5. 5.Part (b) alternative: 1 mark separation of variables; 2 marks for splitting 2x/(x−1)2x/(x-1) and integrating to ln⁡y=2x+2ln⁡∣x−1∣+C\ln y = 2x + 2 \ln|x-1| + C; 1 mark applying y(2)=1y(2)=1 to get C=−4C = -4; 1 mark final solution y=(x−1)2e2x−4y = (x-1)^2 e^{2x-4}.

Hint

For (a): first put p0=19p_0 = 19 into the demand law to get the quantity x_0, then CS=∫0x0f(x)dx−p0⋅x0CS = \int_0^{x_0} f(x) dx - p_0 \cdot x_0. For (b): it is variable-separable — write dy/y=[2x/(x−1)]dxdy/y = [2x/(x-1)] dx and split 2x/(x−1)=2+2/(x−1)2x/(x-1) = 2 + 2/(x-1).

Quick oral answer

Consumer's surplus is the extra benefit buyers get because the market price is below what they were willing to pay — computed as the definite integral of the demand curve minus price times quantity; here it works out to 22/322/3, about Rs 7.33.

Common mistakes

  1. 1.In (a), keeping the negative root x=−3x = -3 as a quantity, or computing p0⋅x0p_0 \cdot x_0 with the wrong x₀.
  2. 2.In (b), failing to split 2x/(x−1)2x/(x-1) as 2+2/(x−1)2 + 2/(x-1) and instead integrating incorrectly.
  3. 3.Omitting the constant of integration C or mis-substituting the initial condition, so the particular solution is wrong.

Analysis and explanation

This question tests two distinct calculus applications through internal choice, and a student should attempt whichever they are more confident in.


Concept — Consumer's Surplus (a)


  • Consumer's surplus is the monetary gain to buyers who were willing to pay more than the actual market price. Geometrically it is the area between the demand curve and the horizontal price line p=p0p = p_0, from x=0x = 0 to the equilibrium quantity x₀.
  • The single formula CS=∫0x0f(x)dx−p0⋅x0CS = \int_0^{x_0} f(x) dx - p_0 \cdot x_0 captures this area; the definite integral is the total willingness-to-pay and p₀·x₀ is the amount actually paid.

Concept — Separable ODE (b)


  • An equation of the form g(y)dy=h(x)dxg(y) dy = h(x) dx is solved by integrating both sides. The algebraic trick of writing 2x/(x−1)=2+2/(x−1)2x/(x-1) = 2 + 2/(x-1) (splitting an improper rational function) is the key step and the most common place students get stuck.

Exam trap


  • In (a), students forget to reject the negative root x=−3x = -3, or subtract p₀·x₀ before integrating.
  • In (b), forgetting the +C or applying the initial condition incorrectly loses the accuracy mark.

Real-world


  • Consumer's surplus is a core idea in welfare economics used to measure how much a market price benefits consumers; governments use it when analysing the impact of taxes and subsidies.

Interesting facts

The concept of consumer's surplus was introduced by French engineer Jules Dupuit in 1844 and popularised by Alfred Marshall in his 1890 'Principles of Economics'.

Separable differential equations are the oldest solved class of ODEs — Leibniz described the separation-of-variables method in 1691, just years after inventing calculus notation.

Consumer's surplus is still used today by regulators and antitrust bodies to estimate the welfare loss from monopolies and price increases.

Question details

Q33 of 44 questions

Board
CBSE
Class
Class 12
Subject
Applied Mathematics
Year
2026
Section
Section D
Marks
5
Type
Long Answer
Chapter
Integration and its Applications / Differential Equations
Topic
Consumer's Surplus / Initial Value Differential Equation
Bloom's level
apply
Syllabus topic
Unit 3: Calculus — Application of Integration (Consumer & Producer Surplus) and Differential Equations
Author
CBSE
Reviewed by
Padhantu Academic Team

Spotted a mistake or something unclear?

Tell us and we will fix it.

Frequently asked questions

Why do we reject x=−3x = -3 when finding the quantity in the consumer's surplus problem?

The variable x represents the quantity demanded, which cannot be negative in an economic context. Solving x2+x−6=0x^2 + x - 6 = 0 gives x=2x = 2 and x=−3x = -3; only x=2x = 2 is physically meaningful, so x0=2x_0 = 2 is used as the upper limit of integration.

How do you integrate 2x/(x−1)2x/(x - 1) in the differential equation part?

Since the numerator's degree is not lower than the denominator's, first do the division: 2x/(x−1)=2+2/(x−1)2x/(x - 1) = 2 + 2/(x - 1). Then integrate term by term to get 2x+2ln⁡∣x−1∣2x + 2 \ln|x - 1|, which is the standard technique for improper rational functions.