Unit 3: Calculus — Class 12 Applied Maths | MCQs, Solved Examples & Case Studies

📐 Unit 3: Calculus

Master Differentiation & Its Applications, Integration & Its Applications, Differential Equations

⚖️ Weightage: 15 Marks in Board Exam

This page covers all topics in Unit 3 — Calculus of CBSE Class 12 Applied Mathematics, the second-highest weightage unit at 15 marks in the board exam. You'll find 6 practice MCQs, 6 step-by-step solved examples, 5 long-answer questions (including CBSE 2023, 2024 & 2025 board questions), and 3 case studies on Manufacturing Cost Optimisation, Water Tank Problems and E-commerce Sales Analysis. Topics covered include Higher-order Differentiation, Applications of Derivatives (maxima & minima, marginal cost & revenue), Indefinite and Definite Integrals, Applications of Integrals (consumer and producer surplus, area under curves), and Differential Equations. All content is aligned to the CBSE 2025–26 syllabus.

15
Board Exam Marks
5
Topics Covered
6
Practice MCQs
11
Solved Examples
3
Case Studies

Topics Covered in Unit 3 — Calculus

Master all 5 key topics for the board exam

1. Differentiation

Higher order derivatives and their applications

2. Applications of Derivatives

Rate of change, increasing-decreasing, maxima-minima, marginal cost and revenue

3. Integrals

Indefinite, definite, substitution, partial fractions, integration by parts

4. Applications of Integrals

Cost-revenue functions, consumer and producer surplus, area under curves

5. Differential Equations

Order and degree, formation and solving differential equations

Key Formulas — Unit 3 Quick Reference

Essential formulas to memorise for the board exam

Power Rule

Derivative of xⁿ with respect to x.

d/dx(xⁿ) = n·xⁿ⁻¹

Product Rule

Differentiating the product of two functions u and v.

d/dx(uv) = u·v' + v·u'

Chain Rule

For composite functions f(g(x)).

dy/dx = (dy/du)·(du/dx)

Integration Power Rule

Indefinite integral of xⁿ (n ≠ −1).

∫xⁿ dx = xⁿ⁺¹/(n+1) + C

Definite Integral

Applying the Fundamental Theorem of Calculus.

∫ₐᵇ f(x)dx = F(b) − F(a)

Maxima & Minima Test

Find x where f'(x) = 0; check sign of f''(x).

f''<0 → max | f''>0 → min

Marginal Cost & Revenue

Derivatives of total cost and revenue functions.

MC = dC/dx MR = dR/dx

Consumer Surplus

Area between demand curve and equilibrium price.

CS = ∫₀^x₀ D(x)dx − p₀·x₀

🎥 Video Tutorials — Class 12 Applied Maths Unit 3

3
Calculus
15 Marks · Series of 7 Videos

Integration & Its Applications — Series (7 Videos)

▶ Watch Series on YouTube →
✓ Indefinite Integrals ✓ Definite Integrals ✓ Area Under the Curve ✓ Consumer & Producer Surplus

🔜 Coming Soon

Differentiation & Applications Differential Equations

Practice MCQs with Answers — Unit 3 Calculus

Click "Show Answer" to reveal step-by-step explanations

Question 1
The derivative of x³ + 3x² + 4x + 5 is:
(a) 3x² + 6x + 4
(b) x³ + 3x² + 4
(c) 3x² + 6x + 5
(d) x² + 6x + 4
✓ Correct Answer: (a) 3x² + 6x + 4
Solution:
Using the power rule d/dx(xⁿ) = nxⁿ⁻¹:
d/dx(x³) = 3x²;   d/dx(3x²) = 6x;   d/dx(4x) = 4;   d/dx(5) = 0
Therefore derivative = 3x² + 6x + 4
Question 2
∫(2x + 3)dx equals:
(a) x² + 3x + C
(b) 2x² + 3x + C
(c) x² + 3 + C
(d) 2x + 3x + C
✓ Correct Answer: (a) x² + 3x + C
Solution:
Using ∫xⁿdx = xⁿ⁺¹/(n+1) + C:
∫2x dx = 2 × x²/2 = x²;   ∫3 dx = 3x
Therefore ∫(2x + 3)dx = x² + 3x + C
Question 3
∫₀² x² dx equals:
(a) 8/3
(b) 4/3
(c) 8
(d) 2/3
✓ Correct Answer: (a) 8/3
Solution:
∫x² dx = x³/3. Applying limits 0 to 2:
[x³/3]₀² = (2³/3) − (0³/3) = 8/3
Question 4
The derivative of e^(3x) is:
(a) 3e^(3x)
(b) e^(3x)
(c) e^(3x)/3
(d) 3eˣ
✓ Correct Answer: (a) 3e^(3x)
Solution:
Using chain rule: d/dx[e^(ax)] = a·e^(ax). Here a = 3.
Therefore d/dx[e^(3x)] = 3e^(3x)
Question 5
The product rule for differentiation of uv is:
(a) u(dv/dx) + v(du/dx)
(b) u(dv/dx) − v(du/dx)
(c) (du/dx)(dv/dx)
(d) u + v
✓ Correct Answer: (a) u(dv/dx) + v(du/dx)
Solution:
The product rule: d/dx(uv) = u(dv/dx) + v(du/dx).
This is one of the fundamental rules of differentiation.
Question 6
The second derivative of x⁴ is:
(a) 12x²
(b) 4x³
(c) 12x
(d) 24x
✓ Correct Answer: (a) 12x²
Solution:
First derivative: d/dx(x⁴) = 4x³
Second derivative: d/dx(4x³) = 12x²

Short Answer Questions — Step-by-Step Solved Examples

2-mark and 3-mark questions from CBSE 2022–2025 papers

Example 1
Find the derivative of f(x) = (2x + 1)(3x² − 4)
Solution:
Using the product rule: d/dx(uv) = u(dv/dx) + v(du/dx)
Let u = 2x + 1 and v = 3x² − 4
du/dx = 2  ;  dv/dx = 6x
f'(x) = (2x + 1)(6x) + (3x² − 4)(2) = 12x² + 6x + 6x² − 8
Answer: f'(x) = 18x² + 6x − 8
Example 2
Evaluate ∫₁³ (4x³ − 2x)dx
Solution:
∫(4x³ − 2x)dx = 4(x⁴/4) − 2(x²/2) + C = x⁴ − x² + C
[x⁴ − x²]₁³ = (81 − 9) − (1 − 1) = 72 − 0
Answer: 72
Example 3
A company's profit function is P(x) = −2x² + 40x − 50, where x is the number of units sold (in hundreds). Find the number of units that maximises profit.
Solution:
dP/dx = −4x + 40. Setting dP/dx = 0: −4x + 40 = 0 → x = 10
d²P/dx² = −4 < 0 → maximum confirmed ✓
Answer: 10 hundred units = 1,000 units maximises profit
Example 4
Find the derivative of y = ln(x² + 5x + 1)
Solution:
Using chain rule: d/dx[ln f(x)] = f'(x)/f(x)
f(x) = x² + 5x + 1;   f'(x) = 2x + 5
Answer: dy/dx = (2x + 5)/(x² + 5x + 1)
Example 5
Find the area bounded by the curve y = x², the x-axis, and the lines x = 1 and x = 3
Solution:
Area = ∫₁³ x² dx = [x³/3]₁³
= (27/3) − (1/3) = 26/3 square units
Answer: Area = 26/3 ≈ 8⅔ square units
Example 6
The revenue function is R(x) = 100x − 0.5x², where x is quantity sold. Find the marginal revenue at x = 40 units.
Solution:
MR = dR/dx = 100 − x
At x = 40: MR = 100 − 40 = 60
Answer: Marginal Revenue at 40 units = ₹60 per unit

Long Answer Questions with Complete Solutions

4-mark and 6-mark questions from CBSE 2023, 2024 & 2025 board papers

Question 1 CBSE 2024
A wire of length 28 cm is to be cut into two pieces. One piece will be bent to form a square and the other to form a circle. Find the length of each piece so that the total area enclosed is minimum.
Complete Solution:
Let wire for square = x cm, for circle = (28 − x) cm
Area of square = (x/4)² = x²/16;   Radius of circle r = (28−x)/(2π)
Total area A = x²/16 + (28−x)²/(4π)
dA/dx = 0 → x/8 = (28−x)/(2π) → x(π+4) = 112 → x = 112/(π+4) ≈ 15.68 cm
d²A/dx² > 0 → confirmed minimum ✓
Length for square = 112/(π+4) cm ≈ 15.68 cm
Length for circle = 28π/(π+4) cm ≈ 12.32 cm
Question 2 CBSE 2024
Evaluate: ∫ (x² + 1)/(x⁴ + x² + 1) dx
Complete Solution:
Divide numerator and denominator by x²: = ∫ (1 + 1/x²)/[(x − 1/x)² + 3] dx
Let t = x − 1/x → dt = (1 + 1/x²) dx
= ∫ dt/(t² + (√3)²) = (1/√3) tan⁻¹(t/√3) + C
= (1/√3) tan⁻¹[(x² − 1)/(x√3)] + C
Question 3 CBSE 2023
The marginal cost function is MC = 3x² − 10x + 3. The total cost of producing 3 units is ₹49. Find the total cost function and the cost of producing 5 units.
Complete Solution:
C = ∫(3x² − 10x + 3) dx = x³ − 5x² + 3x + k
C(3) = 49 → 27 − 45 + 9 + k = 49 → k = 58
C(5) = 125 − 125 + 15 + 58 = 73
Total cost function: C(x) = x³ − 5x² + 3x + 58
Cost of producing 5 units = ₹73
Question 4 CBSE 2025
Find the area of the region bounded by the parabola y² = 4x and the line y = 2x − 4.
Complete Solution:
Intersection: y²/4 = (y+4)/2 → y² − 2y − 8 = 0 → y = 4 or y = −2
Area = ∫₋₂⁴ [(y+4)/2 − y²/4] dy = (1/4)∫₋₂⁴ (8 + 2y − y²) dy
= (1/4)[8y + y² − y³/3]₋₂⁴ = (1/4) × 60 = 15
Area = 15 square units
Question 5 CBSE 2023
Solve the differential equation: (x + y) dy/dx = 1, given y = 0 when x = 1.
Complete Solution:
Let t = x + y → integrating: y = log|t + 1| + C = log|x + y + 1| + C
Given y = 0, x = 1: 0 = log 2 + C → C = −log 2
y = log|(x + y + 1)/2|

Case Studies — Real-World Application Questions

4-mark case-based questions as per latest CBSE Class 12 Applied Maths pattern

Case Study 1: Manufacturing Cost Optimisation

Context: A company manufactures electronic components. The total cost function (in thousands of rupees) for producing x units (in hundreds) is:

C(x) = 2x³ − 15x² + 36x + 10

The selling price per unit is fixed at ₹200. The company wants to optimise production to minimise average cost and maximise profit.

(i) Find the marginal cost function dC/dx
C(x) = 2x³ − 15x² + 36x + 10
MC(x) = dC/dx = 6x² − 30x + 36
(ii) Find the value of x for which marginal cost is minimum
d(MC)/dx = 12x − 30 = 0 → x = 2.5
d²(MC)/dx² = 12 > 0 → confirms minimum ✓
Marginal cost is minimum at x = 2.5 hundred units = 250 units
(iii)(a) Find the minimum marginal cost
MC(2.5) = 6(6.25) − 30(2.5) + 36 = 37.5 − 75 + 36 = −1.5
Minimum Marginal Cost = ₹−1,500 per hundred units (₹15 per unit in context of fixed overheads)
(iii)(b) OR: If R(x) = 200x − 2x², find the production level that maximises profit
P(x) = R(x) − C(x) = −2x³ + 13x² + 164x − 10
dP/dx = −6x² + 26x + 164 = 0 → x ≈ 6.73 hundred units
Optimal production ≈ 673 units for maximum profit
Case Study 2: Water Tank Filling Problem

Context: A cylindrical water tank has a radius of 5 metres. The height h (in metres) at time t (in minutes) is:

h(t) = 0.1t² + 0.5t

Management needs to analyse the rate of filling and total volume filled over time. (Use π = 3.14)

(i) Find the rate of change of height at t = 10 minutes
dh/dt = 0.2t + 0.5
At t = 10: dh/dt = 2 + 0.5 = 2.5 m/min
Rate of height increase at t = 10 min: 2.5 metres per minute
(ii) Calculate the height after 20 minutes
h(20) = 0.1(400) + 0.5(20) = 40 + 10 = 50 metres
Height after 20 minutes: 50 metres
(iii)(a) Find the volume of water filled in the first 10 minutes
h(10) = 0.1(100) + 5 = 15 m; V = πr²h = 25π × 15 = 375π
Volume = 375 × 3.14 = 1,177.5 m³
(iii)(b) OR: Find the rate of change of volume at t = 15 minutes
dV/dt = 25π · dh/dt = 25π(0.2t + 0.5)
At t = 15: dV/dt = 25π(3 + 0.5) = 87.5 × 3.14 = 274.75 m³/min
Rate of volume change at t = 15 min: 274.75 m³/min
Case Study 3: E-commerce Sales Analysis

Context: An e-commerce company models its daily sales revenue (₹ thousands) as a function of hours x after business start:

S(x) = −x³ + 9x² + 15x + 20,   0 ≤ x ≤ 8

The company wants to identify peak sales hours and calculate total revenue for optimal resource allocation.

(i) Find the rate of change of sales with respect to time
dS/dx = −3x² + 18x + 15 (₹ thousand per hour)
(ii) Find the time at which sales rate is maximum
d²S/dx² = −6x + 18 = 0 → x = 3 hours
d³S/dx³ = −6 < 0 → confirms maximum ✓
Sales rate is maximum at 3 hours after opening
(iii)(a) Calculate the total sales revenue from hour 2 to hour 6
S(6) = −216 + 324 + 90 + 20 = 218; S(2) = −8 + 36 + 30 + 20 = 78
Difference = 218 − 78 = 140
Revenue from hour 2 to 6: ₹140 thousand
(iii)(b) OR: Find when the sales revenue is exactly ₹100 thousand
−x³ + 9x² + 15x + 20 = 100 → x³ − 9x² − 15x + 80 = 0
By numerical methods: x ≈ 4.37 hours
Sales reach ₹100 thousand at approximately 4 hours 22 minutes

How to Score Full Marks in Unit 3 — Exam Tips

Examiner-tested strategies for CBSE Class 12 Applied Maths

📌 Always Verify Maxima / Minima

After finding the critical point (where f'(x) = 0), always check the sign of f''(x) and write "f''(x) < 0 → maximum" or "f''(x) > 0 → minimum" explicitly. Skipping this loses 1 mark.

∫ Add the Constant of Integration

Never forget "+ C" in indefinite integrals. Many students lose easy marks on 1-mark MCQs by omitting the constant. Make it a habit — always write + C.

💰 Define MC and MR clearly

In business application problems, write "MC = dC/dx" and "MR = dR/dx" at the start. Explicit definitions earn method marks even if arithmetic slips.

📐 Show All Integration Steps

For definite integrals, always write the antiderivative in brackets [F(x)]ₐᵇ and then substitute limits separately. Full marks require showing both the antiderivative and the substitution step.

📋 Verify Case Study Answers

In case studies, substitute your answer back into the original condition. A 30-second check can save 2 marks — case study errors are the most common reason for score loss in Calculus.

🔗 Chain Rule — Label Each Step

For composite functions, write the substitution explicitly: "Let u = inner function, then dy/dx = (dy/du)·(du/dx)". Examiners award partial marks for correct method even with a wrong answer.

Frequently Asked Questions — Unit 3 Calculus

Common questions students ask about Class 12 Applied Maths Unit 3

Unit 3 — Calculus carries 15 marks in the CBSE Class 12 Applied Mathematics board exam. Questions appear as 1-mark MCQs, 3-mark short answers, 4-mark long answers, and 4-mark case studies covering differentiation, integration and differential equations.
Unit 3 covers: (1) Higher-order Differentiation, (2) Applications of Derivatives — rate of change, increasing/decreasing functions, maxima & minima, marginal cost and revenue, (3) Indefinite and Definite Integrals — substitution, partial fractions, integration by parts, (4) Applications of Integrals — consumer and producer surplus, area under curves, (5) Differential Equations — order, degree, formation and solutions.
Marginal Cost (MC) = dC/dx, the derivative of the total cost function. Marginal Revenue (MR) = dR/dx, the derivative of the revenue function. To recover the total cost function from MC, integrate MC with respect to x and use a given boundary condition to find the constant of integration.
Consumer Surplus = ∫₀^x₀ D(x) dx − p₀·x₀, where D(x) is the demand function, x₀ is the equilibrium quantity, and p₀ is the equilibrium price. It represents the total benefit consumers gain by paying the market price rather than the maximum they were willing to pay.
Order is the highest derivative present. Degree is the power of that highest derivative, provided the equation is a polynomial in derivatives. Important: if the equation contains sin(dy/dx) or e^(dy/dx), the degree is not defined — state this explicitly in the exam for full marks.
Based on CBSE 2022–2025 papers: Maxima & Minima (every year as a long answer), Marginal Cost & Revenue (case study or short answer), Definite Integrals (MCQ and long answer), Area under curves, and Differential Equations (long answer). The second derivative test for verifying the nature of extrema appears in almost every paper.
The product rule: d/dx(uv) = u·(dv/dx) + v·(du/dx). Always define u and v clearly at the start of your solution. This is one of the most-tested differentiation rules in MCQs and short-answer questions.