Class 12 Applied Maths Unit 3 — Calculus Study Material
This page covers all topics in Unit 3 of CBSE Class 12 Applied Mathematics — carrying 15 marks in the board exam, making it the second-highest weightage unit.
You'll find 6 MCQs with solutions, 6 short-answer and 5 long-answer questions with complete step-by-step workings, and 3 case studies based on Manufacturing, Water Tanks, and E-commerce.
Topics include Differentiation, Applications of Derivatives (maxima, minima, marginal cost), Definite and Indefinite Integrals, Applications of Integrals (consumer surplus, area), and Differential Equations.
All content is aligned to the CBSE 2026–27 syllabus.
Attempt each question, then click Show Solution to reveal the explanation
Question 1
The derivative of \(x^3 + 3x^2 + 4x + 5\) is:
(a) \(3x^2 + 6x + 4\)
(b) \(x^3 + 3x^2 + 4\)
(c) \(3x^2 + 6x + 5\)
(d) \(x^2 + 6x + 4\)
✓ Correct Answer: (a) \(3x^2 + 6x + 4\)
Using the power rule \(\dfrac{d}{dx}(x^n)=nx^{n-1}\):
\(\dfrac{d}{dx}(x^3)=3x^2\), \(\dfrac{d}{dx}(3x^2)=6x\), \(\dfrac{d}{dx}(4x)=4\), \(\dfrac{d}{dx}(5)=0\)
Adding: derivative \(= 3x^2+6x+4\)
Question 2
\(\displaystyle\int(2x+3)\,dx\) equals:
(a) \(x^2+3x+C\)
(b) \(2x^2+3x+C\)
(c) \(x^2+3+C\)
(d) \(2x+3x+C\)
✓ Correct Answer: (a) \(x^2+3x+C\)
\(\displaystyle\int 2x\,dx = 2\cdot\dfrac{x^2}{2} = x^2\) and \(\displaystyle\int 3\,dx = 3x\)
Adding constant of integration: \(x^2+3x+C\)
The product rule: \(\dfrac{d}{dx}(uv) = u\dfrac{dv}{dx} + v\dfrac{du}{dx}\)
Always define \(u\) and \(v\) clearly at the start — examiners award method marks for this.
Question 6
The second derivative of \(x^4\) is:
(a) \(12x^2\)
(b) \(4x^3\)
(c) \(12x\)
(d) \(24x\)
✓ Correct Answer: (a) \(12x^2\)
First derivative: \(\dfrac{d}{dx}(x^4) = 4x^3\)
Second derivative: \(\dfrac{d}{dx}(4x^3) = 12x^2\)
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Step 3 — Apply the limits using \(\displaystyle\int_a^b f(x)\,dx = F(b) - F(a)\):
At \(x=3\): \(F(3) = 3^4 - 3^2 = 81 - 9 = 72\)
At \(x=1\): \(F(1) = 1^4 - 1^2 = 1 - 1 = 0\)
Step 4 — Subtract: \(F(3) - F(1) = 72 - 0 = 72\)
Answer: \(\mathbf{72}\)
Question 3
A company's profit function is \(P(x) = -2x^2 + 40x - 50\), where \(x\) is units sold (hundreds). Find the number of units that maximises profit.
Step 1 — Differentiate \(P(x) = -2x^2 + 40x - 50\) with respect to \(x\):
\(\dfrac{dP}{dx} = -4x + 40\)
Step 2 — Set the derivative to zero to find the critical point:
\(-4x + 40 = 0 \Rightarrow 4x = 40 \Rightarrow x = 10\)
Step 3 — Second derivative test to confirm whether this is a maximum or minimum:
\(\dfrac{d^2P}{dx^2} = -4\)
Since \(\dfrac{d^2P}{dx^2} = -4 < 0\), the function has a maximum at \(x = 10\). ✓
Step 4 — Calculate the maximum profit by substituting \(x = 10\):
\(P(10) = -2(100) + 40(10) - 50 = -200 + 400 - 50\) = ₹150 (hundreds)
Profit is maximised at \(x = 10\) hundred units = 1,000 units Maximum profit = ₹150 thousand
Question 4
Find the derivative of \(y = \ln(x^2+5x+1)\).
Identify the rule to use: This is a composite function — a logarithm of another function. Use the chain rule for logarithms:
\(\dfrac{d}{dx}[\ln f(x)] = \dfrac{f'(x)}{f(x)}\)
Identify \(f(x)\):
\(f(x) = x^2 + 5x + 1\)
Differentiate \(f(x)\) using the power rule:
\(f'(x) = 2x + 5\)
Apply the formula:
\(\dfrac{dy}{dx} = \dfrac{f'(x)}{f(x)} = \dfrac{2x+5}{x^2+5x+1}\)
\(\dfrac{dy}{dx} = \dfrac{2x+5}{x^2+5x+1}\)
Question 5
Find the area bounded by \(y = x^2\), the \(x\)-axis, and the lines \(x=1\) and \(x=3\).
Area \(= \displaystyle\int_1^3 x^2\,dx = \left[\dfrac{x^3}{3}\right]_1^3\)
4-mark and 5-mark practice questions — click Show Solution to reveal full working
Question 1
A wire of length 28 cm is cut into two pieces. One piece is bent into a square, the other into a circle. Find the length of each piece so that the total enclosed area is minimum.
Let wire for square \(= x\) cm; wire for circle \(= (28-x)\) cm.
Area of square: \(\left(\dfrac{x}{4}\right)^2 = \dfrac{x^2}{16}\). Radius of circle: \(r = \dfrac{28-x}{2\pi}\).
Total area: \(A = \dfrac{x^2}{16} + \dfrac{(28-x)^2}{4\pi}\)
Differentiate and set to zero: \(\dfrac{dA}{dx} = \dfrac{x}{8} - \dfrac{28-x}{2\pi} = 0\)
Solving: \(\pi x = 4(28-x) \Rightarrow x(\pi+4) = 112 \Rightarrow x = \dfrac{112}{\pi+4}\approx 15.68\) cm.
The marginal cost function is \(MC = 3x^2 - 10x + 3\). The total cost of producing 3 units is ₹49. Find the total cost function and the cost of producing 5 units.
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Case Studies — Real-World Application Questions
4-mark case-based questions as per the latest CBSE Class 12 Applied Maths pattern — click Show Solution under each part
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:
\[C(x) = 2x^3 - 15x^2 + 36x + 10\]
The selling price per unit is fixed at ₹200. The company wants to optimise production.
(i) Find the marginal cost function \(dC/dx\)
\(C(x) = 2x^3-15x^2+36x+10\)
\(MC = \dfrac{dC}{dx} = 6x^2-30x+36\)
(ii) Find the value of \(x\) for which marginal cost is minimum
\(\dfrac{d(MC)}{dx} = 12x-30 = 0 \Rightarrow x = 2.5\)
(iii)(b) OR: If \(R(x) = 200x-2x^2\), find the production level that maximises profit
\(P(x) = R(x)-C(x) = -2x^3+13x^2+164x-10\)
\(\dfrac{dP}{dx} = -6x^2+26x+164 = 0 \Rightarrow x \approx 6.73\)
Optimal production ≈ 673 units for maximum profit
Case Study 2: Water Tank Filling Problem
Context: A cylindrical water tank has radius 5 metres. The height \(h\) (in metres) at time \(t\) (in minutes) is \(h(t) = 0.1t^2+0.5t\). (Use \(\pi = 3.14\))
(i) Find the rate of change of height at \(t = 10\) minutes
\(\dfrac{dh}{dt} = 0.2t+0.5\)
At \(t=10\): \(\dfrac{dh}{dt} = 2+0.5 = 2.5\) m/min
Rate of height increase at \(t=10\) min = 2.5 m/min
(ii) Calculate the height after 20 minutes
\(h(20) = 0.1(400)+0.5(20) = 40+10 = 50\) m
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
\(V = \pi r^2 h = 25\times 3.14\times 15\)
\(V = \mathbf{1{,}177.5}\) m³
(iii)(b) OR: Find the rate of change of volume at \(t = 15\) minutes
At \(t=15\): \(\dfrac{dV}{dt} = 78.5\times(3+0.5) = 78.5\times 3.5\)
Rate of volume change = 274.75 m³/min
Case Study 3: E-Commerce Sales Analysis
Context: An e-commerce company models daily sales revenue (₹ thousands) as a function of hours \(x\) after business start:
\[S(x) = -x^3+9x^2+15x+20, \quad 0\le x\le 8\]
(i) Find the rate of change of sales with respect to time
\(\dfrac{dS}{dx} = -3x^2+18x+15\) (₹ thousand per hour)
(ii) Find the time at which the sales rate is maximum
\(\dfrac{d^2S}{dx^2} = -6x+18 = 0 \Rightarrow x = 3\) hours
\(\dfrac{d^3S}{dx^3} = -6 < 0\) → confirms maximum ✓
Sales rate is maximum at 3 hours after opening
(iii)(a) Calculate the total sales revenue from hour 2 to hour 6
\(S(6) = -216+324+90+20 = 218\)
\(S(2) = -8+36+30+20 = 78\)
Revenue increase \(= 218-78 = 140\)
Revenue from hour 2 to 6 = ₹140 thousand
(iii)(b) OR: Find when the sales revenue is exactly ₹100 thousand
Set \(-x^3+9x^2+15x+20 = 100\), i.e. \(x^3-9x^2-15x+80 = 0\).
By numerical methods: \(x\approx 4.37\) hours.
Sales reach ₹100 thousand at approximately 4 hours 22 minutes
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How to Score Full Marks in Unit 3 — Exam Tips
Common mistakes examiners flag every year in CBSE Class 12 Applied Maths
✅ Tip 1
Always verify maxima or minima using the second derivative test. Finding where \(f'(x)=0\) is only half the answer. You must then compute \(f''(x)\) at that point and write explicitly: "Since \(f''(x) < 0\), this is a maximum" or "Since \(f''(x) > 0\), this is a minimum." Skipping this verification step costs 1 mark in almost every long-answer question on maxima and minima.
🔒 More tips — including integration step-writing, consumer surplus setup, differential equations conclusion format, and MCQ shortcuts — are all in the Question Bank.
Answers to questions students commonly ask about Class 12 Applied Maths Unit 3
Unit 3 — Calculus carries 15 marks in the CBSE Class 12 Applied Mathematics board exam, making it the second-highest weightage unit. Questions can appear as 1-mark MCQs, 3-mark short answers, 4-mark long answers, and 4-mark case studies.
Unit 3 covers five topics: Higher-order Differentiation, Applications of Derivatives (rate of change, maxima and minima, marginal cost and revenue), Indefinite and Definite Integrals (substitution, partial fractions, integration by parts), Applications of Integrals (consumer and producer surplus, area under curves), and Differential Equations (order, degree, formation and solutions).
Marginal Cost \((MC) = \dfrac{dC}{dx}\) — the derivative of the total cost function. Marginal Revenue \((MR) = \dfrac{dR}{dx}\).
To recover total cost from MC, integrate it and use the given boundary condition to find the constant of integration.
Consumer Surplus \(= \displaystyle\int_0^{x_0} D(x)\,dx - p_0 x_0\), where \(D(x)\) is the demand function, \(x_0\) is the equilibrium quantity, and \(p_0\) is the equilibrium price. It represents the total benefit consumers gain by paying the market price rather than the maximum they were individually willing to pay.
Order = the highest derivative present in the equation. Degree = the power of that highest derivative, provided the equation is polynomial in derivatives.
Important: if the equation contains \(\sin\!\left(\dfrac{dy}{dx}\right)\) or \(e^{dy/dx}\), the degree is not defined — state this explicitly in the exam for full marks.
Based on CBSE papers, the most frequently asked topics are:
• Maxima and Minima — every year as a long answer
• Marginal Cost and Revenue — case study or short answer
• Definite Integrals — MCQ and long answer
• Area under curves — long answer
• Differential Equations — long answer most years
The product rule: \(\dfrac{d}{dx}(uv) = u\dfrac{dv}{dx} + v\dfrac{du}{dx}\).
Always define \(u\) and \(v\) clearly at the start of your solution — examiners award partial marks for correct method even if arithmetic slips.
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