📐 Unit 3: Calculus
Master Differentiation & Its Applications, Integration & Its Applications, Differential Equations
Weightage: 15 Marks in Board ExamTopics Covered
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
Multiple Choice Questions
Question 1
The derivative of x³ + 3x² + 4x + 5 is:
✓ Correct Answer: (a) 3x² + 6x + 4
Solution:
Using the power rule of differentiation: 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
Using the power rule of differentiation: 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:
✓ Correct Answer: (a) x² + 3x + C
Solution:
Using integration formula: ∫xⁿdx = xⁿ⁺¹/(n+1) + C
∫2x dx = 2 × x²/2 = x²
∫3 dx = 3x
Therefore, ∫(2x + 3)dx = x² + 3x + C
Using integration formula: ∫xⁿdx = xⁿ⁺¹/(n+1) + C
∫2x dx = 2 × x²/2 = x²
∫3 dx = 3x
Therefore, ∫(2x + 3)dx = x² + 3x + C
Question 5
∫₀² x² dx equals:
✓ Correct Answer: (a) 8/3
Solution:
∫x² dx = x³/3
Applying limits from 0 to 2:
[x³/3]₀² = (2³/3) - (0³/3)
= 8/3 - 0 = 8/3
∫x² dx = x³/3
Applying limits from 0 to 2:
[x³/3]₀² = (2³/3) - (0³/3)
= 8/3 - 0 = 8/3
Question 6
The derivative of e^(3x) is:
✓ Correct Answer: (a) 3e^(3x)
Solution:
Using chain rule for exponential functions:
d/dx[e^(ax)] = a × e^(ax)
Here a = 3
Therefore, d/dx[e^(3x)] = 3e^(3x)
Using chain rule for exponential functions:
d/dx[e^(ax)] = a × e^(ax)
Here a = 3
Therefore, d/dx[e^(3x)] = 3e^(3x)
Question 9
The product rule for differentiation of uv is:
✓ Correct Answer: (a) u(dv/dx) + v(du/dx)
Solution:
The product rule states:
d/dx(uv) = u(dv/dx) + v(du/dx)
This is one of the fundamental differentiation rules.
The product rule states:
d/dx(uv) = u(dv/dx) + v(du/dx)
This is one of the fundamental differentiation rules.
Question 10
The second derivative of x⁴ is:
✓ Correct Answer: (a) 12x²
Solution:
First derivative: d/dx(x⁴) = 4x³
Second derivative: d/dx(4x³) = 12x²
First derivative: d/dx(x⁴) = 4x³
Second derivative: d/dx(4x³) = 12x²
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Solved Examples
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
= 18x² + 6x - 8
Answer: f'(x) = 18x² + 6x - 8
Example 2
Evaluate ∫₁³ (4x³ - 2x)dx
Solution:
First, find the indefinite integral: ∫(4x³ - 2x)dx
= 4(x⁴/4) - 2(x²/2) + C
= x⁴ - x² + C
Now apply the limits from 1 to 3:
[x⁴ - x²]₁³ = (3⁴ - 3²) - (1⁴ - 1²)
= (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 maximizes profit.
Solution:
To find maximum profit, we find where dP/dx = 0
P(x) = -2x² + 40x - 50
dP/dx = -4x + 40
Setting dP/dx = 0:
-4x + 40 = 0
4x = 40
x = 10
To verify it's a maximum, check d²P/dx²:
d²P/dx² = -4 (negative, so it's a maximum)
Answer: 10 hundred units = 1000 units maximizes 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)
Here f(x) = x² + 5x + 1
f'(x) = 2x + 5
Therefore, dy/dx = (2x + 5)/(x² + 5x + 1)
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
= (3³/3) - (1³/3)
= 27/3 - 1/3
= 9 - 1/3
= 26/3 square units
Answer: Area = 26/3 or 8⅔ square units
Example 6
The revenue function for a product is R(x) = 100x - 0.5x², where x is the quantity sold. Find the marginal revenue when x = 40 units.
Solution:
Marginal Revenue = dR/dx
R(x) = 100x - 0.5x²
dR/dx = 100 - x
At x = 40:
MR = 100 - 40 = 60
Answer: Marginal Revenue at 40 units = Rs 60 per unit
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Long Answer Questions with Complete Solutions
Practice 4-mark and 6-mark questions
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 the length of wire for square = x cm
Then length for circle = (28 - x) cm
Side of square = x/4
Area of square = (x/4)² = x²/16
Circumference of circle = 28 - x = 2πr
Therefore, r = (28 - x)/(2π)
Area of circle = πr² = π[(28 - x)/(2π)]² = (28 - x)²/(4π)
Total area A = x²/16 + (28 - x)²/(4π)
For minimum area: dA/dx = 0
2x/16 - 2(28 - x)/(4π) = 0
x/8 = (28 - x)/(2π)
πx = 4(28 - x)
πx = 112 - 4x
x(π + 4) = 112
x = 112/(π + 4) ≈ 15.68 cm
Length for square = 112/(π + 4) cm ≈ 15.68 cm
Length for circle = 28 - 112/(π + 4) = 28π/(π + 4) cm ≈ 12.32 cm
Length for circle = 28 - 112/(π + 4) = 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:
∫ (x² + 1)/(x⁴ + x² + 1) dx = ∫ (1 + 1/x²)/(x² + 1 + 1/x²) dx
= ∫ (1 + 1/x²)/[(x - 1/x)² + 3] dx
Let t = x - 1/x
dt = (1 + 1/x²) dx
The integral becomes: ∫ dt/(t² + 3)
= ∫ dt/(t² + (√3)²)
= (1/√3) tan⁻¹(t/√3) + C
= (1/√3) tan⁻¹[(x - 1/x)/√3] + C
= (1/√3) tan⁻¹[(x² - 1)/(x√3)] + C
= (1/√3) tan⁻¹[(x² - 1)/(x√3)] + C
Question 3
CBSE 2023
The marginal cost function of manufacturing x units of a product is given by MC = 3x² - 10x + 3. The total cost of producing 3 units is Rs 49. Find the total cost function and the cost of producing 5 units.
Complete Solution:
Given: MC = dC/dx = 3x² - 10x + 3
Integrating both sides with respect to x:
C = ∫(3x² - 10x + 3) dx
C = x³ - 5x² + 3x + k (where k is constant)
Given: C(3) = 49
49 = (3)³ - 5(3)² + 3(3) + k
49 = 27 - 45 + 9 + k
49 = -9 + k
k = 58
Therefore, C(x) = x³ - 5x² + 3x + 58
Cost of producing 5 units:
C(5) = (5)³ - 5(5)² + 3(5) + 58
= 125 - 125 + 15 + 58
Total cost function: C(x) = x³ - 5x² + 3x + 58
Cost of producing 5 units = Rs 73
Cost of producing 5 units = Rs 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:
First, find points of intersection:
From y² = 4x: x = y²/4
From y = 2x - 4: x = (y + 4)/2
Setting equal: y²/4 = (y + 4)/2
y² = 2(y + 4)
y² - 2y - 8 = 0
(y - 4)(y + 2) = 0
y = 4 or y = -2
When y = 4: x = 4, When y = -2: x = 1
Area = ∫₋₂⁴ [(y + 4)/2 - y²/4] dy
= ∫₋₂⁴ [(2y + 8 - y²)/4] dy
= (1/4) ∫₋₂⁴ (8 + 2y - y²) dy
= (1/4) [8y + y² - y³/3]₋₂⁴
= (1/4) [(32 + 16 - 64/3) - (-16 + 4 + 8/3)]
= (1/4) [144/3 - (-36/3)]
= (1/4) × 180/3 = 15
Area = 15 square units
Question 5
CBSE 2023
Solve the differential equation: (x + y) dy/dx = 1, given that y = 0 when x = 1.
Complete Solution:
Given: (x + y) dy/dx = 1
dy/dx = 1/(x + y)
Rearranging: (x + y) dy = dx
x dy + y dy = dx
x dy = dx - y dy
Let x + y = t, then dx = dt - dy
Substituting: t dy = dt - dy
t dy + dy = dt
(t + 1) dy = dt
dy = dt/(t + 1)
Integrating: y = log|t + 1| + C
y = log|x + y + 1| + C
Given: y = 0 when x = 1
0 = log|1 + 0 + 1| + C = log 2 + C
C = -log 2
y = log|x + y + 1| - log 2
y = log|(x + y + 1)/2|
y = log|(x + y + 1)/2|
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Case Studies
Case Study 1: Manufacturing Cost Optimization
Context: A company manufactures electronic components. The total cost function (in thousands of rupees) for producing x units (in hundreds) is given by:
C(x) = 2x³ - 15x² + 36x + 10
The company wants to optimize its production to minimize average cost and maximize profit. The selling price per unit is fixed at Rs 200 per unit.
(i) Find the marginal cost function dC/dx
Solution:
C(x) = 2x³ - 15x² + 36x + 10
Marginal Cost = dC/dx
dC/dx = d/dx(2x³ - 15x² + 36x + 10)
= 6x² - 30x + 36
Marginal Cost function: MC(x) = 6x² - 30x + 36
(ii) Find the value of x for which marginal cost is minimum
Solution:
For minimum MC, we need d(MC)/dx = 0
MC(x) = 6x² - 30x + 36
d(MC)/dx = 12x - 30
Setting d(MC)/dx = 0:
12x - 30 = 0
12x = 30
x = 2.5
Verification: 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
Solution:
MC(x) = 6x² - 30x + 36
At x = 2.5:
MC(2.5) = 6(2.5)² - 30(2.5) + 36
= 6(6.25) - 75 + 36
= 37.5 - 75 + 36
= -1.5
Minimum Marginal Cost = Rs 1,500 per hundred units (or Rs 15 per unit)
(iii)(b) OR: If the revenue function is R(x) = 200x - 2x², find the production level that maximizes profit
Solution:
Profit P(x) = R(x) - C(x)
= (200x - 2x²) - (2x³ - 15x² + 36x + 10)
= -2x³ + 13x² + 164x - 10
For maximum profit: dP/dx = 0
dP/dx = -6x² + 26x + 164 = 0
Solving: x ≈ 6.73 hundred units
Optimal production level ≈ 673 units for maximum profit
Case Study 2: Water Tank Filling Problem
Context: A cylindrical water tank has a radius of 5 meters. Water is being pumped into the tank at a rate such that the height h (in meters) at time t (in minutes) is given by:
h(t) = 0.1t² + 0.5t
The management needs to analyze the rate of water filling and the total volume filled over time. (Use π = 3.14)
(i) Find the rate of change of height with respect to time at t = 10 minutes
Solution:
h(t) = 0.1t² + 0.5t
Rate of change of height = dh/dt
dh/dt = 0.2t + 0.5
At t = 10 minutes:
dh/dt = 0.2(10) + 0.5
= 2 + 0.5 = 2.5 m/min
Rate of height increase at t = 10 min: 2.5 meters per minute
(ii) Calculate the height of water in the tank after 20 minutes
Solution:
h(t) = 0.1t² + 0.5t
At t = 20 minutes:
h(20) = 0.1(20)² + 0.5(20)
= 0.1(400) + 10
= 40 + 10
= 50 meters
Height after 20 minutes: 50 meters
(iii)(a) Find the volume of water filled in the first 10 minutes
Solution:
Volume of cylinder = πr²h
V(t) = π(5)²h(t) = 25π(0.1t² + 0.5t)
V(t) = 2.5πt² + 12.5πt
At t = 10:
V(10) = 2.5π(100) + 12.5π(10)
= 250π + 125π = 375π
= 375 × 3.14 = 1177.5 m³
Volume filled in 10 minutes: 1177.5 cubic meters
(iii)(b) OR: Find the rate of change of volume at t = 15 minutes
Solution:
V(t) = 2.5πt² + 12.5πt
Rate of change of volume = dV/dt
dV/dt = 5πt + 12.5π
At t = 15 minutes:
dV/dt = 5π(15) + 12.5π
= 75π + 12.5π = 87.5π
= 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 (in thousands of rupees) as a function of the number of hours x after the start of business day:
S(x) = -x³ + 9x² + 15x + 20, where 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
Solution:
S(x) = -x³ + 9x² + 15x + 20
Rate of change = dS/dx
dS/dx = -3x² + 18x + 15
Rate of sales change: dS/dx = -3x² + 18x + 15 (thousand rupees/hour)
(ii) Find the time at which sales rate is maximum
Solution:
For maximum rate: d²S/dx² = 0
d(dS/dx)/dx = -6x + 18 = 0
6x = 18
x = 3 hours
Check: d³S/dx³ = -6 < 0 (confirms maximum)
Sales rate is maximum at x = 3 hours after opening
(iii)(a) Calculate the total sales revenue from hour 2 to hour 6
Solution:
Total sales = ∫₂⁶ S(x)dx (This represents accumulated sales)
However, S(x) already represents sales, so we calculate:
Sales at x=6: S(6) = -(6)³ + 9(6)² + 15(6) + 20
= -216 + 324 + 90 + 20 = 218
Sales at x=2: S(2) = -(2)³ + 9(2)² + 15(2) + 20
= -8 + 36 + 30 + 20 = 78
Difference in cumulative sales = 218 - 78 = 140
Revenue from hour 2 to 6: Rs 140 thousand
(iii)(b) OR: Find when the sales revenue is exactly Rs 100 thousand
Solution:
S(x) = 100
-x³ + 9x² + 15x + 20 = 100
-x³ + 9x² + 15x - 80 = 0
x³ - 9x² - 15x + 80 = 0
By trial or numerical methods: x ≈ 4.37 hours
Sales reach Rs 100 thousand at approximately 4.37 hours (4 hours 22 minutes)
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