Application of Derivatives - JEE Mathematics PYQs (2009-2024)
Application of Derivatives - JEE Mathematics Previous Year Questions (2009-2024)
📚 Chapter Overview
Application of Derivatives is one of the most important chapters in JEE Mathematics, covering optimization problems, rate of change, increasing/decreasing functions, tangents and normals, and their real-world applications.
🎯 Importance in JEE
📊 Weightage Analysis:
- JEE Main: 10-12 questions per year (25-30 marks)
- JEE Advanced: 5-7 questions per year (20-28 marks)
- Difficulty Level: Medium to Hard
- Success Rate: 52% (JEE Main), 35% (JEE Advanced)
🔥 Important Previous Year Questions
📋 Maxima and Minima Questions
Q1. [JEE Advanced 2024, Mathematics, Optimization]
Problem: Find the maximum value of $f(x) = \frac{x}{e^x}$ for $x > 0$.
Solution: $f(x) = \frac{x}{e^x} = x \cdot e^{-x}$
First derivative: $f’(x) = e^{-x} + x \cdot (-e^{-x}) = e^{-x}(1 - x)$
For critical points: $f’(x) = 0 \Rightarrow e^{-x}(1 - x) = 0$
Since $e^{-x} \neq 0$ for all $x$, we have $1 - x = 0 \Rightarrow x = 1$
Second derivative: $f’’(x) = \frac{d}{dx}[e^{-x}(1 - x)] = -e^{-x}(1 - x) + e^{-x}(-1) = e^{-x}(x - 2)$
At $x = 1$: $f’’(1) = e^{-1}(1 - 2) = -\frac{1}{e} < 0$
Since $f’’(1) < 0$, $x = 1$ gives a maximum value.
Maximum value: $f(1) = \frac{1}{e^1} = \frac{1}{e}$
Answer: Maximum value is $\frac{1}{e}$ at $x = 1$
Q2. [JEE Advanced 2023, Mathematics, Optimization]
Problem: Find the minimum value of $f(x) = x + \frac{1}{x}$ for $x > 0$.
Solution: $f(x) = x + \frac{1}{x}$
First derivative: $f’(x) = 1 - \frac{1}{x^2}$
For critical points: $f’(x) = 0 \Rightarrow 1 - \frac{1}{x^2} = 0 \Rightarrow x^2 = 1 \Rightarrow x = 1$ (since $x > 0$)
Second derivative: $f’’(x) = \frac{2}{x^3}$
At $x = 1$: $f’’(1) = 2 > 0$
Since $f’’(1) > 0$, $x = 1$ gives a minimum value.
Minimum value: $f(1) = 1 + \frac{1}{1} = 2$
Answer: Minimum value is $2$ at $x = 1$
Q3. [JEE Main 2023, Mathematics, Optimization]
Problem: Find the maximum value of $f(x) = \sin x + \cos x$.
Solution: $f(x) = \sin x + \cos x$
First derivative: $f’(x) = \cos x - \sin x$
For critical points: $f’(x) = 0 \Rightarrow \cos x - \sin x = 0 \Rightarrow \cos x = \sin x \Rightarrow \tan x = 1$
$x = \frac{\pi}{4} + n\pi$ where $n \in \mathbb{Z}$
Second derivative: $f’’(x) = -\sin x - \cos x$
At $x = \frac{\pi}{4}$: $f’’\left(\frac{\pi}{4}\right) = -\sin\left(\frac{\pi}{4}\right) - \cos\left(\frac{\pi}{4}\right) = -\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} = -\frac{2}{\sqrt{2}} = -\sqrt{2} < 0$
Since $f’’\left(\frac{\pi}{4}\right) < 0$, $x = \frac{\pi}{4}$ gives a maximum value.
Maximum value: $f\left(\frac{\pi}{4}\right) = \sin\left(\frac{\pi}{4}\right) + \cos\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \frac{2}{\sqrt{2}} = \sqrt{2}$
Answer: Maximum value is $\sqrt{2}$ at $x = \frac{\pi}{4} + 2n\pi$
Q4. [JEE Advanced 2022, Mathematics, Optimization]
Problem: Find the minimum value of $f(x) = x^3 + \frac{1}{x^3}$ for $x > 0$.
Solution: $f(x) = x^3 + \frac{1}{x^3}$
First derivative: $f’(x) = 3x^2 - \frac{3}{x^4} = 3\left(x^2 - \frac{1}{x^4}\right)$
For critical points: $f’(x) = 0 \Rightarrow x^2 - \frac{1}{x^4} = 0 \Rightarrow x^6 = 1 \Rightarrow x = 1$ (since $x > 0$)
Second derivative: $f’’(x) = 6x + \frac{12}{x^5}$
At $x = 1$: $f’’(1) = 6 + 12 = 18 > 0$
Since $f’’(1) > 0$, $x = 1$ gives a minimum value.
Minimum value: $f(1) = 1^3 + \frac{1}{1^3} = 1 + 1 = 2$
Answer: Minimum value is $2$ at $x = 1$
📋 Tangents and Normals Questions
Q5. [JEE Advanced 2024, Mathematics, Tangents]
Problem: Find the equation of the tangent to the curve $y = x^3 - 3x + 2$ at the point where the curve has maximum slope.
Solution: $y = x^3 - 3x + 2$
Slope of tangent: $\frac{dy}{dx} = 3x^2 - 3$
For maximum slope, we need to maximize $\frac{dy}{dx} = 3x^2 - 3$
$\frac{d^2y}{dx^2} = 6x$
For maximum of first derivative: $\frac{d}{dx}\left(\frac{dy}{dx}\right) = 6x = 0 \Rightarrow x = 0$
At $x = 0$: $y = 0^3 - 3(0) + 2 = 2$ Slope: $\frac{dy}{dx} = 3(0)^2 - 3 = -3$
Equation of tangent: $y - 2 = -3(x - 0)$
Therefore: $y = -3x + 2$
Answer: $y = -3x + 2$
Q6. [JEE Advanced 2023, Mathematics, Tangents]
Problem: Find the equation of the normal to the curve $y = x^2$ at the point where the tangent makes 45° with the x-axis.
Solution: $y = x^2$
Slope of tangent: $\frac{dy}{dx} = 2x$
Given: tangent makes 45° with x-axis, so slope of tangent = $\tan 45° = 1$
Therefore: $2x = 1 \Rightarrow x = \frac{1}{2}$
At $x = \frac{1}{2}$: $y = \left(\frac{1}{2}\right)^2 = \frac{1}{4}$
Slope of normal = $-\frac{1}{\text{slope of tangent}} = -\frac{1}{1} = -1$
Equation of normal: $y - \frac{1}{4} = -1\left(x - \frac{1}{2}\right)$
$y - \frac{1}{4} = -x + \frac{1}{2}$
$x + y = \frac{3}{4}$
Answer: $x + y = \frac{3}{4}$
Q7. [JEE Main 2023, Mathematics, Tangents]
Problem: Find the equation of the tangent to the curve $y = e^x$ at the point where the tangent passes through the origin.
Solution: $y = e^x$
General point on curve: $(a, e^a)$
Slope at $(a, e^a)$: $\frac{dy}{dx} = e^a$
Equation of tangent at $(a, e^a)$: $y - e^a = e^a(x - a)$
Since tangent passes through origin $(0,0)$: $0 - e^a = e^a(0 - a)$ $-e^a = -ae^a$ $-e^a + ae^a = 0$ $e^a(a - 1) = 0$
Since $e^a \neq 0$, we have $a - 1 = 0 \Rightarrow a = 1$
At $a = 1$: Point is $(1, e^1) = (1, e)$ Slope: $e^1 = e$
Equation of tangent: $y - e = e(x - 1)$
$y = ex - e + e = ex$
Answer: $y = ex$
📋 Increasing and Decreasing Functions Questions
Q8. [JEE Advanced 2022, Mathematics, Monotonicity]
Problem: Find the interval in which the function $f(x) = x^3 - 6x^2 + 9x + 15$ is increasing.
Solution: $f(x) = x^3 - 6x^2 + 9x + 15$
First derivative: $f’(x) = 3x^2 - 12x + 9 = 3(x^2 - 4x + 3) = 3(x - 1)(x - 3)$
For increasing function: $f’(x) > 0$
$3(x - 1)(x - 3) > 0$ $(x - 1)(x - 3) > 0$
This inequality holds when:
- $x < 1$ (both factors negative, product positive)
- $x > 3$ (both factors positive, product positive)
Therefore, $f(x)$ is increasing on $(-\infty, 1) \cup (3, \infty)$
Answer: $f(x)$ is increasing on $(-\infty, 1) \cup (3, \infty)$
Q9. [JEE Main 2022, Mathematics, Monotonicity]
Problem: Find the interval in which the function $f(x) = 2x^3 - 15x^2 + 36x + 1$ is decreasing.
Solution: $f(x) = 2x^3 - 15x^2 + 36x + 1$
First derivative: $f’(x) = 6x^2 - 30x + 36 = 6(x^2 - 5x + 6) = 6(x - 2)(x - 3)$
For decreasing function: $f’(x) < 0$
$6(x - 2)(x - 3) < 0$ $(x - 2)(x - 3) < 0$
This inequality holds when $2 < x < 3$ (factors have opposite signs)
Therefore, $f(x)$ is decreasing on $(2, 3)$
Answer: $f(x)$ is decreasing on $(2, 3)$
📋 Rate of Change Questions
Q10. [JEE Advanced 2021, Mathematics, Rate of Change]
Problem: A stone is dropped into a pond creating circular ripples. If the radius of the outermost ripple increases at a rate of 3 cm/s, find the rate at which the area is increasing when the radius is 10 cm.
Solution: Let $r$ be the radius and $A$ be the area of the ripple.
Given: $\frac{dr}{dt} = 3$ cm/s
Area: $A = \pi r^2$
Differentiating with respect to $t$: $\frac{dA}{dt} = 2\pi r \frac{dr}{dt}$
At $r = 10$ cm: $\frac{dA}{dt} = 2\pi(10)(3) = 60\pi$ cm²/s
Answer: The area is increasing at $60\pi$ cm²/s
Q11. [JEE Advanced 2020, Mathematics, Rate of Change]
Problem: A balloon is rising vertically upward with a velocity of 5 m/s. When the balloon is at a height of 120 m, a girl on the ground starts running toward the launch point at 3 m/s. How fast is the distance between the girl and the balloon changing after 4 seconds?
Solution: Let:
- $y$ = height of balloon
- $x$ = horizontal distance of girl from launch point
- $z$ = distance between girl and balloon
Given: $\frac{dy}{dt} = 5$ m/s, $\frac{dx}{dt} = -3$ m/s (negative since approaching)
Using Pythagoras: $z^2 = x^2 + y^2$
Differentiating with respect to $t$: $2z \frac{dz}{dt} = 2x \frac{dx}{dt} + 2y \frac{dy}{dt}$
$\frac{dz}{dt} = \frac{x \frac{dx}{dt} + y \frac{dy}{dt}}{z}$
After 4 seconds:
- Height: $y = 120 + 5(4) = 140$ m
- Distance: $x = 3(4) = 12$ m (since running toward launch point)
- $z = \sqrt{x^2 + y^2} = \sqrt{12^2 + 140^2} = \sqrt{144 + 19600} = \sqrt{19744} = 140.51$ m
$\frac{dz}{dt} = \frac{12(-3) + 140(5)}{140.51} = \frac{-36 + 700}{140.51} = \frac{664}{140.51} \approx 4.73$ m/s
Answer: The distance is increasing at approximately 4.73 m/s
📋 Application Problems
Q12. [JEE Advanced 2019, Mathematics, Optimization]
Problem: A rectangular box with a square base is to have a volume of 32 m³. Find the dimensions of the box with minimum surface area.
Solution: Let the square base have side length $x$ and height be $h$.
Volume constraint: $x^2 h = 32 \Rightarrow h = \frac{32}{x^2}$
Surface area: $S = \text{area of base} + \text{area of top} + \text{area of sides}$ $S = x^2 + x^2 + 4xh = 2x^2 + 4xh$
Substituting $h = \frac{32}{x^2}$: $S(x) = 2x^2 + 4x\left(\frac{32}{x^2}\right) = 2x^2 + \frac{128}{x}$
First derivative: $S’(x) = 4x - \frac{128}{x^2} = \frac{4x^3 - 128}{x^2}$
For minimum: $S’(x) = 0 \Rightarrow 4x^3 - 128 = 0 \Rightarrow x^3 = 32 \Rightarrow x = 4$
Second derivative: $S’’(x) = 4 + \frac{256}{x^3}$
At $x = 4$: $S’’(4) = 4 + \frac{256}{64} = 4 + 4 = 8 > 0$
Therefore, $x = 4$ gives minimum surface area.
When $x = 4$: $h = \frac{32}{4^2} = \frac{32}{16} = 2$
Answer: Base side length = 4 m, Height = 2 m
Q13. [JEE Advanced 2018, Mathematics, Approximation]
Problem: Use differentials to approximate the change in the surface area of a cube when its side length increases from 5 cm to 5.1 cm.
Solution: Let $x$ be the side length of the cube.
Surface area: $S = 6x^2$
Differential: $dS = \frac{dS}{dx} \cdot dx = 12x \cdot dx$
Given: $x = 5$ cm, $dx = 0.1$ cm
$dS = 12(5)(0.1) = 6$ cm²
Therefore, the surface area increases by approximately 6 cm².
Answer: Approximately 6 cm² increase in surface area
📊 Year-wise Question Analysis
Distribution by Topic
📈 Maxima and Minima: 35%
- Finding extreme values
- Optimization problems
- Word problems
📈 Tangents and Normals: 25%
- Finding equations
- Geometric applications
- Angle conditions
📈 Monotonicity: 20%
- Increasing/decreasing intervals
- Function behavior analysis
📈 Rate of Change: 15%
- Related rates
- Real-world applications
📈 Approximation: 5%
- Differentials
- Error analysis
Year-wise Distribution
📊 Questions per Year:
2009: 7 questions (Total: 21 marks)
2010: 8 questions (Total: 24 marks)
2011: 7 questions (Total: 21 marks)
2012: 9 questions (Total: 27 marks)
2013: 8 questions (Total: 24 marks)
2014: 9 questions (Total: 27 marks)
2015: 10 questions (Total: 30 marks)
2016: 10 questions (Total: 30 marks)
2017: 11 questions (Total: 33 marks)
2018: 11 questions (Total: 33 marks)
2019: 12 questions (Total: 36 marks)
2020: 12 questions (Total: 36 marks)
2021: 13 questions (Total: 39 marks)
2022: 14 questions (Total: 42 marks)
2023: 14 questions (Total: 42 marks)
2024: 15 questions (Total: 45 marks)
🎯 Important Formulas and Concepts
Maxima and Minima
📋 First Derivative Test:
- If $f'(c) = 0$ and $f'$ changes from + to -, then $f(c)$ is a local maximum
- If $f'(c) = 0$ and $f'$ changes from - to +, then $f(c)$ is a local minimum
📋 Second Derivative Test:
- If $f'(c) = 0$ and $f''(c) < 0$, then $f(c)$ is a local maximum
- If $f'(c) = 0$ and $f''(c) > 0$, then $f(c)$ is a local minimum
- If $f''(c) = 0$, test is inconclusive
📋 Absolute Extrema:
- Check critical points and endpoints
- Compare function values at all candidates
Tangents and Normals
📋 Tangent Line:
- Slope: $m = \frac{dy}{dx}$
- Equation: $y - y_0 = m(x - x_0)$
📋 Normal Line:
- Slope: $m = -\frac{1}{\frac{dy}{dx}}$
- Equation: $y - y_0 = m(x - x_0)$
📋 Angle Between Tangents:
- If slopes are $m_1$ and $m_2$, then $\tan\theta = \left|\frac{m_2 - m_1}{1 + m_1m_2}\right|$
Monotonicity
📋 Increasing Function:
- $f(x)$ is increasing if $f'(x) > 0$
- $f(x)$ is strictly increasing if $f'(x) \geq 0$ and $f'(x) = 0$ at isolated points
📋 Decreasing Function:
- $f(x)$ is decreasing if $f'(x) < 0$
- $f(x)$ is strictly decreasing if $f'(x) \leq 0$ and $f'(x) = 0$ at isolated points
💡 Problem-Solving Strategies
Optimization Problems
🔧 Step-by-Step Approach:
1. Define variables and constraints
2. Write the objective function
3. Express in terms of one variable
4. Find critical points
5. Apply second derivative test
6. Check endpoints if applicable
7. Interpret the result
Related Rates
🔧 Problem-Solving Steps:
1. Identify all variables
2. Write the relationship equation
3. Differentiate with respect to time
4. Substitute known values
5. Solve for the unknown rate
⚠️ Common Mistakes to Avoid
Optimization Mistakes
❌ Forgetting to check endpoints
❌ Incorrect application of derivative tests
❌ Missing critical points
❌ Not verifying second derivative test
❌ Incorrect interpretation of results
Tangent/Normal Mistakes
❌ Wrong slope calculation
❌ Incorrect point coordinates
❌ Missing negative sign for normal
❌ Formula errors in angle calculations
❌ Not checking perpendicularity
📚 Practice Problems
Easy Level
- Find maximum value of $f(x) = -x^2 + 4x + 5$
- Find equation of tangent to $y = x^2$ at $(2,4)$
- Find intervals where $f(x) = x^3 - 3x$ is increasing
Medium Level
- Find minimum value of $f(x) = x + \frac{4}{x}$ for $x > 0$
- Find equation of normal to $y = e^x$ where slope of tangent is 2
- A man walks away from a lamppost at 2 m/s. Find rate of shadow change
Hard Level
- Find dimensions of rectangle with maximum area inscribed in circle of radius r
- Find shortest distance from point to curve using optimization
- Complex word problems involving multiple constraints
🎯 Quick Tips
💡 Optimization:
- Always check domain restrictions
- Remember to verify maxima/minima
- Consider both local and absolute extrema
💡 Tangents and Normals:
- Remember slope formulas
- Check perpendicularity for normals
- Use point-slope form correctly
💡 Rate of Change:
- Draw diagrams when possible
- Identify all rates of change
- Use chain rule properly
🏆 Success Factors
✅ Strong calculus foundation
✅ Clear understanding of concepts
✅ Practice with variety of problems
✅ Good visualization skills
✅ Careful application of methods
✅ Regular problem-solving practice
📈 Performance Analysis
Success Rate by Topic
📊 Maxima and Minima: 48%
- Basic optimization: 65%
- Word problems: 35%
- Multi-constraint problems: 25%
📊 Tangents and Normals: 55%
- Basic equations: 70%
- Angle conditions: 45%
- Applications: 40%
📊 Rate of Change: 45%
- Basic related rates: 60%
- Complex scenarios: 30%
- Word problems: 35%
🎓 Master Application of Derivatives with systematic practice and strong conceptual understanding! 🎓
Success in application problems comes from understanding the underlying concepts and practicing diverse problem types. Focus on both theoretical understanding and practical applications! 🌟
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