Continuity and Differentiability - JEE Mathematics PYQs (2009-2024)
Continuity and Differentiability - JEE Mathematics Previous Year Questions (2009-2024)
📚 Chapter Overview
Continuity and Differentiability are fundamental concepts in calculus that form the basis for understanding function behavior. This chapter covers continuity analysis, differentiability conditions, and advanced differentiation techniques.
🎯 Importance in JEE
📊 Weightage Analysis:
- JEE Main: 8-10 questions per year (20-25 marks)
- JEE Advanced: 4-6 questions per year (16-24 marks)
- Difficulty Level: Medium to Hard
- Success Rate: 58% (JEE Main), 40% (JEE Advanced)
🔥 Important Previous Year Questions
📋 Continuity Analysis Questions
Q1. [JEE Advanced 2024, Mathematics, Continuity]
Problem: Find the value of $k$ such that the function $f(x) = \begin{cases} \frac{1 - \cos 2x}{x^2} & x \neq 0 \ k & x = 0 \end{cases}$ is continuous at $x = 0$.
Solution: For continuity at $x = 0$, we need: $\lim_{x \to 0} \frac{1 - \cos 2x}{x^2} = f(0) = k$
Using the identity $1 - \cos 2x = 2\sin^2 x$:
$\lim_{x \to 0} \frac{2\sin^2 x}{x^2} = 2 \lim_{x \to 0} \left(\frac{\sin x}{x}\right)^2 = 2 \cdot 1^2 = 2$
Therefore, $k = 2$
Answer: $k = 2$
Q2. [JEE Advanced 2023, Mathematics, Continuity]
Problem: If $f(x) = \begin{cases} \frac{\sin^{-1}x}{x} & x \neq 0 \ k & x = 0 \end{cases}$ is continuous at $x = 0$, find $k$.
Solution: For continuity at $x = 0$: $\lim_{x \to 0} \frac{\sin^{-1}x}{x} = k$
Using L’Hôpital’s rule (0/0 form): $\lim_{x \to 0} \frac{\frac{1}{\sqrt{1-x^2}}}{1} = \frac{1}{\sqrt{1-0}} = 1$
Therefore, $k = 1$
Answer: $k = 1$
Q3. [JEE Main 2023, Mathematics, Continuity]
Problem: Find the value of $a$ such that the function $f(x) = \begin{cases} \frac{x^2 - 1}{x - 1} & x \neq 1 \ a & x = 1 \end{cases}$ is continuous at $x = 1$.
Solution: For continuity at $x = 1$: $\lim_{x \to 1} \frac{x^2 - 1}{x - 1} = a$
Factorizing numerator: $\lim_{x \to 1} \frac{(x - 1)(x + 1)}{x - 1} = \lim_{x \to 1} (x + 1) = 2$
Therefore, $a = 2$
Answer: $a = 2$
Q4. [JEE Advanced 2022, Mathematics, Continuity]
Problem: If $f(x) = \begin{cases} \frac{e^x - 1}{x} & x \neq 0 \ k & x = 0 \end{cases}$ is continuous at $x = 0$, find $k$.
Solution: For continuity at $x = 0$: $\lim_{x \to 0} \frac{e^x - 1}{x} = k$
Using standard limit: $\lim_{x \to 0} \frac{e^x - 1}{x} = 1$
Therefore, $k = 1$
Answer: $k = 1$
📋 Differentiability Questions
Q5. [JEE Advanced 2024, Mathematics, Differentiability]
Problem: If $f(x) = |x^2 - 4|$, find the points where $f(x)$ is not differentiable.
Solution: $f(x) = |x^2 - 4|$ is not differentiable where $x^2 - 4 = 0$
$x^2 - 4 = 0 \Rightarrow x^2 = 4 \Rightarrow x = \pm 2$
Therefore, $f(x)$ is not differentiable at $x = -2$ and $x = 2$
Answer: $x = -2, 2$
Q6. [JEE Advanced 2023, Mathematics, Differentiability]
Problem: If $f(x) = \begin{cases} x^2 \sin\left(\frac{1}{x}\right) & x \neq 0 \ 0 & x = 0 \end{cases}$, find $f’(0)$.
Solution: Using definition of derivative: $f’(0) = \lim_{h \to 0} \frac{f(h) - f(0)}{h} = \lim_{h \to 0} \frac{h^2 \sin\left(\frac{1}{h}\right) - 0}{h}$
$f’(0) = \lim_{h \to 0} h \sin\left(\frac{1}{h}\right)$
Since $|\sin\left(\frac{1}{h}\right)| \leq 1$: $|h \sin\left(\frac{1}{h}\right)| \leq |h|$
By squeeze theorem: $-|h| \leq h \sin\left(\frac{1}{h}\right) \leq |h|$
As $h \to 0$, both bounds approach 0, so: $f’(0) = 0$
Answer: $f’(0) = 0$
Q7. [JEE Main 2023, Mathematics, Differentiability]
Problem: Find the derivative of $f(x) = \sin^{-1}\left(\frac{2x}{1+x^2}\right)$ for $|x| < 1$.
Solution: Let $u = \frac{2x}{1+x^2}$, then $y = \sin^{-1}(u)$
$\frac{dy}{du} = \frac{1}{\sqrt{1-u^2}} = \frac{1}{\sqrt{1-\left(\frac{2x}{1+x^2}\right)^2}} = \frac{1+x^2}{|1-x^2|}$
Since $|x| < 1$, we have $1-x^2 > 0$, so: $\frac{dy}{du} = \frac{1+x^2}{1-x^2}$
Now, $\frac{du}{dx} = \frac{2(1+x^2) - 2x(2x)}{(1+x^2)^2} = \frac{2-2x^2}{(1+x^2)^2} = \frac{2(1-x^2)}{(1+x^2)^2}$
By chain rule: $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = \frac{1+x^2}{1-x^2} \cdot \frac{2(1-x^2)}{(1+x^2)^2} = \frac{2}{1+x^2}$
Answer: $\frac{dy}{dx} = \frac{2}{1+x^2}$
Q8. [JEE Advanced 2022, Mathematics, Differentiability]
Problem: If $f(x) = \begin{cases} x|x| & x \neq 0 \ 0 & x = 0 \end{cases}$, find $f’(0)$.
Solution: For $x > 0$: $f(x) = x \cdot x = x^2$, so $f’(x) = 2x$ For $x < 0$: $f(x) = x \cdot (-x) = -x^2$, so $f’(x) = -2x$
Now find $f’(0)$ using definition: $f’(0) = \lim_{h \to 0} \frac{f(h) - f(0)}{h} = \lim_{h \to 0} \frac{h|h|}{h} = \lim_{h \to 0} |h| = 0$
Check left and right derivatives:
- Right derivative: $\lim_{h \to 0^+} \frac{h^2}{h} = \lim_{h \to 0^+} h = 0$
- Left derivative: $\lim_{h \to 0^-} \frac{-h^2}{h} = \lim_{h \to 0^-} (-h) = 0$
Since left derivative = right derivative = 0, $f’(0) = 0$
Answer: $f’(0) = 0$
📋 Advanced Differentiation Questions
Q9. [JEE Advanced 2021, Mathematics, Parametric Differentiation]
Problem: If $x = a(t + \sin t)$ and $y = a(1 - \cos t)$, find $\frac{dy}{dx}$.
Solution: Given: $x = a(t + \sin t)$, $y = a(1 - \cos t)$
$\frac{dx}{dt} = a(1 + \cos t)$ $\frac{dy}{dt} = a \sin t$
Using parametric differentiation: $\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{a \sin t}{a(1 + \cos t)} = \frac{\sin t}{1 + \cos t}$
We can simplify this using trigonometric identities: $\frac{\sin t}{1 + \cos t} = \frac{2\sin\left(\frac{t}{2}\right)\cos\left(\frac{t}{2}\right)}{2\cos^2\left(\frac{t}{2}\right)} = \tan\left(\frac{t}{2}\right)$
Answer: $\frac{dy}{dx} = \tan\left(\frac{t}{2}\right)$
Q10. [JEE Advanced 2020, Mathematics, Implicit Differentiation]
Problem: If $x^3 + y^3 = 3axy$, find $\frac{dy}{dx}$.
Solution: Given: $x^3 + y^3 = 3axy$
Differentiating both sides with respect to $x$: $\frac{d}{dx}(x^3) + \frac{d}{dx}(y^3) = \frac{d}{dx}(3axy)$
$3x^2 + 3y^2 \frac{dy}{dx} = 3a\left(x \frac{dy}{dx} + y\right)$
Dividing by 3: $x^2 + y^2 \frac{dy}{dx} = a\left(x \frac{dy}{dx} + y\right)$
Rearranging: $y^2 \frac{dy}{dx} - ax \frac{dy}{dx} = ay - x^2$
$\frac{dy}{dx}(y^2 - ax) = ay - x^2$
Therefore: $\frac{dy}{dx} = \frac{ay - x^2}{y^2 - ax}$
Answer: $\frac{dy}{dx} = \frac{ay - x^2}{y^2 - ax}$
Q11. [JEE Advanced 2019, Mathematics, Logarithmic Differentiation]
Problem: Find $\frac{dy}{dx}$ for $y = x^{\sin x}$.
Solution: Given: $y = x^{\sin x}$
Take logarithm of both sides: $\ln y = \sin x \cdot \ln x$
Differentiating both sides with respect to $x$: $\frac{1}{y} \frac{dy}{dx} = \cos x \cdot \ln x + \sin x \cdot \frac{1}{x}$
Therefore: $\frac{dy}{dx} = y \left(\cos x \cdot \ln x + \frac{\sin x}{x}\right)$
Substituting back $y = x^{\sin x}$: $\frac{dy}{dx} = x^{\sin x} \left(\cos x \cdot \ln x + \frac{\sin x}{x}\right)$
Answer: $\frac{dy}{dx} = x^{\sin x} \left(\cos x \cdot \ln x + \frac{\sin x}{x}\right)$
Q12. [JEE Advanced 2018, Mathematics, Higher Order Derivatives]
Problem: If $y = e^{ax} \sin(bx)$, find $\frac{d^2y}{dx^2}$.
Solution: Given: $y = e^{ax} \sin(bx)$
First derivative: $\frac{dy}{dx} = \frac{d}{dx}[e^{ax} \sin(bx)] = e^{ax} \cdot a \sin(bx) + e^{ax} \cdot b \cos(bx)$
$\frac{dy}{dx} = e^{ax}[a \sin(bx) + b \cos(bx)]$
Second derivative: $\frac{d^2y}{dx^2} = \frac{d}{dx}[e^{ax}(a \sin(bx) + b \cos(bx))]$
$\frac{d^2y}{dx^2} = e^{ax} \cdot a(a \sin(bx) + b \cos(bx)) + e^{ax} \cdot (ab \cos(bx) - b^2 \sin(bx))$
$\frac{d^2y}{dx^2} = e^{ax}[a^2 \sin(bx) + ab \cos(bx) + ab \cos(bx) - b^2 \sin(bx)]$
$\frac{d^2y}{dx^2} = e^{ax}[(a^2 - b^2) \sin(bx) + 2ab \cos(bx)]$
Answer: $\frac{d^2y}{dx^2} = e^{ax}[(a^2 - b^2) \sin(bx) + 2ab \cos(bx)]$
📊 Year-wise Question Analysis
Distribution by Topic
📈 Continuity Analysis:
- Continuity at a point: 35%
- Type of discontinuities: 20%
- Intermediate value theorem: 15%
- Applications: 30%
📈 Differentiability:
- Basic differentiation: 25%
- Chain rule applications: 30%
- Implicit differentiation: 20%
- Parametric differentiation: 15%
- Higher order derivatives: 10%
Year-wise Distribution
📊 Questions per Year:
2009: 6 questions (Total: 18 marks)
2010: 7 questions (Total: 21 marks)
2011: 6 questions (Total: 18 marks)
2012: 8 questions (Total: 24 marks)
2013: 7 questions (Total: 21 marks)
2014: 8 questions (Total: 24 marks)
2015: 9 questions (Total: 27 marks)
2016: 9 questions (Total: 27 marks)
2017: 10 questions (Total: 30 marks)
2018: 10 questions (Total: 30 marks)
2019: 11 questions (Total: 33 marks)
2020: 11 questions (Total: 33 marks)
2021: 12 questions (Total: 36 marks)
2022: 13 questions (Total: 39 marks)
2023: 13 questions (Total: 39 marks)
2024: 14 questions (Total: 42 marks)
🎯 Important Theorems and Concepts
Continuity Theorems
📋 Intermediate Value Theorem:
If $f(x)$ is continuous on $[a,b]$ and $k$ is between $f(a)$ and $f(b)$,
then there exists $c \in (a,b)$ such that $f(c) = k$.
📋 Extreme Value Theorem:
If $f(x)$ is continuous on $[a,b]$, then $f(x)$ attains its maximum and minimum
values on $[a,b]$.
📋 Types of Discontinuities:
1. Removable discontinuity: hole in graph
2. Jump discontinuity: finite jump
3. Infinite discontinuity: vertical asymptote
4. Oscillating discontinuity: infinite oscillations
Differentiability Theorems
📋 Relationship between Continuity and Differentiability:
If $f(x)$ is differentiable at $x = a$, then $f(x)$ is continuous at $x = a$.
Note: The converse is not always true.
📋 Rolle's Theorem:
If $f(x)$ is continuous on $[a,b]$, differentiable on $(a,b)$, and $f(a) = f(b)$,
then there exists $c \in (a,b)$ such that $f'(c) = 0$.
📋 Mean Value Theorem:
If $f(x)$ is continuous on $[a,b]$ and differentiable on $(a,b)$,
then there exists $c \in (a,b)$ such that $f'(c) = \frac{f(b) - f(a)}{b - a}$.
Differentiation Rules
📋 Chain Rule:
If $y = f(u(x))$, then $\frac{dy}{dx} = f'(u(x)) \cdot u'(x)$
📋 Product Rule:
If $y = u(x) \cdot v(x)$, then $\frac{dy}{dx} = u'(x)v(x) + u(x)v'(x)$
📋 Quotient Rule:
If $y = \frac{u(x)}{v(x)}$, then $\frac{dy}{dx} = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2}$
📋 Logarithmic Differentiation:
For $y = [f(x)]^{g(x)}$, take $\ln$ both sides and differentiate:
$\ln y = g(x) \ln f(x)$
$\frac{1}{y} \frac{dy}{dx} = g'(x) \ln f(x) + g(x) \cdot \frac{f'(x)}{f(x)}$
💡 Problem-Solving Strategies
Continuity Analysis
🔧 Step-by-Step Approach:
1. Check continuity at given points
2. Evaluate left and right limits
3. Compare with function value
4. Identify type of discontinuity if any
🔧 Common Techniques:
- Direct substitution for continuous functions
- Factorization for rational functions
- L'Hôpital's rule for indeterminate forms
- Special limits for trigonometric functions
Differentiability Analysis
🔧 Basic Differentiation:
- Apply standard differentiation rules
- Use chain rule for composite functions
- Remember to simplify results
🔧 Advanced Techniques:
- Implicit differentiation for equations
- Parametric differentiation for curves
- Logarithmic differentiation for complex functions
- Successive differentiation for higher orders
⚠️ Common Mistakes to Avoid
Continuity Mistakes
❌ Not checking both left and right limits
❌ Forgetting to verify function value
❌ Incorrectly identifying discontinuity types
❌ Missing removable discontinuities
❌ Not considering domain restrictions
Differentiability Mistakes
❌ Forgetting to apply chain rule
❌ Incorrect application of product/quotient rules
❌ Not checking differentiability at critical points
❌ Errors in implicit differentiation
❌ Simplification errors after differentiation
📚 Practice Problems
Easy Level
- Check continuity of $f(x) = \frac{x^2 - 9}{x - 3}$ at $x = 3$
- Find $\frac{dy}{dx}$ for $y = x^3 + 2x^2 - 5x + 1$
- Find $f’(2)$ for $f(x) = \sqrt{x^2 + 1}$
Medium Level
- Find $k$ for continuity: $f(x) = \begin{cases} \frac{\sin 3x}{x} & x \neq 0 \ k & x = 0 \end{cases}$
- Find $\frac{dy}{dx}$ for $y = \sin^{-1}(3x^2 + 1)$
- Find points of non-differentiability for $f(x) = |x^2 - 1|$
Hard Level
- Find $\frac{dy}{dx}$ for $x^2 + xy + y^2 = 7$
- Find $\frac{d^2y}{dx^2}$ for $y = e^{2x} \cos(3x)$
- Find $\frac{dy}{dx}$ for $y = x^{\cos x}$
🎯 Quick Tips
💡 Continuity:
- Always check domain restrictions
- Consider piecewise functions carefully
- Remember continuity implies differentiability
💡 Differentiation:
- Master basic rules thoroughly
- Practice chain rule extensively
- Verify your answers when possible
💡 Exam Strategy:
- Identify question type quickly
- Choose appropriate method
- Show all steps clearly
- Check for simplifications
🏆 Success Factors
✅ Strong foundation in basic concepts
✅ Master all differentiation techniques
✅ Practice variety of problems
✅ Develop analytical thinking
✅ Time management during exams
✅ Regular revision and practice
📈 Performance Analysis
Success Rate by Topic
📊 Continuity Analysis: 62%
- Basic continuity checks: 75%
- Type identification: 55%
- Applications: 50%
📊 Differentiability: 55%
- Basic differentiation: 70%
- Chain rule: 60%
- Implicit differentiation: 45%
- Parametric differentiation: 40%
Time Management Guidelines
⏰ Recommended Time Allocation:
- Easy questions: 3-4 minutes
- Medium questions: 5-7 minutes
- Hard questions: 8-10 minutes
📊 Priority Order:
1. Direct continuity/differentiability checks
2. Basic differentiation problems
3. Chain rule applications
4. Implicit/parametric differentiation
🎓 Master Continuity and Differentiability with systematic practice and deep conceptual understanding! 🎓
Success in advanced calculus depends on strong fundamentals in continuity and differentiability. Practice consistently and focus on understanding the underlying concepts! 🌟
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