Revision Checklist JEE Calculus

Monotonicity & Maxima/Minima: Complete Formula Sheet

All essential theorems, formulas, and concepts for JEE Main & Advanced

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Basic Definitions Monotonicity Tests First Derivative Test Second Derivative Test Absolute Extrema Applications Special Cases JEE Tips

1. Basic Definitions & Concepts

Monotonic Increasing Function

A function $f(x)$ is monotonically increasing on an interval I if for any $x_1, x_2 \in I$ with $x_1 < x_2$, we have:

$f(x_1) \leq f(x_2)$

Strictly increasing if $f(x_1) < f(x_2)$

Monotonic Decreasing Function

A function $f(x)$ is monotonically decreasing on an interval I if for any $x_1, x_2 \in I$ with $x_1 < x_2$, we have:

$f(x_1) \geq f(x_2)$

Strictly decreasing if $f(x_1) > f(x_2)$

Local Maximum & Minimum

Local Maximum: $f(c)$ is a local maximum if $f(c) \geq f(x)$ for all $x$ in some open interval containing $c$

Local Minimum: $f(c)$ is a local minimum if $f(c) \leq f(x)$ for all $x$ in some open interval containing $c$

Critical Point: A point $c$ where $f'(c) = 0$ or $f'(c)$ does not exist

Absolute Maximum & Minimum

Absolute Maximum: $f(c)$ is the absolute maximum on interval I if $f(c) \geq f(x)$ for all $x \in I$

Absolute Minimum: $f(c)$ is the absolute minimum on interval I if $f(c) \leq f(x)$ for all $x \in I$

2. Monotonicity Tests & Theorems

First Derivative Test for Monotonicity

Let $f$ be continuous on $[a,b]$ and differentiable on $(a,b)$:

If $f'(x) > 0$ for all $x \in (a,b)$, then $f$ is strictly increasing on $[a,b]$
If $f'(x) < 0$ for all $x \in (a,b)$, then $f$ is strictly decreasing on $[a,b]$
If $f'(x) = 0$ for all $x \in (a,b)$, then $f$ is constant on $[a,b]$

Proof Insight: By Mean Value Theorem, for any $x_1 < x_2$, there exists $c \in (x_1, x_2)$ such that $f(x_2) - f(x_1) = f'(c)(x_2 - x_1)$

Critical Points Identification

To find intervals of monotonicity:

Find $f'(x)$
Solve $f'(x) = 0$ to find critical points
Also identify points where $f'(x)$ does not exist
Test sign of $f'(x)$ in each interval between critical points

💡 Quick Tip: Create a sign chart for $f'(x)$ to visualize increasing/decreasing intervals

3. First Derivative Test for Local Extrema

First Derivative Test Statement

Let $c$ be a critical point of a continuous function $f$, and $f$ be differentiable near $c$ (except possibly at $c$):

If $f'(x)$ changes from positive to negative at $c$, then $f$ has a local maximum at $c$
If $f'(x)$ changes from negative to positive at $c$, then $f$ has a local minimum at $c$
If $f'(x)$ does not change sign at $c$, then $f$ has no local extremum at $c$

Step-by-Step Procedure

Find all critical points by solving $f'(x) = 0$
For each critical point $c$, examine sign of $f'(x)$ in intervals:
- Immediately to the left of $c$: $(c-\delta, c)$
- Immediately to the right of $c$: $(c, c+\delta)$
Apply the sign change rule to classify each critical point

4. Second Derivative Test

Second Derivative Test Statement

Suppose $f''(x)$ is continuous near $c$ and $f'(c) = 0$:

If $f''(c) > 0$, then $f$ has a local minimum at $c$
If $f''(c) < 0$, then $f$ has a local maximum at $c$
If $f''(c) = 0$, the test is inconclusive - use first derivative test

Geometric Insight: $f''(c) > 0$ means the function is concave up (shaped like ∪), so critical point is minimum. $f''(c) < 0$ means concave down (shaped like ∩), so critical point is maximum.

Concavity & Points of Inflection

Concave Up: $f''(x) > 0$ - Graph lies above its tangent lines

Concave Down: $f''(x) < 0$ - Graph lies below its tangent lines

Point of Inflection: A point where concavity changes

To find inflection points:

Find $f''(x)$
Solve $f''(x) = 0$
Verify sign change of $f''(x)$ around these points

5. Absolute Extrema on Closed Intervals

Extreme Value Theorem

If $f$ is continuous on a closed interval $[a,b]$, then:

$f$ attains an absolute maximum value $f(c)$ for some $c \in [a,b]$
$f$ attains an absolute minimum value $f(d)$ for some $d \in [a,b]$

Procedure for Finding Absolute Extrema

To find absolute maximum and minimum of $f$ on $[a,b]$:

Find all critical points of $f$ in $(a,b)$
Evaluate $f$ at all critical points
Evaluate $f$ at endpoints $a$ and $b$
The largest value is absolute maximum, smallest is absolute minimum

⚠️ Important: This method works ONLY for continuous functions on closed intervals

6. Applications & Problem Solving

Optimization Problems Framework

Understand the problem: Identify what to maximize/minimize
Draw diagram: Visualize the situation
Define variables: Introduce notation
Write objective function: Express quantity to optimize in terms of variables
Identify constraints: Write equations for given conditions
Reduce variables: Use constraints to express objective in one variable
Find critical points: Differentiate and solve $f'(x) = 0$
Verify: Use first or second derivative test
Answer the question: State complete solution in context

Common JEE Problem Types

Geometry Problems:

Maximum area with fixed perimeter
Minimum surface area with fixed volume
Optimal dimensions for containers

Physics Applications:

Maximum range of projectile
Minimum time problems
Optimal angles

7. Special Cases & Important Functions

Quadratic Functions

For $f(x) = ax^2 + bx + c$:

Vertex at $x = -\frac{b}{2a}$
If $a > 0$: Minimum at vertex, range $[f(-\frac{b}{2a}), \infty)$
If $a < 0$: Maximum at vertex, range $(-\infty, f(-\frac{b}{2a})]$

Trigonometric Functions

$\sin x$ on $[0, 2\pi]$:

Max: 1 at $x = \frac{\pi}{2}$

Min: -1 at $x = \frac{3\pi}{2}$

$\cos x$ on $[0, 2\pi]$:

Max: 1 at $x = 0, 2\pi$

Min: -1 at $x = \pi$

When Second Derivative Test Fails

Cases where $f''(c) = 0$:

$f(x) = x^4$ has minimum at $x = 0$ but $f''(0) = 0$
$f(x) = x^3$ has inflection point at $x = 0$, $f''(0) = 0$
Solution: Always use first derivative test when $f''(c) = 0$

8. JEE Problem Solving Strategies

Time-Saving Techniques

For Monotonicity:

Use sign chart method
Test one point in each interval
Remember standard function behaviors

For Maxima/Minima:

Try second derivative test first
If inconclusive, use first derivative test
Check endpoints for absolute extrema

Common Pitfalls to Avoid

Forgetting to check endpoints for absolute extrema on closed intervals
Assuming all critical points are extrema - some are inflection points
Misapplying second derivative test when $f''(c) = 0$
Not verifying if critical points lie in the domain
Overlooking points where derivative doesn't exist

Must-Know Formulas

Critical Points:

Solve $f'(x) = 0$ or $f'(x)$ DNE

Inflection Points:

Solve $f''(x) = 0$ with sign change

Quadratic Vertex:

$x = -\frac{b}{2a}$

Closed Interval Method:

Check critical points + endpoints

Final Revision Notes

🎯 Key Concepts to Remember

Monotonicity determined by sign of $f'(x)$
Local extrema occur at critical points with sign change of $f'(x)$
Second derivative test faster but sometimes inconclusive
Absolute extrema on closed intervals need endpoint check
Optimization problems follow systematic framework

🚀 Exam Day Strategy

Read optimization problems carefully - identify what to maximize/minimize
For monotonicity, create quick sign chart
Verify your answers with quick reasoning
Practice common JEE problem patterns

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