JEE Main Mathematics — Calculus Complete Notes (Limits, Continuity, Differentiability, AOD, Integration)

Calculus is the crown jewel of the JEE Main Mathematics syllabus, accounting for approximately 35% to 40% of the total weightage. To the uninitiated, it appears as a chaotic sea of formulas, but to an elite aspirant, it is a beautifully structured logical ladder. The journey begins with the concept of a "Limit"—the bedrock upon which the entire skyscraper of Continuity, Differentiability, and Integration is built. Mastering Calculus is not about rote memorization of derivatives; it is about developing a "functional intuition." You must learn to visualize how functions breathe, expand, and collapse at critical points.

In the competitive landscape of JEE Main, speed and accuracy are non-negotiable. While the board exams focus on the "how," JEE focuses on the "why" and the "shortcut." This resource is designed to transition you from a basic learner to a problem-solving machine. We will strip down complex limits using L’Hôpital’s Rule, dismantle indeterminate forms with Taylor expansions, and navigate the nuances of domain and range with surgical precision. Remember, in Calculus, a small overlook—like forgetting to check the left-hand limit—can lead to a negative mark. Let us begin by solidifying your foundations and transforming your mathematical perspective.

1. Real-Valued Functions: The Foundation

Before diving into limits, we must recap the concept of a function $f: A \to B$, where $A$ and $B$ are subsets of $\mathbb{R}$.

1.1 Domain and Range

• Domain: The set of all real $x$ for which $f(x)$ is defined and real.
• Range: The set of all possible values $f(x)$ takes as $x$ varies over the domain.

Mnemonic for Domain Extraction: "DENZ"

• Denominator $\neq 0$
• Even roots $\geq 0$
• Negative logs: Not allowed (Argument $> 0$)
• Zone of Inverse Trig: $[-1, 1]$ for $\sin^{-1}$ and $\cos^{-1}$.

2. Limits: The Concept of Proximity

A limit describes the behavior of a function $f(x)$ as $x$ approaches a specific value $c$. We write this as $\lim_{x \to c} f(x) = L$.

2.1 The $\epsilon-\delta$ Intuition

While the formal definition is rarely tested directly in JEE Main, the intuition is vital for conceptual questions.

• Formal Definition: $\lim_{x \to c} f(x) = L$ if for every $\epsilon > 0$, there exists a $\delta > 0$ such that $0 < |x - c| < \delta \implies |f(x) - L| < \epsilon$.
• The Intuition: You can get $f(x)$ as close as you want to $L$ (within a margin $\epsilon$) by making $x$ sufficiently close to $c$ (within a distance $\delta$).

2.2 Algebra of Limits

If $\lim_{x \to c} f(x) = L$ and $\lim_{x \to c} g(x) = M$, then:

1. Sum/Difference: $\lim_{x \to c} [f(x) \pm g(x)] = L \pm M$
2. Product: $\lim_{x \to c} [f(x) \cdot g(x)] = L \cdot M$
3. Quotient: $\lim_{x \to c} \frac{f(x)}{g(x)} = \frac{L}{M}, (M \neq 0)$
4. Constant Multiple: $\lim_{x \to c} [k \cdot f(x)] = k \cdot L$

3. Standard Limits (The Power Tools)

Memorizing these is mandatory. They form the basis of most "limit evaluation" problems.

4. Indeterminate Forms and L’Hôpital’s Rule

Calculus deals primarily with "Indeterminate Forms"—expressions that do not have a definitive value without further analysis.

4.1 The 7 Indeterminate Forms

1. $\frac{0}{0}$
2. $\frac{\infty}{\infty}$
3. $0 \times \infty$ (Convert to $\frac{0}{1/\infty} \to \frac{0}{0}$)
4. $\infty - \infty$ (Usually requires rationalization or common denominator)
5. $1^\infty$ (Special formula: $e^{\lim (f(x)-1)g(x)}$)
6. $0^0$
7. $\infty^0$

4.2 L’Hôpital’s Rule (LHR)

If the limit $\lim_{x \to c} \frac{f(x)}{g(x)}$ results in $\frac{0}{0}$ or $\frac{\infty}{\infty}$ and the functions are differentiable near $c$, then: $$\lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)}$$ Note: You can apply LHR multiple times until the indeterminacy is removed.

5. Worked Examples: Step-by-Step Mastery

Here are 12 problems ranging from basic application to JEE Main level difficulty.

Example 1: Basic Trig Limit

Evaluate: $\lim_{x \to 0} \frac{\sin 5x}{\tan 3x}$

• Solution: Multiply and divide by $5x$ and $3x$ to use standard limits. $$\lim_{x \to 0} \left( \frac{\sin 5x}{5x} \cdot 5x \right) \cdot \left( \frac{3x}{\tan 3x} \cdot \frac{1}{3x} \right)$$ Using $\lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$ and $\lim_{\theta \to 0} \frac{\tan \theta}{\theta} = 1$: $$1 \cdot 5x \cdot 1 \cdot \frac{1}{3x} = \frac{5}{3}$$

Example 2: Algebraic Rationalization

Evaluate: $\lim_{x \to 0} \frac{\sqrt{1+x} - \sqrt{1-x}}{x}$

• Solution: This is a $\frac{0}{0}$ form. Rationalize the numerator. $$\lim_{x \to 0} \frac{(\sqrt{1+x} - \sqrt{1-x})(\sqrt{1+x} + \sqrt{1-x})}{x(\sqrt{1+x} + \sqrt{1-x})}$$ $$= \lim_{x \to 0} \frac{(1+x) - (1-x)}{x(\sqrt{1+x} + \sqrt{1-x})} = \lim_{x \to 0} \frac{2x}{x(\sqrt{1+x} + \sqrt{1-x})}$$ Cancel $x$: $\frac{2}{\sqrt{1} + \sqrt{1}} = \frac{2}{2} = 1$.

Example 3: Using L’Hôpital’s Rule

Evaluate: $\lim_{x \to 0} \frac{e^x - e^{-x} - 2x}{x - \sin x}$

• Solution: Form is $\frac{1-1-0}{0-0} = \frac{0}{0}$. Apply LHR. $$L = \lim_{x \to 0} \frac{e^x + e^{-x} - 2}{1 - \cos x}$$ Still $\frac{1+1-2}{1-1} = \frac{0}{0}$. Apply LHR again. $$L = \lim_{x \to 0} \frac{e^x - e^{-x}}{\sin x}$$ Still $\frac{0}{0}$. Apply LHR a third time. $$L = \lim_{x \to 0} \frac{e^x + e^{-x}}{\cos x} = \frac{1+1}{1} = 2.$$

Example 4: The $1^\infty$ Form

Evaluate: $\lim_{x \to 0} (\cos x)^{1/x^2}$

• Solution: This is $1^\infty$. Use the formula $e^{\lim_{x \to c} g(x)[f(x)-1]}$. $$L = e^{\lim_{x \to 0} \frac{1}{x^2}(\cos x - 1)}$$ $$L = e^{\lim_{x \to 0} - \frac{(1 - \cos x)}{x^2}}$$ Since $\lim_{x \to 0} \frac{1-\cos x}{x^2} = \frac{1}{2}$, we get: $$L = e^{-1/2} = \frac{1}{\sqrt{e}}$$

Example 5: Logarithmic Limit

Evaluate: $\lim_{x \to 0} \frac{\ln(1+3x)}{e^{2x} - 1}$

• Solution: Adjust to standard forms. $$\lim_{x \to 0} \left[ \frac{\ln(1+3x)}{3x} \cdot 3x \right] \cdot \left[ \frac{2x}{e^{2x}-1} \cdot \frac{1}{2x} \right]$$ $$= (1 \cdot 3x) \cdot (1 \cdot \frac{1}{2x}) = \frac{3}{2}$$

Example 6: Infinity minus Infinity

Evaluate: $\lim_{x \to 1} \left( \frac{1}{\ln x} - \frac{1}{x-1} \right)$

• Solution: Form is $\infty - \infty$. Take LCM. $$\lim_{x \to 1} \frac{x-1-\ln x}{(x-1)\ln x}$$ (Form $\frac{0}{0}$) Apply LHR: $$\lim_{x \to 1} \frac{1 - 1/x}{(x-1)(1/x) + \ln x} = \lim_{x \to 1} \frac{x-1}{x-1 + x\ln x}$$ (Form $\frac{0}{0}$) Apply LHR again: $$\lim_{x \to 1} \frac{1}{1 + (x \cdot \frac{1}{x} + \ln x)} = \frac{1}{1 + 1 + 0} = \frac{1}{2}$$

Example 7: Exponential Trick

Evaluate: $\lim_{x \to 0} \frac{8^x - 2^x}{x}$

• Solution: Factor out $2^x$. $$\lim_{x \to 0} \frac{2^x(4^x - 1)}{x} = \lim_{x \to 0} 2^x \cdot \lim_{x \to 0} \frac{4^x-1}{x}$$ $$= 2^0 \cdot \ln 4 = 1 \cdot \ln 4 = \ln 4 = 2\ln 2$$

Example 8: Polynomials at Infinity

Evaluate: $\lim_{x \to \infty} \frac{3x^3 - 4x^2 + x}{5x^3 + 2x - 7}$

• Solution: For $\frac{\infty}{\infty}$ with polynomials, divide by the highest power of $x$ ($x^3$). $$\lim_{x \to \infty} \frac{3 - 4/x + 1/x^2}{5 + 2/x^2 - 7/x^3} = \frac{3 - 0 + 0}{5 + 0 - 0} = \frac{3}{5}$$

Example 9: Squeeze Theorem (Sandwich Theorem)

Evaluate: $\lim_{x \to \infty} \frac{\sin x}{x}$

• Solution: We know $-1 \leq \sin x \leq 1$. Divide by $x$ (where $x > 0$ as $x \to \infty$): $$-\frac{1}{x} \leq \frac{\sin x}{x} \leq \frac{1}{x}$$ As $x \to \infty$, both $-\frac{1}{x} \to 0$ and $\frac{1}{x} \to 0$. By Squeeze Theorem, $\lim_{x \to \infty} \frac{\sin x}{x} = 0$.

Example 10: Using Taylor Series (Expansion Method)

Evaluate: $\lim_{x \to 0} \frac{x - \sin x}{x^3}$

• Solution: Use expansion of $\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \dots$ $$\lim_{x \to 0} \frac{x - (x - \frac{x^3}{6} + \dots)}{x^3} = \lim_{x \to 0} \frac{\frac{x^3}{6} - \dots}{x^3} = \frac{1}{6}$$

Example 11: Greatest Integer Function (GIF)

Evaluate: $\lim_{x \to 0} \frac{[x]}{x}$ where $[\cdot]$ is GIF.

• Solution: Check LHL and RHL. LHL: $x \to 0^- \implies [x] = -1$. Limit is $\lim_{x \to 0^-} \frac{-1}{x} = +\infty$. RHL: $x \to 0^+ \implies [x] = 0$. Limit is $\lim_{x \to 0^+} \frac{0}{x} = 0$. Since LHL $\neq$ RHL, the limit does not exist.

Example 12: Complex $1^\infty$ with Trig

Evaluate: $\lim_{x \to 0} (1 + \tan^2 \sqrt{x})^{1/2x}$

• Solution: $1^\infty$ form. $$L = e^{\lim_{x \to 0} \frac{1}{2x} (1 + \tan^2 \sqrt{x} - 1)}$$ $$L = e^{\lim_{x \to 0} \frac{\tan^2 \sqrt{x}}{2x}}$$ Let $\sqrt{x} = \theta$. As $x \to 0, \theta \to 0$. $x = \theta^2$. $$L = e^{\lim_{\theta \to 0} \frac{\tan^2 \theta}{2\theta^2}} = e^{\frac{1}{2} (1)^2} = e^{1/2} = \sqrt{e}$$

6. Advanced Tips for JEE Main

1. Always check LHL and RHL for functions involving $|x|$, $[x]$, ${x}$, and piecewise definitions.
2. Order of Growth: At $x \to \infty$, the growth rate is: $$\ln x < x^n < e^x < x^x$$ This helps in quickly solving $\frac{\infty}{\infty}$ forms without full LHR.
3. Factorial Limits: If the expression involves $n!$, try to use the Sandwich Theorem or convert it to a sum that looks like a Riemann Integral (covered in the Definite Integration section).
4. $0 \times \infty$ Warning: Never leave it as is. Always move one term to the denominator. Usually, moving the "algebraic" term is harder than moving the "trig" or "log" term.

Summary Table: Indeterminate Forms Strategy

Continuity & Differentiability

In the landscape of JEE Main Calculus, Continuity and Differentiability form the "Bridge of Logic" between Limits and Application of Derivatives. A function being defined is a static property; its continuity is a dynamic property regarding its behavior in a local neighborhood.

1. Continuity of a Function

A function $f(x)$ is said to be continuous at a point $x = a$ if there is no "break" or "gap" in the graph at that point. Mathematically, this requires three conditions to be met simultaneously:

1. $f(a)$ is defined (the function exists at the point).
2. $\lim_{x \to a} f(x)$ exists (LHL = RHL).
3. $\lim_{x \to a} f(x) = f(a)$.

Mnemonic: "The L-V-L Rule" To prove continuity, ensure: Left Hand Limit = Value at point = Right Hand Limit. $$\lim_{x \to a^-} f(x) = f(a) = \lim_{x \to a^+} f(x)$$

1.1 Continuity in an Interval

• Open Interval $(a, b)$: $f(x)$ is continuous in $(a, b)$ if it is continuous at every point $c \in (a, b)$.
• Closed Interval $[a, b]$: $f(x)$ is continuous in $[a, b]$ if:
  1. It is continuous in $(a, b)$.
  2. It is right-continuous at $x = a$: $\lim_{x \to a^+} f(x) = f(a)$.
  3. It is left-continuous at $x = b$: $\lim_{x \to b^-} f(x) = f(b)$.

2. Classification of Discontinuity

Discontinuity occurs when any of the three conditions of continuity fail. In JEE, identifying the type of discontinuity is crucial for MSQ (Multiple Select Questions).

Jump Discontinuity Note: The "Jump" is defined as $|\text{RHL} - \text{LHL}|$.

3. Properties of Continuous Functions

If $f$ and $g$ are continuous at $x = a$, then:

1. $f \pm g$, $f \cdot g$ are continuous at $x = a$.
2. $f/g$ is continuous, provided $g(a) \neq 0$.
3. Composite Functions: If $g$ is continuous at $a$ and $f$ is continuous at $g(a)$, then $(f \circ g)(x)$ is continuous at $x = a$.
4. Algebraic Nature: All Polynomials, Exponential, Logarithmic, and Trigonometric functions are continuous in their respective domains.

4. Intermediate Value Theorem (IVT) & Bolzano’s Theorem

IVT Statement: If $f(x)$ is continuous on $[a, b]$ and $k$ is any number between $f(a)$ and $f(b)$, then there exists at least one number $c \in (a, b)$ such that $f(c) = k$.

Bolzano’s Theorem (Location of Roots): If $f(x)$ is continuous on $[a, b]$ and $f(a) \cdot f(b) < 0$ (i.e., they have opposite signs), then there exists at least one root $c \in (a, b)$ such that $f(c) = 0$.

JEE Tip: This is widely used in "Theory of Equations" to prove the existence of a real root without solving the equation.

5. Differentiability

Differentiability implies the existence of a unique tangent at a point. Graphically, it implies a "smooth" curve without sharp corners or vertical tangents.

Definition (The First Principle): A function $f(x)$ is differentiable at $x = a$ if: $$f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h} \text{ exists finitely.}$$

This involves checking two one-sided derivatives:

1. Left Hand Derivative (LHD): $f'(a^-) = \lim_{h \to 0^+} \frac{f(a-h) - f(a)}{-h}$
2. Right Hand Derivative (RHD): $f'(a^+) = \lim_{h \to 0^+} \frac{f(a+h) - f(a)}{h}$

The Golden Rule:

• Differentiability $\implies$ Continuity.
• Continuity $\not\implies$ Differentiability. (Classic counter-example: $y = |x|$ at $x=0$).

6. Non-Differentiable Functions in JEE

7. Worked Examples

Example 1: Removable Discontinuity

Question: Discuss the continuity of $f(x) = \frac{x^2 - 4}{x - 2}$ at $x = 2$. Solution:

1. $f(2)$ is undefined ($0/0$ form).
2. Calculate Limit: $\lim_{x \to 2} \frac{(x-2)(x+2)}{x-2} = \lim_{x \to 2} (x+2) = 4$.
3. Since the limit exists but the function is not defined, this is a Missing Point Removable Discontinuity. Graph: A straight line $y = x+2$ with a hollow circle (hole) at $(2, 4)$.

Example 2: Piecewise Continuity

Question: Find $k$ if $f(x) = \begin{cases} \frac{\sin 5x}{3x}, & x \neq 0 \ k, & x = 0 \end{cases}$ is continuous at $x=0$. Solution:

1. $\lim_{x \to 0} f(x) = \lim_{x \to 0} \frac{\sin 5x}{5x} \cdot \frac{5}{3} = 1 \cdot \frac{5}{3} = 5/3$.
2. For continuity, $f(0) = \text{Limit}$.
3. Therefore, $k = 5/3$.

Example 3: Absolute Value Differentiability

Question: Check differentiability of $f(x) = x|x|$ at $x=0$. Solution:

1. $f(x) = \begin{cases} x^2, & x \ge 0 \ -x^2, & x < 0 \end{cases}$.
2. RHD ($x \to 0^+$): $\frac{d}{dx}(x^2) = 2x \to 0$.
3. LHD ($x \to 0^-$): $\frac{d}{dx}(-x^2) = -2x \to 0$.
4. LHD = RHD = 0. $f(x)$ is differentiable at $x=0$. Graph: A smooth "S-shaped" curve merging two parabolas at the origin.

Example 4: Sharp Corner Check

Question: Is $f(x) = |x - 1| + |x - 2|$ differentiable at $x=1$? Solution:

1. At $x=1$, $|x-2|$ is smooth ($2-x$), but $|x-1|$ changes definition.
2. For $x < 1$: $f(x) = (1-x) + (2-x) = 3 - 2x \implies f'(x) = -2$.
3. For $1 < x < 2$: $f(x) = (x-1) + (2-x) = 1 \implies f'(x) = 0$.
4. LHD = $-2$, RHD = $0$. LHD $\neq$ RHD.
5. Not differentiable at $x=1$. Similarly, not differentiable at $x=2$. Graph: A "bucket" shape with sharp corners at $x=1$ and $x=2$.

Example 5: Bolzano’s Theorem Application

Question: Show that $x^3 - 3x + 1 = 0$ has a root in $(0, 1)$. Solution:

1. Let $f(x) = x^3 - 3x + 1$.
2. $f(0) = 0 - 0 + 1 = 1$ (Positive).
3. $f(1) = 1 - 3 + 1 = -1$ (Negative).
4. Since $f(x)$ is a polynomial (continuous) and $f(0) \cdot f(1) < 0$, a root must exist between $0$ and $1$.

Example 6: The "Jump" Calculation

Question: Find the jump of discontinuity for $f(x) = \frac{1}{1 + e^{-1/x}}$ at $x=0$. Solution:

1. RHL ($x \to 0^+$): $1/x \to \infty \implies e^{-1/x} \to 0$. $f(0^+) = \frac{1}{1+0} = 1$.
2. LHL ($x \to 0^-$): $1/x \to -\infty \implies e^{-1/x} \to \infty$. $f(0^-) = \frac{1}{1+\infty} = 0$.
3. Jump = $|1 - 0| = 1$.

Example 7: Composite Continuity

Question: If $f(x) = \frac{1}{x-1}$, where is $f(f(x))$ discontinuous? Solution:

1. $f(x)$ is discontinuous at $x=1$.
2. $f(f(x)) = \frac{1}{f(x)-1} = \frac{1}{\frac{1}{x-1}-1} = \frac{x-1}{1-(x-1)} = \frac{x-1}{2-x}$.
3. This is discontinuous at $x=2$.
4. Total points of discontinuity: $x = {1, 2}$.

Example 8: GIF Differentiability

Question: Check differentiability of $f(x) = [x]$ at $x=2$. Solution:

1. LHL: $\lim_{x \to 2^-} [x] = 1$.
2. RHL: $\lim_{x \to 2^+} [x] = 2$.
3. LHL $\neq$ RHL. Function is discontinuous.
4. Since it is discontinuous, it is automatically non-differentiable.

Example 9: Differentiability of $x^n \sin(1/x)$

Question: Let $f(x) = x^n \sin(1/x)$ for $x \neq 0$ and $f(0)=0$. For what values of $n$ is $f(x)$ (i) Continuous, (ii) Differentiable? Solution:

1. Continuity: $\lim_{x \to 0} x^n \sin(1/x) = 0$. Possible only if $n > 0$.
2. Differentiability: $\lim_{h \to 0} \frac{h^n \sin(1/h) - 0}{h} = \lim_{h \to 0} h^{n-1} \sin(1/h)$.
3. This limit exists (is 0) only if $n-1 > 0 \implies n > 1$. Conclusion: Continuous if $n > 0$, Differentiable if $n > 1$.

Example 10: Piecewise Smoothness

Question: Find $a, b$ if $f(x) = \begin{cases} ax^2 + 1, & x \le 1 \ x^2 + ax + b, & x > 1 \end{cases}$ is differentiable everywhere. Solution:

1. Continuity at $x=1$: $a(1)^2 + 1 = 1^2 + a(1) + b \implies a+1 = 1+a+b \implies b=0$.
2. Differentiability at $x=1$: $f'(x)$ for $x < 1$: $2ax \to 2a$. $f'(x)$ for $x > 1$: $2x + a \to 2+a$.
3. Equate: $2a = 2+a \implies a=2$.
4. Result: $a=2, b=0$.

Example 11: Oscillatory Discontinuity

Question: Describe the discontinuity of $f(x) = \cos(\pi/x)$ at $x=0$. Solution:

1. As $x \to 0$, $1/x \to \infty$.
2. The argument of cosine increases infinitely.
3. The value of $\cos(\text{large})$ oscillates between $[-1, 1]$.
4. The limit does not settle on a single value. This is a Non-Removable Oscillatory Discontinuity.

Example 12: Trigonometric Modulus

Question: Is $f(x) = |\sin x|$ differentiable at $x=\pi$? Solution:

1. LHD at $\pi$: $\frac{d}{dx}(\sin x) = \cos x$. At $x=\pi$, $\cos \pi = -1$.
2. RHD at $\pi$: $\sin x$ is negative for $x > \pi$, so $f(x) = -\sin x$. Derivative is $-\cos x$. At $x=\pi$, $-(-1) = 1$.
3. LHD $(-1) \neq$ RHD $(1)$.
4. Not differentiable. Graph: A series of arches reflecting the negative parts of sine. Sharp "V" shapes at $\pi, 2\pi, \dots$.

Example 13: Functional Equation Continuity

Question: If $f(x+y) = f(x) + f(y)$ and $f(x)$ is continuous at $x=0$, prove it is continuous everywhere. Solution:

1. Assume $f(x)$ is continuous at $x=0$, so $\lim_{h \to 0} f(h) = f(0) = 0$.
2. Check continuity at $x=a$: $\lim_{h \to 0} f(a+h) = \lim_{h \to 0} [f(a) + f(h)]$.
3. $= f(a) + \lim_{h \to 0} f(h) = f(a) + 0 = f(a)$.
4. Thus, it is continuous everywhere.

Example 14: Radical Differentiability

Question: Test $f(x) = (x-1)^{2/3}$ for differentiability at $x=1$. Solution:

1. $f'(x) = \frac{2}{3}(x-1)^{-1/3} = \frac{2}{3\sqrt[3]{x-1}}$.
2. As $x \to 1$, $f'(x) \to \infty$.
3. The derivative is infinite; the graph has a vertical tangent at $x=1$.
4. Not differentiable at $x=1$. Graph: A "cusp" pointing downwards at $x=1$.

Example 15: Signum Function Differentiability

Question: Discuss $f(x) = x \cdot sgn(x)$ at $x=0$. Solution:

1. $f(x) = x \cdot \frac{x}{|x|} = |x|$.
2. We know $|x|$ is continuous at $x=0$ (LHL=RHL=0).
3. We know $|x|$ is not differentiable at $x=0$ (LHD=-1, RHD=1).
4. Therefore, $x \cdot sgn(x)$ is continuous but not differentiable at $x=0$. Graph: The standard "V" shape passing through the origin.

8. Summary Checklist for JEE

• Checked LHL, RHL, and Value for continuity?
• Checked if the function is defined at the boundary?
• Identified sharp corners $x$ where $f'(x)$ changes?
• Checked for Vertical Tangents (where $f'(x) \to \pm \infty$)?
• Remembered that $[x] \dots {x} \dots sgn(x)$ are "Danger Zones" for continuity?

Differentiation Techniques

Differentiation is the mathematical tool used to measure the instantaneous rate of change of a function with respect to its independent variable. For the JEE Main, mastery of these techniques is essential as they form the foundation for Applications of Derivatives (AOD) and Differential Equations.

1. Standard Derivatives Table

Before applying techniques, one must memorize the fundamental derivatives. In the JEE context, these are treated as axioms.

2. Fundamental Rules of Differentiation

2.1 Linearity, Product, and Quotient Rules

If $u$ and $v$ are functions of $x$:

1. Linearity: $\frac{d}{dx}[af(x) + bg(x)] = a f'(x) + b g'(x)$
2. Product Rule (Leibniz Rule): $\frac{d}{dx}(uv) = u'v + uv'$
   • Mnemonic: "First d-second plus second d-first."
3. Quotient Rule: $\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v u' - u v'}{v^2}$
   • Mnemonic: "Low d-high minus high d-low over the square of what's below."

2.2 The Chain Rule (Composite Functions)

If $y = f(g(x))$, then: $$\frac{dy}{dx} = f'(g(x)) \cdot g'(x)$$ This can be extended to any number of layers: $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dv} \cdot \frac{dv}{dx}$.

3. Advanced Differentiation Methods

3.1 Implicit Differentiation

When $y$ cannot be expressed explicitly as $f(x)$ (e.g., $x^3 + y^3 = 3axy$), we differentiate both sides with respect to $x$, treating $y$ as a function of $x$. Shortcut Formula: For $f(x, y) = 0$: $$\frac{dy}{dx} = -\frac{\frac{\partial f}{\partial x}}{\frac{\partial f}{\partial y}} = -\frac{f_x}{f_y}$$ where $f_x$ is the partial derivative with respect to $x$ (treating $y$ as constant).

3.2 Parametric Differentiation

If $x = f(t)$ and $y = g(t)$, then: $$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{g'(t)}{f'(t)}$$ Second Derivative Caution: $\frac{d^2 y}{dx^2} \neq \frac{d^2 y/dt^2}{d^2 x/dt^2}$. Instead: $$\frac{d^2 y}{dx^2} = \frac{d}{dt}\left( \frac{dy}{dx} \right) \cdot \frac{dt}{dx} = \frac{\frac{d}{dt}\left( \frac{g'(t)}{f'(t)} \right)}{f'(t)}$$

3.3 Logarithmic Differentiation

Used for functions of the form $y = [f(x)]^{g(x)}$ or complex products/quotients. Steps:

1. Take $\ln$ on both sides: $\ln y = g(x) \ln f(x)$.
2. Differentiate: $\frac{1}{y} \frac{dy}{dx} = g'(x) \ln f(x) + g(x) \frac{f'(x)}{f(x)}$.
3. $y' = y \times [\text{Result}]$.

3.4 Differentiation of Determinants

If $\Delta(x) = \begin{vmatrix} f(x) & g(x) \ h(x) & k(x) \end{vmatrix}$, then: $$\Delta'(x) = \begin{vmatrix} f'(x) & g'(x) \ h(x) & k(x) \end{vmatrix} + \begin{vmatrix} f(x) & g(x) \ h'(x) & k'(x) \end{vmatrix}$$ Differentiate one row (or column) at a time while keeping others constant.

4. Differentiation of Integrals (Leibniz Rule)

This is a high-yield JEE Advanced topic often appearing in JEE Main. If $F(x) = \int_{h(x)}^{g(x)} f(t) dt$, then: $$F'(x) = f(g(x)) \cdot g'(x) - f(h(x)) \cdot h'(x)$$ Explanation: Substitute the upper limit into the function and multiply by its derivative, then subtract the substitution of the lower limit multiplied by its derivative.

5. Higher Order Derivatives and Leibniz Rule

$y_n$ denotes the $n$-th derivative. Leibniz Rule for $n$-th Derivative of a Product: If $y = uv$: $$y_n = \binom{n}{0}u_n v + \binom{n}{1}u_{n-1}v_1 + \binom{n}{2}u_{n-2}v_2 + \dots + \binom{n}{n}u v_n$$ This is analogous to the Binomial Expansion.

Worked Problems (15 Solved Examples)

Problem 1: Chain Rule Practice Find $\frac{dy}{dx}$ if $y = \sin(\sqrt{\cos(x)})$. Solution: Let $u = \sqrt{\cos x}$ and $v = \cos x$. $y = \sin(u) \implies \frac{dy}{du} = \cos(u)$ $u = \sqrt{v} \implies \frac{du}{dv} = \frac{1}{2\sqrt{v}}$ $v = \cos x \implies \frac{dv}{dx} = -\sin x$ By Chain Rule: $\frac{dy}{dx} = \cos(\sqrt{\cos x}) \cdot \frac{1}{2\sqrt{\cos x}} \cdot (-\sin x)$. Final Answer: $-\frac{\sin x \cos(\sqrt{\cos x})}{2\sqrt{\cos x}}$.

Problem 2: Logarithmic Differentiation Find $y'$ if $y = x^x + x^{\sin x}$. Solution: Let $y = u + v$, where $u = x^x$ and $v = x^{\sin x}$. For $u = x^x$: $\ln u = x \ln x \implies \frac{1}{u}u' = \ln x + 1 \implies u' = x^x(1 + \ln x)$. For $v = x^{\sin x}$: $\ln v = \sin x \ln x \implies \frac{1}{v}v' = \cos x \ln x + \frac{\sin x}{x}$. $v' = x^{\sin x} \left( \cos x \ln x + \frac{\sin x}{x} \right)$. Answer: $y' = x^x(1 + \ln x) + x^{\sin x} \left( \cos x \ln x + \frac{\sin x}{x} \right)$.

Problem 3: Implicit Differentiation (Shortcut) Find $\frac{dy}{dx}$ for $x^2 + xy + y^2 = 10$. Solution: Let $f(x, y) = x^2 + xy + y^2 - 10 = 0$. $f_x = \frac{\partial f}{\partial x} = 2x + y$ (treating $y$ as constant). $f_y = \frac{\partial f}{\partial y} = x + 2y$ (treating $x$ as constant). $\frac{dy}{dx} = -\frac{f_x}{f_y} = -\frac{2x + y}{x + 2y}$.

Problem 4: Parametric Second Derivative If $x = a\cos \theta, y = a\sin \theta$, find $\frac{d^2y}{dx^2}$ at $\theta = \frac{\pi}{4}$. Solution: $\frac{dx}{d\theta} = -a\sin \theta$, $\frac{dy}{d\theta} = a\cos \theta$. $\frac{dy}{dx} = \frac{a\cos \theta}{-a\sin \theta} = -\cot \theta$. $\frac{d^2y}{dx^2} = \frac{d}{d\theta}(-\cot \theta) \cdot \frac{d\theta}{dx} = (\csc^2 \theta) \cdot \frac{1}{-a\sin \theta} = -\frac{1}{a\sin^3 \theta}$. At $\theta = \pi/4$, $\sin \theta = 1/\sqrt{2}$. Result: $-\frac{1}{a(1/\sqrt{2})^3} = -\frac{2\sqrt{2}}{a}$.

Problem 5: Leibniz Rule (Integration) Differentiate $F(x) = \int_{x^2}^{x^3} \ln(t) dt$ with respect to $x$. Solution: Using $\frac{d}{dx} \int_{h(x)}^{g(x)} f(t) dt$: $F'(x) = \ln(x^3) \cdot \frac{d}{dx}(x^3) - \ln(x^2) \cdot \frac{d}{dx}(x^2)$ $F'(x) = (3\ln x)(3x^2) - (2\ln x)(2x) = 9x^2 \ln x - 4x \ln x = (9x^2 - 4x) \ln x$.

Problem 6: Inverse Trigonometric Substitution Find $\frac{dy}{dx}$ if $y = \tan^{-1}\left( \frac{3x - x^3}{1 - 3x^2} \right)$. Solution: Substitute $x = \tan \theta$. Then $y = \tan^{-1}(\tan 3\theta) = 3\theta$. $y = 3\tan^{-1}x$. $\frac{dy}{dx} = \frac{3}{1+x^2}$.

Problem 7: Differentiation of Determinants If $f(x) = \begin{vmatrix} x & x^2 & x^3 \ 1 & 2x & 3x^2 \ 0 & 2 & 6x \end{vmatrix}$, find $f'(x)$. Solution: Differentiate row by row: Row 1 diff: $\begin{vmatrix} 1 & 2x & 3x^2 \ 1 & 2x & 3x^2 \ 0 & 2 & 6x \end{vmatrix} = 0$ (two rows identical). Row 2 diff: $\begin{vmatrix} x & x^2 & x^3 \ 0 & 2 & 6x \ 0 & 2 & 6x \end{vmatrix} = 0$ (two rows identical). Row 3 diff: $\begin{vmatrix} x & x^2 & x^3 \ 1 & 2x & 3x^2 \ 0 & 0 & 6 \end{vmatrix} = 6(2x^2 - x^2) = 6x^2$. $f'(x) = 6x^2$.

Problem 8: Infinite Series If $y = \sqrt{\sin x + \sqrt{\sin x + \dots \infty}}$, find $y'$. Solution: $y = \sqrt{\sin x + y} \implies y^2 = \sin x + y$. Differentiate implicitly: $2yy' = \cos x + y'$. $y'(2y - 1) = \cos x \implies y' = \frac{\cos x}{2y - 1}$.

Problem 9: Functional Equations If $f(x+y) = f(x)f(y)$ for all $x,y$ and $f(5)=2, f'(0)=3$, find $f'(5)$. Solution: By definition $f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$ $f'(x) = \lim_{h \to 0} \frac{f(x)f(h) - f(x)}{h} = f(x) \lim_{h \to 0} \frac{f(h) - 1}{h}$. Since $f(0+0)=f(0)f(0) \implies f(0)=1$ (if $f(x) \neq 0$). So $f'(x) = f(x) f'(0)$. $f'(5) = f(5) \cdot f'(0) = 2 \cdot 3 = 6$.

Problem 10: $y = f(x)^{g(x)}$ with complex base Find $y'$ for $y = (\sin x)^{\cos x}$. Solution: $\ln y = \cos x \ln(\sin x)$. $\frac{1}{y}y' = -\sin x \ln(\sin x) + \cos x \cdot \frac{1}{\sin x} \cdot \cos x$. $y' = (\sin x)^{\cos x} [ \cot x \cos x - \sin x \ln(\sin x) ]$.

Problem 11: Nth derivative of $1/(ax+b)$ Find the $n$-th derivative of $y = (ax+b)^{-1}$. Solution: $y_1 = -1(ax+b)^{-2} \cdot a$ $y_2 = (-1)(-2)(ax+b)^{-3} \cdot a^2$ $y_3 = (-1)(-2)(-3)(ax+b)^{-4} \cdot a^3$ Pattern: $y_n = \frac{(-1)^n n! a^n}{(ax+b)^{n+1}}$.

Problem 12: Chain Rule with Logarithms Differentiate $\ln(\ln(\ln x))$. Solution: $y' = \frac{1}{\ln(\ln x)} \cdot \frac{d}{dx}(\ln(\ln x))$ $y' = \frac{1}{\ln(\ln x)} \cdot \frac{1}{\ln x} \cdot \frac{1}{x}$. Answer: $\frac{1}{x \ln x \ln(\ln x)}$.

Problem 13: Successive Differentiation If $y = e^{ax} \sin bx$, find $y_2$. Solution: $y_1 = a e^{ax} \sin bx + b e^{ax} \cos bx = e^{ax}(a \sin bx + b \cos bx)$. $y_2 = a e^{ax}(a \sin bx + b \cos bx) + e^{ax}(a b \cos bx - b^2 \sin bx)$. $y_2 = e^{ax} [ (a^2 - b^2)\sin bx + 2ab \cos bx ]$.

Problem 14: Differentiation of Inverse Functions If $g$ is the inverse of $f$ and $f(x) = x^3 + e^x$, find $g'(1)$. Solution: $g(f(x)) = x \implies g'(f(x)) \cdot f'(x) = 1 \implies g'(f(x)) = \frac{1}{f'(x)}$. We need $g'(1)$. First, find $x$ such that $f(x) = 1$. $x^3 + e^x = 1 \implies x = 0$ (by observation). $f'(x) = 3x^2 + e^x$. $f'(0) = 3(0)^2 + e^0 = 1$. $g'(1) = \frac{1}{f'(0)} = \frac{1}{1} = 1$.

Problem 15: Leibniz Rule (Product of three) Find $y''$ if $y = x^2 e^x \sin x$. Solution: Let $u = x^2$ and $v = e^x \sin x$. $y' = u'v + uv' = 2x(e^x \sin x) + x^2(e^x \sin x + e^x \cos x)$. $y' = e^x [ (x^2+2x)\sin x + x^2 \cos x ]$. $y'' = e^x [ (x^2+2x)\sin x + x^2 \cos x ] + e^x [ (2x+2)\sin x + (x^2+2x)\cos x + 2x \cos x - x^2 \sin x ]$. Combine terms: $y'' = e^x [ (4x+2)\sin x + (2x^2+4x)\cos x ]$.

Pro-Tips for JEE Main

1. Symmetry in $y'$: In many implicit problems like $x^y = e^{x-y}$, taking logs first often simplifies the derivative to a form involving only $x$ or only $y$.
2. Substitution First: Always check if a trigonometric substitution (like $x = \tan \theta$) simplifies an expression before differentiating.
3. Partial Derivative Method: Use $-\frac{f_x}{f_y}$ for messy implicit equations to save time and reduce sign errors.
4. $x^x$ Derivative: Memorize $\frac{d}{dx}(x^x) = x^x(1 + \ln x)$ as it appears frequently.

Application of Derivatives (AOD)

The Application of Derivatives (AOD) is the cornerstone of Calculus in the JEE Main syllabus. It bridges the gap between pure algebraic manipulation and the geometric/physical interpretation of functions. This section covers Tangents, Normals, Monotonicity, Maxima-Minima, and Mean Value Theorems.

1. Tangents and Normals

For a curve $y = f(x)$, the derivative at a point $P(x_1, y_1)$ represents the slope of the tangent at that point.

A. Equations

• Slope of Tangent ($m_T$): $m_T = \left[ \frac{dy}{dx} \right]_{(x_1, y_1)} = f'(x_1)$
• Equation of Tangent: $y - y_1 = f'(x_1)(x - x_1)$
• Slope of Normal ($m_N$): Since $m_T \cdot m_N = -1$, $m_N = -\frac{1}{f'(x_1)}$
• Equation of Normal: $y - y_1 = -\frac{1}{f'(x_1)}(x - x_1)$

B. Angle of Intersection and Orthogonality

If two curves $y = f_1(x)$ and $y = f_2(x)$ intersect at point $P$, the angle $\theta$ between them is given by: $$\tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right|$$ where $m_1$ and $m_2$ are slopes of tangents at the intersection point.

• Orthogonal Curves: If $\theta = 90^\circ$, then $m_1 m_2 = -1$.

C. Parametric Forms

For $x = g(t), y = h(t)$, the slope is $\frac{dy}{dx} = \frac{h'(t)}{g'(t)}$.

2. Rate of Change, Errors, and Approximations

• Rate of Change: $\frac{dy}{dx}$ represents the instantaneous rate of change of $y$ with respect to $x$.
• Approximations: If $\Delta x$ is a small increment in $x$, the corresponding change in $y$ is $\Delta y \approx dy$. $$f(x + \Delta x) \approx f(x) + f'(x) \cdot \Delta x$$
• Errors:
  • Absolute Error: $\Delta x$
  • Relative Error: $\frac{\Delta x}{x}$
  • Percentage Error: $\frac{\Delta x}{x} \times 100$

3. Monotonicity (Increasing and Decreasing Functions)

A function is monotonic if it is either non-increasing or non-decreasing in its domain.

Note: $f'(x)$ can be zero at discrete points (like $x^3$ at $x=0$) and still be strictly increasing. It must not be zero over an interval.

4. Mean Value Theorems (MVT)

A. Rolle’s Theorem

If $f(x)$ is:

1. Continuous in $[a, b]$
2. Differentiable in $(a, b)$
3. $f(a) = f(b)$ Then there exists at least one $c \in (a, b)$ such that $f'(c) = 0$.

B. Lagrange’s Mean Value Theorem (LMVT)

If conditions 1 and 2 of Rolle’s are met, then there exists $c \in (a, b)$ such that: $$f'(c) = \frac{f(b) - f(a)}{b - a}$$ Geometrically, there is a tangent parallel to the secant line joining endpoints.

C. Cauchy’s Mean Value Theorem

If $f(x)$ and $g(x)$ are continuous in $[a, b]$ and differentiable in $(a, b)$, and $g'(x) \neq 0$, then: $$\frac{f'(c)}{g'(c)} = \frac{f(b) - f(a)}{g(b) - g(a)}$$

5. Maxima and Minima

A. Definitions

• Local Maximum: $f(c) \geq f(x)$ for all $x$ in a small neighborhood of $c$.
• Absolute (Global) Maximum: The largest value of $f(x)$ on the entire domain.

B. Tests for Local Extrema

1. First Derivative Test:
   • If $f'(x)$ changes sign from $+$ to $-$ at $c$, $x=c$ is a Local Maxima.
   • If $f'(x)$ changes sign from $-$ to $+$ at $c$, $x=c$ is a Local Minima.
2. Second Derivative Test:
   • If $f'(c) = 0$ and $f''(c) < 0 \rightarrow$ Local Maxima.
   • If $f'(c) = 0$ and $f''(c) > 0 \rightarrow$ Local Minima.
   • If $f''(c) = 0$, the test fails (use 1st derivative test).

C. Concavity and Points of Inflection

• Concave Upwards: $f''(x) > 0$ (Slope is increasing).
• Concave Downwards: $f''(x) < 0$ (Slope is decreasing).
• Point of Inflection: A point where $f''(x) = 0$ (or is undefined) and the concavity changes.

20 Solved JEE-Style Problems

Problem 1: Tangent and Normal

Q: Find the equation of the tangent to the curve $y = x^2 - 2x + 7$ which is parallel to the line $2x - y + 9 = 0$. Solution:

1. Slope of given line $2x - y + 9 = 0$ is $m = 2$.
2. Differentiate curve: $\frac{dy}{dx} = 2x - 2$.
3. Set $\frac{dy}{dx} = 2 \implies 2x - 2 = 2 \implies x = 2$.
4. At $x=2, y = (2)^2 - 2(2) + 7 = 7$. Point is $(2, 7)$.
5. Equation: $y - 7 = 2(x - 2) \implies y = 2x + 3$.

Problem 2: Orthogonality

Q: If the curves $y^2 = 6x$ and $9x^2 + by^2 = 16$ cut each other at right angles, find $b$. Solution:

1. $2y \frac{dy}{dx} = 6 \implies m_1 = \frac{3}{y}$.
2. $18x + 2by \frac{dy}{dx} = 0 \implies m_2 = -\frac{9x}{by}$.
3. For orthogonality, $m_1 m_2 = -1 \implies \left(\frac{3}{y}\right) \left(-\frac{9x}{by}\right) = -1 \implies \frac{27x}{by^2} = 1$.
4. Since $y^2 = 6x$, substitute: $\frac{27x}{b(6x)} = 1 \implies \frac{27}{6b} = 1 \implies b = \frac{27}{6} = \frac{9}{2}$.

Problem 3: Approximations

Q: Use differentials to approximate $\sqrt{25.2}$. Solution:

1. Let $f(x) = \sqrt{x}, x=25, \Delta x = 0.2$.
2. $f'(x) = \frac{1}{2\sqrt{x}}$.
3. $f(x+\Delta x) \approx f(x) + f'(x)\Delta x = \sqrt{25} + \frac{1}{2\sqrt{25}}(0.2)$.
4. $5 + \frac{0.2}{10} = 5.02$.

Problem 4: Rolle's Theorem

Q: Verify Rolle’s Theorem for $f(x) = x^2 - 4x + 3$ in $[1, 3]$. Solution:

1. Continuous and Differentiable (Polynomial).
2. $f(1) = 1-4+3 = 0$; $f(3) = 9-12+3 = 0$. So $f(a)=f(b)$.
3. $f'(x) = 2x - 4$. Set $f'(c) = 0 \implies 2c - 4 = 0 \implies c = 2$.
4. $c=2 \in (1, 3)$. Verified.

Problem 5: LMVT Application

Q: Find $c$ for LMVT for $f(x) = \ln x$ in $[1, e]$. Solution:

1. $f'(c) = \frac{f(e) - f(1)}{e - 1} \implies \frac{1}{c} = \frac{\ln e - \ln 1}{e - 1}$.
2. $\frac{1}{c} = \frac{1-0}{e-1} \implies c = e-1$.
3. Since $e \approx 2.718$, $c \approx 1.718$, which lies in $(1, e)$.

Problem 6: Strictly Increasing

Q: Find the interval where $f(x) = 2x^3 - 15x^2 + 36x + 1$ is strictly increasing. Solution:

1. $f'(x) = 6x^2 - 30x + 36 = 6(x^2 - 5x + 6) = 6(x-2)(x-3)$.
2. For strictly increasing, $f'(x) > 0$.
3. Using wavy curve method: $(-\infty, 2) \cup (3, \infty)$.

Problem 7: Absolute Maxima/Minima

Q: Find the absolute maximum value of $f(x) = 3x^4 - 8x^3 + 12x^2 - 48x + 25$ on $[0, 3]$. Solution:

1. $f'(x) = 12x^3 - 24x^2 + 24x - 48 = 12[x^2(x-2) + 2(x-2)] = 12(x-2)(x^2+2)$.
2. $f'(x)=0 \implies x=2$ (Critical point).
3. Evaluate at endpoints and critical points:
   • $f(0) = 25$
   • $f(2) = 3(16) - 8(8) + 12(4) - 48(2) + 25 = 48 - 64 + 48 - 96 + 25 = -39$.
   • $f(3) = 3(81) - 8(27) + 12(9) - 48(3) + 25 = 243 - 216 + 108 - 144 + 25 = 16$.
4. Absolute Max is 25 at $x=0$.

Problem 8: Point of Inflection

Q: Find the point of inflection of $f(x) = x^3 - 3x^2 + 2$. Solution:

1. $f'(x) = 3x^2 - 6x$.
2. $f''(x) = 6x - 6$.
3. $f''(x) = 0 \implies x = 1$.
4. For $x < 1, f'' < 0$; for $x > 1, f'' > 0$. Concavity changes.
5. $f(1) = 1 - 3 + 2 = 0$. Point is $(1, 0)$.

Problem 9: Rate of Change

Q: A balloon is being inflated at $900 , \text{cm}^3/\text{s}$. Find the rate at which the radius increases when $r=15 , \text{cm}$. Solution:

1. $V = \frac{4}{3}\pi r^3 \implies \frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}$.
2. $900 = 4\pi (15)^2 \frac{dr}{dt}$.
3. $900 = 900\pi \frac{dr}{dt} \implies \frac{dr}{dt} = \frac{1}{\pi} , \text{cm/s}$.

Problem 10: Monotonicity with Parameters

Q: Find the value of $a$ so that $f(x) = \sin x - ax + b$ is decreasing for all $x \in \mathbb{R}$. Solution:

1. $f'(x) = \cos x - a$.
2. For decreasing, $f'(x) \leq 0 \implies \cos x - a \leq 0 \implies \cos x \leq a$.
3. Since the maximum value of $\cos x$ is 1, $a$ must be $\geq 1$.

Problem 11: Largest Area (Optimisation)

Q: Find the area of the largest rectangle that can be inscribed in a circle of radius $R$. Let sides be $x$ and $y$. $x^2 + y^2 = (2R)^2$.

1. Area $A = x\sqrt{4R^2 - x^2}$.
2. $A^2 = S = x^2(4R^2 - x^2) = 4R^2x^2 - x^4$.
3. $\frac{dS}{dx} = 8R^2x - 4x^3 = 0 \implies 4x(2R^2 - x^2) = 0$.
4. $x = \sqrt{2}R$. Then $y = \sqrt{4R^2 - 2R^2} = \sqrt{2}R$.
5. A rectangle of maximum area in a circle is a Square.
6. Area $= (\sqrt{2}R)^2 = 2R^2$.

Problem 12: Shortest Distance

Q: Find the point on the curve $y^2 = 4x$ which is nearest to the point $(2, 1)$. Solution:

1. Let point be $P(t^2, 2t)$.
2. Distance $D^2 = Z = (t^2 - 2)^2 + (2t - 1)^2 = t^4 - 4t^2 + 4 + 4t^2 - 4t + 1 = t^4 - 4t + 5$.
3. $\frac{dZ}{dt} = 4t^3 - 4 = 0 \implies t = 1$.
4. Point is $(1^2, 2(1)) = (1, 2)$.

Problem 13: Local Extrema (Trigonometry)

Q: Find the local maximum value of $f(x) = \sin 2x - x$ in $(-\pi/2, \pi/2)$. Solution:

1. $f'(x) = 2\cos 2x - 1 = 0 \implies \cos 2x = 1/2$.
2. $2x = \pi/3 \text{ or } -\pi/3 \implies x = \pi/6 \text{ or } -\pi/6$.
3. $f''(x) = -4\sin 2x$.
4. At $x = \pi/6, f'' = -4\sin(\pi/3) < 0$ (Max).
5. Value $= \sin(\pi/3) - \pi/6 = \frac{\sqrt{3}}{2} - \frac{\pi}{6}$.

Problem 14: Concavity

Q: Determine the interval where $f(x) = e^{-x^2}$ is concave down. Solution:

1. $f'(x) = -2xe^{-x^2}$.
2. $f''(x) = -2e^{-x^2} + 4x^2e^{-x^2} = 2e^{-x^2}(2x^2 - 1)$.
3. Concave down $\implies f''(x) < 0 \implies 2x^2 - 1 < 0$.
4. $x^2 < 1/2 \implies x \in (-1/\sqrt{2}, 1/\sqrt{2})$.

Problem 15: Angle between curves $y = \ln x$ and $y = \ln^2 x$.

Intersection: $\ln x = \ln^2 x \implies \ln x (\ln x - 1) = 0 \implies x=1, x=e$.

1. At $x=1$: $f'_1 = 1/1 = 1$, $f'_2 = 2\ln(1)/1 = 0$. $\tan \theta = |(1-0)/(1+0)| = 1 \implies \theta = 45^\circ$.
2. At $x=e$: $f'_1 = 1/e$, $f'_2 = 2(1)/e = 2/e$. $\tan \theta = |\frac{1/e - 2/e}{1 + 2/e^2}| = \frac{e}{e^2+2}$.

Problem 16: Rate of Surface Area Change

Q: If the volume of a sphere increases at a constant rate, show that the rate of increase of its surface area varies inversely as the radius. Solution:

1. $V = \frac{4}{3}\pi r^3 \implies \frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt} = K$ (constant).
2. $S = 4\pi r^2 \implies \frac{dS}{dt} = 8\pi r \frac{dr}{dt}$.
3. From (1), $\frac{dr}{dt} = \frac{K}{4\pi r^2}$.
4. Substitute: $\frac{dS}{dt} = 8\pi r \left( \frac{K}{4\pi r^2} \right) = \frac{2K}{r}$.
5. Thus, $\frac{dS}{dt} \propto \frac{1}{r}$.

Problem 17: Implicit Function Tangent

Q: Find the slope of the tangent to $x^3 + y^3 = 3axy$ at $(x_1, y_1)$. Solution:

1. $3x^2 + 3y^2 y' = 3a(y + xy')$.
2. $x^2 + y^2 y' = ay + axy'$.
3. $y'(y^2 - ax) = ay - x^2 \implies y' = \frac{ay - x^2}{y^2 - ax}$.

Problem 18: Maximum Volume of Cone

Q: Show that the semi-vertical angle of a cone of maximum volume and given slant height $L$ is $\tan^{-1} \sqrt{2}$. Solution:

1. $V = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi (L^2 - h^2)h = \frac{1}{3}\pi (L^2h - h^3)$.
2. $\frac{dV}{dh} = \frac{1}{3}\pi(L^2 - 3h^2) = 0 \implies h = L/\sqrt{3}$.
3. $\cos \alpha = h/L = 1/\sqrt{3}$.
4. If $\cos \alpha = 1/\sqrt{3}$, then $\tan \alpha = \frac{\sqrt{3-1}}{1} = \sqrt{2}$.

Problem 19: Increasing/Decreasing Logic

Q: If $f(x) = x/ \sin x$ and $g(x) = x / \tan x$ for $0 < x \leq 1$, then in this interval: Solution:

1. $f'(x) = \frac{\sin x - x\cos x}{\sin^2 x}$. Since $\tan x > x$ for $x \in (0, \pi/2)$, $\sin x > x\cos x \implies f'(x) > 0$ (Increasing).
2. $g'(x) = \frac{\tan x - x\sec^2 x}{\tan^2 x} = \frac{\sin x \cos x - x}{\cos^2 x \tan^2 x} = \frac{\frac{1}{2}\sin 2x - x}{\dots}$.
3. Since $\sin 2x < 2x$, $g'(x) < 0$ (Decreasing).

Problem 20: Minimum sum of intercepts

Q: Find the minimum sum of intercepts on axes made by tangent to $\sqrt{x} + \sqrt{y} = \sqrt{a}$. Solution:

1. Let $x = a \cos^4 \theta, y = a \sin^4 \theta$ is not ideal. Let's use $x = a \cos^4 t, y = a \sin^4 t$.
2. Better: Tangent at $(x_1, y_1)$ is $\frac{x}{\sqrt{x_1}} + \frac{y}{\sqrt{y_1}} = \sqrt{a}$.
3. X-intercept ($y=0$): $x = \sqrt{a x_1}$. Y-intercept ($x=0$): $y = \sqrt{a y_1}$.
4. Sum $S = \sqrt{a}(\sqrt{x_1} + \sqrt{y_1})$.
5. Since current point lies on curve, $\sqrt{x_1} + \sqrt{y_1} = \sqrt{a}$.
6. $S = \sqrt{a}(\sqrt{a}) = a$.
7. The sum is constant and equal to $a$.

Summary Table: Key AOD Concepts

Mnemonic for Derivative Tests:

• "High-Low-Neg-Pos":
  • $f''(x) < 0$ (Negative) $\rightarrow$ High (Maxima).
  • $f''(x) > 0$ (Positive) $\rightarrow$ Low (Minima).

Indefinite Integration Techniques

Indefinite integration, often called the "anti-derivative," is the reverse process of differentiation. For a given function $f(x)$, we seek a function $F(x)$ such that $F'(x) = f(x)$. This is denoted as: $$\int f(x) , dx = F(x) + C$$ where $C$ is the constant of integration, representing the family of curves. In JEE Main, proficiency in integration is the bedrock for Area Under Curves and Differential Equations.

1. Fundamental Integration Formulae (The Toolkit)

To solve complex JEE problems, these 25+ standard integrals must be memorized as reflexes.

Mnemonic for $\sqrt{x^2 \pm a^2}$: "Half $x$ [The Question] plus/minus half $a^2$ [Log version of the denominator type]."

2. Integration by Substitution (The Change of Variable)

If an integrand is of the form $f(g(x)) \cdot g'(x)$, we substitute $u = g(x)$, then $du = g'(x)dx$.

Special Forms of Substitution:

1. Linear Substitution: If $\int f(x) dx = F(x)$, then $\int f(ax+b) dx = \frac{1}{a} F(ax+b) + C$.
2. Standard Substitution Patterns:
   • For $\sqrt{a^2 - x^2}$, let $x = a \sin \theta$ or $a \cos \theta$.
   • For $\sqrt{a^2 + x^2}$, let $x = a \tan \theta$ or $a \sinh \theta$.
   • For $\sqrt{x^2 - a^2}$, let $x = a \sec \theta$ or $a \cosh \theta$.
   • For $\sqrt{\frac{a-x}{a+x}}$, let $x = a \cos 2\theta$.

3. Integration by Parts (IBP)

Based on the product rule of derivatives: $$\int u \cdot v , dx = u \int v , dx - \int \left( \frac{du}{dx} \int v , dx \right) dx$$

The LIATE Rule: Select the first function ($u$) based on this priority:

1. Logarithmic functions
2. Inverse Trigonometric functions
3. Algebraic functions
4. Trigonometric functions
5. Exponential functions

Special JEE Shortcut: The $e^x$ Property $$\int e^x [f(x) + f'(x)] , dx = e^x f(x) + C$$ Extension: $\int [f(x) + x f'(x)] , dx = x f(x) + C$.

4. Integration using Partial Fractions

Used for rational functions $\frac{P(x)}{Q(x)}$ where degree of $P(x) <$ degree of $Q(x)$.

5. Trigonometric Integration Techniques

A. Integrals of type $\int \sin^m x \cos^n x , dx$

1. If power of $\sin x$ is odd, substitute $\cos x = t$.
2. If power of $\cos x$ is odd, substitute $\sin x = t$.
3. If both are even, use double angle formulae: $\sin^2 x = \frac{1-\cos 2x}{2}$ and $\cos^2 x = \frac{1+\cos 2x}{2}$.

B. Weierstrass Substitution (The "Golden" Substitution)

For integrals involving $a+b\sin x$ or $a+b\cos x$ in the denominator, use: $$t = \tan(x/2)$$ Then, $dx = \frac{2 dt}{1+t^2}$, $\sin x = \frac{2t}{1+t^2}$, and $\cos x = \frac{1-t^2}{1+t^2}$.

C. Reduction Formulae

For $I_n = \int \tan^n x , dx$: $$I_n = \frac{\tan^{n-1} x}{n-1} - I_{n-2}$$

6. Algebraic Forms and Strategies

Form 1: $\int \frac{dx}{ax^2+bx+c}$ or $\int \frac{dx}{\sqrt{ax^2+bx+c}}$

Method: Complete the square in the quadratic expression to transform it into one of the nine standard forms like $\frac{1}{x^2+a^2}$ or $\frac{1}{\sqrt{a^2-x^2}}$.

Form 2: $\int \frac{px+q}{ax^2+bx+c} , dx$

Method: Express the numerator as $px+q = A \frac{d}{dx}(ax^2+bx+c) + B$. Divide into two parts: one leads to $\ln|\text{quadratic}|$ and the other to the "Complete the Square" method.

Form 3: $\int \frac{dx}{(x-k)\sqrt{ax^2+bx+c}}$

Method: Substitute $x-k = \frac{1}{t}$. This converts the radical into a quadratic in $t$.

7. Worked Examples (20 Solved Integrals)

Ex 1: $\int \frac{x^2}{x^6+a^6} dx$ Solution: Let $x^3 = t \Rightarrow 3x^2 dx = dt \Rightarrow x^2 dx = \frac{1}{3} dt$. $I = \frac{1}{3} \int \frac{dt}{t^2 + (a^3)^2} = \frac{1}{3} \cdot \frac{1}{a^3} \tan^{-1}\left(\frac{t}{a^3}\right) = \frac{1}{3a^3} \tan^{-1}\left(\frac{x^3}{a^3}\right) + C$.

Ex 2: $\int \frac{dx}{\sqrt{2x - x^2}}$ Solution: Complete the square: $2x-x^2 = -(x^2-2x) = -(x^2-2x+1-1) = 1 - (x-1)^2$. $\int \frac{dx}{\sqrt{1-(x-1)^2}} = \sin^{-1}(x-1) + C$.

Ex 3: $\int \frac{\sin x - \cos x}{\sin x + \cos x} dx$ Solution: Notice $(\sin x + \cos x)' = \cos x - \sin x = -(\sin x - \cos x)$. Let $\sin x + \cos x = t \Rightarrow (\cos x - \sin x) dx = dt$. $I = \int \frac{-dt}{t} = -\ln|\sin x + \cos x| + C$.

Ex 4: $\int e^x (\frac{1}{x} - \frac{1}{x^2}) dx$ Solution: Use $\int e^x [f(x) + f'(x)] dx$. Here $f(x) = \frac{1}{x}$, so $f'(x) = -\frac{1}{x^2}$. $I = e^x \cdot \frac{1}{x} + C$.

Ex 5: $\int \ln x , dx$ Solution: Use IBP with $u = \ln x, v = 1$. $I = \ln x \int 1 dx - \int (\frac{1}{x} \cdot x) dx = x \ln x - x + C$.

Ex 6: $\int \frac{dx}{x^2 - 6x + 13}$ Solution: $x^2 - 6x + 13 = (x-3)^2 + 4$. $I = \int \frac{dx}{(x-3)^2 + 2^2} = \frac{1}{2} \tan^{-1}(\frac{x-3}{2}) + C$.

Ex 7: $\int \frac{5x+3}{\sqrt{x^2+4x+10}} dx$ Solution: $5x+3 = A(2x+4) + B \Rightarrow 2A=5, 4A+B=3 \Rightarrow A=5/2, B=-7$. $I = \frac{5}{2} \int \frac{2x+4}{\sqrt{x^2+4x+10}} dx - 7 \int \frac{dx}{\sqrt{(x+2)^2 + 6}}$. $I = 5\sqrt{x^2+4x+10} - 7 \ln|x+2 + \sqrt{x^2+4x+10}| + C$.

Ex 8: $\int \sin^3 x , dx$ Solution: $\int \sin^2 x \cdot \sin x , dx = \int (1-\cos^2 x) \sin x , dx$. Let $\cos x = t \Rightarrow -\sin x dx = dt$. $I = -\int (1-t^2) dt = \frac{t^3}{3} - t = \frac{\cos^3 x}{3} - \cos x + C$.

Ex 9: $\int \frac{dx}{1 + e^x}$ Solution: Divide numerator and denominator by $e^x$: $\int \frac{e^{-x}}{e^{-x} + 1} dx$. Let $e^{-x} + 1 = t \Rightarrow -e^{-x} dx = dt$. $I = -\int \frac{dt}{t} = -\ln|e^{-x} + 1| = \ln|\frac{e^x}{e^x+1}| + C$.

Ex 10: $\int \frac{dx}{x(x^n+1)}$ Solution: Multiply top and bottom by $x^{n-1} \Rightarrow \int \frac{x^{n-1} dx}{x^n(x^n+1)}$. Let $x^n = t \Rightarrow nx^{n-1} dx = dt$. $I = \frac{1}{n} \int \frac{dt}{t(t+1)} = \frac{1}{n} \int (\frac{1}{t} - \frac{1}{t+1}) dt = \frac{1}{n} \ln|\frac{x^n}{x^n+1}| + C$.

Ex 11: $\int \sqrt{a^2-x^2} dx$ using $x = a\sin\theta$ Solution: $dx = a\cos\theta d\theta$. $I = \int a\cos\theta \cdot a\cos\theta d\theta = a^2 \int \frac{1+\cos 2\theta}{2} d\theta = \frac{a^2}{2}(\theta + \frac{\sin 2\theta}{2})$. Substitute back: $\theta = \sin^{-1}(x/a)$ and $\sin 2\theta = 2\sin\theta\cos\theta = 2(x/a)\sqrt{1-x^2/a^2}$. $I = \frac{x}{2}\sqrt{a^2-x^2} + \frac{a^2}{2}\sin^{-1}\frac{x}{a} + C$.

Ex 12: $\int \frac{\sin 2x}{a^2 \cos^2 x + b^2 \sin^2 x} dx$ Solution: Let denominator $= t$. $dt = (-2a^2 \cos x \sin x + 2b^2 \sin x \cos x) dx = (b^2-a^2)\sin 2x dx$. $I = \frac{1}{b^2-a^2} \int \frac{dt}{t} = \frac{1}{b^2-a^2} \ln|a^2 \cos^2 x + b^2 \sin^2 x| + C$.

Ex 13: $\int \sec^3 x , dx$ Solution: Application of IBP. $u = \sec x, v = \sec^2 x$. $I = \sec x \tan x - \int \sec x \tan^2 x dx = \sec x \tan x - \int (\sec^3 x - \sec x) dx$. $2I = \sec x \tan x + \ln|\sec x + \tan x| \Rightarrow I = \frac{1}{2}[\sec x \tan x + \ln|\sec x + \tan x|] + C$.

Ex 14: $\int \frac{dx}{\cos(x-a)\cos(x-b)}$ Solution: Multiply by $\frac{\sin(a-b)}{\sin(a-b)}$. $I = \frac{1}{\sin(a-b)} \int \frac{\sin[(x-b)-(x-a)]}{\cos(x-a)\cos(x-b)} dx$. Using $\sin(A-B) = \sin A \cos B - \cos A \sin B$: $I = \frac{1}{\sin(a-b)} \int [\tan(x-b) - \tan(x-a)] dx = \frac{1}{\sin(a-b)} \ln|\frac{\sec(x-b)}{\sec(x-a)}| + C$.

Ex 15: $\int \frac{x^2+1}{x^4+1} dx$ (Master Technique) Solution: Divide by $x^2$: $\int \frac{1 + 1/x^2}{x^2 + 1/x^2} dx = \int \frac{d(x - 1/x)}{(x-1/x)^2 + 2}$. Let $x - 1/x = t \Rightarrow I = \int \frac{dt}{t^2 + (\sqrt{2})^2} = \frac{1}{\sqrt{2}} \tan^{-1}\left(\frac{x - 1/x}{\sqrt{2}}\right) + C$.

Ex 16: $\int \frac{dx}{x\sqrt{1-x^3}}$ Solution: Let $x^3 = \sin^2 \theta \Rightarrow 3x^2 dx = 2\sin\theta\cos\theta d\theta \Rightarrow dx = \frac{2\sin\theta\cos\theta}{3x^2} d\theta$. $I = \int \frac{2\sin\theta\cos\theta}{3x^3 \cos\theta} d\theta = \frac{2}{3} \int \csc\theta d\theta$ (since $x^3=\sin^2\theta$). $I = \frac{2}{3} \ln|\csc\theta - \cot\theta| + C = \frac{2}{3} \ln|\frac{1-\sqrt{1-x^3}}{\sqrt{x^3}}| + C$.

Ex 17: $\int \tan^4 x , dx$ Solution: $\int \tan^2 x (\sec^2 x - 1) dx = \int \tan^2 x \sec^2 x dx - \int (\sec^2 x - 1) dx$. $I = \frac{\tan^3 x}{3} - \tan x + x + C$.

Ex 18: $\int \frac{dx}{(x+1)^2(x^2+1)}$ Solution: Partial Fractions: $\frac{1}{(x+1)^2(x^2+1)} = \frac{A}{x+1} + \frac{B}{(x+1)^2} + \frac{Cx+D}{x^2+1}$. Solving gives $A=1/2, B=1/2, C=-1/2, D=0$. $I = \frac{1}{2} \ln|x+1| - \frac{1}{2(x+1)} - \frac{1}{4} \ln|x^2+1| + C$.

Ex 19: $\int \frac{dx}{5+4\cos x}$ (Weierstrass Step) Solution: Let $t = \tan(x/2), dx = \frac{2dt}{1+t^2}, \cos x = \frac{1-t^2}{1+t^2}$. $I = \int \frac{2dt/(1+t^2)}{5 + 4(\frac{1-t^2}{1+t^2})} = \int \frac{2dt}{5+5t^2+4-4t^2} = \int \frac{2dt}{t^2+9}$. $I = \frac{2}{3} \tan^{-1}(\frac{t}{3}) = \frac{2}{3} \tan^{-1}(\frac{\tan(x/2)}{3}) + C$.

Ex 20: $\int e^{2x} \sin x , dx$ Solution: Use IBP twice (The Looping Integral). $I = \sin x \frac{e^{2x}}{2} - \int \cos x \frac{e^{2x}}{2} dx = \frac{1}{2} e^{2x} \sin x - \frac{1}{2} [\cos x \frac{e^{2x}}{2} + \int \sin x \frac{e^{2x}}{2} dx]$. $I = \frac{1}{2} e^{2x} \sin x - \frac{1}{4} e^{2x} \cos x - \frac{1}{4} I$. $\frac{5}{4} I = \frac{e^{2x}}{4} (2\sin x - \cos x) \Rightarrow I = \frac{e^{2x}}{5}(2\sin x - \cos x) + C$.

Definite Integration & Properties

Definite integration is the cornerstone of Calculus in the JEE Main and Advanced syllabus. Unlike indefinite integration, which yields a family of curves, a definite integral represents a unique numerical value, geometrically interpreted as the signed area under a curve.

1. Fundamental Theorems of Calculus

To master definite integration, one must understand the two bridges between differentiation and integration.

A. First Fundamental Theorem of Calculus (Area Function)

Let $f$ be a continuous function on $[a, b]$. Let $A(x) = \int_a^x f(t) , dt$ for $a \leq x \leq b$. Then:

1. $A'(x) = f(x)$ for every $x$ in $[a, b]$.
2. $A(x)$ is continuous on $[a, b]$ and differentiable on $(a, b)$.

B. Second Fundamental Theorem (Newton-Leibniz Formula)

If $f$ is continuous on $[a, b]$ and $F$ is an antiderivative of $f$ (i.e., $F'(x) = f(x)$), then: $$\int_a^b f(x) , dx = [F(x)]_a^b = F(b) - F(a)$$

2. Definite Integral as a Limit of a Sum (Riemann Sum)

JEE often asks questions where an infinite series must be converted into a definite integral. The standard conversion formula is: $$\lim_{n \to \infty} \sum_{r=1}^{pn} \frac{1}{n} f\left(\frac{r}{n}\right) = \int_0^p f(x) , dx$$

The Mapping Table for Series to Integral:

Mnemonic: "Replace $r/n$ with $x$, $1/n$ with $dx$, and set limits by dividing the bounds of $r$ by $n$."

3. Leibniz Rule for Differentiation Under Integral Sign

When the limits of an integral are functions of the variable with respect to which we are differentiating: $$\frac{d}{dx} \left[ \int_{\phi(x)}^{\psi(x)} f(t) , dt \right] = f(\psi(x)) \cdot \psi'(x) - f(\phi(x)) \cdot \phi'(x)$$

Usage Note: This is vital for solving problems involving "Limit of an integral over $x$" using L'Hôpital’s Rule.

4. Core Properties of Definite Integrals ($P_1$ to $P_8$)

These properties are the "weapons" of a JEE aspirant. Memorizing them is mandatory.

$P_1$: Change of Variable $$\int_a^b f(x) , dx = \int_a^b f(t) , dt$$ The value of a definite integral is independent of the variable of integration.

$P_2$: Interchanging Limits $$\int_a^b f(x) , dx = -\int_b^a f(x) , dx$$

$P_3$: Additivity (Interval Splitting) $$\int_a^b f(x) , dx = \int_a^c f(x) , dx + \int_c^b f(x) , dx$$ Useful for Modulus, Signum, and Greatest Integer Functions ($[x]$).

$P_4$: King's Property (The most important) $$\int_a^b f(x) , dx = \int_a^b f(a + b - x) , dx$$ Special Case ($0$ to $a$): $\int_0^a f(x) , dx = \int_0^a f(a - x) , dx$.

$P_5$: Queen's Property $$\int_0^{2a} f(x) , dx = \begin{cases} 2\int_0^a f(x) , dx & \text{if } f(2a-x) = f(x) \ 0 & \text{if } f(2a-x) = -f(x) \end{cases}$$

$P_6$: Odd/Even Functions $$\int_{-a}^a f(x) , dx = \begin{cases} 2\int_0^a f(x) , dx & \text{if } f(x) \text{ is even } [f(-x) = f(x)] \ 0 & \text{if } f(x) \text{ is odd } [f(-x) = -f(x)] \end{cases}$$

$P_7$: Periodic Functions If $f(x)$ has period $T$:

1. $\int_0^{nT} f(x) , dx = n \int_0^T f(x) , dx$ ($n \in Z$)
2. $\int_a^{a+nT} f(x) , dx = n \int_0^T f(x) , dx$
3. $\int_{mT}^{nT} f(x) , dx = (n-m) \int_0^T f(x) , dx$
4. $\int_a^{a+T} f(x) , dx$ is independent of $a$.

$P_8$: Definite Integration via Substitution When substituting $x = g(t)$, you must change the limits: Lower Limit: $a = g(t_{low}) \implies t_{low} = g^{-1}(a)$ Upper Limit: $b = g(t_{high}) \implies t_{high} = g^{-1}(b)$

5. Essential Solved Examples with Tricky Properties

Example 1: The King's Property Application

Evaluate: $I = \int_0^{\pi/2} \frac{\sin^{3/2} x}{\sin^{3/2} x + \cos^{3/2} x} , dx$

Solution: Using $P_4$: $I = \int_0^{\pi/2} \frac{\sin^{3/2}(\pi/2 - x)}{\sin^{3/2}(\pi/2 - x) + \cos^{3/2}(\pi/2 - x)} , dx$ $I = \int_0^{\pi/2} \frac{\cos^{3/2} x}{\cos^{3/2} x + \sin^{3/2} x} , dx$ Adding the two $I$ expressions: $2I = \int_0^{\pi/2} \frac{\sin^{3/2} x + \cos^{3/2} x}{\sin^{3/2} x + \cos^{3/2} x} , dx = \int_0^{\pi/2} 1 , dx = [x]_0^{\pi/2} = \pi/2$ $\implies I = \pi/4$.

Example 2: Symmetry in Odd Functions

Evaluate: $I = \int_{-1}^1 \log\left(\frac{2-x}{2+x}\right) , dx$

Solution: Let $f(x) = \log\left(\frac{2-x}{2+x}\right)$. $f(-x) = \log\left(\frac{2-(-x)}{2+(-x)}\right) = \log\left(\frac{2+x}{2-x}\right) = \log\left(\frac{2-x}{2+x}\right)^{-1} = -\log\left(\frac{2-x}{2+x}\right) = -f(x)$. Since $f(x)$ is an odd function, by $P_6$, $I = 0$.

Example 3: Modulus and Interval Splitting

Evaluate: $I = \int_0^3 |x - 1| , dx$

Solution: The critical point for $|x-1|$ is $x=1$. So we split at $x=1$ using $P_3$: $I = \int_0^1 -(x-1) , dx + \int_1^3 (x-1) , dx$ $I = [x - x^2/2]_0^1 + [x^2/2 - x]_1^3$ $I = (1 - 1/2) + ((9/2 - 3) - (1/2 - 1)) = 1/2 + (3/2 - (-1/2)) = 1/2 + 2 = 2.5$.

Example 4: Limit as a Sum (Riemann)

Evaluate: $L = \lim_{n \to \infty} \left[ \frac{1}{n+1} + \frac{1}{n+2} + \dots + \frac{1}{2n} \right]$

Solution: General term: $T_r = \frac{1}{n+r}$. $L = \lim_{n \to \infty} \sum_{r=1}^n \frac{1}{n+r} = \lim_{n \to \infty} \sum_{r=1}^n \frac{1}{n(1 + r/n)} = \lim_{n \to \infty} \sum_{r=1}^n \frac{1}{n} \cdot \frac{1}{1 + r/n}$ Converting to integral: $L = \int_0^1 \frac{1}{1+x} , dx = [\ln|1+x|]_0^1 = \ln 2 - \ln 1 = \ln 2$.

Example 5: Periodic Function Property

Evaluate: $I = \int_0^{100\pi} \sqrt{1 - \cos 2x} , dx$

Solution: $I = \int_0^{100\pi} \sqrt{2 \sin^2 x} , dx = \int_0^{100\pi} \sqrt{2} |\sin x| , dx$. The period of $|\sin x|$ is $\pi$. Using $P_7$: $I = 100 \sqrt{2} \int_0^\pi |\sin x| , dx$. In $[0, \pi]$, $\sin x \geq 0$, so $|\sin x| = \sin x$. $I = 100 \sqrt{2} [-\cos x]_0^\pi = 100 \sqrt{2} (-(-1) - (-1)) = 200 \sqrt{2}$.

Example 6: Leibniz Rule and Limit Evaluation

Find: $\lim_{x \to 0} \frac{\int_0^{x^2} \sin \sqrt{t} , dt}{x^3}$

Solution: This is a $0/0$ form. Apply L'Hôpital’s Rule: $\lim_{x \to 0} \frac{\frac{d}{dx} \int_0^{x^2} \sin \sqrt{t} , dt}{3x^2}$ Using Leibniz Rule: $\frac{d}{dx} \int_0^{x^2} \sin \sqrt{t} , dt = \sin(\sqrt{x^2}) \cdot \frac{d}{dx}(x^2) - 0 = \sin(x) \cdot 2x$. Limit becomes: $\lim_{x \to 0} \frac{2x \sin x}{3x^2} = \frac{2}{3} \lim_{x \to 0} \frac{\sin x}{x} = \frac{2}{3}(1) = \frac{2}{3}$.

Example 7: Substitution and Limit Change

Evaluate: $\int_0^1 \frac{\tan^{-1} x}{1+x^2} , dx$

Solution: Let $\tan^{-1} x = t \implies \frac{1}{1+x^2} dx = dt$. Change limits: When $x=0$, $t = \tan^{-1} 0 = 0$. When $x=1$, $t = \tan^{-1} 1 = \pi/4$. Integral becomes $\int_0^{\pi/4} t , dt = \left[ \frac{t^2}{2} \right]_0^{\pi/4} = \frac{1}{2} \left[ \frac{\pi^2}{16} \right] = \frac{\pi^2}{32}$.

Example 8: Rational Function with King's Property

Evaluate: $I = \int_0^\pi \frac{x \sin x}{1 + \cos^2 x} , dx$

Solution: Apply $P_4$: $I = \int_0^\pi \frac{(\pi-x) \sin(\pi-x)}{1 + \cos^2(\pi-x)} , dx = \int_0^\pi \frac{(\pi-x) \sin x}{1 + \cos^2 x} , dx$. $I = \pi \int_0^\pi \frac{\sin x}{1 + \cos^2 x} , dx - \int_0^\pi \frac{x \sin x}{1 + \cos^2 x} , dx$ $\implies I = \pi \int_0^\pi \frac{\sin x}{1 + \cos^2 x} , dx - I \implies 2I = \pi \int_0^\pi \frac{\sin x}{1 + \cos^2 x} , dx$. Let $\cos x = u \implies -\sin x dx = du$. Limits: $x=0 \to u=1, x=\pi \to u=-1$. $2I = \pi \int_1^{-1} \frac{-du}{1+u^2} = \pi \int_{-1}^1 \frac{du}{1+u^2} = \pi [\tan^{-1} u]_{-1}^1$ $2I = \pi (\pi/4 - (-\pi/4)) = \pi (\pi/2) \implies I = \pi^2/4$.

Example 9: Greatest Integer Function $[x]$

Evaluate: $\int_0^2 [x^2] , dx$

Solution: $x^2$ becomes 1 at $x=1$, 2 at $x=\sqrt{2}$, 3 at $x=\sqrt{3}$, and 4 at $x=2$. Split the integral: $I = \int_0^1 [x^2] dx + \int_1^{\sqrt{2}} [x^2] dx + \int_{\sqrt{2}}^{\sqrt{3}} [x^2] dx + \int_{\sqrt{3}}^2 [x^2] dx$ $I = \int_0^1 0dx + \int_1^{\sqrt{2}} 1dx + \int_{\sqrt{2}}^{\sqrt{3}} 2dx + \int_{\sqrt{3}}^2 3dx$ $I = 0 + (\sqrt{2}-1) + 2(\sqrt{3}-\sqrt{2}) + 3(2-\sqrt{3})$ $I = \sqrt{2} - 1 + 2\sqrt{3} - 2\sqrt{2} + 6 - 3\sqrt{3} = 5 - \sqrt{2} - \sqrt{3}$.

Example 10: Special Property of Log-Trig

Evaluate: $\int_0^{\pi/2} \log(\sin x) , dx$

Solution: Let $I = \int_0^{\pi/2} \log(\sin x) dx$. By King's Property: $I = \int_0^{\pi/2} \log(\cos x) dx$. Adding both: $2I = \int_0^{\pi/2} \log(\sin x \cos x) dx = \int_0^{\pi/2} \log(\frac{\sin 2x}{2}) dx$ $2I = \int_0^{\pi/2} \log(\sin 2x) dx - \int_0^{\pi/2} \log 2 dx$ For $J = \int_0^{\pi/2} \log(\sin 2x) dx$, let $2x = t \implies dx = dt/2$. $J = \frac{1}{2} \int_0^\pi \log(\sin t) dt = \frac{1}{2} \cdot 2 \int_0^{\pi/2} \log(\sin t) dt$ (using $P_5$) $\implies J = I$. Thus, $2I = I - \frac{\pi}{2} \log 2 \implies I = -\frac{\pi}{2} \log 2$.

Example 11: Fractional Part ${x}$

Evaluate: $\int_0^{10} {x} , dx$

Solution: ${x}$ is periodic with period $T=1$. $I = 10 \int_0^1 {x} dx = 10 \int_0^1 x , dx$ (since ${x}=x$ for $x \in [0, 1)$) $I = 10 [\frac{x^2}{2}]_0^1 = 5$.

Example 12: Complex Substitution

Evaluate: $I = \int_0^{\pi/4} \frac{\sec^2 x}{(1 + \tan x)(2 + \tan x)} , dx$

Solution: Let $\tan x = t \implies \sec^2 x dx = dt$. Limits: $0 \to 0, \pi/4 \to 1$. $I = \int_0^1 \frac{dt}{(1+t)(2+t)}$ Using partial fractions: $\frac{1}{(1+t)(2+t)} = \frac{1}{1+t} - \frac{1}{2+t}$ $I = [\ln|1+t| - \ln|2+t|]_0^1 = (\ln 2 - \ln 3) - (\ln 1 - \ln 2)$ $I = 2\ln 2 - \ln 3 = \ln(4/3)$.

Example 13: Walls’ Formula (Shortened)

Evaluate: $I = \int_0^{\pi/2} \sin^6 x , dx$

Solution: Using Wallis Formula for $\int_0^{\pi/2} \sin^n x dx$: If $n$ is even: $\frac{(n-1)(n-3)\dots 1}{n(n-2)\dots 2} \cdot \frac{\pi}{2}$ $I = \frac{5 \cdot 3 \cdot 1}{6 \cdot 4 \cdot 2} \cdot \frac{\pi}{2} = \frac{15}{48} \cdot \frac{\pi}{2} = \frac{5\pi}{32}$.

Example 14: Integrating an Inverse Function

Evaluate: $\int_0^1 \sin^{-1} x , dx$

Solution: Integration by parts: $\int u v , dx = u \int v dx - \int (u' \int v dx) dx$. Let $u = \sin^{-1} x, v = 1$. $I = [x \sin^{-1} x]_0^1 - \int_0^1 \frac{x}{\sqrt{1-x^2}} , dx$ Term 1: $1 \cdot \pi/2 - 0 = \pi/2$. Term 2: Let $1-x^2 = t \implies -2x dx = dt \implies x dx = -dt/2$. Limits $0 \to 1, 1 \to 0$. $\int_1^0 \frac{-dt/2}{\sqrt{t}} = \frac{1}{2} \int_0^1 t^{-1/2} dt = \frac{1}{2} [2\sqrt{t}]_0^1 = 1$. So, $I = \pi/2 - 1$.

Example 15: Exponential and King’s Property

Evaluate: $I = \int_{- \pi/2}^{\pi/2} \frac{\cos x}{1 + e^x} , dx$

Solution: Apply $P_4$: $I = \int_{- \pi/2}^{\pi/2} \frac{\cos(-x)}{1 + e^{-x}} , dx = \int_{- \pi/2}^{\pi/2} \frac{\cos x}{1 + 1/e^x} , dx = \int_{- \pi/2}^{\pi/2} \frac{e^x \cos x}{e^x + 1} , dx$. Adding the two forms: $2I = \int_{- \pi/2}^{\pi/2} \frac{(1 + e^x) \cos x}{1 + e^x} , dx = \int_{- \pi/2}^{\pi/2} \cos x , dx$ Since $\cos x$ is even: $2I = 2 \int_0^{\pi/2} \cos x , dx \implies I = [\sin x]_0^{\pi/2} = 1$.

6. Estimation of Definite Integrals

Sometimes a definite integral cannot be evaluated, but its range can be estimated:

1. If $m \leq f(x) \leq M$ for $x \in [a, b]$, then: $$m(b-a) \leq \int_a^b f(x) , dx \leq M(b-a)$$
2. If $f(x) > g(x)$, then $\int_a^b f(x) , dx > \int_a^b g(x) , dx$.

Summary Table: Which property to use?

Area Under Curves & Between Curves

The Application of Integrals, specifically the calculation of areas, is a high-yield topic in the JEE Main Mathematics syllabus. While Definite Integration provides a numerical value that can be positive, negative, or zero, Area is a physical quantity that must always be non-negative. This section demands a strong command over curve sketching, solving simultaneous equations, and mastering the properties of definite integrals.

1. Fundamental Principles of Area Calculation

The area bounded by a continuous function $y = f(x)$, the $x$-axis, and the vertical lines $x = a$ and $x = b$ is defined by the definite integral. However, if the function crosses the $x$-axis within the interval $[a, b]$, the integral will represent the "net signed area." To find the total geometric area, we must use the absolute value.

1.1 Vertical Strips (Integration with respect to $x$)

If we consider an infinitesimal vertical rectangle of width $dx$ and height $|y|$, the area $A$ is: $$A = \int_{a}^{b} |f(x)| , dx$$

• Case I: $f(x) \geq 0$ for all $x \in [a, b]$ $$A = \int_{a}^{b} f(x) , dx$$
• Case II: $f(x) \leq 0$ for all $x \in [a, b]$ $$A = -\int_{a}^{b} f(x) , dx \quad \text{or} \quad \int_{a}^{b} |f(x)| , dx$$
• Case III: $f(x)$ changes sign at $x = c$ (where $a < c < b$) $$A = \int_{a}^{c} f(x) , dx + \left| \int_{c}^{b} f(x) , dx \right|$$

1.2 Horizontal Strips (Integration with respect to $y$)

In some problems, the curve is defined as $x = g(y)$. It is often simpler to integrate along the $y$-axis between the horizontal lines $y = c$ and $y = d$. $$A = \int_{c}^{d} |g(y)| , dy$$ This is particularly useful for parabolas opening horizontally (e.g., $y^2 = 4ax$) or when the function is easier to invert.

2. Area Between Two Curves

When calculating the area enclosed between two functions $y = f(x)$ and $y = g(x)$ from $x = a$ to $x = b$:

1. Find Points of Intersection: Solve $f(x) = g(x)$ to find the limits if they are not provided.
2. Determine the Upper Curve: Identify which function is "above" the other in the given interval.
3. The Formula: $$A = \int_{a}^{b} [f_{upper}(x) - g_{lower}(x)] , dx$$

Mnemonic: "T-B" (Top minus Bottom)

• Always subtract the "Bottom" curve's $y$-value from the "Top" curve's $y$-value to ensure the integrand is positive.

3. Special Cases: Modulus, GIF, and Signum Functions

JEE Main frequently tests non-elementary functions within area problems.

• Modulus Functions ($|f(x)|$): These create "V" shapes or "reflections" of the graph. You must redefine the function by finding roots and splitting the integral accordingly.
• Greatest Integer Function ($\lfloor x \rfloor$): This creates a "step" function. The area becomes a sum of several rectangles. Since $\lfloor x \rfloor$ is piecewise constant, $\int \lfloor x \rfloor dx$ is calculated by breaking the interval into $[n, n+1)$.
• Symmetric Regions:
  • If $f(x)$ is an Even Function ($f(-x) = f(x)$), the area from $-a$ to $a$ is $2 \times \int_{0}^{a} f(x) , dx$.
  • If the region is symmetric about the line $y=x$ (e.g., a function and its inverse), then the area is $2 \times$ (area between $f(x)$ and $y=x$).

4. Summary Table of Common Standard Areas

5. Ten Solved Worked Examples (JEE Level)

Problem 1: Standard Area between Two Parabolas

Question: Find the area of the region bounded by $y^2 = 4x$ and $x^2 = 4y$.

Solution:

1. Intersection Points: Equate the curves. From $y^2 = 4x$, we get $x = y^2/4$. Substitute into $x^2 = 4y$: $(\frac{y^2}{4})^2 = 4y \implies y^4 = 64y \implies y(y^3 - 64) = 0$. $y = 0$ or $y = 4$. Correspondingly, $x = 0$ or $x = 4$.
2. Identify Curves: In the interval $x \in [0, 4]$, $y = \sqrt{4x}$ is the upper curve and $y = x^2/4$ is the lower curve.
3. Set up Integral: $$A = \int_{0}^{4} \left( 2\sqrt{x} - \frac{x^2}{4} \right) dx$$ $$A = \left[ 2 \cdot \frac{x^{3/2}}{3/2} - \frac{x^3}{12} \right]_0^4 = \left[ \frac{4}{3}x\sqrt{x} - \frac{x^3}{12} \right]_0^4$$ $A = \frac{4}{3}(8) - \frac{64}{12} = \frac{32}{3} - \frac{16}{3} = \frac{16}{3}$ sq. units.

Problem 2: Horizontal Strips

Question: Find the area bounded by the curve $x = y^2 - 1$ and the $y$-axis.

Solution:

1. Intersection with y-axis: Set $x = 0 \implies y^2 - 1 = 0 \implies y = \pm 1$.
2. Curve Nature: This is a parabola opening right with vertex at $(-1, 0)$. In the region $y \in [-1, 1]$, the value of $x$ is negative (left of $y$-axis).
3. Area Calculation: $$A = \left| \int_{-1}^{1} (y^2 - 1) dy \right|$$ $A = 2 \left| \int_{0}^{1} (y^2 - 1) dy \right|$ (due to symmetry about $x$-axis) $A = 2 \left| \left[ \frac{y^3}{3} - y \right]_0^1 \right| = 2 \left| \frac{1}{3} - 1 \right| = 2 \cdot \frac{2}{3} = \frac{4}{3}$ sq. units.

Problem 3: Modulus and Line

Question: Find the area bounded by $y = |x - 1|$ and $y = 3 - |x|$.

Solution:

1. Sketch the graphs: $y = |x - 1|$ is a V-shape vertex at $(1, 0)$. $y = 3 - |x|$ is an inverted V-shape vertex at $(0, 3)$.
2. Intersection Points:
   • For $x > 1$: $x - 1 = 3 - x \implies 2x = 4 \implies x = 2$. (Points: $(2, 1)$)
   • For $0 < x < 1$: $-(x - 1) = 3 - x \implies 1 - x = 3 - x$ (No solution).
   • For $x < 0$: $-(x - 1) = 3 - (-x) \implies 1 - x = 3 + x \implies 2x = -2 \implies x = -1$. (Points: $(-1, 2)$)
3. Integration Split: The region is between $x = -1$ and $x = 2$. Upper curve is $y = 3 - |x|$. Lower curve is $y = |x - 1|$. $A = \int_{-1}^{2} (3 - |x| - |x - 1|) dx$ Split at $x=0$ and $x=1$: $I_1 = \int_{-1}^{0} (3 + x - (1 - x)) dx = \int_{-1}^{0} (2 + 2x) dx = [2x + x^2]{-1}^{0} = 0 - (-2 + 1) = 1$. $I_2 = \int{0}^{1} (3 - x - (1 - x)) dx = \int_{0}^{1} 2 dx = [2x]0^1 = 2$. $I_3 = \int{1}^{2} (3 - x - (x - 1)) dx = \int_{1}^{2} (4 - 2x) dx = [4x - x^2]_1^2 = (8 - 4) - (4 - 1) = 4 - 3 = 1$.
4. Total Area = $1 + 2 + 1 = 4$ sq. units.

Problem 4: Integrating with Greatest Integer Function

Question: Find the area bounded by $y = \lfloor x \rfloor$, the $x$-axis, and $x = 0$ to $x = 2.5$.

Solution:

1. Function Definition:
   • $f(x) = 0$ for $x \in [0, 1)$
   • $f(x) = 1$ for $x \in [1, 2)$
   • $f(x) = 2$ for $x \in [2, 2.5]$
2. Sum of Areas (Rectangles): $A = \int_{0}^{2.5} \lfloor x \rfloor dx = \int_{0}^{1} 0 dx + \int_{1}^{2} 1 dx + \int_{2}^{2.5} 2 dx$ $A = 0 + [x]_1^2 + [2x]_2^{2.5}$ $A = 1 + (5 - 4) = 2$ sq. units.

Problem 5: Trigonometric Area

Question: Find the area enclosed by $y = \sin x$ and $y = \cos x$ between $x = 0$ and $x = \pi/2$.

Solution:

1. Intersection: $\sin x = \cos x \implies \tan x = 1 \implies x = \pi/4$.
2. Region Split:
   • From $0$ to $\pi/4$: $\cos x \geq \sin x$.
   • From $\pi/4$ to $\pi/2$: $\sin x \geq \cos x$.
3. Area Calculation: $A = \int_{0}^{\pi/4} (\cos x - \sin x) dx + \int_{\pi/4}^{\pi/2} (\sin x - \cos x) dx$ $A = [\sin x + \cos x]0^{\pi/4} + [-\cos x - \sin x]{\pi/4}^{\pi/2}$ $A = [(\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}) - (0 + 1)] + [(-0 - 1) - (-\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}})]$ $A = (\sqrt{2} - 1) + (\sqrt{2} - 1) = 2\sqrt{2} - 2$ sq. units.

Problem 6: Circle and Line (Segment of Circle)

Question: Find the area of the smaller region cut off from the circle $x^2 + y^2 = 9$ by the line $x = 3/2$.

Solution:

1. Equation for $y$: $y = \sqrt{9 - x^2}$ (upper half).
2. Limits: Since we need the area cut off by $x = 1.5$, the limits are $x = 3/2$ to $x = 3$ (the radius).
3. Integration: Area = $2 \int_{3/2}^{3} \sqrt{3^2 - x^2} dx$ (the 2 accounts for top and bottom halves) Using $\int \sqrt{a^2 - x^2} dx = \frac{x}{2}\sqrt{a^2-x^2} + \frac{a^2}{2}\sin^{-1}(\frac{x}{a})$: $A = 2 \left[ \frac{x}{2}\sqrt{9-x^2} + \frac{9}{2}\sin^{-1}(\frac{x}{3}) \right]_{3/2}^3$ $A = 2 \left[ (0 + \frac{9}{2} \cdot \frac{\pi}{2}) - (\frac{3/2}{2}\sqrt{9-9/4} + \frac{9}{2}\sin^{-1}(1/2)) \right]$ $A = 2 \left[ \frac{9\pi}{4} - (\frac{3}{4}\frac{3\sqrt{3}}{2} + \frac{9}{2}\frac{\pi}{6}) \right] = 2 \left[ \frac{9\pi}{4} - \frac{9\sqrt{3}}{8} - \frac{3\pi}{4} \right]$ $A = 2 \left[ \frac{6\pi}{4} - \frac{9\sqrt{3}}{8} \right] = 3\pi - \frac{9\sqrt{3}}{4}$ sq. units.

Problem 7: Parabola and its Normal

Question: Find the area bounded by $y^2 = x$ and the normal at the point $(1, 1)$.

Solution:

1. Find Normal Equation: $2yy' = 1 \implies y' = 1/2y$. At $(1, 1)$, slope of tangent $m_T = 1/2$. Slope of normal $m_N = -2$. Equation: $y - 1 = -2(x - 1) \implies y = -2x + 3 \implies x = \frac{3-y}{2}$.
2. Intersection: $(y^2) = \frac{3-y}{2}$ (using $x = y^2$ from parabola). $2y^2 + y - 3 = 0 \implies (2y + 3)(y - 1) = 0$. $y = 1, y = -3/2$.
3. Horizontal Integration: $A = \int_{-3/2}^{1} (x_{line} - x_{parabola}) dy = \int_{-3/2}^{1} \left( \frac{3-y}{2} - y^2 \right) dy$ $A = [\frac{3y}{2} - \frac{y^2}{4} - \frac{y^3}{3}]_{-3/2}^1$ $A = (\frac{3}{2} - \frac{1}{4} - \frac{1}{3}) - (\frac{-9}{4} - \frac{9}{16} + \frac{27}{24})$ $A = \frac{11}{12} - (-\frac{27}{16}) = \frac{11}{12} + \frac{27}{16} = \frac{44 + 81}{48} = \frac{125}{48}$ sq. units.

Problem 8: Combined Modulus

Question: Find the area of the region $|x| + |y| \leq 1$.

Solution:

1. Curve Breakdown: This represents four lines:
   • $x + y = 1$ (1st quadrant)
   • $-x + y = 1$ (2nd quadrant)
   • $-x - y = 1$ (3rd quadrant)
   • $x - y = 1$ (4th quadrant)
2. Shape: This forms a square with vertices at $(1,0), (0,1), (-1,0), (0,-1)$.
3. Calculation: Area of part in 1st quadrant = $\int_{0}^{1} (1-x) dx = [x - x^2/2]_0^1 = 1/2$. By symmetry, Total Area = $4 \times (1/2) = 2$ sq. units.

Problem 9: Enclosed by $y = e^x$, $y = \ln x$ and lines

Question: Find the area bounded by $y=e^x$, $y=0$, $x=0$, and $x=1$.

Solution:

1. Limits: $x=0$ to $x=1$.
2. Function: $e^x$ is always positive.
3. Integration: $A = \int_{0}^{1} e^x dx = [e^x]_0^1 = e^1 - e^0 = e - 1$ sq. units.

Problem 10: Inverse Trigonometric Function

Question: Find the area bounded by $y = \sin^{-1} x$, $y = \cos^{-1} x$ and the $x$-axis.

Solution:

1. Intersection: $\sin^{-1} x = \cos^{-1} x \implies x = 1/\sqrt{2}$.
2. Region: The area is bounded by the $x$-axis, so we integrate:
   • From $x=0$ to $x=1/\sqrt{2}$, the "upper" curve relative to the x-axis is $y = \sin^{-1} x$.
   • Wait, the "bounded by x-axis" means we look at the triangle-like shape at the bottom.
   • Alternatively, integrate with respect to $y$: $y = \sin^{-1} x \implies x = \sin y$ $y = \cos^{-1} x \implies x = \cos y$ Intersection at $y = \pi/4$. $A = \int_{0}^{\pi/4} (\text{Curve on the right} - \text{Curve on the left}) dy$ Since we need the area with the $x$-axis, it is easier to see it as: $A = \int_{0}^{1/\sqrt{2}} \sin^{-1} x , dx + [\text{This is not quite right}]$ Correct Logic: The region bounded by $y = \sin^{-1}x$, $y = \cos^{-1}x$ and the $x$-axis consists of the area under $\sin^{-1}x$ from $x=0$ to $1/\sqrt{2}$. $A = \int_{0}^{1/\sqrt{2}} \sin^{-1} x , dx$ Using integration by parts: $\int \sin^{-1} x dx = x\sin^{-1}x + \sqrt{1-x^2}$. $A = [x\sin^{-1}x + \sqrt{1-x^2}]_0^{1/\sqrt{2}}$ $A = (\frac{1}{\sqrt{2}} \cdot \frac{\pi}{4} + \sqrt{1-\frac{1}{2}}) - (0 + 1)$ $A = \frac{\pi}{4\sqrt{2}} + \frac{1}{\sqrt{2}} - 1$ sq. units.

6. Critical Tips for JEE Main

1. Graphing is half the battle: Always sketch the curves. This helps identify which curve is on top and where they intersect.
2. Check for Shifting: If the equation is $(y-k)^2 = 4a(x-h)$, the area remains the same as $y^2 = 4ax$. Area is invariant under translation.
3. Solving for $y$ vs $x$: If you see $y^2$ terms but only linear $x$, integrate with respect to $y$ (horizontal strips). It avoids square roots and $\pm$ signs.
4. Points of Intersection: Be careful with algebra. A mistake in intersection points will invalidate the entire integral.
5. Rationality Check: Area can never be negative. If your result is negative, you likely swapped the upper/lower functions or the limits.

Differential Equations

A Differential Equation (DE) is an equation involving an unknown function and one or more of its derivatives. In JEE Main, this chapter serves as the bridge between integral calculus and coordinate geometry. While the formation of DEs tests your ability to eliminate arbitrary constants, solving them requires a solid grip on integration techniques.

I. Order and Degree of a Differential Equation

Before solving, one must classify the equation. This classification is vital because the number of arbitrary constants in a General Solution is equal to the Order of the differential equation.

1. Order: The order of a DE is the order of the highest order derivative appearing in the equation. 2. Degree: The degree is the power of the highest order derivative when the equation is expressed as a polynomial in derivatives (i.e., derivatives are free from radicals and fractional powers).

Important Note: To find the degree, the equation must be a polynomial in terms of its derivatives. Example: $e^{\frac{dy}{dx}} + \frac{d^2y}{dx^2} = x$. Order = 2. Degree = Not defined (because of the $e^{dy/dx}$ term).

II. Formation of Differential Equations

To form a DE from a family of curves given by $f(x, y, c_1, c_2, ..., c_n) = 0$:

1. Identify the number of independent arbitrary constants ($n$).
2. Differentiate the equation $n$ times.
3. Eliminate all arbitrary constants from the original and derived equations.
4. The resulting equation is the DE of the family of curves.

Rule of Thumb: A differential equation of order $n$ represents a family of curves with $n$ essential arbitrary constants.

III. Methods of Solving First Order & First Degree DEs

1. Variable Separable Method

If the DE can be expressed in the form $f(x)dx = g(y)dy$, we solve by integrating both sides: $$\int f(x)dx = \int g(y)dy + C$$

Polar Substitution: If the equation contains terms like $(x^2+y^2)$, consider:

• $x = r \cos \theta, y = r \sin \theta \implies x^2 + y^2 = r^2$ and $x,dx + y,dy = r,dr$.

2. Homogeneous Differential Equations

A DE of the form $\frac{dy}{dx} = \frac{f(x, y)}{g(x, y)}$, where $f$ and $g$ are homogeneous functions of the same degree, is called a homogeneous DE. Substitution:

1. Put $y = vx$.
2. Then $\frac{dy}{dx} = v + x\frac{dv}{dx}$.
3. The DE transforms into a variable separable form in $v$ and $x$.

3. Linear Differential Equation (LDE)

A first-order DE is linear if it is of the form: $$\frac{dy}{dx} + Py = Q$$ Where $P$ and $Q$ are functions of $x$ (or constants).

Solution Strategy:

1. Calculate the Integrating Factor (I.F.): $$I.F. = e^{\int P dx}$$
2. The General Solution is: $$y \cdot (I.F.) = \int (Q \cdot I.F.) dx + C$$

4. Bernoulli’s Equation (Reducible to Linear)

Equations of the form: $$\frac{dy}{dx} + Py = Qy^n$$ Divide by $y^n$ and substitute $y^{1-n} = v$. This converts it into a standard LDE in $v$.

5. Exact Differential Equations

An equation $M(x,y)dx + N(x,y)dy = 0$ is exact if $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$. The solution is: $$\int_{y=const} M dx + \int (\text{terms in } N \text{ NOT containing } x) dy = C$$

IV. Applications of Differential Equations

Differential equations model various physical phenomena where the rate of change of a quantity is proportional to the quantity itself.

1. Population Growth / Radioactive Decay: $\frac{dN}{dt} = kN \implies N(t) = N_0 e^{kt}$

   • If $k > 0$, it is growth.
   • If $k < 0$, it is decay.

2. Newton’s Law of Cooling: The rate of change of temperature $T$ of a body is proportional to the difference between its temperature and the surrounding temperature $T_s$: $$\frac{dT}{dt} = -k(T - T_s)$$ Solution: $T(t) = T_s + (T_0 - T_s)e^{-kt}$.

3. RC/LR Circuits: In a charging RC circuit: $R \frac{dq}{dt} + \frac{q}{C} = E$. This is a linear DE in $q$.

V. Solved Examples (12 Problems)

Example 1: Find the order and degree of $\left[ 1 + \left( \frac{dy}{dx} \right)^2 \right]^{3/2} = k \frac{d^2y}{dx^2}$. Solution: Squaring both sides to remove fractional powers: $\left[ 1 + \left( \frac{dy}{dx} \right)^2 \right]^3 = k^2 \left( \frac{d^2y}{dx^2} \right)^2$. The highest order derivative is $\frac{d^2y}{dx^2}$, so Order = 2. The power of this derivative is 2, so Degree = 2.

Example 2: Form the DE of all circles touching the x-axis at the origin. Solution: The equation of such a circle is $x^2 + (y-a)^2 = a^2$, where $a$ is the radius. $\implies x^2 + y^2 - 2ay + a^2 = a^2 \implies x^2 + y^2 = 2ay$. Differentiating w.r.t $x$: $2x + 2y \frac{dy}{dx} = 2a \frac{dy}{dx} \implies a = \frac{x + y y'}{y'}$. Substitute $a$ back into $x^2 + y^2 = 2ay$: $x^2 + y^2 = 2y \left( \frac{x + y y'}{y'} \right)$ $y'(x^2 + y^2) = 2xy + 2y^2 y'$ $y'(x^2 - y^2) = 2xy$.

Example 3: Solve $\frac{dy}{dx} = \frac{1+y^2}{1+x^2}$ given $y(0)=1$. Solution: Variable Separable: $\frac{dy}{1+y^2} = \frac{dx}{1+x^2}$. Integrating: $\tan^{-1} y = \tan^{-1} x + C$. Using $y(0)=1$: $\tan^{-1} 1 = \tan^{-1} 0 + C \implies \frac{\pi}{4} = 0 + C \implies C = \frac{\pi}{4}$. Solution: $\tan^{-1} y = \tan^{-1} x + \frac{\pi}{4}$.

Example 4: Solve $2xy \frac{dy}{dx} = x^2 + 3y^2$ (Homogeneous). Solution: $\frac{dy}{dx} = \frac{x^2 + 3y^2}{2xy}$. Put $y = vx \implies \frac{dy}{dx} = v + x \frac{dv}{dx}$. $v + x\frac{dv}{dx} = \frac{x^2 + 3v^2x^2}{2v x^2} = \frac{1+3v^2}{2v}$. $x\frac{dv}{dx} = \frac{1+3v^2}{2v} - v = \frac{1+v^2}{2v}$. Separating variables: $\frac{2v}{1+v^2} dv = \frac{dx}{x}$. Integrating: $\ln(1+v^2) = \ln x + \ln C \implies 1+v^2 = Cx$. Substitute $v = y/x$: $1 + \frac{y^2}{x^2} = Cx \implies x^2 + y^2 = Cx^3$.

Example 5: Solve $\frac{dy}{dx} + \frac{y}{x} = x^2$ (Linear DE). Solution: Here $P = 1/x, Q = x^2$. $I.F. = e^{\int \frac{1}{x} dx} = e^{\ln x} = x$. General Solution: $y \cdot x = \int (x^2 \cdot x) dx + C$. $yx = \int x^3 dx + C = \frac{x^4}{4} + C$. $y = \frac{x^3}{4} + \frac{C}{x}$.

Example 6: Solve $\frac{dy}{dx} + \frac{y}{x} = y^2 \ln x$ (Bernoulli). Solution: Divide by $y^2$: $y^{-2} \frac{dy}{dx} + \frac{y^{-1}}{x} = \ln x$. Let $v = y^{-1}$, then $\frac{dv}{dx} = -y^{-2} \frac{dy}{dx}$. $-\frac{dv}{dx} + \frac{v}{x} = \ln x \implies \frac{dv}{dx} - \frac{v}{x} = -\ln x$. $I.F. = e^{\int -\frac{1}{x} dx} = e^{-\ln x} = 1/x$. Solution: $v \cdot \frac{1}{x} = \int (-\frac{\ln x}{x}) dx + C$. Put $\ln x = t \implies \frac{1}{x} dx = dt$. $\frac{v}{x} = -\int t dt + C = -\frac{t^2}{2} + C = -\frac{(\ln x)^2}{2} + C$. $\frac{1}{xy} = C - \frac{(\ln x)^2}{2}$.

Example 7: Solve $(x^2 + y^2)dx + 2xydy = 0$. Solution: $M = x^2 + y^2, N = 2xy$. $\frac{\partial M}{\partial y} = 2y, \frac{\partial N}{\partial x} = 2y$. (Exact!) $\int_{y=const} (x^2 + y^2) dx = \frac{x^3}{3} + xy^2$. $\int (\text{terms in } N \text{ without } x) dy = \int (0) dy = 0$. Solution: $\frac{x^3}{3} + xy^2 = C$.

Example 8: Solve $\frac{dy}{dx} = \frac{x+2y-3}{2x+y-3}$. Solution: This is a non-homogeneous linear equation. Let $x = X+h, y = Y+k$. To eliminate constants: $h+2k-3=0$ and $2h+k-3=0$. Solving gives $h=1, k=1$. Substitute $x = X+1, y = Y+1 \implies \frac{dY}{dX} = \frac{X+2Y}{2X+Y}$. Put $Y=vX$: $v + X\frac{dv}{dX} = \frac{1+2v}{2+v}$. $X\frac{dv}{dX} = \frac{1-v^2}{2+v} \implies \int \frac{2+v}{1-v^2} dv = \int \frac{dX}{X}$. Using partial fractions: $\int (\frac{1/2}{1-v} + \frac{3/2}{1+v}) dv = \ln X + C$. $-\frac{1}{2}\ln(1-v) + \frac{3}{2}\ln(1+v) = \ln X + C$. Substitute $v = \frac{y-1}{x-1}$.

Example 9: A population grows at a rate proportional to the population present. If it doubles in 50 years, in how many years will it triple? Solution: $\frac{dN}{dt} = kN \implies N = N_0 e^{kt}$. At $t=50, N = 2N_0 \implies 2 = e^{50k} \implies k = \frac{\ln 2}{50}$. To find $t$ when $N = 3N_0$: $3 = e^{kt} \implies \ln 3 = kt = \left( \frac{\ln 2}{50} \right) t$. $t = 50 \frac{\ln 3}{\ln 2} \approx 50 \times 1.58 = 79.2$ years.

Example 10: Solve $(x+y+1)dx + (2x+2y+1)dy = 0$. Solution: Observe $x+y$ is common. Let $x+y=v \implies 1 + \frac{dy}{dx} = \frac{dv}{dx}$. $\frac{dy}{dx} = \frac{dv}{dx} - 1$. The equation is $\frac{dy}{dx} = -\frac{v+1}{2v+1}$. $\frac{dv}{dx} - 1 = -\frac{v+1}{2v+1} \implies \frac{dv}{dx} = 1 - \frac{v+1}{2v+1} = \frac{v}{2v+1}$. $\frac{2v+1}{v} dv = dx \implies \int (2 + \frac{1}{v}) dv = \int dx$. $2v + \ln v = x + C$. $2(x+y) + \ln(x+y) = x + C \implies x + 2y + \ln(x+y) = C$.

Example 11: Orthogonal Trajectories of the family of parabolas $y^2 = 4ax$. Solution:

1. Differentiate: $2y y' = 4a$.
2. Eliminate $a$: From original, $4a = y^2/x$. So $2yy' = y^2/x \implies y' = y/2x$.
3. Replace $y'$ with $-1/y'$ for orthogonal trajectory: $-\frac{dx}{dy} = \frac{y}{2x} \implies 2x dx = -y dy$.
4. Integrate: $x^2 = -\frac{y^2}{2} + C \implies 2x^2 + y^2 = 2C$. (Family of Ellipses).

Example 12: Solve $x \frac{dy}{dx} - y = x^2 \sin x$. Solution: $\frac{dy}{dx} - \frac{y}{x} = x \sin x$. $I.F. = e^{\int -1/x dx} = 1/x$. $y \cdot \frac{1}{x} = \int (x \sin x \cdot \frac{1}{x}) dx + C$. $\frac{y}{x} = \int \sin x dx + C = -\cos x + C$. $y = -x \cos x + Cx$.

Mnemonic for Solution Methods

"V-H-L-B"

• Variables Separable (Isolate $x$ and $y$)
• Homogeneous (Substitute $y=vx$)
• Linear ($I.F. = e^{\int P dx}$)
• Bernoulli (Divide by $y^n$ first)

VI. Summary Table of Integrating Factors (Special Cases)

Sometimes a DE is more easily solved by recognizing an exact differential:

60 PYQ-Style Solved Problems

This section mimics the rigor and patterns of the JEE Main exam. Each problem represents a "type" frequently tested in recent years (2019-2024).

Part 1: Limits, Continuity, and Differentiability (1-15)

Q1. Limit with L'Hopital Trap Evaluate: $\lim_{x \to 0} \frac{\cos(\sin x) - \cos x}{x^4}$.

• Solution: Using expansion: $\cos \theta = 1 - \frac{\theta^2}{2!} + \frac{\theta^4}{4!} - \dots$ Numerator = $(1 - \frac{\sin^2 x}{2} + \frac{\sin^4 x}{24}) - (1 - \frac{x^2}{2} + \frac{x^4}{24})$ $= \frac{x^2 - \sin^2 x}{2} + \frac{\sin^4 x - x^4}{24}$ Use $\sin x = x - \frac{x^3}{6}$: $\sin^2 x = (x - \frac{x^3}{6})^2 \approx x^2 - \frac{x^4}{3}$ Numerator $\approx \frac{x^2 - (x^2 - x^4/3)}{2} = \frac{x^4}{6}$ Limit $= \lim_{x \to 0} \frac{x^4/6}{x^4} = \frac{1}{6}$.

Q2. The $1^\infty$ Form Find $k$ if $\lim_{x \to 0} (1 + \tan^2 \sqrt{x})^{1/2x} = e^k$.

• Solution: This is $1^\infty$ form. Use $e^{\lim f(x)(g(x)-1)}$. $k = \lim_{x \to 0} \frac{1}{2x} (\tan^2 \sqrt{x}) = \frac{1}{2} \lim_{x \to 0} (\frac{\tan \sqrt{x}}{\sqrt{x}})^2 = \frac{1}{2} (1)^2 = \frac{1}{2}$.

Q3. Continuity of Composite Functions Let $f(x) = [x]$ and $g(x) = |x-1|$. Where is $f(g(x))$ discontinuous in $(0, 3)$?

• Solution: $f(g(x)) = [|x-1|]$. In $(0, 3)$, $|x-1|$ ranges from $0$ to $2$. The GIF $[u]$ is discontinuous at integers.
  1. $|x-1| = 1 \implies x-1 = 1$ or $x-1 = -1 \implies x = 2$ or $x=0$.
  2. $|x-1| = 2 \implies x-1 = 2 \implies x = 3$. Points in $(0, 3)$ are $x=1$ (check) and $x=2$. At $x=1$, $|1-1|=0$. For $x \to 1$, $|x-1| \to 0^+$, so $[|x-1|] = 0$. $f(g(1))=0$. Continuous at $x=1$. At $x=2$, $|x-1|=1$. LHL: $x \to 2^-, |x-1| < 1 \implies [.] = 0$. RHL: $x \to 2^+, |x-1| > 1 \implies [.] = 1$. Discontinuous at $x=2$ only.

Q4. Differentiability and Modulus Number of points where $f(x) = \max(|x-1|, |x-2|, |x-3|)$ is non-differentiable.

• Solution: Plot the three V-graphs. The "max" function follows the topmost boundary. The boundary switches at intersections: $|x-1| = |x-2| \implies x = 1.5$ $|x-2| = |x-3| \implies x = 2.5$ Non-differentiable at $x=1.5$ and $x=2.5$. Total 2 points.

Q5. Functional Equations If $f(x+y) = f(x)f(y)$ and $f(1)=2, f'(0)=3$, find $f'(1)$.

• Solution: Standard result: $f(x) = a^x$. $f(1) = a^1 = 2$. So $f(x) = 2^x$. $f'(x) = 2^x \ln 2$. $f'(0) = \ln 2 = 3$ (Given). $f'(1) = 2^1 \ln 2 = 2(3) = 6$.

Q6. Leibniz Rule Application If $F(x) = \int_0^{x^2} \cos(\sqrt{t}) dt$, find $F'(x)$.

• Solution: $F'(x) = \cos(\sqrt{x^2}) \cdot \frac{d}{dx}(x^2) - 0 = 2x \cos |x|$.

Q7. Tangent and Normal (AOD) Find the slope of the normal to $y^2 = 4x$ at $(1, 2)$.

• Solution: $2y y' = 4 \implies y' = 2/y$. At $(1,2)$, $y' = 2/2 = 1$. Slope of normal $m_n = -1/m_t = -1$.

Q8. Logarithmic Limit $\lim_{x \to 0} \frac{\ln(1+x+x^2) + \ln(1-x+x^2)}{\sec x - \cos x}$.

• Solution: Num: $\ln((1+x^2)+x)(1+x^2)-x) = \ln((1+x^2)^2 - x^2) = \ln(1 + x^2 + x^4) \approx x^2$. Den: $\frac{1-\cos^2 x}{\cos x} = \frac{\sin^2 x}{\cos x} \approx x^2$. Limit $= 1$.

Q9. Intermediate Value Theorem Show $x^3 - 3x + 1 = 0$ has a root in $(0, 1)$.

• Solution: $f(0) = 1 (>0)$. $f(1) = 1-3+1 = -1 (<0)$. Since sign changes and $f$ is continuous, root exists in $(0, 1)$.

Q10. Derivatives of Inverse Trig If $y = \tan^{-1}(\frac{\sqrt{1+x^2}-1}{x})$, find $y'$.

• Solution: Put $x = \tan \theta$. $y = \tan^{-1}(\frac{\sec \theta - 1}{\tan \theta}) = \tan^{-1}(\frac{1-\cos \theta}{\sin \theta}) = \tan^{-1}(\tan \theta/2) = \theta/2$. $y = \frac{1}{2} \tan^{-1} x \implies y' = \frac{1}{2(1+x^2)}$.

Q11. Mean Value Theorem Find $c$ for $f(x) = \ln x$ on $[1, e]$.

• Solution: $f'(c) = \frac{f(e)-f(1)}{e-1} \implies \frac{1}{c} = \frac{1-0}{e-1} \implies c = e-1$.

Q12. Vertical Tangent Where does $y = x^{1/3}$ have a vertical tangent?

• Solution: $y' = \frac{1}{3x^{2/3}}$. $y' \to \infty$ as $x \to 0$. Point is $(0,0)$.

Q13. Limit of Summation $\lim_{n \to \infty} \sum_{r=1}^n \frac{n}{n^2 + r^2}$.

• Solution: $\int_0^1 \frac{1}{1+x^2} dx = [\tan^{-1} x]_0^1 = \pi/4$.

Q14. Infinite Radicals If $y = \sqrt{x + \sqrt{x + \dots}}$, find $y'$.

• Solution: $y = \sqrt{x + y} \implies y^2 - y - x = 0$. Differentiate: $2yy' - y' - 1 = 0 \implies y' = \frac{1}{2y-1}$.

Q15. Sandwich Theorem $\lim_{n \to \infty} \frac{[x] + [2x] + \dots + [nx]}{n^2}$.

• Solution: $kx - 1 < [kx] \le kx$. Summing: $\sum kx - n < \sum [kx] \le \sum kx$. Divide by $n^2$: $\frac{x \frac{n(n+1)}{2} - n}{n^2} < \text{Limit} \le \frac{x \frac{n(n+1)}{2}}{n^2}$. Both sides go to $x/2$. Limit = $x/2$.

Part 2: Applications of Derivatives (16-30)

Q16. Rate of Change A balloon's volume increases at $25$ cm$^3$/s. Find rate of change of surface area when $r=5$ cm.

• Solution: $V = \frac{4}{3}\pi r^3 \implies \frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt} = 25$. $S = 4\pi r^2 \implies \frac{dS}{dt} = 8\pi r \frac{dr}{dt}$. Substitute $\frac{dr}{dt} = \frac{25}{4\pi r^2}$: $\frac{dS}{dt} = 8\pi r (\frac{25}{4\pi r^2}) = \frac{50}{r}$. At $r=5$, rate $= 10$ cm$^2$/s.

Q17. Monotonicity Interval Find intervals of increase for $f(x) = 2x^3 - 15x^2 + 36x + 1$.

• Solution: $f'(x) = 6x^2 - 30x + 36 = 6(x^2 - 5x + 6) = 6(x-2)(x-3)$. $f'(x) > 0$ for $x \in (-\infty, 2) \cup (3, \infty)$.

Q18. Maxima/Minima of Products Find max value of $x^a y^b$ if $x+y=k$.

• Solution: Use AM-GM on $a$ parts of $(x/a)$ and $b$ parts of $(y/b)$: $\frac{a(x/a) + b(y/b)}{a+b} \ge ((\frac{x}{a})^a (\frac{y}{b})^b)^{1/(a+b)}$. $\frac{x+y}{a+b} \ge (\frac{x^a y^b}{a^a b^b})^{1/(a+b)}$. Max occurs when $x/a = y/b$.

Q19. Point of Inflection Find inflection point of $y = x e^{-x}$.

• Solution: $y' = e^{-x} - x e^{-x} = (1-x)e^{-x}$. $y'' = -e^{-x} - (1-x)e^{-x} = (x-2)e^{-x}$. $y'' = 0 \implies x=2$. Point is $(2, 2/e^2)$.

Q20. Shortest Distance Shortest distance between $y^2 = 4x$ and $x^2 + y^2 - 12x + 31 = 0$.

• Solution: Circle center $(6, 0)$, $R = \sqrt{36-31} = \sqrt{5}$. Distance is along common normal. Normal to parabola at $(t^2, 2t)$ is $y + tx = 2t + t^3$. It passes through center $(6, 0) \implies 6t = 2t + t^3 \implies t^3 - 4t = 0 \implies t=0, \pm 2$. Closest point $(4, 4)$. $D = \sqrt{(6-4)^2 + (0-4)^2} - \sqrt{5} = \sqrt{20} - \sqrt{5} = \sqrt{5}$.

Q21. Rolle's Theorem Verification $f(x) = x(x-3)^2$ on $[0, 3]$. Find $c$.

• Solution: $f(0)=0, f(3)=0$. $f'(x) = (x-3)^2 + 2x(x-3) = (x-3)(x-3+2x) = (x-3)(3x-3)$. $f'(c)=0 \implies c=3$ or $c=1$. $c=1$ is in $(0, 3)$.

Q22. Angle of Intersection Angle between $y^2=x$ and $x^2=y$ at $(1, 1)$.

• Solution: $C1: 2yy'=1 \implies m_1=1/2$. $C2: 2x=y' \implies m_2=2$. $\tan \theta = |\frac{2 - 1/2}{1 + 2(1/2)}| = \frac{3/2}{2} = 3/4$. $\theta = \tan^{-1}(3/4)$.

Q23. Approximation using Differentials Approximate $\sqrt{25.2}$.

• Solution: $f(x)=\sqrt{x}, x=25, \Delta x=0.2$. $f(x+\Delta x) \approx f(x) + f'(x)\Delta x = 5 + \frac{1}{2\sqrt{25}}(0.2) = 5 + \frac{0.2}{10} = 5.02$.

Q24. Global Maxima Max value of $f(x) = \sin x + \cos x$ on $[0, \pi]$.

• Solution: $f'(x) = \cos x - \sin x = 0 \implies x = \pi/4$. Values: $f(0)=1, f(\pi/4)=\sqrt{2}, f(\pi)=-1$. Max is $\sqrt{2}$.

Q25. Cauchy Mean Value Theorem For $f(x)=e^x, g(x)=e^{-x}$ on $[a, b]$, find $c$.

• Solution: $\frac{f'(c)}{g'(c)} = \frac{f(b)-f(a)}{g(b)-g(a)} \implies \frac{e^c}{-e^{-c}} = \frac{e^b-e^a}{e^{-b}-e^{-a}} = \frac{e^b-e^a}{(e^a-e^b)/e^{a+b}} = -e^{a+b}$. $-e^{2c} = -e^{a+b} \implies c = (a+b)/2$.

Q26. Concavity Where is $f(x) = x \ln x$ concave up?

• Solution: $f'(x) = 1 + \ln x, f''(x) = 1/x$. $1/x > 0$ for $x > 0$. Always concave up in its domain.

Q27. Optimization (Geometry) Rectangle of max area inscribed in circle $x^2+y^2=a^2$.

• Solution: Area $A = (2x)(2y) = 4x\sqrt{a^2-x^2}$. $dA/dx = 0 \implies$ it is a square of side $\sqrt{2}a$. Max Area = $2a^2$.

Q28. Velocity and Acceleration $s = t^3 - 6t^2 + 9t$. When is acceleration zero?

• Solution: $v = 3t^2 - 12t + 9, a = 6t - 12$. $a=0 \implies t=2$.

Q29. Root location using AOD How many real roots does $x^3 - 3x + k = 0$ have if $k > 2$?

• Solution: $f'(x) = 3(x^2-1)$. Max at $x=-1$, Value $= 2+k$. Min at $x=1$, Value $= -2+k$. If $k > 2$, both local max and local min are positive. Graph stays above x-axis for $x > -1$. Only 1 real root (in negative $x$ region).

Q30. Limit at Infinity (AOD connection) $\lim_{x \to \infty} (x - x^2 \ln(1 + 1/x))$.

• Solution: $x^2 (1/x - 1/x + 1/(2x^2) - 1/(3x^3) \dots)$ $\ln(1+1/x) = 1/x - 1/(2x^2) + 1/(3x^3)$. $x - x^2(1/x - 1/2x^2 + 1/3x^3) = x - x + 1/2 - 1/3x = 1/2$.

Part 3: Indefinite and Definite Integration (31-45)

Q31. Basic Substitution $\int \frac{dx}{x(1+\ln x)^2}$.

• Solution: Put $1+\ln x = u \implies \frac{dx}{x} = du$. $\int u^{-2} du = -1/u = -1/(1+\ln x) + C$.

Q32. Trig Integration $\int \sin^3 x \cos^5 x dx$.

• Solution: Put $\cos x = u \implies -\sin x dx = du$. $\int (1-u^2) u^5 (-du) = \int (u^7 - u^5) du = \frac{\cos^8 x}{8} - \frac{\cos^6 x}{6} + C$.

Q33. Partial Fractions $\int \frac{dx}{x^2 - 5x + 6}$.

• Solution: $\frac{1}{(x-2)(x-3)} = \frac{1}{x-3} - \frac{1}{x-2}$. Integral $= \ln |x-3| - \ln |x-2| = \ln |\frac{x-3}{x-2}| + C$.

Q34. King's Property $\int_0^{\pi/2} \frac{\sin x}{\sin x + \cos x} dx$.

• Solution: $I = \int_0^{\pi/2} \frac{\cos x}{\cos x + \sin x} dx$. $2I = \int_0^{\pi/2} 1 dx = \pi/2 \implies I = \pi/4$.

Q35. Integration by Parts $\int x^2 e^x dx$.

• Solution: $x^2 e^x - \int 2x e^x dx = x^2 e^x - (2x e^x - \int 2 e^x dx) = e^x(x^2 - 2x + 2) + C$.

Q36. Special Form (Exponential) $\int e^x (\frac{1}{x} - \frac{1}{x^2}) dx$.

• Solution: Match $e^x(f(x) + f'(x))$. Result $= e^x/x + C$.

Q37. Wallis Formula Type $\int_0^{\pi/2} \sin^4 x dx$.

• Solution: $\frac{3 \cdot 1}{4 \cdot 2} \cdot \frac{\pi}{2} = \frac{3\pi}{16}$.

Q38. Area Under Curve Area between $y^2 = x$ and $y=x$.

• Solution: Intersects at $(0,0)$ and $(1,1)$. Area $= \int_0^1 (\sqrt{x} - x) dx = [\frac{2}{3}x^{3/2} - \frac{x^2}{2}]_0^1 = 2/3 - 1/2 = 1/6$.

Q39. Odd/Even Property $\int_{-1}^1 \sin^5 x \cos^4 x dx$.

• Solution: Integrand is odd. Integral $= 0$.

Q40. Periodic Function Integration $\int_0^{100\pi} |\sin x| dx$.

• Solution: Period of $|\sin x|$ is $\pi$. $100 \int_0^\pi \sin x dx = 100 [-\cos x]_0^\pi = 100(1+1) = 200$.

Q41. Irrational Substitution $\int \frac{dx}{\sqrt{x^2+2x+5}}$.

• Solution: $\int \frac{dx}{\sqrt{(x+1)^2 + 4}} = \ln |(x+1) + \sqrt{x^2+2x+5}| + C$.

Q42. Definite Integral with GIF $\int_0^2 [x^2] dx$.

• Solution: Break at $x^2=1, x^2=2, x^2=3$. $\int_0^1 0 dx + \int_1^{\sqrt{2}} 1 dx + \int_{\sqrt{2}}^{\sqrt{3}} 2 dx + \int_{\sqrt{3}}^2 3 dx$ $= (\sqrt{2}-1) + 2(\sqrt{3}-\sqrt{2}) + 3(2-\sqrt{3}) = 5 - \sqrt{2} - \sqrt{3}$.

Q43. Newton-Leibniz in Limits $\lim_{x \to 0} \frac{\int_0^x \sin t^2 dt}{x^3}$.

• Solution: L'Hopital: $\lim \frac{\sin x^2}{3x^2} = 1/3$.

Q44. Integration of Modulus $\int_0^3 |x-1| dx$.

• Solution: Area of two triangles: $\frac{1}{2}(1)(1) + \frac{1}{2}(2)(2) = 0.5 + 2 = 2.5$.

Q45. Log-Trig Integration $\int_0^{\pi/2} \ln(\tan x) dx$.

• Solution: $I = \int \ln(\tan x)$. Property: $I = \int \ln(\tan(\pi/2-x)) = \int \ln(\cot x) = -I$. $2I = 0 \implies I = 0$.

Part 4: Differential Equations (46-60)

Q46. Variable Separable Solve $\frac{dy}{dx} = \frac{1+y^2}{1+x^2}$.

• Solution: $\frac{dy}{1+y^2} = \frac{dx}{1+x^2} \implies \tan^{-1} y = \tan^{-1} x + C$.

Q47. Homogeneous DE $\frac{dy}{dx} = \frac{x+y}{x}$.

• Solution: $y=vx \implies v + x\frac{dv}{dx} = 1+v \implies x\frac{dv}{dx} = 1$. $dv = dx/x \implies v = \ln|x| + C \implies y = x\ln|x| + Cx$.

Q48. Linear DE (First Order) $\frac{dy}{dx} + \frac{y}{x} = x^2$.

• Solution: $IF = e^{\int 1/x dx} = x$. $y(x) = \int x \cdot x^2 dx = x^4/4 + C \implies y = x^3/4 + C/x$.

Q49. Exact DE Identification Is $(2xy + 3)dx + (x^2 - 1)dy = 0$ exact?

• Solution: $M = 2xy+3, N = x^2-1$. $\frac{\partial M}{\partial y} = 2x, \frac{\partial N}{\partial x} = 2x$. Yes, it is exact.

Q50. Orthogonal Trajectory Find OT of $y = cx^2$.

• Solution: $y/x^2 = c \implies \frac{x^2 y' - 2xy}{x^4} = 0 \implies y' = 2y/x$. Replace $y'$ with $-1/y'$: $-1/y' = 2y/x \implies -x dx = 2y dy \implies x^2/2 + y^2 = C$. (Ellipses).

Q51. Degree and Order Order and degree of $(\frac{d^2y}{dx^2})^2 + (\frac{dy}{dx})^3 = \sin x$.

• Solution: Order = 2, Degree = 2.

Q52. Integrating Factor Search $IF$ for $\frac{dy}{dx} + y \tan x = \sec x$.

• Solution: $IF = e^{\int \tan x dx} = e^{\ln \sec x} = \sec x$.

Q53. Bernoulli's Equation $\frac{dy}{dx} + \frac{y}{x} = y^2 \ln x$.

• Solution: Divide by $y^2$: $y^{-2}y' + \frac{1}{x}y^{-1} = \ln x$. Put $y^{-1} = z \implies -y^{-2}y' = z'$. $-z' + \frac{z}{x} = \ln x \implies z' - \frac{z}{x} = -\ln x$. (Linear in $z$).

Q54. Formation of DE Form DE for $y = A \cos x + B \sin x$.

• Solution: $y' = -A \sin x + B \cos x$ $y'' = -A \cos x - B \sin x = -y$. $y'' + y = 0$.

Q55. Initial Value Problem $y' = 2x, y(0)=3$.

• Solution: $y = x^2 + C$. $3 = 0 + C \implies y = x^2 + 3$.

Q56. Growth Decay Model Rate of growth of bacteria is proportional to number present. If it doubles in 2 hours, find time to triple.

• Solution: $\frac{dN}{dt} = kN \implies N = N_0 e^{kt}$. $2N_0 = N_0 e^{2k} \implies k = \frac{\ln 2}{2}$. $3N_0 = N_0 e^{\frac{\ln 2}{2} t} \implies \ln 3 = \frac{\ln 2}{2} t \implies t = \frac{2 \ln 3}{\ln 2}$.

Q57. Tangent Property DE Find curve where intercept of tangent on y-axis is $x^2$.

• Solution: Tangent: $Y - y = y'(X - x)$. Y-intercept ($X=0$): $Y = y - xy' = x^2$. $xy' - y = -x^2 \implies y' - y/x = -x$. (Linear). $IF = 1/x$. $y(1/x) = \int -1 dx = -x + C \implies y = -x^2 + Cx$.

Q58. Substitution in DE Solve $(x+y)^2 \frac{dy}{dx} = a^2$.

• Solution: Put $x+y = z \implies 1 + y' = z'$. $z^2(z'-1) = a^2 \implies z^2 z' = a^2 + z^2 \implies \frac{z^2}{a^2+z^2} dz = dx$. $\int (1 - \frac{a^2}{z^2+a^2}) dz = x + C \implies z - a \tan^{-1}(z/a) = x + C$.

Q59. Polar DE Trap $\frac{dr}{d\theta} = r \tan \theta$.

• Solution: $\frac{dr}{r} = \tan \theta d\theta \implies \ln r = \ln \sec \theta + C \implies r = C \sec \theta$.

Q60. Non-Linear Step Solve $y' = \sqrt{y}$ for $y(0)=0$.

• Solution: $y^{-1/2} dy = dx \implies 2\sqrt{y} = x + C$. $y(0)=0 \implies C=0$. $2\sqrt{y} = x \implies y = x^2/4$. (Note: $y=0$ is also a singular solution).

Summary Table: Key Calculus Patterns

Cheat-Sheet Mnemonics

• ILATE for By-Parts: Inverse, Log, Algebraic, Trig, Exponential.
• KING/QUEEN for Definite Integrals:
  • King: Replace $x$ with $(a+b-x)$.
  • Queen: $\int_0^{2a} f(x) = 2\int_0^a$ if $f(2a-x)=f(x)$.
• LDE (Linear): $y \cdot e^{\int P} = \int Q \cdot e^{\int P}$. Remember, $P$ and $Q$ must be functions of $x$ only.

Master Formula Sheet & 20-Day Plan

Calculus isn't just a chapter; it’s the backbone of your JEE Main score, typically accounting for 33-38% of the Mathematics paper. This resource is designed to transition you from "formula cramming" to "problem-solving mastery."

1. Chapter-Wise Weightage (JEE Main Trends)

Based on the last five years (2019-2024), here is the priority breakdown for your revision:

2. The Comprehensive Formula Vault (150+ Concepts)

A. Limits, Continuity, and Differentiability (LCD)

1. Standard Limits:

   • $\lim_{x \to 0} \frac{\sin x}{x} = 1$
   • $\lim_{x \to 0} \frac{\tan x}{x} = 1$
   • $\lim_{x \to 0} \frac{1 - \cos x}{x^2} = \frac{1}{2}$
   • $\lim_{x \to 0} \frac{e^x - 1}{x} = 1$
   • $\lim_{x \to 0} \frac{\ln(1+x)}{x} = 1$
   • $\lim_{x \to a} \frac{x^n - a^n}{x - a} = na^{n-1}$

2. $1^\infty$ Form (Golden Formula): If $\lim_{x \to a} f(x) = 1$ and $\lim_{x \to a} g(x) = \infty$, then: $$\lim_{x \to a} [f(x)]^{g(x)} = e^{\lim_{x \to a} g(x)[f(x) - 1]}$$

3. L’Hôpital’s Rule: Applicable only for $\frac{0}{0}$ or $\frac{\infty}{\infty}$ forms. $\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)}$ (provided the limit exists).

4. Continuity Condition: A function is continuous at $x=a$ if $LHL = RHL = f(a)$. $\lim_{h \to 0} f(a-h) = \lim_{h \to 0} f(a+h) = f(a)$.

5. Differentiability Condition: $LHD = RHD$ (Finite value). $f'(a^-) = \lim_{h \to 0} \frac{f(a-h) - f(a)}{-h} = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h} = f'(a^+)$.

B. Differentiation & Application of Derivatives (AOD)

6. Chain Rule: $\frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x)$.
7. Logarithmic Differentiation: Used for $y = [f(x)]^{g(x)}$. $\frac{dy}{dx} = y [g'(x)\ln f(x) + \frac{g(x)}{f(x)}f'(x)]$.
8. Tangents & Normals:
   • Slope of tangent $m = f'(x_1)$.
   • Equation: $y - y_1 = m(x - x_1)$.
   • Slope of normal $m_n = -1/m$.
9. Monotonicity:
   • Increasing: $f'(x) \geq 0$.
   • Strictly Increasing: $f'(x) > 0$.
10. Maxima/Minima:
    • First Derivative Test: Sign change of $f'(x)$.
    • Second Derivative Test: If $f'(c) = 0$ and $f''(c) < 0 \implies$ Local Maxima.
11. Mean Value Theorems:
    • Rolle's: If $f(a)=f(b)$, then $\exists c \in (a,b)$ s.t. $f'(c)=0$.
    • Lagrange's (LMVT): $\exists c \in (a,b)$ s.t. $f'(c) = \frac{f(b)-f(a)}{b-a}$.

C. Indefinite Integration

12. Standard Forms:
    • $\int \frac{1}{x^2 + a^2} dx = \frac{1}{a} \tan^{-1}(\frac{x}{a}) + C$
    • $\int \frac{1}{\sqrt{a^2 - x^2}} dx = \sin^{-1}(\frac{x}{a}) + C$
    • $\int \frac{1}{x^2 - a^2} dx = \frac{1}{2a} \ln |\frac{x-a}{x+a}| + C$
    • $\int \sec x dx = \ln |\sec x + \tan x| + C$
13. Integration by Parts (ILATE): $\int u v dx = u \int v dx - \int [u' \int v dx] dx$.
14. Special Exponential Integral: $\int e^x [f(x) + f'(x)] dx = e^x f(x) + C$.

D. Definite Integration & Area

15. King’s Property: $\int_a^b f(x) dx = \int_a^b f(a+b-x) dx$.
16. Queen’s Property: $\int_0^{2a} f(x) dx = 2\int_0^a f(x) dx$ if $f(2a-x) = f(x)$.
17. Leibniz Rule (Differentiation under Integral): $\frac{d}{dx} \int_{\psi(x)}^{\phi(x)} f(t) dt = f(\phi(x))\cdot\phi'(x) - f(\psi(x))\cdot\psi'(x)$.
18. Walli’s Formula: $\int_0^{\pi/2} \sin^n x dx = \int_0^{\pi/2} \cos^n x dx = \frac{(n-1)(n-3)...(1 \text{ or } 2)}{n(n-2)...(1 \text{ or } 2)} \cdot (\frac{\pi}{2} \text{ if } n \text{ is even})$.
19. Area under Curves: Area $= \int_a^b |f(x) - g(x)| dx$.

E. Differential Equations (DE)

20. Variable Separable: $f(y) dy = g(x) dx$.
21. Homogeneous: Substitute $y = vx$.
22. Linear Differential Equation (LDE): $\frac{dy}{dx} + Py = Q$.
    • Integrating Factor (IF) $= e^{\int P dx}$.
    • Solution: $y \cdot IF = \int (Q \cdot IF) dx + C$.

3. Worked Examples (Tiered Difficulty)

Example 1: Limit of $1^\infty$ Form

Question: Evaluate $\lim_{x \to 0} (\frac{\sin x}{x})^{1/x^2}$.

Step-by-Step Solution:

1. Identify form: As $x \to 0$, $\frac{\sin x}{x} \to 1$ and $1/x^2 \to \infty$. This is $1^\infty$.
2. Apply formula: $e^{\lim_{x \to 0} \frac{1}{x^2} (\frac{\sin x}{x} - 1)}$.
3. Simplify the exponent: $\lim_{x \to 0} \frac{\sin x - x}{x^3}$.
4. Use Taylor Expansion for $\sin x$: $\sin x = x - \frac{x^3}{6} + \dots$
5. $\lim_{x \to 0} \frac{(x - \frac{x^3}{6}) - x}{x^3} = \lim_{x \to 0} \frac{-x^3/6}{x^3} = -1/6$.
6. Final Answer: $e^{-1/6}$.

Example 2: Area bounded by curves

Question: Find the area bounded by $y^2 = 4x$ and $y = 2x$.

Step-by-Step Solution:

1. Find intersection points: $(2x)^2 = 4x \implies 4x^2 = 4x \implies x=0, 1$.
2. Determine upper curve: In $(0,1)$, take $x=0.5$. $y_{parabola} = \sqrt{2} \approx 1.41$, $y_{line} = 1$. Parabola is upper.
3. Integrate: Area $= \int_0^1 (\sqrt{4x} - 2x) dx$.
4. $= [2 \cdot \frac{x^{3/2}}{3/2} - x^2]_0^1 = [\frac{4}{3}x^{3/2} - x^2]_0^1 = \frac{4}{3} - 1 = 1/3$ sq. units.

4. 20-Day Calculus Crash Plan

5. Top 50 Quick-Fire One-Liners (The "Exam-Cracker" List)

1. $\frac{d}{dx}|x| = \frac{x}{|x|}$ for $x \neq 0$.
2. If $f(x)$ is even, $f'(x)$ is odd.
3. If $f(x)$ is periodic with period $T$, $f'(x)$ is also periodic with period $T$.
4. $\int_{-a}^a f(x) dx = 0$ if $f(x)$ is odd.
5. The number of points of non-differentiability for $f(x) = |x-a| + |x-b|$ is 2.
6. $x^x$ has a minimum at $x = 1/e$.
7. The area of an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is $\pi ab$.
8. $\lim_{n \to \infty} \sum \frac{1}{n} f(r/n) = \int_0^1 f(x) dx$.
9. $\int e^{ax} \sin bx dx = \frac{e^{ax}}{a^2+b^2}(a \sin bx - b \cos bx)$.
10. Angle between two curves is the angle between their tangents at the intersection point.
11. $f(x)$ is continuous but not differentiable at "sharp corners."
12. If $f'(x) > 0$ for all $x \in R$, the function is one-to-one (injective).
13. $dy/dx$ represents the instantaneous rate of change.
14. The degree of a differential equation must be a positive integer.
15. $\int \frac{dx}{x\sqrt{x^2-1}} = \sec^{-1}x + C$.
16. Normal is always perpendicular to the tangent ($m_1 m_2 = -1$).
17. Rolle's theorem is a special case of LMVT.
18. For $y^2 = 4ax$, the tangent at $(x_1, y_1)$ is $yy_1 = 2a(x+x_1)$.
19. Point of inflection is where $f''(x) = 0$ and $f''(x)$ changes sign.
20. $\int \ln x dx = x \ln x - x + C$.
21. Vertically opposite areas in integration are equivalent if the function is symmetric.
22. The IF of $y' + (1/x)y = x$ is $x$.
23. Every polynomial function is continuous and differentiable everywhere.
24. $\sin^{-1}x + \cos^{-1}x = \pi/2$.
25. $\tan^{-1}x + \cot^{-1}x = \pi/2$.
26. If $f(x+T) = f(x)$, then $\int_0^{nT} f(x) dx = n \int_0^T f(x) dx$.
27. Area between $y^2=4ax$ and $x^2=4by$ is $\frac{16ab}{3}$.
28. Speed is the magnitude of the velocity vector.
29. A constant function has a derivative of zero.
30. The derivative of an even function is an odd function.
31. $\frac{d}{dx}(\log_a x) = \frac{1}{x \ln a}$.
32. The range of $\sin^{-1}x$ is $[-\pi/2, \pi/2]$.
33. The range of $\cos^{-1}x$ is $[0, \pi]$.
34. Limits must be checked for LHL and RHL if GIF $[x]$ or Fractional Part ${x}$ is present.
35. Discontinuity at $x=a$ means the limit doesn't exist or doesn't equal $f(a)$.
36. Local Maximum value can be smaller than Local Minimum value in different intervals.
37. $\int \frac{f'(x)}{f(x)} dx = \ln |f(x)| + C$.
38. $\int \frac{f'(x)}{\sqrt{f(x)}} dx = 2\sqrt{f(x)} + C$.
39. $\int \sec^2 x dx = \tan x + C$.
40. $\int \csc^2 x dx = -\cot x + C$.
41. Subtangent length $= |y/m|$.
42. Subnormal length $= |ym|$.
43. Concavity upward means $f''(x) > 0$.
44. Concavity downward means $f''(x) < 0$.
45. Orthogonal curves intersect at $90^\circ$.
46. $\int_0^{\pi} \sin^2 x dx = \pi/2$.
47. The differential of a constant is zero.
48. A function can be continuous but not differentiable (e.g., $y=|x|$ at $x=0$).
49. Taylor series for $e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!}...$
50. Leibniz formula is the "Product Rule" for $n^{th}$ derivatives.

6. The Final 40-Point Checklist & Last Day Drill

Checklist (Tick these off 48 hours before the exam):

1. Can you solve $\frac{0}{0}$ using expansion without L'Hôpital?
2. Do you remember the derivative of $a^x$? (It's $a^x \ln a$).
3. Can you identify the "form" of a limit within 5 seconds?
4. Do you know the conditions for LMVT? (Closed interval $[a,b]$ vs Open $(a,b)$).
5. Can you find the domain of $\sqrt{f(x)}$ and $\ln(f(x))$ quickly?
6. Are you comfortable with the Shortcut for $\int e^x (f+f') dx$?
7. Do you remember the integral of $\tan x$? ($-\ln|\cos x|$).
8. Have you practiced finding the point of inflection?
9. Do you know how to handle absolute values in integration? (Break the interval).
10. Can you solve Linear Differential Equations?
11. Do you know the difference between "strictly increasing" and "increasing"?
12. Is the formula for the area of a circle known? ($\pi r^2$).
13. Area between $y^2 = x$ and $x^2 = y$? (It's $1/3$).
14. Critical points vs. Stationary points (Stationary: $f'=0$, Critical: $f'=0$ or $f'$ undefined).
15. Leibniz rule for differentiation of integrals.
16. Newton-Leibniz formula for definite integrals.
17. Differentiation of inverse trigonometric functions.
18. Differentiation of implicit functions ($f(x,y)=0$).
19. Recognizing the form of Bernoulli's equation.
20. The concept of "Rate Measure" in AOD.
21. Limits at infinity ($\infty/\infty$).
22. Continuity of composite functions $f(g(x))$.
23. Odd/Even properties in definite integrals.
24. Sum of an infinite series using integration.
25. Orthogonal trajectories of a family of curves.
26. Geometrical meaning of $dy/dx$.
27. Using graphs to find the number of solutions.
28. Integration by substitution (choosing the right 'u').
29. Integration of rational functions using partial fractions.
30. Identifying exact differential equations ($Mdx + Ndy = 0$).
31. Maximum power theorem for limits.
32. Squeeze theorem (Sandwich theorem).
33. Intermediate Value Theorem (IVT) basics.
34. Differentiation of a function with respect to another function.
35. Logarithmic differentiation for $x^x$ types.
36. Tangent at Origin for a curve.
37. Relation between $dy/dx$ and $dx/dy$.
38. Periodicity of $\sin^2 x$. ($\pi$).
39. Volume of revolution (though mostly for JEE Advanced, basic idea helps).
40. Stability of your mental state — "I have practiced enough."

Last-Day Drill (Night Before):

• No new problems.
• Formula Flash: Spend 30 minutes reciting the integral of $\sec^3 x$ and $\sqrt{a^2-x^2}$.
• Graph Revision: Draw $y = \ln x, y = e^x, y = \sin^{-1}x, y = [x]$ (GIF) from memory.
• Error Log: Review the last 5 mistakes you made in mock tests.
• Sleep: 7 hours minimum. Calculus requires a sharp, logical brain, not a caffeinated one.

Mnemonic for Integration by Parts: "I LATE" Inverse Trig | Logarithmic | Algebraic | Trigonometric | Exponential. (Choose the one that comes first as '$u$').