Calculus for JEE Mains and Advanced Preparation

Calculus forms the backbone of modern mathematics and is a high-weightage topic in both JEE Mains and JEE Advanced. With a significant number of questions asked every year, mastering calculus is not just important — it’s essential.

1.0Why is Calculus Important for JEE?

In JEE Main, calculus contributes approximately 25–30% of the total marks in Mathematics. In JEE Advanced, calculus often forms complex multi-conceptual problems that test your problem-solving depth.

2.0Important List of Topics in Calculus for JEE

Here is a chapter-wise important list of topics you must master:

1. Limits and Continuity

• Definition of limits
• L'Hôpital's Rule
• Left-hand & Right-hand limits
• Discontinuity types

2. Differentiability

• First Principle of Derivatives
• Chain Rule
• Derivatives of inverse functions
• Parametric, implicit, logarithmic differentiation

3. Application of Derivatives

• Tangents and Normals
• Increasing/Decreasing functions
• Maxima and Minima (1st and 2nd derivative test)
• Rate of change

4. Indefinite Integration

• Standard integrals
• Substitution method
• Integration by parts
• Partial fractions

5. Definite Integration

• Properties of definite integrals
• Area under curves
• Symmetry in integration

6. Differential Equations

• Order and Degree
• Variable separable
• Homogeneous equations
• Linear differential equations

3.0Important Questions Asked in JEE Main and Advanced

Below questions are based on JEE Mains 2024 Exam.

Topic: Application of Derivatives (Maxima minima)

1. Let be a real valued function. If a and b are respectively the   minimum and the maximum values of f, then a² + 2b² is equal to 

(1) 44 (2) 42 (3) 24 (4) 38

Ans. (2)

Sol.

$\begin{array}{rl}& \begin{array}{rl}& f(x)=3\sqrt{x-2}+\sqrt{4-x}\\ & x-2\ge 0and4-x\ge 0\therefore x\in [2,4]\end{array}\\ & \text{ Let }x=2{\sin }^2\theta +4{\cos }^2\theta \\ & \begin{array}{rl}& \therefore \mathrm{f}(\mathrm{x})=3\sqrt{2}|\cos \theta |+\sqrt{2}|\sin \theta |\\ & \therefore \sqrt{2}\le 3\sqrt{2}|\cos \theta |+\sqrt{2}|\sin \theta |\le \sqrt{9\times 2+2}\\ & \sqrt{2}\le 3\sqrt{2}|\cos \theta |+\sqrt{2}|\sin \theta |\le \sqrt{20}\\ & \therefore \mathrm{a}=\sqrt{2}\mathrm{b}=\sqrt{20}\\ & {\mathrm{a}}^2+2{\mathrm{b}}^2=2+40=42\end{array}\end{array}$

2. The number of critical points of the function f(x) = (x – 2)^2/3 (2x + 1) is:

(1) 2 (2) 0 (3) 1 (4) 3

Ans. (1)

Sol.

$\begin{array}{rl}& f(x)=(x-2{)}^{2/3}(2x+1)\\ & f^{\prime }(x)=\frac{2}{3}(x-2{)}^{-1/3}(2x+1)+(x-2{)}^{2/3}(2)\\ & f^{\prime }(x)=2\times \frac{(2x+1)+(x-2)}{3(x-2{)}^{1/3}}\\ & \frac{3x-1}{(x-2{)}^{1/3}}=0\end{array}$

Critical points x = 1/3 and x = 2

3. Let the set of all values of p, for which f(x) = (p² –6p + 8) (sin²2x – cos²2x) + 2(2 – p)x    + 7 does not have any critical point, be the interval (a, b). Then 16ab is equal to _____ .

Ans. (252)

Sol. f(x) = – (p² –6p + 8) cos 4n + 2(2–p)n + 7

f¹(x) = +4(p²– 6p + 8) sin 4x + (4–2p) ≠ 0

$\begin{array}{rl}& \sin 4x\ne \frac{2p-4}{4(p-4)(p-2)}\\ & \sin 4x\ne \frac{2(p-2)}{4(p-4)(p-2)}\\ & p\ne 2\\ & \sin 4x\ne \frac{1}{2(p-4)}\\ & \Rightarrow |\frac{1}{2(p-4)\mid }>1\\ & \text{ on solving we get }\\ & p\in \left(\frac{7}{2},\frac{9}{2}\right)\\ & \therefore \\ & \text{ Hence }a=\frac{7}{2},b=\frac{9}{2}\\ & \therefore 16\text{ ab }=252\end{array}$

Topic: Definite Integration

1. Let $f(x)={\int }_0^x\left(t+\sin \left(1-e^t\right)\right)dt,x\in \mathbb{Z}.Then{\lim }_{x\to 0}\frac{f(x)}{x^3}$ is equal to 

(1) 1/6    (2) -1/6 (3) -2/3   (4) 2/3

Ans. (2)

Sol. ${\lim }_{x\to 0}\frac{f(x)}{x^3}$

Using L'Hospital's Rule.

${\lim }_{x\to 0}\frac{f^{\prime }(x)}{3x^2}={\lim }_{x\to 0}\frac{x+\sin \left(1-e^x\right)}{3x^2}$  (Again, L Hopital)

Using L.H. Rule

$\begin{array}{rl}& =\underset{x\to 0}{\lim }\frac{-\left[\sin \left(1-e^x\right)\left(-e^x\right)\cdot e^x+\cos \left(1-e^x\right)\cdot e^x\right]}{6}\\ & =-\frac{1}{6}\end{array}$

2. If the value of the integral . Then, a value of α is 

1. $\frac{\pi }{6}$              (2) $\frac{\pi }{2}$ (3) $\frac{\pi }{3}$ (4) $\frac{\pi }{4}$

Ans. (2)

Sol. Let $I={\int }_{-1}^{+1}\frac{\cos \alpha x}{1+3^x}dx\dots (I)$

$\begin{array}{rl}& I={\int }_{-1}^{+1}\frac{\cos \alpha x}{1+3^{-x}}dx\\ & \left(\text{ using }{\int }_a^{b}f(x)dx={\int }_a^{b}f(a+b-x)dx\right)\end{array}$  …(II)

Add (1) and (II)

$\begin{array}{rl}& 2I={\int }_{-1}^{+1}\cos (\alpha x)dx=2{\int }_0^1\cos (\alpha x)dx\\ & I=\frac{\sin \alpha }{\alpha }=\frac{2}{\pi }(\text{ given })\\ & \therefore \alpha =\frac{\pi }{2}\end{array}$

3. If ${\int }_0^{\frac{\pi }{4}}\frac{{\sin }^{2}x}{1+\sin x\cos x}dx=\frac{1}{a}{\log }_e\left(\frac{a}{3}\right)+\frac{\pi }{b\sqrt{3}}$, where a, b ∈ N, then a + b is equal to _____.

Ans. (8)

Sol.

=$\begin{array}{rl}& {\int }_0^{\frac{\pi }{2}}\frac{{\sin }^{2}x}{1+\frac{1}{2}\sin 2x}dx={\int }_0^{\frac{\pi }{4}}\frac{1-\cos 2x}{2+\sin 2x}dx\\ & =\int \frac{1}{2+\sin 2x}-\int \frac{\cos 2x}{2+\sin 2x}\\ & =\left(I_1\right)-\left(I_2\right)\end{array}$

$\begin{array}{rl}& Here,\left(I_1\right)=\int \frac{dx}{2+\frac{2\tan x}{1+{\tan }^{2}x}}=\frac{\pi }{4}\\ & {\int }_0^{\pi }\frac{{\sec }^{2}xdx}{2{\tan }^{2}x+2\tan x+2}\\ & put\tan x=t\end{array}$

$\begin{array}{rl}& \frac{1}{2}{\int }_0^1\frac{dt}{{\left(t+\frac{1}{2}\right)}^2+\frac{3}{4}}=\frac{\pi }{6\sqrt{3}}\\ & \begin{array}{rl}& I_2={\int }_0^{\pi /4}\frac{\cos 2x}{2+\sin 2x}dx=\frac{1}{2}\left(\ln \frac{3}{2}\right)\\ & I_1-I_2=\frac{1}{\sqrt{3}}\frac{\pi }{6}+\frac{1}{2}\ln \frac{2}{3}\\ & \Rightarrow a=2,b=6\end{array}\end{array}$

Ans. 8

Topic: Differential Equations

1. Let y = y(x) be the solution of the differential equation (x² + 4)²dy + (2x³y + 8xy – 2)dx = 0. If y(0) = 0, then y(2) is equal to  

1. $\frac{\pi }{8}(2)\frac{\pi }{16}(3)\frac{\pi }{2}(4)\frac{\pi }{32}$

Ans. (4)

Sol.

$\begin{array}{rl}& \frac{dy}{dx}+y\left(\frac{2x^3+8x}{{\left(x^2+4\right)}^2}\right)=\frac{2}{{\left(x^2+4\right)}^2}\\ & \frac{dy}{dx}+y\left(\frac{2x}{x^2+4}\right)=\frac{2}{{\left(x^2+4\right)}^2}\\ & IF=e^{\int \frac{2x}{x^2+4}dx}\end{array}$

$\begin{array}{rl}\text{ IF }=& x^2+4\\ & y\times \left(x^2+4\right)=\int \frac{2}{{\left(x^2+4\right)}^2}\times \left(x^2+4\right)\\ & y\left(x^2+4\right)=2\int \frac{dx}{x^2+2^2}\\ & y\left(x^2+4\right)=\frac{2}{2}{\tan }^{-1}\left(\frac{x}{2}\right)+c\\ & 0=0+c=c=0\\ & y\left(x^2+4\right)={\tan }^{-1}\left(\frac{x}{2}\right)\\ & y\text{ at }x=2\\ & y(4+4)={\tan }^{-1}(1)\\ & y(2)=\frac{\pi }{32}\end{array}$

Option (4) is correct

2. Let y = y(x) be the solution of the differential equation (x + y + 2)² dx = dy, y(0) = –2. Let the maximum and minimum values of the function y = y(x) in $\left[0,\frac{\pi }{3}\right]$ be α and β, respectively. If $(3\alpha +\pi {)}^2+{\beta }^2=\gamma +\delta \sqrt{3},\gamma ,\delta \in \mathbb{Z}$, then γ + δ equals …..

Ans. (31)

Sol. $\frac{dy}{dx}=(x+y+2{)}^2$  ...(1),     y(0) = –2

Let x + y + 2 = v

$1+\frac{dy}{dx}=\frac{dv}{dx}$

from (1) $\frac{dv}{dx}=1+v^2$

$\int \frac{dv}{1+v^2}=\int dx$

tan^–1(v) = x + C

tan^–1(x + y + 2) = x + C

at x = 0  y = – 2  ⇒ C = 0

⇒ tan^–1(x + y + 2) = x

y = tan x – x – 2

f(x) = tan x – x – 2, x ∈ $\left[0,\frac{\pi }{3}\right]$

f '(x) = sec²x – 1 > 0 ⇒ f(x) ↑ 

f_min = f(0) = –2 = β

f_max = $f\left(\frac{\pi }{3}\right)=\sqrt{3}-\frac{\pi }{3}-2=\alpha$

now (3α + π)² + β² = γ + δ $\sqrt{3}$

⇒ (3α + π)² + β² = $(3\sqrt{3}–6)2$ + 4

γ + $\delta \sqrt{3}$ = 67 – 36 $\sqrt{3}$

⇒ γ = 67 and δ = –36 ⇒ γ + δ = 31  

Topic: Area under curves

1. One of the points of intersection of the curves y = 1 + 3x – 2x² and y = 1/x  is (½,2). Let the area of the region enclosed by these curves be $\frac{1}{24}(\ell \sqrt{5}+m)$ – nlog_e $(1+\sqrt{5})$, where l, m, n ∈ N. Then l + m + n is equal to

(1) 32 (2) 30 (3) 29 (4) 31

Ans. (2)

Sol.

$\begin{array}{rl}& A={\int }_{\frac{1}{2}}^{\frac{1+\sqrt{5}}{2}}\left(1+3x-2x^2-\frac{1}{x}\right)dx\\ & A={\left[x+\frac{3x^2}{2}-\frac{2x^3}{3}-\ln x\right]}_{\frac{1}{2}}^{\frac{1+\sqrt{5}}{2}}\\ & A=\frac{1+\sqrt{5}}{2}+\frac{3}{2}{\left(\frac{1+\sqrt{5}}{2}\right)}^2-\frac{2}{3}{\left(\frac{1+\sqrt{5}}{2}\right)}^3-\ln \left(\frac{1+\sqrt{5}}{2}\right)\\ & -\frac{1}{2}-\frac{3}{2}\left(\frac{1}{4}\right)+\frac{2}{3}\left(\frac{1}{8}\right)+\ln \left(\frac{1}{2}\right)\\ & A=\frac{1}{2}+\frac{\sqrt{5}}{2}+\frac{3}{8}+\frac{3}{4}\sqrt{5}+\frac{15}{8}-\frac{4}{3}-\frac{2}{3}\sqrt{5}\end{array}$

$\begin{array}{rl}& -\frac{1}{2}-\frac{3}{8}+\frac{1}{12}-\ln (1+\sqrt{5})\\ & \begin{array}{rl}& =\sqrt{5}\left(\frac{1}{2}+\frac{3}{4}-\frac{2}{3}\right)+\frac{15}{8}-\frac{4}{3}+\frac{1}{12}-\ln (1+\sqrt{5})\\ & =\frac{14}{24}\sqrt{5}+\frac{15}{24}-\ln (1+\sqrt{5})\end{array}\end{array}$

2. The area enclosed between the curves y = x|x| and y = x – |x| is :

(1) 8/3 (2) 2/3 (3) 1 (4) 4/3

Ans. (4)

Sol.

$A={\int }_{-2}^0-x^2-2x=\frac{4}{3}$

3. The area of the region enclosed by the parabolas y = x² – 5x and y = 7x – x² is __________.

Ans. (72)

NTA Ans. (198)

Sol. y = x² – 5x and y = 7x – x²

$\begin{array}{rl}& {\int }_0^6(g(x)-f(x))dx\\ & {\int }_0^6\left(\left(7x-x^2\right)-\left(x^2-5x\right)\right)dx\\ & {\int }_0^6\left(12x-2x^2\right)dx={\left[12\frac{x^2}{2}-\frac{2x^3}{3}\right]}_0^6\\ & \Rightarrow 6(6{)}^2-\frac{2}{3}(6{)}^3\\ & =216-144=72{\text{ unit }}^2\end{array}$

4.0Preparation Tips for Calculus

1. Master Basics First: Especially continuity, differentiability, and standard derivatives/integrals.
2. Practice Problems Daily: Focus on both objective (MCQ) and subjective (integer, matrix match).
3. Use Graphical Understanding: For increasing/decreasing and area under curves.
4. Solve PYQs: This gives you a clear idea of the pattern and level of difficulty.
5. Make Formula Sheets: For integration, differentiation rules, and DE shortcuts.

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