Top 10 Most Repeated Maths Topics for JEE Mains 2026 - High-Scoring Chapters

Most Repeated Maths Topics for JEE Mains 2026

By now you already know how competitive and challenging the JEE Mains exam is. But don’t worry, every exam is crackable and there's someone who always tops the exam. The important thing to know is that strategy and smart preparation will get you to the top. Hence, in this section we will be looking at the top 10 most repeated topics in maths for JEE Mains 2026. We will be looking at the deeply researched and analyzed data of the last 10 years of JEE Mains (all slots included). Each topic in the below table is marked with how many questions were asked from it in the last 10 years. This data about most repeated maths topics for JEE Mains will help you recognize the high weighted topics needed to ace the exam.

Liner Differential equation is by far the most important topic with 102 questions more than any other topic ever repeated! The second is Area bound by two curves with 79 questions. The important thing to note is that the most important topic is clearly visible here with a huge marginal gap therefore you must lay a lot of importance on strategizing which topics to master! As an extra piece of information, Sum of n terms of an AP and Quadratic Equation come in the list after the Multiplication of two matrices but couldn’t make it to the top 10 most repeated topics in maths for JEE Mains. However, they can’t be ignored either.

Let’s go through the questions from each of the top 10 most repeated topics in Maths for JEE mains:

1. Linear Differential Equation

If $x=f(y)$ is the solution of the differential equation
$$(1+y^2)+(x-2e^{{\tan }^{-1}y})\frac{dy}{dx}=0,y\in (-\frac{\pi }{2},\frac{\pi }{2}),$$
with $f(0)=1$, then $f\left(\frac{1}{\sqrt{3}}\right)$ is equal to:

$$\text{Option 1:}{e}^{\pi /4},\text{Option 2:}{e}^{\pi /12},\text{Option 3:}{e}^{\pi /3},\text{Option 4:}{e}^{\pi /6}$$

\noindent \textbf{Solution:}
$$\begin{array}{rl}& \frac{dx}{dy}+\frac{x}{1+y^2}=\frac{2e^{{\tan }^{-1}y}}{1+y^2},\\ & \text{I.F.}=e^{{\tan }^{-1}y},\\ & x{e}^{{\tan }^{-1}y}=\int \frac{2e^{2{\tan }^{-1}y}}{1+y^2}dy,\\ & \text{Put}{\tan }^{-1}y=t,\frac{dy}{1+y^2}=dt,\\ & x{e}^{{\tan }^{-1}y}=\int 2e^{2t}dt=e^{2{\tan }^{-1}y}+c,\\ & x=e^{{\tan }^{-1}y}+c{e}^{-{\tan }^{-1}y},\\ & \text{Since}y=0,x=1\Rightarrow 1=1+c\Rightarrow c=0,\\ & \text{For}y=\frac{1}{\sqrt{3}},x=e^{\pi /6}.\end{array}$$

Hence, the answer is option (4).

2. Area Bounded by Two Curves

If the area of the larger portion bounded between the curves $x^2+y^2=25$ and $y=|x-1|$ is $\frac{1}{4}(b\pi +c),b,c\in \mathbb{N}$, then $b+c$ is equal to \_\_\_.

\noindent \textbf{Solution:}
$$A=25\pi -{\int }_{-3}^4\sqrt{25-x^2}dx+\frac{1}{2}(4\cdot 4)+\frac{1}{2}(3\cdot 3).$$

$$A=25\pi +\frac{25}{2}-{[\frac{x}{2}\sqrt{25-x^2}+\frac{25}{2}{\sin }^{-1}\left(\frac{x}{5}\right)]}_{-3}^4$$

$$A=\frac{75\pi }{4}+\frac{1}{2}=\frac{1}{4}(75\pi +2).$$

Thus, $b=75,c=2\Rightarrow b+c=77$.

% ---------------------------
3. Dispersion (Variance and Standard Deviation)

Mean and Variance of $1,3,a,7,b$ are $5$ and $10$ respectively. Find the variance of $2,5,a+3,11,b+5$.

\noindent \textbf{Solution:}
$$5=\frac{1+3+a+7+b}{5}\Rightarrow a+b=14,$$
$$\frac{1+9+a^2+49+b^2}{5}-25=10\Rightarrow a^2+b^2=116.$$

So $a=10,b=4$. New data: $2,5,13,11,9$.

$$\text{Var}=\frac{4+25+169+121+81}{5}-64=80.2-64=16.2\approx 16.$$

Hence, option (4).

4. General Term of Binomial Expansion

If coefficients of $x^4,x^5,x^6$ in $(1+x{)}^n$ are in A.P., then max $n$ is?

$$2(\genfrac{}{}{0ex}{}{n}{5})=(\genfrac{}{}{0ex}{}{n}{4})+(\genfrac{}{}{0ex}{}{n}{6}).$$

Solving:
$$2=\frac{5}{n-4}+\frac{n-5}{6}\Rightarrow n^2-21n+98=0,$$
$$n=14\text{or}7,n_{max}=14.$$

Hence, option (1).

% ---------------------------
5. Cramer’s law

Coefficient of $x^{2/3}$ and $x^{-2/5}$ in $(x^{2/3}+\frac{1}{2}{x}^{-2/5}{)}^9$.

$$T_{r+1}=(\genfrac{}{}{0ex}{}{9}{r}){\left(x^{2/3}\right)}^{9-r}{\left(\frac{x^{-2/5}}{2}\right)}^r.$$

For $x^{2/3}$ exponent:
$$\frac{2}{3}(9-r)-\frac{2}{5}r=\frac{2}{3}\Rightarrow r=5.$$

Coefficient: $(\genfrac{}{}{0ex}{}{9}{5}){\left(\frac{1}{2}\right)}^5$.

For $x^{-2/5}$ exponent:
$$\frac{2}{3}(9-r)-\frac{2}{5}r=-\frac{2}{5}\Rightarrow r=6.$$

Coefficient: $(\genfrac{}{}{0ex}{}{9}{6}){\left(\frac{1}{2}\right)}^6$.

$$\text{Sum}=\frac{21}{4}.$$

Hence, option (1).

% ---------------------------
6. Vector (or Cross) Product of Two Vectors

Given $\overrightarrow{a}=\hat{i}+2\hat{j}-3\hat{k},\overrightarrow{b}=2\hat{i}+\hat{j}-\hat{k}$. Condition: $\overrightarrow{a}\times \overrightarrow{r}=\overrightarrow{a}\times \overrightarrow{b},\overrightarrow{a}\cdot \overrightarrow{r}=0$.

After solving:
$$\overrightarrow{r}=\frac{3}{2}\hat{i}+\frac{1}{2}\hat{k},|\overrightarrow{r}|=\sqrt{\frac{9}{4}+\frac{1}{4}}=\sqrt{10}/2\ne \sqrt{10}.$$

So Statement (I) false.
Statement (II): $\cos 2A+\cos 2B+\cos 2C\ge -\frac{3}{2}$ is true.

Hence, option (2).

7. Maxima and Minima of a Function

Roots of $2x^2+(\cos \theta )x-1=0$ are $\alpha ,\beta$. Required: ${\alpha }^4+{\beta }^4$.

$${\alpha }^4+{\beta }^4=({\alpha }^2+{\beta }^2{)}^2-2(\alpha \beta {)}^2.$$

$$\alpha +\beta =-\frac{\cos \theta }{2},\alpha \beta =-\frac{1}{2}.$$

So expression $={(\frac{{\cos }^2\theta }{4}+1)}^2-\frac{1}{2}$.

Max $=\frac{17}{16}$, min $=\frac{1}{2}$.

$$16(M+m)=25.$$

Hence, option (2).

% ---------------------------
8. Shortest Distance between Two Lines

Between parabola $y^2=8x$ and circle $x^2+y^2+12y+35=0$.

Centre $C=(0,-6)$, radius $=1$. Normal meets circle at $P(2,-4)$.

$$CP-r=\sqrt{(2-0{)}^2+(-4+6{)}^2}-1=\sqrt{8}-1=2\sqrt{2}-1.$$

Hence, option (4).

9. Application Of Inequality In Definite Integration

$${\int }_0^{\pi /4}\frac{{\cos }^{2}x{\sin }^{2}x}{({\cos }^{3}x+{\sin }^{3}x{)}^2}dx.$$

Divide numerator/denominator by $\cos x$:

$$={\int }_0^{\pi /4}\frac{{\tan }^{2}x{\sec }^{2}x}{(1+{\tan }^{3}x{)}^2}dx.$$

Let $t=1+{\tan }^{3}x$, then $dt=3{\tan }^{2}x{\sec }^{2}xdx$.

$$=\frac{1}{3}{\int }_1^2\frac{dt}{t^2}=\frac{1}{6}.$$

Hence, option (3).

10. Multiplication of two matrices

$A=\left[\begin{array}{cc}1+i& 1\\ -i& 0\end{array}\right]$.

$$A^2=\left[\begin{array}{cc}i& 1+i\\ 1-i& -i\end{array}\right],A^4=I.$$

So $A^{4p+1}=A$. Valid $n=1,5,9,\dots ,97$. Total $25$ values.

Answer: $25$.

\end{document}

We can say that we have a clear understanding of topics to prioritize and which ones have made it to the most repeated maths topics for JEE Mains. Now, lets strategize from a broader perspective and see the most important chapters for JEE mains maths 2026.

Top 10 Most Important Chapters For JEE Mains Maths 2026

After seeing the most repeated maths topics for JEE Mains, it’s time to dive into chapter! Our experts have curated a similar list for you, except this time we will list the most important chapters for JEE mains maths 2026. This list is created by Maths experts and will help you prioritize Maths chapters. The following table has the top 10 most important chapters for Maths JEE Mains on one side and the total number of questions on the other side (total questions asked in the last 10 years). This breakdown will give you clarity on the chapters that hold most marks and you can't miss them at all!

While every chapter is significant, there are some chapters that will give you the boon for higher marks coverage as they have the most repeated questions (as per past data). Now, lets understand the weightage associated with each and every chapter.

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This table will help you understand the JEE Mains maths weightage chapter wise 2026 . The weightage is given in percentage and is according to all the question papers of JEE Mains 2025 (latest). We have also given how many questions from each chapter were difficult, easy or moderate so that you get extra insights for your preparation of JEE Mains 2026.

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