Top 10 Most Repeated Maths Topics for JEE Main 2026: If you are preparing for JEE Main 2026, then just hard work is not enough; you also need some smart work, and one of the easiest ways is to find the topics that are asked frequently in the exam. The JEE Main exam has three main subjects, out of which Maths is an important subject that follows a pattern where some chapters carry high weightage and are repeatedly asked. By focusing on these most-scoring topics in JEE Main Maths paper 2026, students can strengthen their preparation. In this article, we have provided the Top 10 most repeated topics for JEE Mains Maths 2026. If you want to appear for JEE Advanced Exam then it is very important to qualify for this exam. The JEE Main Session 1 examination was conducted successfully from 21 to 30 January 2026. The JEE Main 2026 Session 2 is scheduled to take place from 02 April to 09 April.
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The topics asked in the JEE Mains exam follow a certain exam pattern where some chapters are asked more frequently than others. Focusing on these JEE Main Maths important topics to strengthen your exam preparation and help in maximising scores. Given below some of the most important and repeated topics that students should prioritise for JEE Mains 2026.
|
Topic Name |
Total Number of Questions |
|
102 | |
|
Area Bounded by Two Curves |
79 |
|
70 | |
|
59 | |
|
56 | |
|
51 | |
|
48 | |
|
40 | |
|
39 | |
|
39 |
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From the above table, it is clear that Linear Differential equation is the most important and repeated topic with 102 questions, the second is Area bounded by two curves with 79 questions. The huge gap between the first and second topics shows why we should give priority to the linear equation. Additionally, some topics like the Sum of n terms of an AP and the Quadratic Equation are often asked repeatedly.
Let’s go through the questions from each of the top 10 most repeated topics in Maths for JEE mains:
Question: If the solution curve of the differential equation $\left(\left(\tan ^{-1} y\right)-x\right) d y=\left(1+y^2\right) d x$ passes through the point $(1,0)$, then the abscissa of the point on the curve whose ordinate is $\tan (1)$, is
(1) 2 e
(2) $\frac{2}{\mathrm{e}}$
(3) 2
(4) $\frac{1}{e}$
Solution:
$
\frac{d x}{d y}+\frac{x}{1+y^2}=\frac{\tan ^1 y}{1+y^2}
$
Linear differential equation
Integral factor
$
\begin{aligned}
& \text { If }=\mathrm{e}^{\int \frac{1}{1+\mathrm{y}^2} \mathrm{dy}}=\mathrm{e}^{\tan ^{-1} \mathrm{y}} \\
& \Rightarrow x \times e^{\tan ^{-1} y}=\int \frac{\tan ^{-1} y e^{\tan ^{-1} y}}{1+y^2} d y
\end{aligned}
$
Let $\tan ^{-1} y=t \rightarrow x \times e^{\tan ^{-1} y}=\int \frac{t e^t}{I I I} d t$
$
\begin{aligned}
& \Rightarrow \mathrm{x} \times \mathrm{e}^{\tan ^{-1} \mathrm{y}}=\mathrm{te}^{\mathrm{t}}-\int \mathrm{e}^{\mathrm{t}} \mathrm{dt} \\
& \Rightarrow \mathrm{x} \times \mathrm{e}^{\tan ^{-1} \mathrm{y}}=\tan ^{-1} \mathrm{ye}^{\tan ^{-1} \mathrm{y}}-\mathrm{e}^{\tan ^{-1} \mathrm{y}}+\mathrm{c} \\
& \mathrm{x}=1, \mathrm{y}=0 \Rightarrow 1 \times \mathrm{e}^0=0-\mathrm{e}^0+\mathrm{c} \Rightarrow \mathrm{c}=2 \\
& y=\tan 1 \Rightarrow x \times e^1=1 \times e^1-e^1+2 \Rightarrow x=2 / e
\end{aligned}
$
Hence, the answer is the option (2).
Question: Consider a curve $y=y(x)$ in the first quadrant as shown in the figure. Let the area $A_1$ is twice the area $A_2$. Then the normal to the curve perpendicular to the line $2 x-12 y=15$ does NOT pass through the point.
(1) (6,21)
(2) (8,9)
(3) (10,−4)
(4) (12,−15)
Answer:
$
\begin{gathered}
A_1+A_2=x y-8 \\
\Rightarrow A_1+\frac{A_1}{2}=x y-8 \\
\Rightarrow A_1=\frac{2}{3}(x y-8) \\
\Rightarrow \int_4^{\mathrm{x}} \mathrm{f}(\mathrm{x}) \mathrm{dx}=\frac{2}{3} \mathrm{x} \cdot \mathrm{f}(\mathrm{x})-\frac{16}{3}
\end{gathered}
$
Differentiate w.r.t. $\times$
$
\begin{aligned}
& \begin{aligned}
\mathrm{f}(\mathrm{x}) & =\frac{2}{3} \mathrm{f}(\mathrm{x})+\frac{2}{3} \mathrm{xf}^{\prime}(\mathrm{x}) \\
\Rightarrow & \frac{\mathrm{f}(\mathrm{x})}{3}=\frac{2}{3} \mathrm{xf}(\mathrm{x}) \\
\Rightarrow & \frac{\mathrm{f}^{\prime}(\mathrm{x})}{\mathrm{f}(\mathrm{x})}=\frac{1}{2 \mathrm{x}} \\
\Rightarrow & 2 \int \frac{\mathrm{f}^{\prime}(\mathrm{x})}{\mathrm{f}(\mathrm{x})} \mathrm{dx}=\int \frac{\mathrm{dx}}{\mathrm{x}}
\end{aligned} \\
& 2 \ln |f(x)|=\ln |x|+C
\end{aligned}
$
Simplify the logarithmic terms:
$
\ln |f(x)|=\frac{1}{2} \ln |x|+\frac{C}{2}
$
Exponentiate both sides:
$
\begin{aligned}
|f(x)| & =e^{\frac{C}{2}} \sqrt{x}
\end{aligned}
$
Let $e^{\overline{2}}=k$, where $k$ is an arbitrary constant:
$
f(x)=k \sqrt{x}, k \in R .
$
and satisfy the $(10,-4)$.
Hence, the correct answer is the option (3).
Question: The outcome of each of 30 items was observed; 10 items gave an outcome $\frac{1}{2}-d$ each, 10 items gave outcome $\frac{1}{2}$ each and remaining 10 items gave outcome $\frac{1}{2}+d$ each. If the variance of this outcome data is $\frac{4}{3}$ then $|d|$ equals:
(1) 2
(2) $\frac{2}{3}$
(3) $\frac{\sqrt{5}}{2}$
(4) $\sqrt{2}$
Answer:
Variance -
In case of discrete data
$
\begin{aligned}
& \sigma^2=\left(\frac{\sum x_i^2}{n}\right)-\left(\frac{\sum x_i}{n}\right)^2 \\
& \sigma^2=\frac{\sum x^2}{N}-\mu^2=\frac{4}{3} \\
& \mu=\frac{10\left(\frac{1}{3}-d\right)+10 \times \frac{1}{2}+10\left(\frac{1}{3}+d\right)}{30} \\
& =\frac{1}{2} \\
& \sigma^2=\frac{10 \times\left(\frac{1}{3}+d\right)^2+10 \times \frac{1}{4}+10\left(\frac{1}{2}-d\right)^2}{30}-\frac{1}{4} \\
& {\left[\left(\frac{1}{2}+d\right)^2+\left(\frac{1}{2}-d\right)^2=2\left(\frac{1}{4}+d^2\right)\right]} \\
& \Rightarrow \sigma^2=\frac{1}{3}\left[2\left(\frac{1}{4}+d^2+\frac{1}{4}\right)\right]-\frac{1}{4} \\
& =\frac{1}{3}\left[\frac{3}{4}+2 d^2\right]-\frac{1}{4} \\
& =\frac{2 d^2}{3}
\end{aligned}
$
Now, $\sigma^2=\frac{4}{3}$
$
\Rightarrow \frac{2 d^2}{3}=\frac{4}{3} \Rightarrow|d|=\sqrt{2}
$
Hence, the answer is the option 4.
Question: A possible value of $x^{\prime}$, for which the ninth term in the expansion of $\left\{3^{\log _3 \sqrt{25^{x-1}+7}}+3\left(-\frac{1}{8}\right) \log _3\left(5^{x-1}+1\right)\right\}^{10}$ in the increasing powers of $3\left(-\frac{1}{8}\right) \log _3\left(5^{x-1}+1\right)$ is equal to 180 , is
(1) 0
(2) -1
(3) 2
(4) 1
Answer:
$\begin{aligned} & 3^{\log _3\left(\sqrt{25^{x-1}}+7\right)}=\sqrt{25^{x-1}+7} \text { and } \\ & 3^{-\frac{1}{8} \log _3\left(5^{x-1}+1\right)}=\left(5^{x-1}+1\right)^{-\frac{1}{8}} \\ & \text { Now } T_9=T_8+1=10 C_8 \cdot\left(\sqrt{25^{x-1}+7}\right)^2\left(5^{x-1}+1\right)^{-\frac{1}{8} \cdot 8} \\ & \Rightarrow 10 C_8 \cdot\left(25^{x-1}+7\right)\left(5^{x-1}+1\right)^{-1}=180 \\ & \quad \Rightarrow \frac{25^{x-1}+7}{5^{x-1}+1}=4 \\ & \quad \Rightarrow \operatorname{Let}^x-1 \\ & \quad \Rightarrow \frac{t^2+7}{t+1}=4 \\ & \quad \Rightarrow t^2-4 t+3=0 \\ & \Rightarrow 5^{x-1}=1, t=5^{\circ} \cdot 5^x-1=3 \\ & \Rightarrow x-1=0, x-1=\log _5 3 \\ & \Rightarrow x=1, x=1+\log _5 3\end{aligned}$
Hence, the answer is the option 4.
Question: If the system of linear equations $2 x-3 y=\gamma+5$ and $\alpha \mathrm{x}+5 \mathrm{y}=\beta+1$, where $\alpha, \beta, \gamma \in R$ has infinitely many solutions, then the value of $|9 \alpha+3 \beta+5 \gamma|$ is equal to $\_\_\_\_$ .
(1) 58
(2) 72
(3) 86
(4) 67
Answer:
$
\begin{aligned}
& 2 x-3 y=\gamma+5 \\
& \alpha x+5 y=\beta+1
\end{aligned}
$
For infinite solutions, these two lines should coincide
$
\begin{aligned}
& \frac{2}{\alpha}=\frac{-3}{5}=\frac{\gamma+5}{\beta+1} \\
& \Rightarrow \frac{2}{\alpha}=-\frac{3}{5} \Rightarrow \alpha=-\frac{10}{3} \text { and } \\
& \frac{\gamma+5}{\beta+1}=\frac{-3}{5} \\
& \Rightarrow 5 \gamma+25=-3 \beta-3 \\
& \Rightarrow 3 \beta+5 \gamma=-28 \\
& \therefore|9 \alpha+3 \beta+5 \gamma|=|-30-28|=58
\end{aligned}
$
Hence, the answer is the option (1)
Question: Let $\vec{a}=\hat{i}+\hat{j}+\hat{k}, \vec{b}$ and $\vec{c}=\hat{j}-\hat{k}$ be three vectors such that $\vec{a} \times \vec{b}=\vec{c}$ and $\vec{a} \cdot \vec{b}=1$ If the length of the projection vector of the vector $\vec{b}$ on the vector $\vec{a} \times \vec{c}$ is $l$, then the value of $3 l^2$ is equal to $\_\_\_\_$ .
(1) 2
(2) 4
(3) 6
(4) 8
Answer:
Projection of $\vec{b}$ on $(\vec{a} \times \vec{c})=\frac{|b(\vec{a} \times \vec{c})|}{\mid \vec{a} \times \vec{d}}=l$ $\_\_\_\_$
Now $\vec{a} \times \vec{b}=\vec{c}$
$
\begin{aligned}
& \vec{c} \cdot(\vec{a} \times \vec{b})=\vec{c} \cdot \vec{c}=|\vec{c}|^2 \\
& \Rightarrow[\vec{c} \vec{a} \vec{b}]=(\sqrt{2})^2=2 \\
& \Rightarrow[\vec{b} \vec{a} \vec{c}]=-2 \\
& \Rightarrow|\vec{b} \cdot(\vec{a} \times \vec{c})|=2
\end{aligned}
$
Also $\vec{a} \times \vec{c}=\left|\begin{array}{ccc}i & j & k \\ 1 & 1 \\ 0 & 1 & -1\end{array}\right|$
$
\begin{aligned}
& =i(-1-1)-j(-1)+k(1) \\
& =-2 i+j+k \\
& |\vec{a} \times \vec{c}|=\sqrt{4+1+1}=\sqrt{6}
\end{aligned}
$
From (i)
$
\begin{aligned}
& l=\frac{2}{\sqrt{6}} \\
& \Rightarrow 3 l^2=3 \cdot \frac{4}{6} \\
& =2
\end{aligned}
$
Hence, the answer is the option (1).
Question: Let the maximum area of the triangle that can be inscribed in the ellipse $\frac{x^2}{a^2}+\frac{y^2}{4}=1, a>2$, having one of its vertices at one end of the major axis of the ellipse and one of its sides parallel to the $y$-axis, be $6 \sqrt{3}$. Then the eccentricity of the ellipse is:
(1) $\frac{\sqrt{3}}{2}$
(2) $\frac{1}{2}$
(3) $\frac{1}{\sqrt{2}}$
(4) $\frac{\sqrt{3}}{4}$
Answer:
Area of $\triangle \mathrm{ABC}=\frac{1}{2} \times 4 \sin \theta \times \mathrm{a}(1-\cos \theta)$
(A)
for the maximum area,
$
\begin{gathered}
\frac{\mathrm{dA}}{\mathrm{~d} \theta}=0 \\
2 \mathrm{a} \cos \theta(1-\cos \theta)+2 \mathrm{a} \sin \theta(\sin \theta)=0 \\
2 \mathrm{a}\left(\sin ^2 \theta-\cos ^2 \theta+\cos \theta\right)=0 \\
2 \mathrm{a}\left(\sin ^2 \theta-\cos ^2 \theta+\cos \theta\right)=0 \\
-2 \cos ^2 \theta+\cos \theta+1=0 \\
\cos \theta=1, \cos \theta=-\frac{1}{2} \\
\theta=0 \theta=\frac{2 \pi}{3}
\end{gathered}
$
(not possible)
$
\begin{aligned}
\frac{\mathrm{d}^2 \mathrm{~A}}{\mathrm{~d} \theta^2} & =2 \mathrm{a}(4 \sin \theta \cos \theta-\sin \theta) \\
\theta & =\frac{2 \pi}{3} ; \frac{\mathrm{d}^2 \mathrm{~A}}{\mathrm{~d} \theta^2}=\text { negative }
\end{aligned}
$
Hence at $\theta=\frac{2 \pi}{3}$, Area is maximum
$
\begin{aligned}
& \text { maximum Area }=\frac{1}{2} \times 4 \times \frac{\sqrt{3}}{2} \times a\left(1+\frac{1}{2}\right)=6 \sqrt{3} \\
& \begin{array}{l}
a \sqrt{3}\left(\frac{3}{2}\right)=6 \sqrt{3} \\
a=4
\end{array} \\
& \therefore b^2=a^2\left(1-e^2\right) \\
& 4=16\left(1-e^2\right)
\end{aligned}
$
$e=\frac{\sqrt{3}}{2}$
Hence, the answer is the option (1)
Question: If the lines $\frac{x-2}{1}=\frac{y-3}{1}=\frac{z-4}{-k}$ and $\frac{x-1}{k}=\frac{y-4}{2}=\frac{z-5}{1}$ are coplanar, then k can have:
(1) exactly three values
(2) any value
(3) exactly one value
(4) exactly two values
Answer:
For coplanar lines we have
$
\begin{aligned}
& \left|\begin{array}{cc}
1-1-1 \\
1-\frac{1}{2}-k & 1
\end{array}\right|=0 \\
& 1(1+2 k)+1\left(1+k^2\right)-1(2-k)=0 \\
& 2 k+1+k^2+1-2+k=0 \\
& k^2+3 k=0 \\
& \mathrm{k}=0,-3
\end{aligned}
$
We have two values.
Hence, the answer is the option (4).
Question: If $l=\int_1^2 \frac{d x}{\sqrt{2 x^2-9 x^2+12 x+4}}$ then:
(1) $\frac{1}{6}<l^2<\frac{1}{2}$
(2) $\frac{1}{8}<I^2<\frac{1}{4}$
(3) $\frac{1}{9}<l^2<\frac{1}{8}$
(4) $\frac{1}{16}<I^2<\frac{1}{9}$
Answer:
$\begin{aligned} & f(x)=\frac{1}{\sqrt{2 x^3-9 x^2+12 x+4}} \\ & f^{\prime}(x)=\frac{-1 \quad\left(6 x^2-18 x+12\right)}{\left(2 x^3-9 x^2+12 x+4\right)^{\frac{3}{2}}} \\ & =\frac{-6(x-1)(x-2)}{2\left(2 x^3-9 x^2+12 x+4\right)^{\frac{3}{2}}} \\ & f(1)=\frac{1}{3} f(2)=\frac{1}{\sqrt{8}} \\ & \frac{1}{3}<I<\frac{1}{\sqrt{8}}\end{aligned}$
Hence, the answer is the option (3).
Question: If $p=\left[\begin{array}{c}\frac{\sqrt{3}}{2} \frac{1}{2} \\ -\frac{1}{2} \frac{\sqrt{3}}{2}\end{array}\right], \mathrm{A}=[11]$ and $\mathrm{Q}=\mathrm{PAP}^{\top}$, then $\mathrm{P}^{\top} \mathrm{Q}^{2015} \mathrm{P}$ is:
(1) $\left[\begin{array}{c}0 & 2015 \\ 0 & 0\end{array}\right]$
(2) $\left[\begin{array}{cc}2015 & 1 \\ 0 & 1\end{array}\right]$
(3) $\left[\begin{array}{ccc}2015 & 0 \\ 1 & 2015\end{array}\right]$
(4) $\left[\begin{array}{c}1 & 2015 \\0 & 1\end{array}\right]$
Answer:
$\begin{gathered}P^T Q^{2015} P \\ =P^T Q \cdot Q \cdot Q \cdot Q \cdot Q \ldots \ldots \ldots Q P \\ 2015 \mathrm{times} \\ \text { Now, if } P=\left[\begin{array}{l}\sqrt{3} / 2 \\ -1 / 2 \sqrt{3} / 2\end{array}\right] \\ \text { so, } P^T P=\left[\begin{array}{r}\sqrt{3} / 2-1 / 2 \\ 1 / 2 \sqrt{3} / 2\end{array}\right] \cdot\left[\begin{array}{l}\sqrt{3} / 2 \\ -1 / 2 \sqrt{3} / 2\end{array}\right] \\ {\left[\begin{array}{l}101 \\ 01\end{array}\right]=I} \\ \rightarrow P^T Q^{2015}=A^{2015} \\ =\left[\begin{array}{ll}11 & \\ 01 & \cdot[11]\end{array} \ldots[01]\right. \\ {\left[\begin{array}{l}1 & 2015 \\ 0 & 1\end{array}\right]}\end{gathered}$
Hence, the correct answer is option (4).
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.
To make your preparation easy, a list of the most important chapters for JEE Mains Maths 2026 has been prepared by our subject experts. This list is created on the basis of past year exams and shows which chapters have been asked the most over the last 10 years. In the table given below, students will find the top 10 chapters with the number of questions asked from each. This table will help you understand which chapters are most important and must be given top priority in your JEE Main 2026 study plan.
|
Chapter |
Number of Questions |
|
Vector Algebra |
1561 |
|
Co-ordinate Geometry |
410 |
|
Integral Calculus |
300 |
|
Limit, Continuity and Differentiability |
290 |
|
Three-Dimensional Geometry |
231 |
|
Matrices and Determinants |
202 |
|
Complex Numbers and Quadratic Equations |
200 |
|
Sets, Relations and Functions |
191 |
|
Statistics and Probability |
177 |
|
Sequence and Series |
173 |
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, let's understand the JEE Main 2026 Maths weightage associated with each and every chapter.
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. We have also given how many questions from each chapter were difficult, easy or moderate so that you get extra insights for your preparation.
|
Chapter |
Percentage |
|
Co-ordinate geometry |
16.80% |
|
Integral Calculus |
11.20% |
|
8.00% | |
|
7.60% | |
|
7.60% | |
|
7.60% | |
|
6.80% | |
|
6.40% | |
|
5.60% | |
|
5.20% | |
|
4.80% | |
|
4.80% | |
|
Trigonometry |
4.40% |
|
3.20% | |
|
Grand Total |
100.00% |
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Now, let’s see some other resources and the best books for the preparation of JEE Mains maths 2026!
These are the best books to refer to for the preparation of JEE Mains Maths 2026. For more insights, we have also given you the best uses and why it is recommended by experts.
|
Topic |
Best Books |
Best Use |
Why It’s Recommended |
|
Linear Differential Equation |
Differential Calculus – Amit M Agarwal (Arihant); Problems in Calculus of One Variable – I.A. Maron |
Concept clarity + advanced problem-solving |
Covers theory systematically (Arihant) and Maron gives tough JEE-style practice. |
|
Area Bounded by Two Curves |
Integral Calculus – Amit M Agarwal (Arihant); Integral Calculus for JEE – G. Tewani (Cengage) |
Application-based practice |
Cengage builds step-by-step problem approach; Arihant balances theory + solved examples. |
|
Dispersion (Variance and Standard Deviation) |
Mathematics for Class 11 & 12 – R.D. Sharma; Objective Mathematics – R.D. Sharma |
Strong basics + exam-level MCQs |
Statistics is often NCERT-driven; Sharma’s problems directly match JEE style. |
|
General Term of Binomial Expansion |
Algebra – Dr. S.K. Goyal (Arihant); Problems in Algebra – V. Govorov et al. |
Practice of tricky expansions |
Goyal is exam-oriented, Govorov adds depth and challenging questions. |
|
Cramer’s Law |
Matrices and Determinants – Amit M Agarwal (Arihant); Higher Algebra – Hall & Knight |
Direct formula-based applications |
Arihant simplifies for JEE; Hall & Knight builds classical algebra foundation. |
|
Vector (Cross Product of Two Vectors) |
Vectors & 3D Geometry – Amit M Agarwal (Arihant); Mathematics for JEE Advanced – G. Tewani (Cengage) |
Concept strengthening + 3D applications |
Excellent mix of theory, solved problems, and geometry-based vector visualization. |
|
Maxima and Minima of a Function |
Differential Calculus – Amit M Agarwal (Arihant); Problems in Calculus of One Variable – I.A. Maron |
Problem-solving drills |
Arihant explains approaches; Maron tests problem-solving speed and accuracy. |
|
Shortest Distance between Two Lines |
Vectors & 3D Geometry – Amit M Agarwal (Arihant); Coordinate Geometry – S.L. Loney |
Application of 3D concepts |
Arihant makes formulas exam-ready, Loney strengthens fundamentals and derivations. |
|
Definite Integration |
Integral Calculus – Amit M Agarwal (Arihant); Problems in Calculus of One Variable – I.A. Maron |
Mastery of standard results |
Arihant gives shortcut formulas, Maron helps with tough definite integrals. |
|
Multiplication of Two Matrices |
Matrices and Determinants – Amit M Agarwal (Arihant); Higher Algebra – Hall & Knight |
Speed + accuracy in matrix problems |
Arihant is concise for JEE; Hall & Knight is deep and concept-rich. |
Frequently Asked Questions (FAQs)
They are chosen based on chapter-wise question frequency in past exam papers, the official syllabus weightage, and their importance in building problem-solving skills.
No, while the top 10 chapters provide maximum weightage and higher chances of scoring, you should not ignore the remaining syllabus since competitive exams often include mixed-difficulty questions.
JEE Mains maths weightage chapter wise 2026 shows that Vector Algebra, Coordinate Geometry, Integral Calculus, and Three-Dimensional Geometry usually hold major importance and carry significant marks in the exam.
First, master the fundamental concepts, then practice a variety of solved examples and previous years’ questions. Consistent problem-solving and revision are key to scoring well.
On Question asked by student community
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