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JEE Main 2026 Maths Sample Paper with Solution - We will let you improve your preparation by giving you the JEE Main 2026 Maths Sample Papers, which are made based on the National Testing Agency (NTA) pattern. These Papers are created to replicate the Level of Difficulty and format for an actual JEE Main exam that you will take on exam day. Doing the practice of JEE Main 2026 Questions every day will condition you to be able to solve the same type of problems on the exam day, leading to higher confidence in your JEE Mains exam preparation.
The National Testing Agency (NTA) is likely to release the JEE Main 2026 Session 2 city intimation slip soon at jeemain.nta.nic.in.
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The JEE Main 2026 Maths Sample Papers, along with an Answer Key, will let you check your performance on the practice questions and also pinpoint your weak spots and work on your Overall Speed and Accuracy. Regular practice using JEE Main 2026 Maths model papers will enhance your confidence and enable you to score well on the exam day. The detailed and step-by-step solutions of JEE Main 2026 Maths sample papers will instruct you on how to best perform each question and prevent you from making common errors. By Practising the JEE Main 2026 Maths sample paper with answer key, you will Improve Your Study Strategies and also learn how to effectively manage your Time while taking the Examination.
Also Read: JEE Main 2026 April Attempt Strategy
Get ready for JEE Main 2026 Mathematics with a detailed sample paper that is made as per the new NTA exam pattern. This JEE Main 2026 Maths sample paper with answer key will help you analyse your performance and strengthen your concepts.
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Students may get access to a wide range of free online resources (including a free JEE Main 2026 Maths sample paper with answer keys) that Careers360 offers, and these include: JEE Main free study materials, chapterwise practice tests, subjectwise practice tests, sample papers, and mock tests.
Question 1: $(\sin x \cos y)(f(2x+2y) - f(2x-2y)) = (\cos x \sin y)(f(2x+2y) + f(2x-2y))$, for all $x, y \in \mathbb{R}$. If $f'(0) = \frac{1}{2}$, then the value of $24f''\left(\frac{5\pi}{3}\right)$ is:
Option 1: 2 $\quad$
Option 2: $-3$ $\quad$
Option 3: 3 $\quad$
Option 4: $-2$
Correct Answer: $-3$
Solution:
$(\sin x \cos y)(f(2x+2y) - f(2x-2y)) = (\cos x \sin y)(f(2x+2y) + f(2x-2y))$
$f(2x+2y)\sin(x-y) = f(2x-2y)\sin(x+y)$
$\dfrac{f(2x+2y)}{\sin(x+y)} = \dfrac{f(2x-2y)}{\sin(x-y)}$
Let $2x+2y = m$, $2x-2y = n$:
$\dfrac{f(m)}{\sin\left(\frac{m}{2}\right)} = \dfrac{f(n)}{\sin\left(\frac{n}{2}\right)} = K$
$\Rightarrow f(x) = K\sin\left(\dfrac{x}{2}\right)$
$f'(x) = \dfrac{K}{2}\cos\left(\dfrac{x}{2}\right)$
At $x = 0$: $\dfrac{1}{2} = \dfrac{K}{2} \Rightarrow K = 1$
$f'(x) = \dfrac{1}{2}\cos\dfrac{x}{2}$
$f''(x) = -\dfrac{1}{4}\sin\dfrac{x}{2}$
$24f''\left(\dfrac{5\pi}{3}\right) = 24 \cdot \left(-\dfrac{1}{4}\sin\left(\dfrac{5\pi}{6}\right)\right) = \dfrac{-24}{8} = -3$
Hence, the correct answer is Option (2).
Question 2: Let $A = \begin{bmatrix} \alpha & -1 \ 6 & \beta \end{bmatrix}$, $\alpha > 0$, such that $\det(A) = 0$ and $\alpha + \beta = 1$. If $I$ denotes the $2 \times 2$ identity matrix, then the matrix $(I+A)^8$ is:
Option 1: $\begin{bmatrix} 4 & -1 \ 6 & -1 \end{bmatrix}$ $\quad$
Option 2: $\begin{bmatrix} 257 & -64 \ 514 & -127 \end{bmatrix}$ $\quad$
Option 3: $\begin{bmatrix} 1025 & -511 \ 2024 & -1024 \end{bmatrix}$ $\quad$
Option 4: $\begin{bmatrix} 766 & -255 \ 1530 & -509 \end{bmatrix}$
Correct Answer: $\begin{bmatrix} 766 & -255 \ 1530 & -509 \end{bmatrix}$
Solution:
$|A| = 0 \Rightarrow \alpha\beta + 6 = 0 \Rightarrow \alpha\beta = -6$
$\alpha + \beta = 1 \Rightarrow \alpha = 3,\ \beta = -2$
$A = \begin{bmatrix} 3 & -1 \ 6 & -2 \end{bmatrix}$
$A^2 = \begin{bmatrix} 3 & -1 \ 6 & -2 \end{bmatrix}\begin{bmatrix} 3 & -1 \ 6 & -2 \end{bmatrix} = \begin{bmatrix} 3 & -1 \ 6 & -2 \end{bmatrix} = A$
$\therefore A^2 = A^3 = A^4 = \cdots = A$
$(I+A)^8 = I + \binom{8}{1}A + \binom{8}{2}A^2 + \cdots + \binom{8}{8}A^8$
$= I + A\left(\binom{8}{1} + \binom{8}{2} + \cdots + \binom{8}{8}\right)$
$= I + A(2^8 - 1)$
$= \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix} + 255\begin{bmatrix} 3 & -1 \ 6 & -2 \end{bmatrix}$
$= \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix} + \begin{bmatrix} 765 & -255 \ 1530 & -510 \end{bmatrix} = \begin{bmatrix} 766 & -255 \ 1530 & -509 \end{bmatrix}$
Hence, the correct answer is Option (4).
Question 3: If $\theta \in [-2\pi, 2\pi]$, then the number of solutions of $2\sqrt{2}\cos^2\theta + (2 - \sqrt{6})\cos\theta - \sqrt{3} = 0$ is equal to:
Option 1: 12 $\quad$
Option 2: 6 $\quad$
Option 3: 8 $\quad$
Option 4: 10
Correct Answer: 8
Solution:
$2\sqrt{2}\cos^2\theta + 2\cos\theta - \sqrt{6}\cos\theta - \sqrt{3} = 0$
$(2\cos\theta - \sqrt{3})(\sqrt{2}\cos\theta + 1) = 0$
$\cos\theta = \dfrac{\sqrt{3}}{2},\quad \cos\theta = \dfrac{-1}{\sqrt{2}}$
Number of solutions $= 8$.
Hence, the correct answer is Option (3).
Question 4: If the system of linear equations
$3x + y + \beta z = 3$
$2x + \alpha y - z = -3$
$x + 2y + z = 4$
has infinitely many solutions, find $22\beta - 9\alpha$.
Option 1: 49 $\quad$
Option 2: 31 $\quad$
Option 3: 43 $\quad$
Option 4: 37
Correct Answer: 31
Solution:
$\Delta = \begin{vmatrix} 3 & 1 & \beta \ 2 & \alpha & -1 \ 1 & 2 & 1 \end{vmatrix} = 0$
$\Rightarrow 3\alpha + 4\beta - \alpha\beta + 3 = 0$
$\Delta_3 = \begin{vmatrix} 3 & 1 & 3 \ 2 & \alpha & -3 \ 1 & 2 & 4 \end{vmatrix} = 0$
$\Rightarrow 9\alpha + 19 = 0 \Rightarrow \alpha = \dfrac{-19}{9},\quad \beta = \dfrac{6}{11}$
$\Rightarrow 22\beta - 9\alpha = 31$
Hence, the correct answer is Option (2).
Question 5: Let $[\cdot]$ denote the greatest integer function. If $\displaystyle\int_0^{e^3} \left[\dfrac{1}{e^{x-1}}\right]dx = \alpha - \log_e 2$, then $\alpha^3$ is equal to ____
Correct Answer: 8
Solution:
When $\dfrac{1}{e^{x-1}} = 2 \Rightarrow x = 1 - \ln 2$
When $\dfrac{1}{e^{x-1}} = 1 \Rightarrow x = 1$
$f(0) = e^1 \approx 2.71$, $\quad f(e^3) = e^{1-e^3} \in (0,1)$
$I = \displaystyle\int_0^{1-\ln 2} 2, dx + \int_{1-\ln 2}^{1} 1, dx + \int_1^{e^3} 0, dx$
$= 2(1 - \ln 2) + (1 - 1 + \ln 2) + 0$
$= 2 - 2\ln 2 + \ln 2 = 2 - \ln 2$
$\therefore\ \alpha - \ln 2 = 2 - \ln 2 \Rightarrow \alpha = 2$
$\alpha^3 = 8$
In order to effectively prepare for JEE Main Mathematics, it is imperative to have a comprehensive understanding of the entire mathematics syllabus distributed by JEE; thus, prior to moving forward with the 2026 JEE Mains Maths Sample Paper and Solutions, we will outline the entire mathematics syllabus.
Improves Speed & Accuracy – The JEE Mains 2026 Maths Sample Paper with Solutions, when practiced on a consistent basis, will help you in developing speed and accuracy in answering questions in this highly time sensitive examination. This type and frequency of practice develops an ability to recall a variety of different formulas and techniques automatically once the student arrives on the examination day.
Build Exam Familiarity – By working through the JEE Mains 2026 Maths Sample Paper with Solutions, students gain a familiarity with the look, and layout of the actual JEE Mains examination. The JEE Mains 2026 Maths Sample Paper with Solutions is created to provide students with exposure to the various question styles and levels of difficulty they may encounter, reducing their test anxiety and increasing their level of confidence going into their JEE Mains examination.
Identifies Weak Areas – Completing the JEE Mains 2026 Maths Sample Paper with Solutions and reviewing the solutions to the sample paper will provide an overview to which topic areas require the most focus for preparation purposes, allowing candidates to direct their studying those particular topics, thus providing candidates with the greatest potential to achieve the highest possible score in JEE Mains 2026.
Improves Problem-Solving Abilities - The JEE Mains sample papers for Mathematics for 2026 provides different types of questions and at the same time an opportunity to develop critical thinking, confidence, and improve upon logical ways to solve problems in various ways.
Management Of Time – The completion of JEE Main Mathematics sample papers under timed conditions prepares candidates to efficiently manage the division of time to each question. Students also get used to giving more importance to the easiest questions when going for the exam to get the maximum score.
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JEE Main Chapter-Wise Weightage
Frequently Asked Questions (FAQs)
Yes, you can refer and start with the sample papers and understand what kind of questions are asked in the exam paper but you have to complete each portion of the syllabus and then solve the sample papers.
Yes, after completing each topic, you can solve the sample paper based on that topic.
On Question asked by student community
The WBJEE 2026 exam will not be held in Maharashtra, as per the details mentioned in the brochure. The exam centres will be available in West Bengal, Assam and Tripura. If you are from any other state, then your options will be limited to WB.
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The chances of getting admission in DAIICT Gandhinagar are tough with 90 percentile in JEE Mains .
NIT Mizoram / Arunachal Pradesh / Manipur / Sikkim (Core branches)
Guru Ghasidas Vishwavidyalaya, Bilaspur (Central University)
Assam University, Silchar (Assam University)
Sant Longowal Institute of Engineering and Technology (SLIET), Punjab
Gurukula Kangri Vishwavidyalaya, Haridwar
It is hard to get a seat in NIT Warangal with a 59.8 JEE Main percentile.
The chances of getting NITs or IIITs are very low. Better to go for state-level counseling.
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