JEE Main 2026 Maths Sample Paper with Solutions (Free PDF)

JEE Main 2026 Mathematics Sample Paper with Answer Key

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.

• JEE Main 2026 Sample Papers
• JEE Main 2026 Maths Sample Paper

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JEE Main 2026 Mathematics Sample Questions

1. Let $f:\mathbf{R}\to \mathbf{R}$ be a twice differentiable function such that $(\sin x\cos y)(f(2x+2y)-f(2x-2y))=(\cos x\sin \mathrm{y})(f(2\mathrm{x}+2\mathrm{y})+f(2\mathrm{x}-2\mathrm{y}))$, for all $\mathrm{x},\mathrm{y}\in \mathbf{R}$. If $f^{\prime }(0)=\frac{1}{2}$, then the value of $24f^{\prime \prime }\left(\frac{5\pi }{3}\right)$ is:

Option 1: 2

Option 2:-3

Option 3: 3

Option 4:-2

Correct Answer:- -3

$

\begin{aligned}

&\begin{aligned}

& \quad(\sin x \cos y)(f(2 x+2 y)-f(2 x-2 y))=(\cos x \sin y) \

& f(2 x+2 y)+f(2 x-2 y)) \

& f(2 x+2 y)(\sin (x-y))=f(2 x-2 y) \sin (x+y) \

& \frac{f(2 x+2 y)}{\sin (x+y)}=\frac{f(2 x-2 y)}{\sin (x-y)} \

& \text { ut } 2 x+2 y=m, 2 x-2 y=n \

& \frac{f(m)}{\sin \left(\frac{\pi}{2}\right)}=\frac{f(n)}{\sin \left(\frac{n}{2}\right)}=K

\end{aligned}\

&\text { CAREEF }\

&\begin{aligned}

& \Rightarrow f(m)=K \sin \left(\frac{m}{2}\right) \

& f(x)=K \sin \left(\frac{x}{2}\right) \

& f^{\prime}(x)=\frac{K}{2} \cos \left(\frac{x}{2}\right)

\end{aligned}\

&\text { ut } x=0 ; \frac{1}{2}=\frac{K}{2} \Rightarrow K=1\

&\begin{aligned}

& \prime(x)=\frac{1}{2} \cos \frac{x}{2} \

& f^{\prime \prime}(x)=-\frac{1}{4} \sin \frac{x}{2}

\end{aligned}\

&\begin{aligned}

& f^{\prime \prime}\left(\frac{5 \pi}{3}\right)=\left(-\frac{1}{4} \sin \left(\frac{5 \pi}{6}\right)\right) 24 \

& =\frac{-24}{8}=-3

\end{aligned}

\end{aligned}

$

Hence, the correct answer is option (2).

2. Let $\mathbf{A}=\left[\begin{array}{cc}\alpha & -1\\ 6& \beta \end{array}\right],\alpha >0$, such that $\det (\mathrm{A})=0$ and $\alpha +\beta =1$. If I denotes $2\times 2$ identity matrix, then the matrix $(1+A{)}^8$ is:

Option 1:

$

\left[\begin{array}{ll}

4 & -1 \

6 & -1

\end{array}\right]

$

Option 2

$

\left[\begin{array}{cc}

257 & -64 \

514 & -127

\end{array}\right]

$

Option 3:

$

\left[\begin{array}{cc}

1025 & -511 \

2024 & -1024

\end{array}\right]

$

Option 4:

$

\left[\begin{array}{cc}

766 & -255 \

1530 & -509

\end{array}\right]

$$

Correct Answer:

$

\left[\begin{array}{cc}

766 & -255 \

1530 & -509

\end{array}\right]

$

Solution:

$

\begin{aligned}

& |A|=0 \

& \alpha \beta+6=0 \

& \alpha \beta=-6 \

& \alpha+\beta=1 \

& \Rightarrow \alpha=3, \beta=-2 \

& A=\left[\begin{array}{ll}

3 & -1 \

6 & -2

\end{array}\right] \

& A^2=\left[\begin{array}{ll}

3 & -1 \

6 & -2

\end{array}\right]\left[\begin{array}{ll}

3 & -1 \

6 & -2

\end{array}\right]=\left[\begin{array}{ll}

3 & -1 \

6 & -2

\end{array}\right] \

& \therefore \mathrm{A}^2=\mathrm{A} \

& A=A^2=A^3=A^4=A^5 \

& (\mathrm{I}+\mathrm{A})^8 \

& =\mathrm{I}+{ }^8 \mathrm{C}_1 \mathrm{~A}^7+{ }^8 \mathrm{C}_2 \mathrm{~A}^6+\ldots \ldots+{ }^8 \mathrm{C}_8 \mathrm{~A}^8 \

& =\mathrm{I}+\mathrm{A}\left({ }^8 \mathrm{C}_1+{ }^8 \mathrm{C}_2+\ldots \ldots+{ }^8 \mathrm{C}_8\right) \

& =\mathrm{I}+\mathrm{A}\left(2^8-1\right) \

& =\left[\begin{array}{ll}

1 & 0 \

0 & 1

\end{array}\right]+\left[\begin{array}{cc}

765 & -255 \

1530 & -510

\end{array}\right] \

& =\left[\begin{array}{cc}

766 & -255 \

1530 & -509

\end{array}\right]

\end{aligned}

$

Hence, the correct answer is option (4).

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

Option 2:

6

Option 3:

8

Option 4:

10

Correct Answer:

8

Solution:

$

\begin{aligned}

& 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=\frac{\sqrt{3}}{2}, \frac{-1}{\sqrt{2}}

\end{aligned}

$

Number of solutions = 8.

Hence, the correct answer is option (3).

4. If the system of linear equations

$

\begin{aligned}

& 3 x+y+\beta z=3 \

& 2 x+\alpha y-z=-3 \

& x+2 y+z=4

\end{aligned}

$

Option 1:

49

Option 2:

31

Option 3:

43

Option 4:

37

Correct Answer:

31

Solution:

$

\begin{aligned}

& \Delta=\left|\begin{array}{ccc}

3 & 1 & \beta \

2 & \alpha & -1 \

1 & 2 & 1

\end{array}\right|=0 \

& 3 \alpha+4 \beta-\alpha \beta+3=0 \

& \Delta_3=\left|\begin{array}{ccc}

3 & 1 & 3 \

2 & \alpha & -3 \

1 & 2 & 4

\end{array}\right|=0 \

& 9 \alpha+19=0 \

& \alpha=\frac{-19}{9}, \beta=\frac{6}{11} \

& \Rightarrow 22 \beta-9 \alpha=31

\end{aligned}

$

5. Let [.] denote the greatest integer function. If ${\int }_0^{{\mathrm{e}}^{\mathrm{e}}}\left[\frac{1}{{\mathrm{e}}^{\mathrm{x}-1}}\right]\mathrm{dx}=\alpha -{\log }_{\mathrm{e}}2$, then ${\alpha }^3$ is equal to $\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}$

Correct Answer: 8

Solution:

$

\begin{aligned}

&\begin{array}{c|c}

f(x)=2 & f(x)=1 \

\frac{1}{e^{x-1}}=2 & x=1 \

x=1-\ln 2 &

\end{array}\

&\begin{aligned}

& \mathrm{f}(0)=\mathrm{e}^1=2.71 \

& f\left(e^3\right)=e^{1-e^3} \in(0,1) \

& \mathrm{I}=\int_0^{1-\ln 2} 2 \mathrm{dx}+\int_{1-\ln 2}^1 1 \mathrm{dx}+\int_1^{\mathrm{e}^3} 0 \mathrm{dx} \

& =2(1-\ell \mathrm{n} 2-0)+1(1-1+\ell \mathrm{n} 2)+0 \

& \alpha-\ln 2=2-\ln 2 \

& \alpha=2 \

& \alpha^3=8

\end{aligned}

\end{aligned}

$

JEE Main Mathematics 2026 Syllabus

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.

Download JEE Main Mathematics Syllabus 2026 PDF

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5. 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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