IIT JAM Mathematics Question Paper 2025 With Solutions PDF: Download Math Paper Here

IIT JAM Mathematics Exam Pattern

Aspirants must familiarise themselves with the official IIT JAM 2026 exam pattern for mathematics to prepare effectively.

IIT JAM Mathematics Question Paper 2025 – Overview and Analysis

The IIT JAM Mathematics 2025 exam, held on 2nd February 2025 by IIT Madras, was of moderate to slightly difficult level. It tested both conceptual understanding and problem-solving skills. The paper had 60 questions in total: MCQs, MSQs, and NATs. Most students found the 1-mark MCQs easier and scoring, while questions in the NAT section were longer and required careful calculation.

Section-wise, the paper was balanced but time management was important. Section A (MCQs) was mostly straightforward if you knew the basics. Section B (MSQs) was a bit tougher and needed good analytical thinking, especially for topics like Group Theory and Multivariable Calculus. Section C (NATs) was the most challenging since it had numerical problems that needed accuracy and careful calculations.

Topic-wise Highlights

• Questions on Group Theory, Linear Algebra, and Differential Equations carried significant weightage.

• A new question type appeared on relations and functions based on set homomorphism.

• Some questions, like those on differential equations, also included ideas from Real Analysis, which meant students had to apply concepts rather than just solve equations.

• 1-mark questions were mostly straightforward and scoring, while NAT questions were challenging and time-consuming.

For students preparing for IIT JAM Mathematics 2026, the important takeaways are clear: focus on Real Analysis, Linear Algebra, Differential Equations, and Group Theory, practice previous year questions, and improve speed and accuracy, especially in Sections B and C, where there is no negative marking.

IIT JAM Mathematics Questions with Solutions (Asked in JAM 2025)

Q 1. The sum of the infinite series $
\displaystyle\displaystyle\sum_{n=1}^{\infty}(-1)^{n+1} \frac{\pi^{2 n+1}}{2^{2 n+1}(2 n)!} $ is equal to

(A) $-\pi$
(B) $\frac{\pi}{4}$
(C) $\frac{\pi}{2}$
(D) $-\frac{\pi}{4}$

Solution:

Step 1: Rewrite the series to resemble a known Maclaurin series

We can rewrite the term inside the summation to make it more recognisable. Let's pull out a factor of $\frac{\pi}{2}$.

$
\displaystyle\sum_{n=1}^{\infty}(-1)^{n+1} \frac{\pi^{2 n+1}}{2^{2 n+1}(2 n)!}=\frac{\pi}{2} \displaystyle\sum_{n=1}^{\infty}(-1)^{n+1} \frac{\left(\frac{\pi}{2}\right)^{2 n}}{(2 n)!}
$
The Maclaurin series for $\cos (x)$ is given by:

$
\cos (x)=\displaystyle\sum_{n=0}^{\infty}(-1)^n \frac{x^{2 n}}{(2 n)!}=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\ldots
$

The sum in our series starts at $n=1$ and has the power of $(-1)$ as $n+1$, which is the negative of the power of ( -1 ) in the cosine series. Let's adjust the series for $\cos (x)$.

$
\cos (x)=(-1)^0 \frac{x^0}{0!}+\displaystyle\sum_{n=1}^{\infty}(-1)^n \frac{x^{2 n}}{(2 n)!}=1+\displaystyle\sum_{n=1}^{\infty}(-1)^n \frac{x^{2 n}}{(2 n)!}
$
Therefore,

$
\displaystyle\sum_{n=1}^{\infty}(-1)^n \frac{x^{2 n}}{(2 n)!}=\cos (x)-1
$
Multiplying by -1 gives:

$
\displaystyle\sum_{n=1}^{\infty}(-1)^{n+1} \frac{x^{2 n}}{(2 n)!}=-(\cos (x)-1)=1-\cos (x)
$

Now we can substitute $x=\frac{\pi}{2}$ into this expression.

Step 2: Evaluate the sum

Using the result from Step 1, we can evaluate our series by substituting $x=\frac{\pi}{2}$.

$
\frac{\pi}{2} \displaystyle\sum_{n=1}^{\infty}(-1)^{n+1} \frac{\left(\frac{\pi}{2}\right)^{2 n}}{(2 n)!}=\frac{\pi}{2}\left(1-\cos \left(\frac{\pi}{2}\right)\right)
$
Since $\cos \left(\frac{\pi}{2}\right)=0$, we have:

$
\frac{\pi}{2}(1-0)=\frac{\pi}{2}
$

The sum of the infinite series is (C) $\frac{\pi}{2}$.

Q 2. For which one of the following choices of $N(x, y)$, is the equation $
\left(e^x \sin y-2 y \sin x\right) \mathrm{d} x+N(x, y) \mathrm{d} y=0
$ an exact differential equation?

(A) $N(x, y)=e^x \sin y+2 \cos x$
(B) $N(x, y)=e^x \cos y+2 \cos x$
(C) $N(x, y)=e^x \cos y+2 \sin x$
(D) $N(x, y)=e^x \sin y+2 \sin x$

Solution:

For an equation of the form $M(x, y) \mathrm{d} x+N(x, y) \mathrm{d} y=0$ to be an exact differential equation, the following condition must be satisfied:

$
\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}
$
In the given equation, $M(x, y)=e^x \sin y-2 y \sin x$.

Step 1: Calculate $\frac{\partial M}{\partial y}$

To find $\frac{\partial M}{\partial y}$, we differentiate $M(x, y)$ with respect to $y$, treating $x$ as a constant:

$
\begin{gathered}
\frac{\partial M}{\partial y}=\frac{\partial}{\partial y}\left(e^x \sin y-2 y \sin x\right) \\
\frac{\partial M}{\partial y}=e^x \cos y-2 \sin x
\end{gathered}
$

Step 2: Check each choice for $N(x, y)$

Now, we must find the choice for $N(x, y)$ for which $\frac{\partial N}{\partial x}$ is equal to the expression found in Step 1.

(A) $N(x, y)=e^x \sin y+2 \cos x$

$\frac{\partial N}{\partial x}=\frac{\partial}{\partial x}\left(e^x \sin y+2 \cos x\right)=e^x \sin y-2 \sin x$

This is not equal to $e^x \cos y-2 \sin x$.

(B) $N(x, y)=e^x \cos y+2 \cos x$

$\frac{\partial N}{\partial x}=\frac{\partial}{\partial x}\left(e^x \cos y+2 \cos x\right)=e^x \cos y-2 \sin x$

This is equal to $e^x \cos y-2 \sin x$.

(C) $N(x, y)=e^x \cos y+2 \sin x$

$\frac{\partial N}{\partial x}=\frac{\partial}{\partial x}\left(e^x \cos y+2 \sin x\right)=e^x \cos y+2 \cos x$

This is not equal to $e^x \cos y-2 \sin x$.

(D) $N(x, y)=e^x \sin y+2 \sin x$

$\frac{\partial N}{\partial x}=\frac{\partial}{\partial x}\left(e^x \sin y+2 \sin x\right)=e^x \sin y+2 \cos x$

This is not equal to $e^x \cos y-2 \sin x$.

The correct choice is (B), as it satisfies the condition for an exact differential equation.
(B) $N(x, y)=e^x \cos y+2 \cos x$

To practice the complete IIT JAM previous year question paper, you can download the IIT JAM Mathematics 2025 Question Paper with Answers (PDF) below and use it as a valuable resource for your upcoming preparation.

Key Topics from IIT JAM Mathematics (Must-Prepare for 2026)

Based on last year’s question distribution, these are the high-weightage and frequently repeated topics expected to appear again in IIT JAM Mathematics 2026:

Students should prioritise conceptual clarity in these areas and practice problems from past papers to ensure a strong command before the 2026 exam.