JEE Advanced Mathematics 2025 - Top Skipped, Wrongly Attempted and Correctly Solved Questions

JEE Advanced 2025 Maths Analysis - Paper 1 and Paper 2 Question-Wise Candidate’s Performance

Students can refer to the table given below that provides information about how students performed in Paper 1 and Paper 2 of JEE Advanced 2025 Maths section. This analysis will help students to determine the difficulty levels, and common mistakes, help them identify their strengths and weaknesses for Maths preparation.

Most Skipped Mathematics Questions (High Non-Attempt %)

The most skipped questions in JEE Advanced Maths were primarily from topics like Matrices, 3D Geometry, Probability, and Integral Calculus. For example, in Paper 1, Q4 (Matrices) had a 72.11% skip rate, while in Paper 2, Q6 (Integral Calculus) was skipped by 78.55% of candidates.

Paper 1

Paper 2

In Paper I, questions with the highest percentage of skipped responses were Q4 , Q5 and Q1 . The least skipped response was for Q2. Paper 2, on the other hand, had even higher skips, with Q6 being the most skipped question, touching 141,728 candidates, followed by Q3 and Q7. From this table it is clear that Paper 2 questions had a higher percentage of non attempts compared to Paper.

Detailed Concept-wise Analysis of JEE Advanced Mathematics 2025 Paper 1 & Paper 2 is given below:

JEE Advanced Maths Paper 2 Question 6

Question: Let $S$ denote the locus of the mid-points of those chords of the parabola $y^2=x$, such that the area of the region enclosed between the parabola and the chord is $\frac{4}{3}$. Let $R$ denote the region lying in the first quadrant, enclosed by the parabola $y^2=x$, the curve $S$, and the lines $x=1$ and $x=4$.

Then which of the following statements is (are) TRUE?
$1-(4, \sqrt{3}) \in S$ (Correct)
$2-(5, \sqrt{2}) \in S$
3- Area of $R$ is $\frac{14}{3}-2 \sqrt{3}$ (Correct)
4- Area of $R$ is $\frac{14}{3}-\sqrt{3}$

Solution:-

Given: The parabola is $y^2=x$
Let a chord of this parabola have midpoint $(h, k)$. The equation of the chord with given midpoint in terms of $T=S_1$ form is:

$
\begin{aligned}
& y k-\frac{y+h}{2}=k^2-h \\
& 2 k y-y^2-h=2 k^2-2 h \\
& \text { Now, } A=\int_{y_1}^{y_2}\left(2 k y-2 k^2+h 0-y^2\right) d y=\frac{4}{3} \\
& \left(k y^2+\left(h-2 k^2\right) y-\frac{y^3}{3}\right)_{y_1}^{y_2}=\frac{4}{3} \\
& \left(y_2-y_1\right)\left[k-2 k+h-2 k^2-\frac{1}{3}\left(4 k^2-2 k^2+h\right)\right]=\frac{4}{3} \\
& \left(h-k^2\right)^{\frac{3}{2}}=1 \\
& h-k^2=1
\end{aligned}
$

Thus, the locus $S$ is:

$x-y^2=1 \Rightarrow y^2=x-1$

We'll check if these points lie on the line, by putting the values,

$(\sqrt{3})^2=4-1 \text {---True }$
But $5^2=\sqrt{2}-1$----False,
Thus, point $(4, \sqrt{3})$ lies on the line, so Option 1 is correct.
Now find the area $R$ enclosed between:
- Parabola $y^2=x \Rightarrow y=\sqrt{x}$
- Curve $S: y=\sqrt{x-1}$
- Vertical lines $x=1$ and $x=4$

The area of region $R$ is:

$\begin{aligned}
& A=\int_1^4(\sqrt{x}-\sqrt{x-1}) d x \\
& =\left[\frac{2}{3} x^{3 / 2}-\frac{2}{3}(x-1)^{3 / 2}\right]_1^4=\frac{2}{3}[8-3 \sqrt{3}-1]=\frac{2}{3}(7-3 \sqrt{3}) \\
& R=\frac{14}{3}-2 \sqrt{3}
\end{aligned}$
Option 3 is also correct

Hence, the correct answers are option 1,3.

Why most students skip this question?

Double Concept

• In this question two ideas are mixed using parameters for the parabola and then solving with definite integration.

Confusion with Region in First Quadrant

• When students read this part, they have to imagine the shapes of the curves ${\mathrm{\$y^2=x\$\; and\; \$x=y^2+1\$}}^{}$$\mathrm{}$ on a graph. Many get confused about which curve lies above or below the other.

Two-Stage Question Format

• Even after finding the locus S, the problem is not finished. It also asks to check which points lie on S.

Multiple Correct Answer-Type

• The question has options (A–D) where more than one can be correct. With so many steps: locus, integration, and checking points, the chances of making a mistake are high, so many skip it.

Indirect Parametrisation Trick

• The simple method is to use parameters for the parabola: $x=p^2, y=p$.

The question looks too lengthy, indirect and too integration-heavy at first sight. It involves two different conceptual tricks like parametrisation of a parabola, simplification of the area formula.

Most Wrongly Attempted Mathematics Questions

The Mathematics JEE Advanced 2025 analysis for Paper 2 shows Question 13 Complex Numbers saw 72.99% wrong attempts. These questions looked straight but were full of concepts like sign errors, misusing formulas, or ignoring principal values.

Paper 1

Paper 2

This table shows that some questions like Q13 (Paper 2) saw 72.99% wrong responses, while Q10 (Paper 2) had 69.77% wrong responses. Similarly, Q11 (Paper 1) and Q8 (Paper 1) recorded 69.48% and 69.04% wrong responses, respectively. The common pattern across these questions was that the % Not Attempted values remained moderate (around 23 to 27%). However, the % Full Marks was very low, only 1.44% for Q10 (Paper 2) and 2.81% for Q13 (Paper 2). While questions like Q12 (Paper 1) with 40.02% not attempted but only 51.42% wrong responses.

JEE Advanced Maths Paper 2 Question 13

Question: For a non-zero complex number $z$, let $\arg (z)$ denote the principal argument of $z$, with $-\pi<\arg (z) \leq \pi$. Let $\omega$ be the cube root of unity for which $0<\arg (\omega)<\pi$. Let

$\alpha=\arg \left(\sum_{n=1}^{2025}(-\omega)^n\right)$
Then the value of $\frac{3 \alpha}{\pi}$ is $\_\_\_\_$ .

Correct Answer:- -2$$

Solution:

Let $r=-\omega$. Then

$S=\sum_{n=1}^{2025}(-\omega)^n=\sum_{n=1}^{2025} r^n=\frac{r\left(1-r^{2025}\right)}{1-r}=\frac{-\omega\left(1-(-\omega)^{2025}\right)}{1+\omega}$
Note $(-\omega)^3=-1$, so

$(-\omega)^{2025}=\left((-\omega)^3\right)^{675}=(-1)^{675}=-1$
Thus $1-(-\omega)^{2025}=1-(-1)=2$, and

$S=\frac{-\omega \cdot 2}{1+\omega}=\frac{-2 \omega}{1+\omega}$$\frac{}{}$

Using $1+\omega+\omega^2=0$ gives $1+\omega=-\omega^2$ Hence

$S=\frac{-2 \omega}{-\omega^2}=2 \frac{\omega}{\omega^2}=2 \omega^2$
Now $\omega^2=e^{4 \pi i / 3}$ has principal argument $\arg \left(\omega^2\right)=-\frac{2 \pi}{3}$. Therefore

$\alpha=\arg (S)=\arg \left(2 \omega^2\right)=\arg \left(\omega^2\right)=-\frac{2 \pi}{3}$
Finally, $\frac{3 \alpha}{\pi}=\frac{3 \cdot\left(-\frac{2 \pi}{3}\right)}{\pi}=-2$
Hence, the correct answer is -2 .

Why many students got it wrong?

• There are two non-real cube roots of unity.

• Some think $(-\omega)^n$ repeats every 3 steps, but it actually takes 6 steps to repeat.

• In the formula $\sum r^n$, many students put the wrong minus sign or denominator

• Even if they get the complex sum correct, they sometimes write the angle as $4 \pi / 3$ instead of the main value $-2 \pi / 3$. This changes $\frac{3 \alpha}{\pi}$ a lot.

• The identity $1+\omega+\omega^2=0$ makes the problem short.

Most Correctly Solved Mathematics Questions

Some questions were the most solved questions in JEE Advanced Maths 2025. For example, Paper 1 Q15 (Maxima & Minima) had 37.69% full marks, while Paper 2 Q9 (Differential Equations) had 28.93%.

Paper 1

Paper 2

JEE Advanced Maths Paper 1 Question 15

Question: Let $R$ denote the set of all real numbers. For a real number x , let $[\mathrm{x}]$ denote the greatest integer less than or equal to x . Let n denote a natural number.
Match each entry in List-I to the correct entry in List-II and choose the correct option.

1) $(\mathrm{P}) \longrightarrow(1)(\mathrm{Q}) \rightarrow(3)(\mathrm{R}) \rightarrow(2)(\mathrm{S}) \rightarrow(5)$
2) $(\mathrm{P}) \rightarrow(2)(\mathrm{Q}) \rightarrow(1)(\mathrm{R}) \rightarrow(4)(\mathrm{S}) \rightarrow(3)$ (Correct)
3) $(\mathrm{P}) \rightarrow(5)(\mathrm{Q}) \rightarrow(1)(\mathrm{R}) \rightarrow(4)(\mathrm{S}) \rightarrow(3)$
4) $(\mathrm{P}) \rightarrow(2)(\mathrm{Q}) \rightarrow(3)(\mathrm{R}) \rightarrow(1)(\mathrm{S}) \rightarrow(5)$$$

Correct Option:- 2

Solution:

Given: $\left\{\begin{array}{cc}(P) & f(x)=\left[\frac{10 x^3-45 x^2+60 x+35}{n}\right], x \in[1,2] \\ (Q) & g(x)=\left(2 n^2-13 n-15\right)\left(x^3+3 x\right), x \in R \\ (R) & h(x)=\left(x^2-9\right)^n\left(x^2+2 x+3\right), n>5 \\ (S) & I(x)=\sum_{k=0}^4\left(\sin |x-k|+\cos \left|x-k+\frac{1}{2}\right|\right), x \in R\end{array}\right.$

$(\mathrm{P})$ : Find minimum $n$ so that $f(x)$ is continuous on $[1,2]$.
Define $g(x)=10 x^3-45 x^2+60 x+35$
Calculate derivative: $g^{\prime}(x)=30 x^2-90 x+60=30\left(x^2-3 x+2\right)=30(x-1)(x-2)$
On $[1,2]$, since $(x-1)(x-2) \leq 0, g(x)$ is decreasing.
Evaluate: $f(1)=\left[\frac{g(1)}{n}\right]=\left[\frac{60}{n}\right], f(2)=\left[\frac{g(2)}{n}\right]=\left[\frac{55}{n}\right]$

For continuity, $f(1)=f(2)$ :

$\left[\frac{60}{n}\right]=\left[\frac{55}{n}\right]$
Try $n=9:\left[\frac{60}{9}\right]=6,\left[\frac{55}{9}\right]=6$
So minimum $n=9$.

$(P) \rightarrow(2)$

(Q): Find minimum $n$ such that $g(x)=\left(2 n^2-13 n-15\right)\left(x^3+3 x\right)$ is increasing on $R$.

$g^{\prime}(x)=\left(2 n^2-13 n-15\right)\left(3 x^2+3\right)=3\left(2 n^2-13 n-15\right)\left(x^2+1\right)$
Since $x^2+1>0$, sign depends on $2 n^2-13 n-15$.
Set: $2 n^2-13 n-15>0$
Solve quadratic inequality:

$2 n^2-13 n-15=0 \Longrightarrow n=\frac{13 \pm \sqrt{169+120}}{4}=\frac{13 \pm 17}{4}$
Roots:

$n=7.5, n=-1$
Since $n$ natural, $n \geq 8$ for positivity.
$(Q) \rightarrow(1)$$$

$(\mathrm{R})$ : Find smallest natural $n>5$ so that $x=3$ is local minimum of $h(x)=\left(x^2-9\right)^n\left(x^2+2 x+3\right)$
Note: $h(3)=(9-9)^n(9+6+3)=0$

1. For local minimum at $x=3, h(x)$ should increase on both sides near 3 for some $n$.

Check behavior near 3:
- For $x>3,\left(x^2-9\right)>0$,
- For $x<3,\left(x^2-9\right)<0$.

Since the factor $\left(x^2-9\right)^n$ changes sign depending on $n$ even or odd:
$n=6$ (even and $>5$ ) ensures local minimum at $x=3$.
$(R) \rightarrow(4)$

$(\mathrm{S})$ : Find number of points $x_0$ where $I(x)=\sum_{k=0}^4\left(\sin |x-k|+\cos \left|x-k+\frac{1}{2}\right|\right)$ is NOT differentiable.
$-\sin |x-a|$ is not differentiable at $x=a$.
$-\cos |x-a|$ is differentiable everywhere.
Non-differentiable points are at $x=k$ for $k=0,1,2,3,4$.
Number of such points $=5$.
$(S) \rightarrow(3)$
(P) : 9 (List-II (2))
(Q) : 8 (List-II (1))
(R) : 6 (List-II (4))
(S) : 5 (List-II (3))

Final matching: $(P) \rightarrow(2),(Q) \rightarrow(1),(R) \rightarrow(4),(S) \rightarrow(3)$.

Hence, the answer is option 2.

Why most students solved it correctly

Each part is straight from the textbook and can be checked in one line.

• (Q) comes down to checking the sign of a quadratic; the derivative is always positive because $3 x^2+3>0$.
• ( R ) is about behaviour near a root: $\left(x^2-9\right)^n$ has factor $(x-3)^n$; if $n$ is even, it makes a local minimum
• ( S ) is counting points where $\sin |x-k|$ is not differentiable at $x=k$ (there are five such $k$ )
• (P) is just checking values of a cubic in a small interval; its range is [55,60] and you only test multiples.

No long algebra or tough integrals are needed

Intervals like [55,60], the set $0,1,2,3,4$, and easy quadratic roots make checking quick and easy.

The question was straight from the textbook ideas. Each part needed only one quick check. So students could solve it fast without long algebra.

Distribution of Total Marks in Aggregate (all, qualified and allotted candidates)

For all candidates, the aggregate marks distribution shows the overall performance in JEE Advanced 2025. Given below this information using charts:

For All Candidates

For Qualified Candidates

For Seat Allotted Candidates

JEE Advanced 2025 Mathematics: Most Difficult Questions

The JEE Advanced 2025 question trends clearly show that the toughest problems were those with:

• Low % full marks,

• Moderate attempts, but

• Very high wrong response percentages.

Paper 1

The most difficult questions in JEE Advanced Maths Paper 1 are Q8, Q9, Q10, Q11, Q12. Questions like Q8 and Q11 had nearly 70% wrong attempts. Q10 stood out with the lowest full marks (1.44%). Q9 and Q12 had higher non-attempt rates (34 to 40%). Q12 still had slightly better full marks.

Paper 2

The toughest questions in JEE Advanced Maths Paper 2 are Q13, Q16, Q10, Q12, Q15). Q13, from Complex Numbers and Quadratic Equations, was the most difficult with only 2.81% full marks and a massive 72.99% wrong response. Q16 (Integral Calculus) also saw only 3.7% full marks and 65.12% wrong responses. Q10 and Q12, from Binomial Theorem and Vector Algebra, respectively, had slightly better full marks (6.64% and 8.6%). Q15 (Trigonometry) showed the highest success of this set with 9.62% full marks, yet more than 64% wrong responses.

Detailed Analysis of JEE Main 2026 and 2025 Paper

Students can refer to the table given below to get the detailed JEE Main analysis of 2025 and 2026 January session. JEE Main paper analysis helps aspirants identify commonly asked topics, understand the difficulty level, and recognise the type of questions asked.

Analysis of JEE Main 2026 Session 1

Analysis of JEE Main 2025