JEE Main 2026 April 5 Shift 2 mathematics question paper is available here with answer key and solutions. NTA conducted the second shift of the day on April 5, 2026, from 3:00 PM to 6:00 PM.
- The JEE Main Mathematics Question Paper contains a total of 25 questions.
- Each correct answer gets you 4 marks while incorrect answers gets you a negative mark of 1.
Candidates can download the JEE Main 2026 April 5 Shift 2 mathematics question paper along with detailed solutions to analyze their performance and understand the exam pattern better.
JEE Main 2026 April 5 Shift 2 Mathematics Question Paper with Solution PDF

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Question 1:
Let \(\alpha, \beta\) be the roots of the equation \(x^2 - x + p = 0\) and \(\gamma, \delta\) be the roots of the equation \(x^2 - 4x + q = 0\); \(p, q \in \mathbb{Z}\). If \(\alpha, \beta, \gamma, \delta\) are in G.P., then \(|p+q|\) equals :
Let \(z_1, z_2 \in \mathbb{C}\) be the distinct solutions of the equation \(z^2 + 4z - (1 + 12i) = 0\). Then \(|z_1|^2 + |z_2|^2\) is equal to :
If \(f: \mathbb{N} \to \mathbb{Z}\) is defined by \[ f(n) = \begin{vmatrix} n & -1 & -5
-2n^2 & 3(2k+1) & 2k+1
-3n^3 & 3k(2k+1) & 3k(k+2)+1 \end{vmatrix}, k \in \mathbb{N}, \]
and \(\sum_{n=1}^k f(n) = 98\), then \(k\) is equal to :
Let M be a \(3 \times 3\) matrix such that \(M \begin{pmatrix} 1
1
0 \end{pmatrix} = \begin{pmatrix} 1
2
3 \end{pmatrix}, M \begin{pmatrix} 0
1
0 \end{pmatrix} = \begin{pmatrix} 0
1
2 \end{pmatrix}\) and \(M \begin{pmatrix} 0
0
1 \end{pmatrix} = \begin{pmatrix} -1
1
1 \end{pmatrix}\). If \(M \begin{pmatrix} x
y
z \end{pmatrix} = \begin{pmatrix} 1
7
11 \end{pmatrix}\), then \(x+y+z\) equals :
If the sum of the first 10 terms of the series \(\frac{1}{1+1^4 \cdot 4} + \frac{2}{1+2^4 \cdot 4} + \frac{3}{1+3^4 \cdot 4} + \dots\) is \(\frac{m}{n}\), \(gcd(m, n) = 1\), then \(m+n\) is equal to :
Let \(A_1, A_2, A_3, \dots, A_{39}\) be 39 arithmetic means between the numbers 59 and 159. Then the mean of \(A_{25}, A_{28}, A_{31}\) and \(A_{36}\) is equal to :
The coefficient of \(x^2\) in the expansion of \(\left( 2x^2 + \frac{1}{x} \right)^{10}, x \neq 0\), is :
The probabilities that players A and B of a team are selected for the captaincy for a tournament are 0.6 and 0.4, respectively. If A is selected the captain, the probability that the team wins the tournament is 0.8 and if B is selected the captain, the probability that the team wins the tournament is 0.7. Then the probability, that the team wins the tournament, is :
A box contains 5 blue, 6 yellow and 4 red balls. The number of ways, of drawing 8 balls containing at least two balls of each colour, is :
A variable X takes values \(0, 0, 2, 6, 12, 20, \dots, n(n-1)\) with frequencies \(\binom{n}{0}, \binom{n}{1}, \binom{n}{2}, \binom{n}{3}, \binom{n}{4}, \binom{n}{5}, \dots, \binom{n}{n}\) respectively. If the mean of this data is 60, then its median is :
Let the point P be the vertex of the parabola \(y = x^2 - 6x + 12\). If a line passing through the point P intersects the circle \(x^2 + y^2 - 2x - 4y + 3 = 0\) at the points R and S, then the maximum value of \((PR + PS)^2\) is :
Let the directrix of the parabola \(P : y^2 = 8x\), cut \(x\)-axis at the point A. Let \(B(\alpha, \beta), \alpha > 1\), be a point on P such that the slope of AB is 3/5. If BC is a focal chord of P, then six times the area of \(\Delta ABC\) is :
Let the eccentricity \(e\) of a hyperbola satisfy the equation \(6e^2 - 11e + 3 = 0\). If the foci of the hyperbola are \((3, 5)\) and \((3, -4)\), then the length of its latus rectum is :
Let a triangle PQR be such that P and Q lie on the line \(\frac{x+3}{8} = \frac{y-4}{2} = \frac{z+1}{2}\) and are at a distance of 6 units from R(1, 2, 3). If \((\alpha, \beta, \gamma)\) is the centroid of \(\Delta PQR\), then \(\alpha + \beta + \gamma\) is equal to :
If the distance of the point \((a, 2, 5)\) from the image of the point \((1, 2, 7)\) in the line \(\frac{x}{1} = \frac{y-1}{1} = \frac{z-2}{2}\) is 4, then the sum of all possible values of \(a\) is equal to :
Let O be the origin, \(\vec{OP} = \vec{a}\) and \(\vec{OQ} = \vec{b}\). If R is the point on \(\vec{OP}\) such that \(\vec{OP} = 5\vec{OR}\), and M is the point such that \(\vec{OQ} = 5\vec{RM}\), then \(\vec{PM}\) is equal to :
Let \(f(x) = \lim_{y \to 0} \frac{(1 - \cos(xy)) \tan(xy)}{y^3}\). Then the number of solutions of the equation \(f(x) = \sin x\), \(x \in \mathbb{R}\) is :
Let \((2^{1-a} + 2^{1+a}), f(a), (3^a + 3^{-a})\) be in A.P. and \(\alpha\) be the minimum value of \(f(a)\). Then the value of the integral \(\int_{\log_e(\alpha-1)}^{\log_e(\alpha)} \frac{dx}{(e^{2x} - e^{-2x})}\) is :
Let \(f: [1, \infty) \to \mathbb{R}\) be a differentiable function defined as \(f(x) = \int_1^x f(t) dt + (1-x)(\log_e x - 1) + e\). Then the value of \(f(f(1))\) is :
Let \(f(x)\) and \(g(x)\) be twice differentiable functions satisfying \(f''(x) = g''(x)\) for all \(x \in \mathbb{R}\), \(f'(1) = 2g'(1) = 4\) and \(g(2) = 3f(2) = 9\). Then \(f(25) - g(25)\) is equal to :
Let \(A = \{1, 4, 7\}\) and \(B = \{2, 3, 8\}\). Then the number of elements in the relation \(R = \{ ((a_1, b_1), (a_2, b_2)) \in (A \times B) \times (A \times B) : a_1+b_2 divides a_2+b_1 \}\) is :
From the point \((-1, -1)\), two rays are sent making angles of \(45^\circ\) with the line \(x+y=0\). These rays get reflected from the mirror \(x+2y=1\). If the equations of the reflected rays are \(ax+by=9\) and \(cx+dy=7\), \(a, b, c, d \in \mathbb{Z}\), then the value of \(ad+bc\) is :
If \(S = \{ \theta \in [-\pi, \pi] : \cos\theta \cos\frac{5\theta}{2} = \cos 7\theta \cos\frac{7\theta}{2} \}\), then \(n(S)\) is equal to :
Let \(f: \mathbb{R} \to \mathbb{R}\) be a function such that \(f(x) + 3f\left( \frac{\pi}{2} - x \right) = \sin x, x \in \mathbb{R}\). Let the maximum value of \(f\) on \(\mathbb{R}\) be \(\alpha\). If the area of the region bounded by the curves \(g(x) = x^2\) and \(h(x) = \beta x^3, \beta > 0\), is \(\alpha^2\), then \(30\beta^3\) is equal to :
Let \(y = y(x)\) be the solution of the differential equation \((\tan x)^{1/2} dy = (\sec^3 x - (\tan x)^{3/2}) dx, 0 < x < \frac{\pi}{2}, y\left( \frac{\pi}{4} \right) = \frac{6\sqrt{2}}{5}\). If \(y\left( \frac{\pi}{3} \right) = \frac{4}{5} \alpha\), then \(\alpha^4\) equals :
JEE Main 2026 Mathematics Exam Pattern
| Particulars | Details |
|---|---|
| Exam Mode | Online (Computer-Based Test) |
| Paper | B.E./B.Tech |
| Medium of Exam | 13 languages: English, Hindi, Gujarati, Bengali, Tamil, Telugu, Kannada, Marathi, Malayalam, Odia, Punjabi, Assamese, Urdu |
| Type of Questions | Multiple Choice Questions (MCQs) + Numerical Value Questions |
| Total Marks | 100 marks |
| Marking Scheme | +4 for correct answer & -1 for incorrect MCQ and Numerical Value-based Questions |
| Total Questions | 25 Questions |









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