JEE Main 2026 April 6 Shift 1 mathematics question paper is available here with answer key and solutions. NTA is conducting the first shift of the day on April 6, 2026, from 9:00 AM to 12:00 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 6 Shift 1 mathematics question paper along with detailed solutions to analyze their performance and understand the exam pattern better.
JEE Main 2026 April 6 Shift 1 Mathematics Question Paper with Solution PDF

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Find \( I = \int_{-\pi/4}^{\pi/4} \frac{32 \cos^4 \theta}{1 + e^{\sin \theta}} \, d\theta \)
Let \( S = \{\theta \in (-2\pi, 2\pi): \cos \theta + 1 = \sqrt{3} \sin \theta\} \). Then \( \sum_{\theta \in S} \theta \) is equal to:
\( a_1, a_2, a_3, \dots, a_n \) are in A.P. and sum of first 10 terms is 160. \( g_1, g_2, g_3, \dots, g_n \) are in G.P. where \( g_1 + g_2 = 8 \). If the first term of A.P. is equal to common ratio of G.P. and first term of G.P. is equal to common difference of A.P., then sum of all possible values of \( g_1 \) is equal to:
If \( P\left( \frac{a}{3}, 0, a+c \right) \) is the image of \( Q(1, 6, a) \) with respect to line \( L: \frac{x}{1} = \frac{y-1}{2} = \frac{z-a+1}{b} \), where \( a > 0, b > 0 \). If \( S(\alpha, \beta, \gamma) \) is at a distance of \( 2\sqrt{14} \) from the foot of the perpendicular of Q on L, then \( \alpha^2 + \beta^2 + \gamma^2 \) is:
Consider the circle \( x^2 + y^2 + 2gx + 2fy + 25 = 0 \) having centre on line \( -2x + y + 4 = 0 \) and radius 6. If the line \( x = 1 \) cuts the circle at A and B then \( AB^2 \) is:
The value of \( 1^3 - 2^3 + 3^3 - 4^3 + \cdots - 14^3 + 15^3 \) is equal to:
Evaluate the limit \( \lim_{x \to 0} \frac{x^2 \sin^2 x}{x^2 - \sin^2 x} \)
If coefficients of middle terms in the expansion \( (1 + \alpha x)^{26} \) \& \( (1 - \alpha x)^{28} \) are equal then \( \alpha \) is:
Let \( f : \{1,2,3,4\} \rightarrow \{1, e, e^2, e^3\} \) is a strictly increasing and bijective function and \( g : \{1, e, e^2, e^3\} \rightarrow \{1, \frac{1}{2}, \frac{1}{3}, \frac{1}{4}\} \) strictly decreasing and bijective function. If \( \phi(x) = [f^{-1}(g^{-1}(1/2))]^x \), then find \( \int_0^1 (\phi(x) - x^2) \, dx \):
The number of four letter words which can be formed using two vowels and two consonants from the word INCONSEQUENTIAL (words can be meaningful or meaningless) is:
If \( \tan^{-1}(1-\alpha) + \tan^{-1}(1-\beta) = \frac{\pi}{4} \) \& \( \alpha = \frac{1}{\beta} \) then find \( |\alpha + \beta| \):
Let the set of all values of \( K \in \mathbb{R} \) such that the equation, \( z(\bar{z} + 2 + i) + K(2 + 3i) = 0 \), \( z \in \mathbb{C} \) has at least one solution, be the interval \( [\alpha, \beta] \). Then \( 9(\alpha + \beta) = \)
If \( x_1, x_2, x_3, \dots, x_{25} \) be 25 observations such that \( \sum_{i=1}^{25} (x_i + 5)^2 = 2500 \) and \( \sum_{i=1}^{25} (x_i - 5)^2 = 1000 \). Then the ratio of mean and standard deviation of the given observation, is:
If there are (n + 1) coins of which n coins are fair and one is double headed. A coin is randomly selected and tossed, if probability that head occurs is \( \frac{9}{16} \) then n is:
Given point P(6, \( 4\sqrt{5} \)) satisfy hyperbola \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \) where eccentricity is root of equation \( 9e^2 - 21e + 10 = 0 \). Then find the length of latus rectum of hyperbola \( \frac{x^2}{a^2} - \frac{y^2}{2(1+b^2)} = 1 \).
Given two lines \( L_1: \frac{x-1}{3} = \frac{y-2}{2} = \frac{z+1}{1} \) and \( L_2: \frac{x+2}{1} = \frac{y-1}{1} = \frac{z}{1} \). Third line \( L_3 \) is perpendicular to both lines \( L_1 \) \& \( L_2 \). Find acute angle between lines \( L_1 \) \& \( L_2 \).
If the ratio of y-ordinates of points on the parabola \( y^2 = 12x \) is 1 : 2 and the length of the chord joining these points is \( 3\sqrt{13} \), then find the angle subtended by the chord at the focus of the parabola:
If domain of \( f(x) = \sin^{-1} \left( \frac{x + |x|}{3} \right) \) is \( [\alpha, \beta) \), then \( (\alpha^2 + \beta^2) \) is:
Find the area bounded by \( 0 \le y \le 6 - x \), \( y^2 + 3 \le 4x \) and \( x > 0 \):
Solution of differential equation \( \frac{dy}{dx} + \frac{y(x - \sqrt{x^2 - 1})}{x^2 - x\sqrt{x^2 - 1}} = \frac{x}{x^2 - x\sqrt{x^2 - 1}} \) satisfies the condition \( y(1) = 1 \), then find \( [y(\sqrt{5})] \). (Here [·] denotes greatest integer function):
Consider \( e_1 \) and \( e_2 \) be roots of the equation \( x^2 - a x + 2 = 0 \). Set of exhaustive values of 'a' for which \( e_1 \) and \( e_2 \) are eccentricities of hyperbolas then \( a \in [\alpha, \beta) \) and set of values of 'a' for which \( e_1 \) and \( e_2 \) are eccentricity of the parabola and ellipse is \( (\gamma, \infty) \) then \( (\alpha^2 + \beta^2 + \gamma^2) \) equal:
Given vectors \( \vec{a} = \hat{i} + \hat{j} + \hat{k} \) and \( \vec{b} = \hat{j} - \hat{k} \). Another vector \( \vec{c} \) satisfy equations \( \vec{a} \cdot \vec{c} = 3 \) and \( \vec{a} \times \vec{c} = \vec{b} \), then find \( \vec{a} \cdot (\vec{c} - 2\vec{b}) \):
Given that quadratic equation \( (k^2 - 15k + 27) x^2 + 9(k - 1)x + 18 = 0 \) has one root twice of other. Then find the length of the latus rectum of the parabola \( y^2 = 6kx \):
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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