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JEE Main 2026 (April 2nd, Shift 1) Practice Questions & Answers

75 Questions·Free Practice·MCQ with Explanations

JEE Main 2026 - Session 2 (April 2nd, Shift 1)

Date: April 2, 2026 | Shift: Morning (09:00 AM - 12:00 PM)
Conducted By: National Testing Agency (NTA)

This is the official Paper 1 (B.E./B.Tech) question paper containing 75 questions (25 per subject). Students are required to attempt 75 questions in total.

Exam Pattern Breakdown:

Sections: Physics, Chemistry, and Mathematics.
Questions per Subject: 20 Multiple Choice Questions (MCQs) in Section A and 10 Numerical Value questions in Section B.

Practice this paper to evaluate your speed, accuracy, and readiness for the actual NTA JEE Main exam.

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Let $\alpha, \alpha + 2, \alpha \in \mathbb{Z}$ , be the roots of the quadratic equation $x(x + 2) + (x + 1)(x + 3) + (x + 2)(x + 4) + \dots + (x + n - 1)(x + n + 1) = 4n$ for some $n \in \mathbb{N}$ . Then $n + \alpha$ is equal to :

View Answer & Explanation

Correct Answer: Option C -

Explanation:

The given equation is $\sum_{k=0}^{n-1} (x+k)(x+k+2) = 4n$ .

$\sum_{k=0}^{n-1} (x^2 + 2kx + 2x + k^2 + 2k) = 4n$
$n x^2 + 2x \frac{(n-1)n}{2} + 2x n + \sum_{k=0}^{n-1} k(k + 2) = 4n$
This simplifies to:
$n x^2 + n(n+1)x + \frac{(n-1)n(2n-1)}{6} + \frac{2(n-1)n}{2} - 4n = 0$
Divide by $n$ : $x^2 + (n+1)x + \frac{(n-1)(2n-1)}{6} + (n-1) - 4 = 0$
Sum of roots = $2\alpha + 2 = -(n+1) \implies 2\alpha = -(n+3)$ . Since $\alpha$ is an integer, $n$ must be an odd integer.
Product of roots = $\alpha(\alpha+2) = \frac{2n^2 - 3n + 1 + 6n - 30}{6} = \frac{2n^2 + 3n - 29}{6}$
From $n = -2\alpha - 3$ , $(n+3)/2 = -\alpha$ . Then $\alpha(\alpha+2) = \frac{-(n+3)}{2} \cdot \frac{-(n-1)}{2} = \frac{n^2+2n-3}{4}$ .
Equating the two expressions for the product:
$\frac{n^2+2n-3}{4} = \frac{2n^2+3n-29}{6}$
$3n^2 + 6n - 9 = 4n^2 + 6n - 58 \implies n^2 = 49 \implies n = 7$ (since $n \in \mathbb{N}$ ).
Then $2\alpha = -10 \implies \alpha = -5$ .
So $n + \alpha = 7 - 5 = 2$ .

Let $x$ and $y$ be real numbers such that $50 \left( \frac{2x}{1 + 3i} - \frac{y}{1 - 2i} \right) = 31 + 17i, i = \sqrt{-1}$ . Then the value of $10(x - 3y)$ is :

View Answer & Explanation

Correct Answer: Option D -

Explanation:

Multiply the numerators and denominators by their respective complex conjugates:
$50 \left( \frac{2x(1 - 3i)}{10} - \frac{y(1 + 2i)}{5} \right) = 31 + 17i$
$10x(1 - 3i) - 10y(1 + 2i) = 31 + 17i$
Equating the real parts: $10x - 10y = 31 \implies x - y = 3.1$
Equating the imaginary parts: $-30x - 20y = 17 \implies 30x + 20y = -17$
We solve the linear system for $x$ and $y$ .
Multiply the first equation by 3: $30x - 30y = 93$ .
Subtract it from the second equation: $(30x + 20y) - (30x - 30y) = -17 - 93 \implies 50y = -110 \implies y = -2.2$ .
Substitute $y$ back: $x - (-2.2) = 3.1 \implies x = 0.9$ .
Then we compute $10(x - 3y)$ :
$10(0.9 - 3(-2.2)) = 10(0.9 + 6.6) = 10(7.5) = 75$ .

Let $\alpha, \beta \in \mathbb{R}$ be such that the system of linear equations
$x + 2y + z = 5$
$2x + y + \alpha z = 5$
$8x + 4y + \beta z = 18$
has no solution. Then $\frac{\beta}{\alpha}$ is equal to :

-4
4
8
-8

View Answer & Explanation

Correct Answer: Option B -

Explanation:

For a system of equations to have no solution, the determinant of the coefficient matrix $\Delta$ must be zero, and at least one of $\Delta_x, \Delta_y, \Delta_z$ must be non-zero.
$\Delta = \begin{vmatrix} 1 & 2 & 1 \\ 2 & 1 & \alpha \\ 8 & 4 & \beta \end{vmatrix} = 1(\beta - 4\alpha) - 2(2\beta - 8\alpha) + 1(8 - 8)$
$\Delta = \beta - 4\alpha - 4\beta + 16\alpha = 12\alpha - 3\beta$
Setting $\Delta = 0$ , we get $12\alpha = 3\beta \implies \beta = 4\alpha$ .
If $\beta = 4\alpha$ , notice that the third equation's LHS is exactly 4 times the second equation's LHS ( $4(2x + y + \alpha z) = 8x + 4y + \beta z$ ).
However, the RHS of the third equation is $18$ , while $4 \times 5 = 20$ . This guarantees the system is inconsistent, and thus has no solution.
Therefore, $\frac{\beta}{\alpha} = 4$ .

Let $A = \begin{bmatrix} 1 & 2 \\ 1 & \alpha \end{bmatrix}$ and $B = \begin{bmatrix} 3 & 3 \\ \beta & 2 \end{bmatrix}$ . If $A^2 - 4A + I = O$ and $B^2 - 5B - 6I = O$ , then among the two statements :
(S1): $[(B - A)(B + A)]^T = \begin{bmatrix} 13 & 15 \\ 7 & 10 \end{bmatrix}$
and
(S2): $\det(\text{adj}(A + B)) = -5$ ,

only (S1) is correct
only (S2) is correct
both (S1) and (S2) are correct
both (S1) and (S2) are wrong

View Answer & Explanation

Correct Answer: Option B -

only (S2) is correct

Explanation:

By the Cayley-Hamilton theorem, a $2 \times 2$ matrix $M$ satisfies $M^2 - \text{tr}(M)M + \det(M)I = O$ .
For matrix $A$ : $A^2 - (1+\alpha)A + (\alpha-2)I = O$ . Given $A^2 - 4A + I = O$ , comparing coefficients gives $\text{tr}(A) = 1+\alpha = 4 \implies \alpha = 3$ .
For matrix $B$ : $B^2 - 5B - 6I = O$ . The trace is $3+2=5$ , which matches. The determinant is $|B| = -6 \implies 6 - 3\beta = -6 \implies 3\beta = 12 \implies \beta = 4$ .
Thus, $A = \begin{bmatrix} 1 & 2 \\ 1 & 3 \end{bmatrix}$ and $B = \begin{bmatrix} 3 & 3 \\ 4 & 2 \end{bmatrix}$ .
Evaluate (S1): $B-A = \begin{bmatrix} 2 & 1 \\ 3 & -1 \end{bmatrix}$ and $B+A = \begin{bmatrix} 4 & 5 \\ 5 & 5 \end{bmatrix}$ .
$(B-A)(B+A) = \begin{bmatrix} 2 & 1 \\ 3 & -1 \end{bmatrix} \begin{bmatrix} 4 & 5 \\ 5 & 5 \end{bmatrix} = \begin{bmatrix} 13 & 15 \\ 7 & 10 \end{bmatrix}$ .
Then $[(B-A)(B+A)]^T = \begin{bmatrix} 13 & 7 \\ 15 & 10 \end{bmatrix} \neq \begin{bmatrix} 13 & 15 \\ 7 & 10 \end{bmatrix}$ . So (S1) is incorrect.
Evaluate (S2): $\det(A+B) = \det \begin{bmatrix} 4 & 5 \\ 5 & 5 \end{bmatrix} = 20 - 25 = -5$ .
For a $2 \times 2$ matrix, $\det(\text{adj}(M)) = \det(M)^{2-1} = \det(M)$ . So $\det(\text{adj}(A+B)) = -5$ . Thus, (S2) is correct.

Let $A$ be the set of first 101 terms of an A.P., whose first term is 1 and the common difference is 5 and let $B$ be the set of first 71 terms of an A.P., whose first term is 9 and the common difference is 7. Then the number of elements in $A \cap B$ , which are divisible by 3, is :

View Answer & Explanation

Correct Answer: Option B -

Explanation:

Set A terms: $1, 6, 11, 16, \dots$ . The $n$ -th term is $5n-4$ , up to the 101st term which is $5(101)-4 = 501$ .
Set B terms: $9, 16, 23, \dots$ . The $m$ -th term is $7m+2$ , up to the 71st term which is $7(71)+2 = 499$ .
We find the first common term is $16$ . The common difference of the new A.P. is the LCM of their common differences, $\text{LCM}(5, 7) = 35$ .
So the common terms $A \cap B$ form the sequence: $16, 51, 86, \dots$ , with the general term $T_k = 16 + 35k$ for $k \ge 0$ .
We require $16 + 35k \le 499 \implies 35k \le 483 \implies k \le 13.8$ , so $k \in \{0, 1, 2, \dots, 13\}$ .
We also need the terms to be divisible by 3.
$16 + 35k \equiv 0 \pmod 3 \implies 1 + 2k \equiv 0 \pmod 3 \implies 2k \equiv 2 \pmod 3 \implies k \equiv 1 \pmod 3$ .
For $k \in \{0, 1, \dots, 13\}$ , the values of $k$ satisfying $k \equiv 1 \pmod 3$ are $1, 4, 7, 10, 13$ .
There are exactly 5 such values.

The number of seven-digit numbers, that can be formed by using the digits 1, 2, 3, 5 and 7 such that each digit is used at least once, is :

15400
17800
16800
29400

View Answer & Explanation

Correct Answer: Option C -

16800

Explanation:

We need to form 7-digit numbers using the 5 given digits (1, 2, 3, 5, 7), with each appearing at least once.
Since there are 7 places and 5 digits, there are exactly 2 "extra" digit uses. This can happen in two scenarios:
Case 1: One digit appears 3 times, and the remaining 4 digits appear 1 time each.
The number of ways to choose which digit repeats 3 times is $\binom{5}{1} = 5$ .
The number of arrangements is $\frac{7!}{3!1!1!1!1!} = \frac{5040}{6} = 840$ .
Total numbers for Case 1 = $5 \times 840 = 4200$ .

Case 2: Two digits appear 2 times each, and the remaining 3 digits appear 1 time each.
The number of ways to choose which two digits repeat 2 times is $\binom{5}{2} = 10$ .
The number of arrangements is $\frac{7!}{2!2!1!1!1!} = \frac{5040}{4} = 1260$ .
Total numbers for Case 2 = $10 \times 1260 = 12600$ .

Total possible 7-digit numbers = $4200 + 12600 = 16800$ .

The number of elements in the set $S = \left\{ (r, k) : k \in \mathbb{Z} \text{ and } {}^{36}C_{r+1} = \frac{6 \left({}^{35}C_r\right)}{k^2 - 3} \right\}$ , is :

View Answer & Explanation

Correct Answer: Option B -

Explanation:

We use the identity ${}^nC_r = \frac{n}{r} {}^{n-1}C_{r-1}$ .
Applying this to ${}^{36}C_{r+1}$ , we get ${}^{36}C_{r+1} = \frac{36}{r+1} {}^{35}C_r$ .
Equating this to the given expression:
$\frac{36}{r+1} {}^{35}C_r = \frac{6}{k^2 - 3} {}^{35}C_r$
Assuming ${}^{35}C_r \neq 0$ (so $0 \le r \le 35$ ), we get:
$\frac{6}{r+1} = \frac{1}{k^2 - 3} \implies k^2 - 3 = \frac{r+1}{6} \implies k^2 = \frac{r+19}{6}$
Since $k$ is an integer, $k^2$ must be a perfect square. Also, $\frac{r+19}{6}$ must be an integer, which implies $r + 19$ is a multiple of 6, or $r \equiv 5 \pmod 6$ .
The possible values of $r$ in the range $0 \le r \le 35$ are 5, 11, 17, 23, 29, 35.
Checking these values for perfect squares $k^2$ :
If $r = 5$ , $k^2 = (5+19)/6 = 24/6 = 4 \implies k = \pm 2$ . Valid pairs: $(5, 2), (5, -2)$ .
If $r = 11$ , $k^2 = 30/6 = 5$ (not a perfect square).
If $r = 17$ , $k^2 = 36/6 = 6$ (not a perfect square).
If $r = 23$ , $k^2 = 42/6 = 7$ (not a perfect square).
If $r = 29$ , $k^2 = 48/6 = 8$ (not a perfect square).
If $r = 35$ , $k^2 = 54/6 = 9 \implies k = \pm 3$ . Valid pairs: $(35, 3), (35, -3)$ .
Total number of pairs in $S$ is 4.

If the mean of the data

Class	$5 - 10$	$10 - 15$	$15 - 20$	$20 - 25$	$25 - 30$	$30 - 35$
Frequency	2	$k$	28	54	$k + 1$	5

is $21$ , then $k$ is one of the roots of the equation :

$2x^2 - 23x - 10 = 0$
$4x^2 - 35x + 24 = 0$
$2x^2 - 19x - 10 = 0$
$2x^2 - 35x + 98 = 0$

View Answer & Explanation

Correct Answer: Option C -

$2x^2 - 19x - 10 = 0$

Explanation:

The class marks ( $x_i$ ) are the midpoints of the classes: $7.5, 12.5, 17.5, 22.5, 27.5, 32.5$ .
The frequencies ( $f_i$ ) are: $2, k, 28, 54, k+1, 5$ .
Total frequency $N = \sum f_i = 2 + k + 28 + 54 + (k + 1) + 5 = 2k + 90$ .
Mean $\bar{x} = \frac{\sum f_i x_i}{N}$ . Let's use the step-deviation method with assumed mean $a = 22.5$ and class size $h = 5$ .
$u_i = \frac{x_i - 22.5}{5} \implies -3, -2, -1, 0, 1, 2$ .
$\sum f_i u_i = 2(-3) + k(-2) + 28(-1) + 54(0) + (k+1)(1) + 5(2) = -6 - 2k - 28 + k + 1 + 10 = -k - 23$ .
$\bar{x} = a + h\left(\frac{\sum f_i u_i}{N}\right) = 22.5 + 5\left(\frac{-k - 23}{2k + 90}\right) = 21$ .
$1.5 + 5\left(\frac{-k - 23}{2k + 90}\right) = 0 \implies 3 = 10\left(\frac{k + 23}{2k + 90}\right)$ .
$3(2k + 90) = 10(k + 23) \implies 6k + 270 = 10k + 230 \implies 4k = 40 \implies k = 10$ .
Now we check which quadratic equation has $10$ as a root:
For (3): $2(10)^2 - 19(10) - 10 = 200 - 190 - 10 = 0$ . This fits exactly.

Let the mid points of the sides of a triangle $ABC$ be $\left(\frac{5}{2}, 7\right)$ , $\left(\frac{5}{2}, 3\right)$ and $(4, 5)$ . If its incentre is $(h, k)$ , then $3h + k$ is equal to :

View Answer & Explanation

Correct Answer: Option C -

Explanation:

Let the midpoints form the medial triangle with vertices $D(5/2, 7)$ , $E(5/2, 3)$ , $F(4, 5)$ .
The side lengths of the medial triangle are:
$DE = \sqrt{(5/2 - 5/2)^2 + (7 - 3)^2} = 4$
$EF = \sqrt{(4 - 5/2)^2 + (5 - 3)^2} = \sqrt{9/4 + 4} = 5/2$
$FD = \sqrt{(4 - 5/2)^2 + (5 - 7)^2} = \sqrt{9/4 + 4} = 5/2$
The sides of $\Delta ABC$ are twice the length of the medial triangle sides: $a = 2(EF) = 5$ , $b = 2(FD) = 5$ , $c = 2(DE) = 8$ .
Vertices of $\Delta ABC$ are found using the property that the midpoints are the averages of the vertices:
$A = E + F - D = (5/2 + 4 - 5/2, 3 + 5 - 7) = (4, 1)$
$B = F + D - E = (4 + 5/2 - 5/2, 5 + 7 - 3) = (4, 9)$
$C = D + E - F = (5/2 + 5/2 - 4, 7 + 3 - 5) = (1, 5)$
The side opposite to $A$ is $BC = a = 5$ , opposite to $B$ is $CA = b = 5$ , opposite to $C$ is $AB = c = 8$ .
Incentre $I(h,k)$ is given by:
$x_I = \frac{ax_A + bx_B + cx_C}{a+b+c} = \frac{5(4) + 5(4) + 8(1)}{5 + 5 + 8} = \frac{20+20+8}{18} = \frac{48}{18} = \frac{8}{3}$
$y_I = \frac{ay_A + by_B + cy_C}{a+b+c} = \frac{5(1) + 5(9) + 8(5)}{18} = \frac{5+45+40}{18} = \frac{90}{18} = 5$
Thus, $h = 8/3$ and $k = 5$ .
$3h + k = 3(8/3) + 5 = 8 + 5 = 13$ .

Let an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1, a < b$ , pass through the point $(4, 3)$ and have eccentricity $\frac{\sqrt{5}}{3}$ . Then the length of its latus rectum is :

$\frac{4\sqrt{5}}{3}$
$2\sqrt{5}$
$\frac{7\sqrt{5}}{3}$
$\frac{8\sqrt{5}}{3}$

View Answer & Explanation

Correct Answer: Option D -

$\frac{8\sqrt{5}}{3}$

Explanation:

Since $a < b$ , the major axis is along the y-axis, and the relation for eccentricity is $a^2 = b^2(1 - e^2)$ .
Given $e = \frac{\sqrt{5}}{3}$ , we have $e^2 = \frac{5}{9}$ .
$a^2 = b^2 \left(1 - \frac{5}{9}\right) = \frac{4b^2}{9}$ .
The ellipse passes through $(4, 3)$ , so we substitute $x=4$ and $y=3$ into the ellipse equation:
$\frac{16}{a^2} + \frac{9}{b^2} = 1$
Substituting $a^2 = \frac{4b^2}{9}$ :
$\frac{16}{(4b^2/9)} + \frac{9}{b^2} = 1 \implies \frac{36}{b^2} + \frac{9}{b^2} = 1 \implies \frac{45}{b^2} = 1 \implies b^2 = 45$ .
Then $a^2 = \frac{4(45)}{9} = 20$ .
The length of the latus rectum for $a < b$ is $\frac{2a^2}{b}$ .
We have $a^2 = 20$ and $b = \sqrt{45} = 3\sqrt{5}$ .
Latus Rectum = $\frac{2 \times 20}{3\sqrt{5}} = \frac{40}{3\sqrt{5}} = \frac{8\sqrt{5}}{3}$ .

JEE Main 2026 (April 2nd, Shift 1) Practice Questions & Answers

JEE Main 2026 - Session 2 (April 2nd, Shift 1)

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