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

75 Questions·Free Practice·MCQ with Explanations

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

Date: April 2, 2026 | Shift: Evening (03:00 PM - 06:00 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.

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Let $\alpha, \beta$ be the roots of the equation $x^2 - 3x + r = 0$ , and $\frac{\alpha}{2}, 2\beta$ be the roots of the equation $x^2 + 3x + r = 0$ . If the roots of the equation $x^2 + 6x = m$ are $2\alpha + \beta + 2r$ and $\alpha - 2\beta - \frac{r}{2}$ , then $m$ is equal to:

$-135$
$-567$
$135$
$567$

View Answer & Explanation

Correct Answer: Option D -

$567$

Explanation:

Let $\alpha, \beta$ be roots of $x^2 - 3x + r = 0 \implies \alpha + \beta = 3$ and $\alpha\beta = r$ .
Also, $\frac{\alpha}{2}, 2\beta$ are roots of $x^2 + 3x + r = 0 \implies \frac{\alpha}{2} + 2\beta = -3$ and $(\frac{\alpha}{2})(2\beta) = r \implies \alpha\beta = r$ , which is consistent.
Solving the linear equations $\alpha + \beta = 3$ and $\alpha + 4\beta = -6$ , we get $3\beta = -9 \implies \beta = -3$ , and $\alpha = 6$ . Then $r = -18$ .
The given roots of $x^2 + 6x - m = 0$ are $r_1 = 2\alpha + \beta + 2r = 12 - 3 - 36 = -27$ and $r_2 = \alpha - 2\beta - \frac{r}{2} = 6 + 6 + 9 = 21$ .
Notice that $r_1 + r_2 = -6$ , matching the sum of roots. The product of roots is $-m = (-27)(21) = -567$ , giving $m = 567$ .

Let the circles $C_1 : |z| = r$ and $C_2 : |z - 3 - 4i| = 5, z \in \mathbb{C}$ , be such that $C_2$ lies within $C_1$ . If $z_1$ moves on $C_1$ , $z_2$ moves on $C_2$ and $\min |z_1 - z_2| = 2$ , then $\max |z_1 - z_2|$ is equal to:

$12$
$17$
$22$
$24$

View Answer & Explanation

Correct Answer: Option C -

$22$

Explanation:

Center of $C_1$ is at origin $(0,0)$ with radius $r$ . Center of $C_2$ is $C(3,4)$ with radius $5$ . Since $C_2$ lies within $C_1$ , the minimum distance between points on the two circles along the common radius is $\min |z_1 - z_2| = r - (d + 5)$ , where $d = |C| = \sqrt{3^2 + 4^2} = 5$ .
Given minimum distance is $2 \implies r - (5 + 5) = 2 \implies r = 12$ .
The maximum distance between points on the circles occurs diametrically opposite through the centers: $\max |z_1 - z_2| = r + d + 5 = 12 + 5 + 5 = 22$ .

If the system of equations
$x + 5y + 6z = 4$ ,
$2x + 3y + 4z = 7$ ,
$x + 6y + az = b$
has infinitely many solutions, then the point $(a, b)$ lies on the line:

$y - x = 3$
$x - y = 3$
$x + y = 11$
$x + y = 12$

View Answer & Explanation

Correct Answer: Option B -

$x - y = 3$

Explanation:

For infinitely many solutions, the determinant of coefficients must be zero:
$\Delta = \begin{vmatrix} 1 & 5 & 6 \\ 2 & 3 & 4 \\ 1 & 6 & a \end{vmatrix} = 0 \implies -7a + 50 = 0 \implies a = \frac{50}{7}$ .
Now using elimination to find $b$ , subtracting $R_1$ from $R_3$ gives $y + (a-6)z = b-4$ . Subtracting $2R_1$ from $R_2$ gives $-7y - 8z = -1 \implies 7y + 8z = 1$ .
For proportional rows: $\frac{1}{7} = \frac{a-6}{8} = \frac{b-4}{1}$ .
$b-4 = \frac{1}{7} \implies b = \frac{29}{7}$ .
So, $(a, b) = (\frac{50}{7}, \frac{29}{7})$ . Checking the options, $a - b = \frac{50}{7} - \frac{29}{7} = \frac{21}{7} = 3$ , so it lies on $x - y = 3$ .

Let $a_1, a_2, a_3, \dots$ be an A.P. and $g_1 = a_1, g_2, g_3, \dots$ be an increasing G.P. If $a_1 = a_2 + g_2 = 1$ and $a_3 + g_3 = 4$ , then $a_{10} + g_5$ is equal to:

$81$
$76$
$62$
$55$

View Answer & Explanation

Correct Answer: Option D -

$55$

Explanation:

From the given, $a_1 = 1$ . Let $d$ be common difference of AP and $r$ be common ratio of GP. Then $a_2 = 1 + d$ and $g_2 = r$ .
We have $a_2 + g_2 = 1 \implies 1 + d + r = 1 \implies d = -r$ .
Also, $a_3 + g_3 = 4 \implies (1 + 2d) + r^2 = 4 \implies 1 - 2r + r^2 = 4 \implies (r-1)^2 = 4$ .
Since the GP is increasing ( $r > 1$ ), $r-1 = 2 \implies r = 3$ . Then $d = -3$ .
$a_{10} = 1 + 9(-3) = -26$ . $g_5 = 1 \cdot 3^4 = 81$ .
$a_{10} + g_5 = -26 + 81 = 55$ .

The sum $\frac{1^3}{1} + \frac{1^3 + 2^3}{1 + 3} + \frac{1^3 + 2^3 + 3^3}{1 + 3 + 5} + \dots$ up to 8 terms, is:

$70$
$71$
$72$
$73$

View Answer & Explanation

Correct Answer: Option B -

$71$

Explanation:

The general term $T_k = \frac{\sum_{i=1}^k i^3}{\sum_{i=1}^k (2i-1)} = \frac{(k(k+1)/2)^2}{k^2} = \frac{(k+1)^2}{4}$ .
Sum to 8 terms = $\sum_{k=1}^8 \frac{(k+1)^2}{4} = \frac{1}{4} (2^2 + 3^2 + \dots + 9^2)$ .
This equals $\frac{1}{4} (\frac{9 \times 10 \times 19}{6} - 1^2) = \frac{1}{4}(285 - 1) = \frac{284}{4} = 71$ .

If for $3 \leq r \leq 30$ , $\binom{30}{30-r} + 3\binom{30}{31-r} + 3\binom{30}{32-r} + \binom{30}{33-r} = \binom{m}{r}$ , then $m$ equals:

$31$
$32$
$33$
$34$

View Answer & Explanation

Correct Answer: Option C -

$33$

Explanation:

Using the identity $\binom{n}{k} = \binom{n}{n-k}$ , we can write the expression as $\binom{30}{r} + 3\binom{30}{r-1} + 3\binom{30}{r-2} + \binom{30}{r-3}$ .
This corresponds to the sum of combinations forming the coefficient of $x^r$ in the expansion of $(1+x)^{30} (1+x)^3 = (1+x)^{33}$ .
Thus, it equals $\binom{33}{r}$ , making $m = 33$ .

Let $p_n$ denote the total number of triangles formed by joining the vertices of an $n$ -sided regular polygon. If $p_{n+1} - p_n = 66$ , then the sum of all distinct prime divisors of $n$ is:

View Answer & Explanation

Correct Answer: Option C -

$5$

Explanation:

The number of triangles formed by joining vertices of an $n$ -sided polygon is $p_n = \binom{n}{3}$ .
$p_{n+1} - p_n = \binom{n+1}{3} - \binom{n}{3} = \binom{n}{2}$ .
Given $\binom{n}{2} = 66 \implies \frac{n(n-1)}{2} = 66 \implies n(n-1) = 132$ . Thus, $n = 12$ .
Distinct prime divisors of $12$ are $2$ and $3$ . Their sum is $2 + 3 = 5$ .

A man throws a fair coin repeatedly. He gets 10 points for each head he throws and 5 points for each tail he throws. If the probability that he gets exactly 30 points is $\frac{m}{n}$ , $\gcd(m, n) = 1$ , then $m + n$ is equal to:

$53$
$55$
$107$
$105$

View Answer & Explanation

Correct Answer: Option C -

$107$

Explanation:

Let $P(k)$ be the probability of getting exactly $5k$ points. On each toss, adding a Head gives $10$ points (two units of $5$ ) and adding a Tail gives $5$ points (one unit of $5$ ).
Therefore, $P(k) = \frac{1}{2} P(k-2) + \frac{1}{2} P(k-1)$ with $P(0) = 1$ and $P(1) = 1/2$ .
Evaluating recursively:
$P(2) = \frac{1}{2}(1) + \frac{1}{2}(\frac{1}{2}) = \frac{3}{4}$
$P(3) = \frac{1}{2}(\frac{1}{2}) + \frac{1}{2}(\frac{3}{4}) = \frac{5}{8}$
$P(4) = \frac{1}{2}(\frac{3}{4}) + \frac{1}{2}(\frac{5}{8}) = \frac{11}{16}$
$P(5) = \frac{1}{2}(\frac{5}{8}) + \frac{1}{2}(\frac{11}{16}) = \frac{21}{32}$
$P(6) = \frac{1}{2}(\frac{11}{16}) + \frac{1}{2}(\frac{21}{32}) = \frac{43}{64}$ .
$30$ points means $k=6$ . So probability is $43/64$ , meaning $m = 43$ and $n = 64$ . $m+n = 107$ .

The mean and variance of $n$ observations are $8$ and $16$ , respectively. If the sum of the first $(n - 1)$ observations is $48$ and the sum of squares of the first $(n - 1)$ observations is $496$ , then the value of $n$ is:

$21$
$16$
$13$
$7$

View Answer & Explanation

Correct Answer: Option D -

$7$

Explanation:

Let total sum of $n$ observations be $S_n$ and sum of squares be $Q_n$ .
Given mean is $8 \implies S_n = 8n$ . Variance is $16 \implies \frac{Q_n}{n} - 8^2 = 16 \implies Q_n = 80n$ .
Let the $n$ -th observation be $x_n$ . $x_n = S_n - 48 = 8n - 48$ .
$x_n^2 = Q_n - 496 = 80n - 496$ .
$(8n-48)^2 = 80n - 496 \implies 64n^2 - 768n + 2304 = 80n - 496 \implies 64n^2 - 848n + 2800 = 0$ .
Dividing by $16 \implies 4n^2 - 53n + 175 = 0$ .
Solving gives $n = \frac{53 \pm \sqrt{2809 - 2800}}{8} = \frac{53 \pm 3}{8}$ . Hence $n=7$ (as $n=50/8$ is not an integer).

Let a circle pass through the origin and its centre be the point of intersection of two mutually perpendicular lines $x + (k - 1)y + 3 = 0$ and $2x + k^2y - 4 = 0$ . If the line $x - y + 2 = 0$ intersects the circle at the points A and B, then $(AB)^2$ is equal to:

$10$
$27$
$18$
$34$

View Answer & Explanation

Correct Answer: Option C -

$18$

Explanation:

Since the two lines are perpendicular, the product of their slopes is $-1$ : $\frac{-1}{k-1} \times \frac{-2}{k^2} = -1 \implies k^3 - k^2 + 2 = 0$ . Using trial and error, $k = -1$ .
The lines become $x - 2y + 3 = 0$ and $2x + y - 4 = 0$ . Solving these gives the intersection center point $C(1, 2)$ .
The circle passes through $(0,0)$ , so its radius squared $R^2 = 1^2 + 2^2 = 5$ .
The perpendicular distance $p$ from $C(1,2)$ to line $x - y + 2 = 0$ is $p = \frac{|1 - 2 + 2|}{\sqrt{2}} = \frac{1}{\sqrt{2}}$ .
The length of chord $AB$ is $2\sqrt{R^2 - p^2} = 2\sqrt{5 - 1/2} = 2\sqrt{9/2} = \sqrt{18}$ . Therefore, $(AB)^2 = 18$ .

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

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

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