NTA JEE Mains 2nd April Shift 2 2026 - Mathematics

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

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 :

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 :

Let $$a_1, a_2, a_3, \ldots$$ be an A.P. and $$g_1 = a_1, g_2, g_3, \ldots$$ 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 :

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

If for $$3 \le r \le 30$$, $$\left({}^{30}C_{30-r}\right) + 3\left( ^{30}C_{31-r} \right) + 3\left( ^{30}C_{32-r}\right) + \left( ^{30}C_{33-r}\right) = {}^{m}C_r$$, then $$m$$ equals :

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

A man throws a fair coin repeatedly. He gets 10 points for each head 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 :

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 :

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 $$2x + k^2 y - 4 = 0$$. If the line $$x - y + 2 = 0$$ intersects the circle at the points $$A$$ and $$B$$, then $$(AB)^2$$ is equal to :

Let $$O$$ be the origin, and $$P$$ and $$Q$$ be two points on the rectangular hyperbola $$xy = 12$$ such that the mid point of the line segment $$PQ$$ is $$\left(\frac{1}{2}, -\frac{1}{2}\right)$$. Then the area of the triangle $$OPQ$$ equals :

Let the parabola $$y = x^2 + px + q$$ passing through the point $$(1, -1)$$ be such that the distance between its vertex and the x-axis is minimum. Then the value of $$p^2 + q^2$$ is :

Let $$P = \{\theta \in [0, 4\pi] : \tan^2\theta \ne 1\}$$ and $$S = \{a \in \mathbb{Z} : 2(\cos^8\theta - \sin^8\theta)\sec 2\theta = a^2, \theta \in P\}$$. Then $$n(S)$$ is :

Let the vectors $$\vec{a} = -\hat{i} + \hat{j} + 3\hat{k}$$ and $$\vec{b} = \hat{i} + 3\hat{j} + \hat{k}$$. For some $$\lambda, \mu \in \mathbb{R}$$, let $$\vec{c} = \lambda\vec{a} + \mu\vec{b}$$. If $$\vec{c} \cdot (3\hat{i} - 6\hat{j} + 2\hat{k}) = 10$$ and $$\vec{c} \cdot (\hat{i} + \hat{j} + \hat{k}) = -2$$, then $$|\vec{c}|^2$$ is equal to :

Let the point $$A$$ be the foot of perpendicular drawn from the point $$P(a, b, 0)$$ on the line $$\frac{x-1}{2} = \frac{y-2}{1} = \frac{z-\alpha}{3}$$. If the midpoint of the line segment $$PA$$ is $$\left(0, \frac{3}{4}, -\frac{1}{4}\right)$$, then the value of $$a^2 + b^2 + \alpha^2$$ is equal to :

Two adjacent sides of a parallelogram $$PQRS$$ are given by $$\vec{PQ} = \hat{j} + \hat{k}$$ and $$\vec{PS} = \hat{i} - \hat{j}$$. If the side $$PS$$ is rotated about the point $$P$$ by an acute angle $$\alpha$$ in the plane of the parallelogram so that it becomes perpendicular to the side $$PQ$$, then $$\sin^2\left(\frac{5\alpha}{2}\right) - \sin^2\left(\frac{\alpha}{2}\right)$$ is equal to :

The value of $$\int_0^{20\pi} (\sin^4 x + \cos^4 x)\, dx$$ is equal to :

Let $$f(x)$$ be a polynomial of degree 5, and have extrema at $$x = 1$$ and $$x = -1$$. If $$\lim_{x \to 0} \left(\frac{f(x)}{x^3}\right ) = -5$$, then $$f(2) - f(-2)$$ is equal to :

Let $$f(x) = \int \left( \frac{16x + 24}{x^2 + 2x - 15}\right) dx$$. If $$f(4) = 14\log_e(3)$$ and $$f(7) = \log_e(2^\alpha \cdot 3^\beta)$$, $$\alpha, \beta \in \mathbb{N}$$, then $$\alpha + \beta$$ is equal to :

Let $$x = x(y)$$ be the solution of the differential equation $$2y^2 \frac{dx}{dy} - 2xy + x^2 = 0$$, $$y > 1$$, $$x(e) = e$$. Then $$x(e^2)$$ is equal to :