NTA JEE Mains 2nd April Shift 2 2026

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 :