NTA JEE Mains 4th April Shift 2 2026

Let  $$H : \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$$ be a  hyperbola such that the distance between its foci equal to $$6$$ and distance between its directrices  is  $$\frac{8}{3}$$. If the line $$x = \alpha$$ intersects the hyperbola $$H$$ at $$A$$ and $$B$$, such that  the area of $$\triangle AOB$$ (where $$O$$ is the origin) is $$4\sqrt{15}$$, then $$\alpha^2$$ is equal to :

$$\displaystyle\max_{0 \leq x \leq \pi}\left(16\sin\frac{x}{2}\cos^3\frac{x}{2}\right)$$ is equal to :

The shortest distance between the lines $$\vec{r} = \left(\frac{1}{3}\hat{i} + 2\hat{j} + \frac{8}{3}\hat{k}\right) + \lambda(2\hat{i} - 5\hat{j} + 6\hat{k})$$ and $$\vec{r} = \left(-\frac{2}{3}\hat{i} - \frac{1}{3}\hat{k}\right) + \mu(\hat{j} - \hat{k})$$, where $$\lambda, \mu \in \mathbb{R}$$, is :

If $$(2\alpha + 1,\; \alpha^2 - 3\alpha,\; \frac{\alpha - 1}{2})$$ is the image of $$(\alpha, 2\alpha, 1)$$ in the line $$\frac{x-2}{3} = \frac{y-1}{2} = \frac{z}{1}$$, then the possible value(s) of $$\alpha$$ is/are :

Let $$\hat{u}$$ and $$\hat{v}$$ be unit vectors inclined at acute angle such that $$|\hat{u} \times \hat{v}| = \frac{\sqrt{3}}{2}$$. If $$\vec{A} = \lambda\hat{u} + \hat{v} + (\hat{u} \times \hat{v})$$, then $$\lambda$$ is equal to :

Let for some $$\alpha \in \mathbb{R}$$ $$f : \mathbb{R} \to \mathbb{R}$$ be a function  satisfying  $$f(x+y) = f(x) + 2y^2 + y + \alpha xy$$ for all $$x, y \in \mathbb{R}$$. If $$f(0) = -1$$ and $$f(1) = 2$$, then the value of $$\displaystyle\sum_{n=1}^{5}(\alpha + f(n))$$ is  :

Let $$A = \{(a, b, c) : a, b, c \text{ are non-negative integers and } a + b + 2c = 22\}$$. Then $$n(A)$$ is equal to :

The area of the region bounded by the curves $$x + 3y^2 = 0$$ and $$x + 4y^2 = 1$$ is :

Let $$y = y(x)$$ be the solution of kthe differential equation :$$\frac{dy}{dx} + \left(\frac{6x^2 + (3x^2 + 2x^3 + 4)e^{-2x}}{(x^3 + 2)(2 + e^{-2x})}\ \right),y = 2 + e^{-2x}$$, $$x \in (-1, 2)$$, $$y(0) = \frac{3}{2}$$. If $$y(1) = \alpha(2 + e^{-2})$$, then $$\alpha$$ is equal to :

The integral $$\displaystyle\int_0^1 \cot^{-1}(1 + x + x^2)\,dx$$ is equal to :