NTA JEE Mains 6th April Shift 2 2026

The eccentricity of an ellipse E with centre at the origin O is $$\dfrac{\sqrt{3}}{2}$$ and its directrices are $$x = \pm \dfrac{4\sqrt{6}}{3}$$. Let $$H: \dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1$$ be a hyperbola whose eccentricity is equal to the length of semi-major axis of E, and whose length of latus rectum is equal to the length of minor axis of E. Then the distance between the foci of H is :

Let $$x = 9$$ be a directrix of an ellipse E, whose centre is at the origin and eccentricity is $$\dfrac{1}{3}$$. Let $$P(\alpha, 0)$$, $$\alpha > 0$$, be a focus of E and AB be a chord passing through P. Then the locus of the mid point of AB is :

If $$\sin(\tan^{-1}(x\sqrt{2})) = \cot(\sin^{-1}\sqrt{1 - x^2})$$, $$x \in (0, 1)$$, then the value of $$x$$ is :

The shortest distance between the lines $$\dfrac{x - 4}{1} = \dfrac{y - 3}{2} = \dfrac{z - 2}{-3}$$ and $$\dfrac{x + 2}{2} = \dfrac{y - 6}{4} = \dfrac{z - 5}{-5}$$ is :

Let $$\vec{a} = 2\hat{i} + 3\hat{j} + 3\hat{k}$$ and $$\vec{b} = 6\hat{i} + 3\hat{j} + 3\hat{k}$$. Then the square of the area of the triangle with adjacent sides determined by the vectors $$(2\vec{a} + 3\vec{b})$$ and $$(\vec{a} - \vec{b})$$ is :

Let $$\displaystyle\lim_{x \to 2} \dfrac{(\tan(x - 2))(rx^2 + (p - 2)x - 2p)}{(x - 2)^2} = 5$$ for some $$r, p \in \mathbb{R}$$. If the set of all possible values of q, such that the roots of the equation $$rx^2 - px + q = 0$$ lie in $$(0, 2)$$, be the interval $$(\alpha, \beta]$$, then $$4(\alpha + \beta)$$ equals :

Let $$A = \begin{bmatrix} 1 & 3 & -1 \\ 2 & 1 & \alpha \\ 0 & 1 & -1 \end{bmatrix}$$ be a singular matrix. Let $$f(x) = \displaystyle\int_0^x (t^2 + 2t + 3)\,dt$$, $$x \in [1, \alpha]$$. If M and m are respectively the maximum and the minimum values of $$f$$ in $$[1, \alpha]$$, then $$3(M - m)$$ is equal to :

Let $$f: \mathbb{R} \to \mathbb{R}$$ be such that $$f(xy) = f(x)f(y)$$, for all $$x, y \in \mathbb{R}$$ and $$f(0) \ne 0$$. Let $$g: [1, \infty) \to \mathbb{R}$$ be a differentiable function such that $$$x^2 g(x) = \int_1^x (t^2 f(t) - tg(t))\,dt.$$$ Then $$g(2)$$ is equal to :

The area of the region $$\{(x, y) : x^2 - 8x \le y \le -x\}$$ is :

The value of the integral $$\displaystyle\int_{-1}^{1} \left(\dfrac{x^3 + |x| + 1}{x^2 + 2|x| + 1}\right) dx$$ is equal to :