NTA JEE Mains 8th April Shift 2 2026

Let O be the vertex of the parabola $$y^2 = 4x$$ and its chords OP and OQ are perpendicular to each other. If the locus of the mid-point of the line segment PQ is a conic C, then the length of its latus rectum is :

Let $$\alpha = 3\sin^{-1}\left(\frac{6}{11}\right)$$ and $$\beta = 3\cos^{-1}\left(\frac{4}{9}\right)$$, where inverse trigonometric functions take only the principal values.
Given below are two statements :
Statement I : $$\cos(\alpha + \beta) > 0$$.
Statement II : $$\cos(\alpha) < 0$$.
In the light of the above statements, choose the correct answer from the options given below :

For the function $$f(x) = e^{\sin|x|} - |x|$$, $$x \in \mathbf{R}$$, consider the following statements :
Statement I : $$f$$ is differentiable for all $$x \in \mathbf{R}$$.
Statement II : $$f$$ is increasing in $$\left(-\pi, -\frac{\pi}{2}\right)$$.
In the light of the above statements, choose the correct answer from the options given below :

Let $$\vec{a} = 4\hat{i} - \hat{j} + 3\hat{k}$$, $$\vec{b} = 10\hat{i} + 2\hat{j} - \hat{k}$$ and a vector $$\vec{c}$$ be such that $$2(\vec{a} \times \vec{b}) + 3(\vec{b} \times \vec{c}) = \vec{0}$$. If $$\vec{a} \cdot \vec{c} = 15$$, then $$\vec{c} \cdot (\hat{i} + \hat{j} - 3\hat{k})$$ is equal to :

Let the foot of perpendicular from the point $$(\lambda, 2, 3)$$ on the line $$\frac{x-4}{1} = \frac{y-9}{2} = \frac{z-5}{1}$$ be the point $$(1, \mu, 2)$$. Then the distance between the lines $$\frac{x-1}{2} = \frac{y-2}{3} = \frac{z+4}{6}$$ and $$\frac{x-\lambda}{2} = \frac{y-\mu}{3} = \frac{z+5}{6}$$ is equal to :

The value of the integral $$\int_0^2 \frac{\sqrt{x(x^2 + x + 1)}}{(\sqrt{x} + 1)(\sqrt{x^4 + x^2 + 1})} \, dx$$ is equal to :

Let $$y = y(x)$$ be the solution of the differential equation $$x\sqrt{1-x^2} \, dy + \left(y\sqrt{1-x^2} - x\cos^{-1}x\right)dx = 0$$, $$x \in (0,1)$$, $$\lim_{x \to 1^-} y(x) = 1$$. Then $$y\left(\frac{1}{2}\right)$$ equals :

Let $$f : (1, \infty) \to \mathbf{R}$$ be a function defined as $$f(x) = \frac{x-1}{x+1}$$. Let $$f^{i+1}(x) = f(f^i(x))$$, $$i = 1, 2, ..., 25$$, where $$f^1(x) = f(x)$$. If $$g(x) + f^{26}(x) = 0$$, $$x \in (1, \infty)$$, then the area of the region bounded by the curves $$y = g(x)$$, $$2y = 2x - 3$$, $$y = 0$$ and $$x = 4$$ is :

Let $$f(x) = \begin{cases} \frac{1}{3}, & x \le \frac{\pi}{2} \\ \frac{b(1 - \sin x)}{(\pi - 2x)^2}, & x > \frac{\pi}{2} \end{cases}$$. If $$f$$ is continuous at $$x = \pi/2$$, then the value of $$\int_0^{3b-6} |x^2 + 2x - 3| \, dx$$ is :

Let $$\frac{x^2}{f(a^2+7a+3)} + \frac{y^2}{f(3a+15)} = 1$$ represent an ellipse with major axis along $$y$$-axis, where $$f$$ is a strictly decreasing positive function on $$\mathbf{R}$$. If the set of all possible values of $$a$$ is $$\mathbf{R} - [\alpha, \beta]$$, then $$\alpha^2 + \beta^2$$ is equal to :