NTA JEE Mains 4th April Shift 1 2026

Let the line $$L_1: x + 3 = 0$$ intersect the lines $$L_2: x - y = 0$$ and $$L_3: 3x + y = 0$$ at the points $$A$$ and $$B$$, respectively. Let the bisector of the obtuse angle between the lines $$L_2$$ and $$L_3$$ intersect the line $$L_1$$ at the point $$C$$. Then $$BC^2 : AC^2$$ is equal to :

Let the vertex $$A$$ of a triangle $$ABC$$ be $$(1, 2)$$, and the mid-point of the side $$AB$$ be $$(5, -1)$$. If the centroid of this triangle is $$(3, 4)$$ and its circumcenter is $$(\alpha, \beta)$$, then $$21(\alpha + \beta)$$ is equal to :

Suppose that two chords, drawn from the point $$(1, 2)$$ on the circle $$x^2 + y^2 + x - 3y = 0$$ are bisected by the $$y$$-axis. If the other ends of these chords are $$R$$ and $$S$$, and the mid point of the line segment $$RS$$ is $$(\alpha, \beta)$$, then $$6(\alpha + \beta)$$ is equal to :

A line with direction ratios $$1, -1, 2$$ intersects the lines $$\frac{x}{2} = \frac{y}{3} = \frac{z+1}{3}$$ and $$\frac{x+1}{-1} = \frac{y-2}{1} = \frac{z}{4}$$ at the points $$P$$ and $$Q$$, respectively. If the length of the line segment $$PQ$$ is $$\alpha$$, then $$225\alpha^2$$ is equal to :

The square of the distance of the point $$(-2, -8, 6)$$ from the line $$\frac{x-1}{1} = \frac{y-1}{2} = \frac{z}{-1}$$ along the line $$\frac{x+5}{1} = \frac{y+5}{-1} = \frac{z}{2}$$ is equal to :

If $$y = \tan^{-1}\left(\frac{3\cos x - 4\sin x}{4\cos x + 3\sin x}\right) + 2\tan^{-1}\left(\frac{x}{1+\sqrt{1-x^2}}\right)$$, then $$\frac{dy}{dx}$$ at $$x = \frac{\sqrt{3}}{2}$$ is equal to :

Let $$f$$ be a real polynomial of degree $$n$$ such that $$f(x) = f'(x) \cdot f''(x)$$, for all $$x \in \mathbb{R}$$. If $$f(0) = 0$$, then $$36\left(f'(2) + f''(2) + \int_0^2 f(x)\,dx\right)$$ is equal to :

The area of the region $$\{(x,y): y \leq \pi - |x|, \; y \leq |x \sin x|, \; y \geq 0\}$$ is :

Let $$\int_{-2}^{2} (|\sin x| + [x \sin x])\,dx = 2(3 - \cos 2) + \beta$$, where $$[\cdot]$$ is the greatest integer function. Then $$\beta \sin\left(\frac{\beta}{2}\right)$$ equals :

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