NTA JEE Mains 6th April Shift 1 2026

If the eccentricity $$e$$ of the hyperbola $$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$$, passing through $$(6, 4\sqrt{3})$$, satisfies $$15(e^2 + 1) = 34e$$, then the length of the latus rectum of the hyperbola $$\frac{x^2}{b^2} - \frac{y^2}{2(a^2 + 1)} = 1$$ is:

Let chord PQ of length $$3\sqrt{13}$$ of the parabola $$y^2 = 12x$$ be such that the ordinates of points P and Q are in the ratio 1:2. If the chord PQ subtends an angle $$\alpha$$ at the focus of the parabola, then $$\sin \alpha$$ is equal to:

Let $$0 < \alpha < 1$$, $$\beta = \frac{1}{3\alpha}$$ and $$\tan^{-1}(1 - \alpha) + \tan^{-1}(1 - \beta) = \frac{\pi}{4}$$. Then $$6(\alpha + \beta)$$ is equal to:

Let $$S = \left\{\theta \in (-2\pi, 2\pi) : \cos\theta + 1 = \sqrt{3}\sin\theta\right\}$$.
Then $$\sum_{\theta \in S} \theta$$ is equal to:

Let the image of the point P(1, 6, a) in the line L: $$\frac{x}{1} = \frac{y - 1}{2} = \frac{z - a + 1}{b}$$, $$b > 0$$, be $$\left(\frac{a}{3}, 0, a + c\right)$$. If S($$\alpha, \beta, \gamma$$), $$\alpha > 0$$, is the point on L such that the distance of S from the foot of perpendicular from the point P on L is $$2\sqrt{14}$$, then $$\alpha + \beta + \gamma$$ is equal to:

Let a line L be perpendicular to both the lines
$$L_1: \frac{x + 1}{3} = \frac{y + 3}{5} = \frac{z + 5}{7}$$ and $$L_2: \frac{x - 2}{1} = \frac{y - 4}{4} = \frac{z - 6}{7}$$.
If $$\theta$$ is the acute angle between the lines L and
$$L_3: \frac{x - \frac{8}{7}}{2} = \frac{y - \frac{4}{7}}{1} = \frac{z}{2}$$, then $$\tan\theta$$ is equal to:

The value of $$\lim_{x \to 0}\left(\frac{x^2 \sin^2 x}{x^2 - \sin^2 x}\right)$$ is:

The value of the integral $$\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\left(\frac{32\cos^4 x}{1 + e^{\sin x}}\right)dx$$ is:

The area of the region $$\{(x, y) : 0 \leq y \leq 6 - x, y^2 \geq 4x - 3, x \geq 0\}$$ is:

Let $$e$$ be the base of natural logarithm and let $$f : \{1, 2, 3, 4\} \to \{1, e, e^2, e^3\}$$ and $$g : \{1, e, e^2, e^3\} \to \left\{1,\frac{1}{2}, \frac{1}{3}, \frac{1}{4}\right\}$$ be two bijective functions such that $$f$$ is strictly decreasing and $$g$$ is strictly increasing. If $$\phi(x) = \left[f^{-1}\left\{g^{-1}\left(\frac{1}{2}\right)\right\}\right]^x$$, then the area of the region R = {(x, y): $$x^2 \leq y \leq \phi(x)$$, $$0 \leq x \leq 1$$} is: