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NTA JEE Main 10th April 2019 Shift 1 - Mathematics

For the following questions answer them individually

If $$\alpha$$ and $$\beta$$ are the roots of the quadratic equation $$x^2 + x\sin\theta - 2\sin\theta = 0$$, $$\theta \in \left(0, \frac{\pi}{2}\right)$$, then $$\frac{\alpha^{12} + \beta^{12}}{(\alpha^{-12} + \beta^{-12})\cdot(\alpha - \beta)^{24}}$$ is equal to:

If a directrix of a hyperbola centered at the origin and passing through the point $$(4, -2\sqrt{3})$$ is $$5x = 4\sqrt{5}$$ and its eccentricity is e, then:

ABC is a triangular park with AB = AC = 100 metres. A vertical tower is situated at the mid-point of BC. If the angles of elevation of the top of the tower at A and B are $$\cot^{-1}(3\sqrt{2})$$ and $$\text{cosec}^{-1}(2\sqrt{2})$$ respectively, then the height of the tower (in metres) is

If $$\Delta_1 = \begin{vmatrix} x & \sin\theta & \cos\theta \\ -\sin\theta & -x & 1 \\ \cos\theta & 1 & x \end{vmatrix}$$ and $$\Delta_2 = \begin{vmatrix} x & \sin 2\theta & \cos 2\theta \\ -\sin 2\theta & -x & 1 \\ \cos 2\theta & 1 & x \end{vmatrix}$$, $$x \neq 0$$; then for all $$\theta \in \left(0, \frac{\pi}{2}\right)$$:

If $$f(x) = \begin{cases} \frac{\sin(p+1)x + \sin x}{x}, & x < 0 \\ q, & x = 0 \\ \frac{\sqrt{x + x^2} - \sqrt{x}}{x^{3/2}}, & x > 0 \end{cases}$$ is continuous at $$x = 0$$, then the ordered pair (p, q) is equal to:

Let $$f(x) = e^x - x$$ and $$g(x) = x^2 - x$$, $$\forall$$ x ∈ R. Then the set of all x ∈ R, where the function $$h(x) = (f \circ g)(x)$$ is increasing, is:

If $$\int \frac{dx}{(x^2 - 2x + 10)^2} = A\left(\tan^{-1}\left(\frac{x-1}{3}\right) + \frac{f(x)}{x^2 - 2x + 10}\right) + C$$, then (where C is a constant of integration)

If $$y = y(x)$$ is the solution of the differential equation $$\frac{dy}{dx} = (\tan x - y)\sec^2 x$$, $$x \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$$, such that $$y(0) = 0$$, then $$y\left(-\frac{\pi}{4}\right)$$ is equal to:

Let A(3, 0, -1), B(2, 10, 6) and C(1, 2, 1) be the vertices of a triangle and M be the mid-point of AC. If G divides BM in the ratio, 2 : 1, then $$\cos(\angle GOA)$$ (O being the origin) is equal to