NTA JEE Main 12th April 2014 Online

Two tangents are drawn from a point $$(-2, -1)$$ to the curve, $$y^2 = 4x$$. If $$\alpha$$ is the angle between them, then $$|\tan\alpha|$$ is equal to:

The minimum area of a triangle formed by any tangent to the ellipse $$\frac{x^2}{16} + \frac{y^2}{81} = 1$$ and the co-ordinate axes is:

Let p, q, r denote arbitrary statements. Then the logically equivalent of the statement $$p \Rightarrow (q \vee r)$$ is:

Let $$\bar{X}$$ and M.D. be the mean and the mean deviation about $$\bar{X}$$ of n observations $$x_i$$, i = 1, 2, n. If each of the observations is increased by 5, then the new mean and the mean deviation about the new mean, respectively, are:

A relation on the set A = {x : |x| < 3, x $$\in$$ Z}, where Z is the set of integers is defined by R = {(x, y) : y = |x|, x $$\neq$$ $$-1$$}. Then the number of elements in the power set of R is:

If $$A = \begin{bmatrix} 1 & 2 & x \\ 3 & -1 & 2 \end{bmatrix}$$ and $$B = \begin{bmatrix} y \\ x \\ 1 \end{bmatrix}$$ be such that AB = $$\begin{bmatrix} 6 \\ 8 \end{bmatrix}$$, then:

If $$\begin{vmatrix} a^2 & b^2 & c^2\\ (a+\lambda)^2 & (b+\lambda)^2 & (c+\lambda)^2 \\ (a-\lambda)^2 & (b-\lambda)^2 & (c-\lambda)^2  \end{vmatrix}$$ = $$k\lambda \begin{vmatrix} a^2 & b^2 & c^2 \\ a & b & c \\ 1 & 1 & 1 \end{vmatrix}$$, $$\lambda \neq 0$$, then k is equal to:

If $$f(\theta) = \begin{vmatrix} 1 & \cos\theta & 1 \\ -\sin\theta & 1 & -\cos\theta \\ -1 & \sin\theta & 1 \end{vmatrix}$$ and A and B are respectively the maximum and the minimum values of $$f(\theta)$$, then (A, B) is equal to:

Statement I: The equation $$(\sin^{-1}x)^3 + (\cos^{-1}x)^3 - a\pi^3 = 0$$ has a solution for all $$a \geq \frac{1}{32}$$. 

Statement II: For any x $$\in$$ R, $$\sin^{-1}x + \cos^{-1}x = \frac{\pi}{2}$$ and $$0 \leq \left(\sin^{-1}x - \frac{\pi}{4}\right)^2 \leq \frac{9\pi^2}{16}$$.

If $$f(x) = x^2 - x + 5$$, $$x > \frac{1}{2}$$, and g(x) is its inverse function, then g'(7) equals: