NTA JEE Main 23rd April 2013 Online

A tangent to the hyperbola $$\frac{x^2}{4} - \frac{y^2}{2} = 1$$ meets x-axis at P and y-axis at Q. Lines PR and QR are drawn such that OPRQ is a rectangle (where O is the origin). Then R lies on :

For integers $$m$$ and $$n$$, both greater than 1, consider the following three statements : P : m divides n Q : m divides $$n^2$$ R : m is prime, then

If the median and the range of four numbers $$\{x, y, 2x + y, x - y\}$$, where $$0 < y < x < 2y$$, are 10 and 28 respectively, then the mean of the numbers is :

If the extremities of the base of an isosceles triangle are the points (2a, 0) and (0, a) and the equation of one of the sides is $$x = 2a$$, then the area of the triangle, in square units, is :

On the sides AB, BC, CA of a $$\triangle ABC$$, 3, 4, 5 distinct points (excluding vertices A, B, C) are respectively chosen. The number of triangles that can be constructed using these chosen points as vertices are :

Let $$R = \{(x, y) : x, y \in N$$ and $$x^2 - 4xy + 3y^2 = 0\}$$, where N is the set of all natural numbers. Then the relation R is :

Let A, other than I or - I, be a $$2 \times 2$$ real matrix such that $$A^2 = I$$, I being the unit matrix. Let Tr(A) be the sum of diagonal elements of A.
Statement-1: Tr(A) = 0
Statement-2: det(A) = -1

Statement-1: The system of linear equations
$$x + (\sin\alpha)y + (\cos\alpha)z = 0$$
$$x + (\cos\alpha)y + (\sin\alpha)z = 0$$
$$x - (\sin\alpha)y - (\cos\alpha)z = 0$$
has a non-trivial solution for only one value of $$\alpha$$ lying in the interval $$(0, \frac{\pi}{2})$$.
Statement-2: The equation in $$\alpha$$
$$\begin{vmatrix} \cos\alpha & \sin\alpha & \cos\alpha \\ \sin\alpha & \cos\alpha & \sin\alpha \\ \cos\alpha & -\sin\alpha & -\cos\alpha \end{vmatrix} = 0$$
has only one solution lying in the interval $$(0, \frac{\pi}{2})$$.

$$S = \tan^{-1}\left(\frac{1}{n^2+n+1}\right) + \tan^{-1}\left(\frac{1}{n^2+3n+3}\right) + \ldots + \tan^{-1}\left(\frac{1}{1+(n+19)(n+20)}\right)$$, then $$\tan S$$ is equal to :

Let $$f$$ be a composite function of $$x$$ defined by $$f(u) = \frac{1}{u^2+u-2}$$, $$u(x) = \frac{1}{x-1}$$. Then the number of points $$x$$ where $$f$$ is discontinuous is :