NTA JEE Main 23rd April 2013 Online

The least integral value $$\alpha$$ of $$x$$ such that $$\frac{x-5}{x^2+5x-14} > 0$$, satisfies :

Let $$a = \text{Im}\left(\frac{1+z^2}{2iz}\right)$$, where z is any non-zero complex number. The set $$A = \{a : |z| = 1$$ and $$z \neq \pm 1\}$$ is equal to:

The sum of the series : $$(2)^2 + 2(4)^2 + 3(6)^2 + \ldots$$ upto 10 terms is :

If $$a_1, a_2, a_3, \ldots, a_n, \ldots$$ are in A.P. such that $$a_4 - a_7 + a_{10} = m$$, then the sum of first 13 terms of this A.P., is :

The sum of the rational terms in the binomial expansion of $$\left(2^{\frac{1}{2}} + 3^{\frac{1}{5}}\right)^{10}$$ is :

The number of solutions of the equation $$\sin 2x - 2\cos x + 4\sin x = 4$$ in the interval $$[0, 5\pi]$$ is :

If two lines $$L_1$$ and $$L_2$$ in space, are defined by
$$L_1 = \{x = \sqrt{\lambda}y + (\sqrt{\lambda} - 1), z = (\sqrt{\lambda} - 1)y + \sqrt{\lambda}\}$$ and
$$L_2 = \{x = \sqrt{\mu}y + (1 - \sqrt{\mu}), z = (1 - \sqrt{\mu})y + \sqrt{\mu}\}$$
then $$L_1$$ is perpendicular to $$L_2$$, for all nonnegative reals $$\lambda$$ and $$\mu$$, such that :

Let $$\theta_1$$ be the angle between two lines $$2x + 3y + c_1 = 0$$ and $$-x + 5y + c_2 = 0$$ and $$\theta_2$$ be the angle between two lines $$2x + 3y + c_1 = 0$$ and $$-x + 5y + c_3 = 0$$, where $$c_1, c_2, c_3$$ are any real numbers :
Statement-1: If $$c_2$$ and $$c_3$$ are proportional, then $$\theta_1 = \theta_2$$.
Statement-2: $$\theta_1 = \theta_2$$ for all $$c_2$$ and $$c_3$$.

If the circle $$x^2 + y^2 - 6x - 8y + (25 - a^2) = 0$$ touches the axis of x, then a equals.

The point of intersection of the normals to the parabola $$y^2 = 4x$$ at the ends of its latus rectum is :