NTA JEE Main 22nd April 2013 Online

For $$a > 0$$, $$t \in \left(0, \frac{\pi}{2}\right)$$, let $$x = \sqrt{a^{\sin^{-1}t}}$$ and $$y = \sqrt{a^{\cos^{-1}t}}$$. Then, $$1 + \left(\frac{dy}{dx}\right)^2$$ equals :

Statement-1: The function $$x^2(e^x + e^{-x})$$ is increasing for all $$x > 0$$.
Statement-2: The functions $$x^2 e^x$$ and $$x^2 e^{-x}$$ are increasing for all $$x > 0$$ and the sum of two increasing functions in any interval $$(a, b)$$ is an increasing function in $$(a, b)$$.

The maximum area of a right angled triangle with hypotenuse $$h$$ is :

If $$\int \frac{x^2 - x + 1}{x^2 + 1}e^{\cot^{-1}x}dx = A(x)e^{\cot^{-1}x} + C$$, then $$A(x)$$ is equal to :

The integral $$\int_{7\pi/4}^{7\pi/3} \sqrt{\tan^2 x} \ dx$$ is equal to :

The area of the region (in sq. units), in the first quadrant bounded by the parabola $$y = 9x^2$$ and the lines $$x = 0$$, $$y = 1$$ and $$y = 4$$, is :

Consider the differential equation :
$$$\frac{dy}{dx} = \frac{y^3}{2(xy^2 - x^2)}$$$
Statement-1: The substitution $$z = y^2$$ transforms the above equation into a first order homogeneous differential equation.
Statement-2: The solution of this differential equation is $$y^2 e^{-y^2/x} = C$$.

If $$\hat{a}$$, $$\hat{b}$$ and $$\hat{c}$$ are unit vectors satisfying $$\hat{a} - \sqrt{3}\hat{b} + \hat{c} = \vec{0}$$, then the angle between the vectors $$\hat{a}$$ and $$\hat{c}$$ is :

Let Q be the foot of perpendicular from the origin to the plane $$4x - 3y + z + 13 = 0$$ and R be a point (-1, -6) on the plane. Then length QR is :

Given two independent events, if the probability that exactly one of them occurs is $$\frac{26}{49}$$ and the probability that none of them occurs is $$\frac{15}{49}$$, then the probability of more probable of the two events is :