NTA JEE Main 25th April 2013 Online

If an equation of a tangent to the curve, $$y - \cos(x + f)$$, $$-1 - 1 \leq x \leq 1 + \pi$$, is $$x + 2y = k$$ then $$k$$ is equal to :

If the integral $$\int \frac{\cos 8x + 1}{\cot 2x - \tan 2x}dx = A\cos 8x + k$$ where k is an arbitrary constant, then A is equal to:

For $$0 \leq x \leq \frac{\pi}{2}$$, the value of $$\int_0^{\sin^2 x} \sin^{-1}(\sqrt{t})dt + \int_0^{\cos^2 x} \cos^{-1}(\sqrt{t})dt$$ equals :

Let $$f : [-2, 3] \rightarrow [0, \infty)$$ be a continuous function such that $$f(1-x) = f(x)$$ for all $$x \in [-2, 3]$$. If $$R_1$$ is the numerical value of the area of the region bounded by $$y = f(x)$$, $$x = -2$$, $$x = 3$$ and the axis of x and $$R_2 = \int_{-2}^{3} xf(x)dx$$, then :

The equation of the curve passing through the origin and satisfying the differential equation $$(1 + x^2)\frac{dy}{dx} + 2xy = 4x^2$$ is

Let $$\vec{a} = 2\hat{i} + \hat{j} - 2\hat{k}$$, $$\vec{b} = \hat{i} + \hat{j}$$. If $$\vec{c}$$ is a vector such that $$\vec{a} \cdot \vec{c} = |\vec{c}|$$, $$|\vec{c} - \vec{a}| = 2\sqrt{2}$$ and the angle between $$\vec{a} \times \vec{b}$$ and $$\vec{c}$$ is 30°, then $$|(\vec{a} \times \vec{b}) \times \vec{c}|$$ equals:

Let A(-3, 2) and B(-2, 1) be the vertices of a triangle ABC. If the centroid of this triangle lies on the line $$3x + 4y + 2 = 0$$, then the vertex C lies on the line :

Let ABC be a triangle with vertices at points A(2, 3, 5), B(-1, 3, 2) and C($$\lambda$$, 5, $$\mu$$) in three dimensional space. If the median through A is equally inclined with the axes, then $$(\lambda, \mu)$$ is equal to:

The equation of a plane through the line of intersection of the planes $$x + 2y = 3$$, $$y - 2z + 1 = 0$$, and perpendicular to the first plane is :

If the events A and B are mutually exclusive events such that $$P(A) = \frac{3x+1}{3}$$ and $$P(B) = \frac{1-x}{4}$$, then the set of possible values of x lies in the interval :