NTA JEE Main 8 April 2018 Offline

If the curves $$y^2 = 6x$$, $$9x^2 + by^2 = 16$$ intersect each other at right angles, then the value of b is:

Let $$f(x) = x^2 + \frac{1}{x^2}$$ and $$g(x) = x - \frac{1}{x}$$, $$x \in R - \{-1, 0, 1\}$$. If $$h(x) = \frac{f(x)}{g(x)}$$, then the local minimum value of h(x) is:

The integral $$\int \frac{\sin^2 x \cos^2 x}{(\sin^5 x + \cos^3 x \sin^2 x + \sin^3 x \cos^2 x + \cos^5 x)^2} dx$$ is equal to
(where C is the constant of integration).

The value of $$\int_{-\pi/2}^{\pi/2} \frac{\sin^2 x}{1+2^x} dx$$ is:

Let $$g(x) = \cos x^2$$, $$f(x) = \sqrt{x}$$, and $$\alpha, \beta (\alpha < \beta)$$ be the roots of the quadratic equation $$18x^2 - 9\pi x + \pi^2 = 0$$. Then the area (in sq. units) bounded by the curve $$y = (gof)(x)$$ and the lines $$x = \alpha$$, $$x = \beta$$ and $$y = 0$$, is:

Let $$y = y(x)$$ be the solution of the differential equation $$\sin x \frac{dy}{dx} + y \cos x = 4x$$, $$x \in (0, \pi)$$. If $$y\left(\frac{\pi}{2}\right) = 0$$, then $$y\left(\frac{\pi}{6}\right)$$ is equal to:

Let $$\vec{u}$$ be a vector coplanar with the vectors $$\vec{a} = 2\hat{i} + 3\hat{j} - \hat{k}$$ and $$\vec{b} = \hat{j} + \hat{k}$$. If $$\vec{u}$$ is perpendicular to $$\vec{a}$$ and $$\vec{u} \cdot \vec{b} = 24$$, then $$|\vec{u}|^2$$ is equal to:

If $$L_1$$ is the line of intersection of the planes $$2x - 2y + 3z - 2 = 0$$, $$x - y + z + 1 = 0$$ and $$L_2$$ is the line of intersection of the planes $$x + 2y - z - 3 = 0$$, $$3x - y + 2z - 1 = 0$$, then the distance of the origin from the plane, containing the lines $$L_1$$ and $$L_2$$ is:

The length of the projection of the line segment joining the points (5, -1, 4) and (4, -1, 3) on the plane, $$x + y + z = 7$$ is:

A bag contains 4 red and 6 black balls. A ball is drawn at random from the bag, its color is observed and this ball along with two additional balls of the same color are returned to the bag. If now a ball is drawn at random from the bag, then the probability that this drawn ball is red, is: