NTA JEE Main 9th April 2019 Shift 2

A water tank has the shape of an inverted right circular cone, whose semi-vertical angle is $$\tan^{-1}\left(\frac{1}{2}\right)$$. Water is poured into it at a constant rate of 5 cubic m/min. Then the rate (in m/min), at which the level of water is rising at the instant when the depth of water in the tank is 10 m; is:

If $$\int e^{\sec x}(\sec x \tan x f(x) + (\sec x \tan x + \sec^2 x))dx = e^{\sec x}f(x) + C$$, then a possible choice of $$f(x)$$ is:

The value of the integral $$\int_0^1 x\cot^{-1}(1 - x^2 + x^4) dx$$ is:

If $$f: R \rightarrow R$$ is a differentiable function and $$f(2) = 6$$, then $$\lim_{x \to 2} \int_6^{f(x)} \frac{2t \, dt}{(x - 2)}$$ is:

The area (in sq. units) of the region $$A = \left\{(x, y) : \frac{y^2}{2} \le x \le y + 4\right\}$$ is:

If $$\cos x \frac{dy}{dx} - y\sin x = 6x$$, $$(0 < x < \frac{\pi}{2})$$ and $$y\left(\frac{\pi}{3}\right) = 0$$, then $$y\left(\frac{\pi}{6}\right)$$ is equal to:

If a unit vector $$\vec{a}$$ makes angles $$\frac{\pi}{3}$$ with $$\hat{i}$$, $$\frac{\pi}{4}$$ with $$\hat{j}$$ and $$\theta \in (0, \pi)$$ with $$\hat{k}$$, then a value of $$\theta$$ is:

The vertices B and C of a $$\triangle ABC$$ lie on the line, $$\frac{x + 2}{3} = \frac{y - 1}{0} = \frac{z}{4}$$ such that $$BC = 5$$ units. Then the area (in sq. units) of this triangle, given the point $$A(1, -1, 2)$$, is:

Let P be the plane, which contains the line of intersection of the planes, $$x + y + z - 6 = 0$$ and $$2x + 3y + z + 5 = 0$$ and it is perpendicular to the xy-plane. Then the distance of the point (0, 0, 256) from P is equal to:

Two newspapers A and B are published in a city. It is known that 25% of the city population reads A and 20% reads B while 8% reads both A and B. Further, 30% of those who read A but not B look into advertisements and 40% of those who read B but not A look into advertisements, while 50% of those who read both A and B look into advertisements. Then the percentage of the population who look into advertisements is: