NTA JEE Mains 8th April 2024 Shift 1

Let $$f(x) = 4\cos^3 x + 3\sqrt{3}\cos^2 x - 10$$. The number of points of local maxima of $$f$$ in interval $$(0, 2\pi)$$ is

The number of critical points of the function $$f(x) = (x - 2)^{2/3}(2x + 1)$$ is

Let $$I(x) = \int \frac{6}{\sin^2 x (1 - \cot x)^2} dx$$. If $$I(0) = 3$$, then $$I\left(\frac{\pi}{12}\right)$$ is equal to

The value of $$k \in \mathbb{N}$$ for which the integral $$I_n = \int_0^1 (1 - x^k)^n dx, n \in \mathbb{N}$$, satisfies $$147I_{20} = 148I_{21}$$ is

Let $$f(x)$$ be a positive function such that the area bounded by $$y = f(x), y = 0$$ from $$x = 0$$ to $$x = a > 0$$ is $$e^{-a} + 4a^2 + a - 1$$. Then the differential equation, whose general solution is $$y = c_1 f(x) + c_2$$, where $$c_1$$ and $$c_2$$ are arbitrary constants, is

Let $$y = y(x)$$ be the solution of the differential equation $$(1 + y^2)e^{\tan x} dx + \cos^2 x(1 + e^{2\tan x}) dy = 0, y(0) = 1$$. Then $$y\left(\frac{\pi}{4}\right)$$ is equal to

The set of all $$\alpha$$, for which the vectors $$\vec{a} = \alpha t\hat{i} + 6\hat{j} - 3\hat{k}$$ and $$\vec{b} = t\hat{i} - 2\hat{j} - 2\alpha t\hat{k}$$ are inclined at an obtuse angle for all $$t \in \mathbb{R}$$, is

If the shortest distance between the lines $$L_1 : \vec{r} = (2 + \lambda)\hat{i} + (1 - 3\lambda)\hat{j} + (3 + 4\lambda)\hat{k}, \lambda \in \mathbb{R}$$ and $$L_2 : \vec{r} = 2(1 + \mu)\hat{i} + 3(1 + \mu)\hat{j} + (5 + \mu)\hat{k}, \mu \in \mathbb{R}$$ is $$\frac{m}{\sqrt{n}}$$, where $$\gcd(m, n) = 1$$, then the value of $$m + n$$ equals

Let $$P(x, y, z)$$ be a point in the first octant, whose projection in the $$xy$$-plane is the point $$Q$$. Let $$OP = \gamma$$; the angle between $$OQ$$ and the positive $$x$$-axis be $$\theta$$; and the angle between $$OP$$ and the positive $$z$$-axis be $$\phi$$, where $$O$$ is the origin. Then the distance of $$P$$ from the $$x$$-axis is

Let the sum of two positive integers be 24. If the probability, that their product is not less than $$\frac{3}{4}$$ times their greatest possible product, is $$\frac{m}{n}$$, where $$\gcd(m, n) = 1$$, then $$n - m$$ equals