NTA JEE Main 26th July 2022 Shift 2

If $$0 < x < \dfrac{1}{\sqrt{2}}$$ and $$\dfrac{\sin^{-1}x}{\alpha} = \dfrac{\cos^{-1}x}{\beta}$$, then a value of $$\sin\dfrac{2\pi\alpha}{\alpha + \beta}$$ is

The value of $$\log_e 2 \cdot \dfrac{d}{dx}(\log_{\cos x} \csc x)$$ at $$x = \dfrac{\pi}{4}$$ is

Let $$P$$ and $$Q$$ be any points on the curves $$(x-1)^2 + (y+1)^2 = 1$$ and $$y = x^2$$, respectively. The distance between $$P$$ and $$Q$$ is minimum for some value of the abscissa of $$P$$ in the interval

If the maximum value of $$a$$, for which the function $$f_a(x) = \tan^{-1}(2x) - 3ax + 7$$ is non-decreasing in $$\left(-\dfrac{\pi}{6}, \dfrac{\pi}{6}\right)$$, is $$\bar{a}$$, then $$f_{\bar{a}}\left(\dfrac{\pi}{8}\right)$$ is equal to

The integral $$\displaystyle\int \dfrac{1 - \dfrac{1}{\sqrt{3}}(\cos x - \sin x)}{1 + \dfrac{2}{\sqrt{3}}\sin 2x} dx$$ is equal to

$$\displaystyle\int_0^{20\pi} (|\sin x| + |\cos x|)^2 dx$$ is equal to:

The area bounded by the curves $$y = |x^2 - 1|$$ and $$y = 1$$ is

Let the solution curve $$y = f(x)$$ of the differential equation $$\dfrac{dy}{dx} + \dfrac{xy}{x^2 - 1} = \dfrac{x^4 + 2x}{\sqrt{1-x^2}}$$, $$x \in (-1, 1)$$ pass through the origin. Then $$\displaystyle\int_{-\frac{\sqrt{3}}{2}}^{\frac{\sqrt{3}}{2}} f(x) dx$$ is equal to

A vector $$\vec{a}$$ is parallel to the line of intersection of the plane determined by the vectors $$\hat{i}$$, $$\hat{i} + \hat{j}$$ and the plane determined by the vectors $$\hat{i} - \hat{j}$$, $$\hat{i} + \hat{k}$$. The obtuse angle between $$\vec{a}$$ and the vector $$\vec{b} = \hat{i} - 2\hat{j} + 2\hat{k}$$ is

Let $$X$$ be a binomially distributed random variable with mean $$4$$ and variance $$\dfrac{4}{3}$$. Then $$54 P(X \le 2)$$ is equal to