NTA JEE Main 30th January 2023 Shift 2

Let $$a_1 = 1, a_2, a_3, a_4, \ldots$$ be consecutive natural numbers. Then $$\tan^{-1}\left(\frac{1}{1+a_1a_2}\right) + \tan^{-1}\left(\frac{1}{1+a_2a_3}\right) + \ldots + \tan^{-1}\left(\frac{1}{1+a_{2021}a_{2022}}\right)$$ is equal to

The range of the function $$f(x) = \sqrt{3-x} + \sqrt{2+x}$$ is

If the functions $$f(x) = \frac{x^3}{3} + 2bx + \frac{ax^2}{2}$$ and $$g(x) = \frac{x^3}{3} + ax + bx^2$$, $$a \neq 2b$$ have a common extreme point, then $$a + 2b + 7$$ is equal to

$$\lim_{n \to \infty} \frac{3}{n}\left\{4 + \left(2 + \frac{1}{n}\right)^2 + \left(2 + \frac{2}{n}\right)^2 + \ldots + \left(3 - \frac{1}{n}\right)^2\right\}$$ is equal to

Let $$q$$ be the maximum integral value of $$p$$ in $$[0, 10]$$ for which the roots of the equation $$x^2 - px + \frac{5}{4}p = 0$$ are rational. Then the area of the region $$\{(x,y) : 0 \leq y \leq (x-q)^2, 0 \leq x \leq q\}$$ is

The solution of the differential equation $$\frac{dy}{dx} = -\left(\frac{x^2+3y^2}{3x^2+y^2}\right)$$, $$y(1) = 0$$ is

Let $$\lambda \in \mathbb{R}$$, $$\vec{a} = \lambda\hat{i} + 2\hat{j} - 3\hat{k}$$, $$\vec{b} = \hat{i} - \lambda\hat{j} + 2\hat{k}$$. If $$\left((\vec{a}+\vec{b}) \times (\vec{a} \times \vec{b})\right) \times (\vec{a}-\vec{b}) = 8\hat{i} - 40\hat{j} - 24\hat{k}$$ then $$\left|\lambda(\vec{a}+\vec{b}) \times (\vec{a}-\vec{b})\right|^2$$ is equal to

Let $$\vec{a}$$ and $$\vec{b}$$ be two vectors. Let $$|\vec{a}| = 1$$, $$|\vec{b}| = 4$$ and $$\vec{a} \cdot \vec{b} = 2$$. If $$\vec{c} = (2\vec{a} \times \vec{b}) - 3\vec{b}$$, then the value of $$\vec{b} \cdot \vec{c}$$ is

A vector $$\vec{v}$$ in the first octant is inclined to the $$x$$ axis at $$60°$$, to the $$y$$-axis at $$45°$$ and to the $$z$$-axis at an acute angle. If a plane passing through the points $$(\sqrt{2}, -1, 1)$$ and $$(a, b, c)$$, is normal to $$\vec{v}$$, then

If a plane passes through the points $$(-1, k, 0)$$, $$(2, k, -1)$$, $$(1, 1, 2)$$ and is parallel to the line $$\frac{x-1}{1} = \frac{2y+1}{2} = \frac{z+1}{-1}$$, then the value of $$\frac{k^2+1}{(k-1)(k-2)}$$ is