NTA JEE Main 1st February 2023 Shift 1

Let $$S$$ denote the set of all real values of $$\lambda$$ such that the system of equations
$$\lambda x + y + z = 1$$
$$x + \lambda y + z = 1$$
$$x + y + \lambda z = 1$$
is inconsistent, then $$\sum_{\lambda \in S} |\lambda|^2 + |\lambda|$$ is equal to

Let $$S$$ be the set of all solutions of the equation $$\cos^{-1}(2x) - 2\cos^{-1}\sqrt{1-x^2} = \pi$$, $$x \in [-\frac{1}{2}, \frac{1}{2}]$$. Then $$\sum_{x \in S} 2\sin^{-1}(x^2 - 1)$$ is equal to

Let $$f(x) = 2x + \tan^{-1}x$$ and $$g(x) = \log_e(\sqrt{1+x^2} + x)$$, $$x \in [0, 3]$$. Then

Let $$f(x) = \begin{vmatrix} 1+\sin^2 x & \cos^2 x & \sin 2x \\ \sin^2 x & 1+\cos^2 x & \sin 2x \\ \sin^2 x & \cos^2 x & 1+\sin 2x \end{vmatrix}$$, $$x \in [\frac{\pi}{6}, \frac{\pi}{3}]$$. If $$\alpha$$ and $$\beta$$ respectively are the maximum and the minimum values of $$f$$, then

$$\lim_{n \to \infty} \frac{1}{1+n} + \frac{1}{2+n} + \frac{1}{3+n} + \ldots + \frac{1}{2n}$$ is equal to:

The area enclosed by the closed curve $$C$$ given by the differential equation $$\frac{dy}{dx} + \frac{x+a}{y-2} = 0$$, $$y(1) = 0$$ is $$4\pi$$. Let $$P$$ and $$Q$$ be the points of intersection of the curve $$C$$ and the y-axis. If normals at $$P$$ and $$Q$$ on the curve $$C$$ intersect x-axis at points $$R$$ and $$S$$ respectively, then the length of the line segment $$RS$$ is

If $$y = yx$$ is the solution curve of the differential equation $$\frac{dy}{dx} + y\tan x = x\sec x$$, $$0 \leq x \leq \frac{\pi}{3}$$, $$y(0) = 1$$, then $$y(\frac{\pi}{6})$$ is equal to

Let the image of the point $$P(2, -1, 3)$$ in the plane $$x + 2y - z = 0$$ be $$Q$$. Then the distance of the plane $$3x + 2y + z + 29 = 0$$ from the point $$Q$$ is

The shortest distance between the lines $$\frac{x-5}{1} = \frac{y-2}{2} = \frac{z-4}{-3}$$ and $$\frac{x+3}{1} = \frac{y+5}{4} = \frac{z-1}{-5}$$ is

In a binomial distribution B(n, p), the sum and product of the mean & variance are 5 and 6 respectively, then find $$6(n + p - q)$$ is equal to :-