NTA JEE Mains 27th Jan 2024 Shift 2

If $$\alpha, \beta$$ are the roots of the equation, $$x^2 - x - 1 = 0$$ and $$S_n = 2023\alpha^n + 2024\beta^n$$, then

Let $$\alpha = \dfrac{(4!)!}{(4!)^{3!}}$$ and $$\beta = \dfrac{(5!)!}{(5!)^{4!}}$$. Then :

The 20th term from the end of the progression $$20, 19\frac{1}{4}, 18\frac{1}{2}, 17\frac{3}{4}, \ldots, -129\frac{1}{4}$$ is :

If $$2\tan^2\theta - 5\sec\theta = 1$$ has exactly 7 solutions in the interval $$\left[0, \frac{n\pi}{2}\right]$$, for the least value of $$n \in \mathbb{N}$$ then $$\sum_{k=1}^{n} \frac{k}{2^k}$$ is equal to :

Let $$A$$ and $$B$$ be two finite sets with $$m$$ and $$n$$ elements respectively. The total number of subsets of the set $$A$$ is 56 more than the total number of subsets of $$B$$. Then the distance of the point $$P(m, n)$$ from the point $$Q(-2, -3)$$ is

Let R be the interior region between the lines $$3x - y + 1 = 0$$ and $$x + 2y - 5 = 0$$ containing the origin. The set of all values of $$a$$, for which the points $$(a^2, a + 1)$$ lie in R, is :

Let $$e_1$$ be the eccentricity of the hyperbola $$\frac{x^2}{16} - \frac{y^2}{9} = 1$$ and $$e_2$$ be the eccentricity of the ellipse $$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$, $$a > b$$, which passes through the foci of the hyperbola. If $$e_1 e_2 = 1$$, then the length of the chord of the ellipse parallel to the x-axis and passing through $$(0, 2)$$ is :

If $$\lim_{x \to 0} \frac{3 + \alpha \sin x + \beta \cos x + \log_e(1 - x)}{3\tan^2 x} = \frac{1}{3}$$, then $$2\alpha - \beta$$ is equal to :

The values of $$\alpha$$, for which $$\begin{vmatrix} 1 & \frac{3}{2} & \alpha + \frac{3}{2} \\ 1 & \frac{1}{3} & \alpha + \frac{1}{3} \\ 2\alpha + 3 & 3\alpha + 1 & 0 \end{vmatrix} = 0$$, lie in the interval

Considering only the principal values of inverse trigonometric functions, the number of positive real values of $$x$$ satisfying $$\tan^{-1}(x) + \tan^{-1}(2x) = \frac{\pi}{4}$$ is :