NTA JEE Main 2025 April 3rd Shift 2

Instructions

For the following questions answer them individually

NTA JEE Main 2025 April 3rd Shift 2 - Question 11


The area of the region $$\{(x, y) : |x - y| \le y \le 4\sqrt{x}\}$$ is

NTA JEE Main 2025 April 3rd Shift 2 - Question 12


If the domain of the function $$f(x) = \log_x(1 - \log_4(x^2 - 9x + 18))$$ is $$(\alpha, \beta) \cup (\gamma, \delta)$$, then $$\alpha + \beta + \gamma + \delta$$ is equal to

NTA JEE Main 2025 April 3rd Shift 2 - Question 13


If the probability that the random variable X takes the value x is given by $$P(X = x) = k(x + 1)3^{-x}$$, $$x = 0, 1, 2, 3, \ldots$$, where k is a constant, then $$P(X \ge 3)$$ is equal to

NTA JEE Main 2025 April 3rd Shift 2 - Question 14


Let $$y = y(x)$$ be the solution of the differential equation $$\dfrac{dy}{dx} + 3(\tan^2 x) \, y + 3y = \sec^2 x$$, $$y(0) = \dfrac{1}{3} + e^3$$. Then $$y\left(\dfrac{\pi}{4}\right)$$ is equal to

NTA JEE Main 2025 April 3rd Shift 2 - Question 15


If $$z_1, z_2, z_3 \in \mathbb{C}$$ are the vertices of an equilateral triangle, whose centroid is $$z_0$$, then $$\displaystyle\sum_{k=1}^{3} (z_k - z_0)^2$$ is equal to

NTA JEE Main 2025 April 3rd Shift 2 - Question 16


The number of solutions of equation $$(4 - \sqrt{3})\sin x - 2\sqrt{3}\cos^2 x = \dfrac{-4}{1 + \sqrt{3}}$$, $$x \in \left[-2\pi, \dfrac{5\pi}{2}\right]$$ is

NTA JEE Main 2025 April 3rd Shift 2 - Question 17


Let C be the circle of minimum area enclosing the ellipse $$E : \dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$$ with eccentricity $$\dfrac{1}{2}$$ and foci $$(\pm 2, 0)$$. Let PQR be a variable triangle, whose vertex P is on the circle C and the side QR of length 29 is parallel to the major axis of E and contains the point of intersection of E with the negative y-axis. Then the maximum area of the triangle PQR is:

NTA JEE Main 2025 April 3rd Shift 2 - Question 18


The shortest distance between the curves $$y^2 = 8x$$ and $$x^2 + y^2 + 12y + 35 = 0$$ is:

NTA JEE Main 2025 April 3rd Shift 2 - Question 19


The distance of the point $$(7, 10, 11)$$ from the line $$\dfrac{x - 4}{1} = \dfrac{y - 4}{0} = \dfrac{z - 2}{3}$$ along the line $$\dfrac{x - 9}{2} = \dfrac{y - 13}{3} = \dfrac{z - 17}{6}$$ is

NTA JEE Main 2025 April 3rd Shift 2 - Question 20


The sum $$1 + \dfrac{1 + 3}{2!} + \dfrac{1 + 3 + 5}{3!} + \dfrac{1 + 3 + 5 + 7}{4!} + \ldots$$ upto $$\infty$$ terms, is equal to

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