NTA JEE Main 2025 April 7th Shift 1

Among the statements
(S1) : The set $$\{z \in \mathbb{C} - \{-i\} : |z| = 1$$ and $$\frac{z-i}{z+i}$$ is purely real$$\}$$ contains exactly two elements, and
(S2) : The set $$\{z \in \mathbb{C} - \{-1\} : |z| = 1$$ and $$\frac{z-1}{z+1}$$ is purely imaginary$$\}$$ contains infinitely many elements.

The mean and standard deviation of 100 observations are 40 and 5.1, respectively. By mistake one observation is taken as 50 instead of 40. If the correct mean and the correct standard deviation are $$\mu$$ and $$\sigma$$ respectively, then $$10(\mu + \sigma)$$ is equal to

Let $$x_1, x_2, x_3, x_4$$ be in a geometric progression. If 2, 7, 9, 5 are subtracted respectively from $$x_1, x_2, x_3, x_4$$ then the resulting numbers are in an arithmetic progression. Then the value of $$\frac{1}{24}(x_1 x_2 x_3 x_4)$$ is :

Let the set of all values of $$p \in \mathbb{R}$$, for which both the roots of the equation $$x^2 - (p+2)x + (2p+9) = 0$$ are negative real numbers, be the interval $$(\alpha, \beta]$$. Then $$\beta - 2\alpha$$ is equal to

Let A be a $$3 \times 3$$ matrix such that $$|\text{adj}(\text{adj}(\text{adj } A))| = 81$$. If $$S = \{n \in \mathbb{Z} : (|\text{adj}(\text{adj } A)|)^{\frac{(n-1)^2}{2}} = |A|^{(3n^2 - 5n - 4)}\}$$, then $$\sum_{n \in S} |A^{(n^2+n)}|$$ is equal to

If the area of the region bounded by the curves $$y = 4 - \frac{x^2}{4}$$ and $$y = \frac{x-4}{2}$$ is equal to $$\alpha$$, then $$6\alpha$$ equals

Let the system of equations : $$2x + 3y + 5z = 9$$, $$7x + 3y - 2z = 8$$, $$12x + 3y - (4 + \lambda)z = 16 - \mu$$, have infinitely many solutions. Then the radius of the circle centred at $$(\lambda, \mu)$$ and touching the line $$4x = 3y$$ is

Let the line L pass through (1, 1, 1) and intersect the lines $$\frac{x-1}{2} = \frac{y+1}{3} = \frac{z-1}{4}$$ and $$\frac{x-3}{1} = \frac{y-4}{2} = \frac{z}{1}$$. Then, which of the following points lies on the line L?

Let the angle $$\theta, 0 < \theta < \frac{\pi}{2}$$ between two unit vectors $$\hat{a}$$ and $$\hat{b}$$ be $$\sin^{-1}\left(\frac{\sqrt{65}}{9}\right)$$. If the vector $$\vec{c} = 3\hat{a} + 6\hat{b} + 9(\hat{a} \times \hat{b})$$, then the value of $$9(\vec{c} \cdot \hat{a}) - 3(\vec{c} \cdot \hat{b})$$ is

Let ABC be the triangle such that the equations of lines AB and AC be $$3y - x = 2$$ and $$x + y = 2$$, respectively, and the points B and C lie on x-axis. If P is the orthocentre of the triangle ABC, then the area of the triangle PBC is equal to