NTA JEE Mains 28th Jan 2025 Shift 2

Let A and B be the two points of intersection of the line $$y + 5 = 0$$ and the mirror image of the parabola $$y^{2}=4x$$ with respect to the line $$x + y + 4 = 0$$. If d denotes the distance between A and B , and a denotes the area of $$\triangle SAB$$ where $$S$$ is the focus of the parabola $$y^{2}=4x$$, then the value of (a + d) is

The number of natural numbers, between 212 and 999 , such that the sum of their digits is 15 , is

If $$y=y(x)$$ is the solution of the differential equation.
$$\sqrt{4 - x^2}\,\frac{dy}{dx} = \left(\left(\sin^{-1}\left(\frac{x}{2}\right)\right)^2 - y\right)\sin^{-1}\left(\frac{x}{2}\right), \quad -2 \le x \le 2,\quad y(2) = \frac{\pi^2 - 8}{4}$$, then $$y^{2}(0)$$ is equal to

The interior angles of a polygon with n sides, are in an A.P. with common difference $$6^{\circ}$$ . If the largest interior angle of the polygon is $$219^{\circ}$$, then n is equal to

Let $$f(x) = \lim_{n \to \infty} \sum_{r=0}^{n}\left(\frac{\tan\left(\frac{x}{2^{r+1}}\right) + \tan^{3}\left(\frac{x}{2^{r+1}}\right)}{1 - \tan^{2}\left(\frac{x}{2^{r+1}}\right)}\right).\quad\text{Then }\lim_{x \to 0} \frac{e^{x} - e^{f(x)}}{x - f(x)}$$ is equal to.

A bar magnet has total length $$2l = 20$$ units and the field point $$P$$ is at a distance d = 10 units from the centre of the magnet. If the relative uncertainty of length measurement is 1%, then uncertainty of the magnetic field at point P is :

A concave mirror produces an image of an object such that the distance between the object and image is 20 cm . If the magnification of the image is ' -3 ', then the magnitude of the radius of curvature of the mirror is :

The velocity-time graph of an object moving along a straight line is shown in figure. What is the distance covered by the object between $$t = 0 to t = 4 s$$ ?

A body of mass 4 kg is placed on a plane at a point P having coordinate (3,4)m. Under the action of force $$\overrightarrow{F} = (2\hat{i}+3\hat{j})N$$, it moves to a new point Q having coordinates (6,10)m in 4 sec . The average power and instanteous power at the end of 4 sec are in the ratio of :

Match List - I with List - II.