NTA JEE Main 2025 April 4th Shift 2

Let the matrix $$A = \begin{bmatrix} 1 & 0 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{bmatrix}$$ satisfy $$A^n = A^{n-2} + A^2 - I$$ for $$n \geq 3$$. Then the sum of all the elements of $$A^{50}$$ is :

If the sum of the first 20 terms of the series $$\frac{4 \cdot 1}{4 + 3 \cdot 1^2 + 1^4} + \frac{4 \cdot 2}{4 + 3 \cdot 2^2 + 2^4} + \frac{4 \cdot 3}{4 + 3 \cdot 3^2 + 3^4} + \frac{4 \cdot 4}{4 + 3 \cdot 4^2 + 4^4} + \ldots$$ is $$\frac{m}{n}$$, where m and n are coprime, then $$m + n$$ is equal to :

If $$1^2 \cdot ({^{15} C_{1}}) + 2^2 \cdot ({^{15} C_{2}}) + 3^2 \cdot ({^{15} C_{3}}) + \ldots + 15^2 \cdot ({^{15} C_{15}}) = 2^m \cdot 3^n \cdot 5^k$$, where $$m, n, k \in \mathbb{N}$$, then $$m + n + k$$ is equal to :

Let for two distinct values of p the lines $$y = x + p$$ touch the ellipse E : $$\frac{x^2}{4^2} + \frac{y^2}{3^2} = 1$$ at the points A and B. Let the line $$y = x$$ intersect E at the points C and D. Then the area of the quadrilateral ABCD is equal to

Consider two sets A and B, each containing three numbers in A.P. Let the sum and the product of the elements of A be 36 and p respectively and the sum and the product of the elements of B be 36 and q respectively. Let d and D be the common differences of A.P.'s in A and B respectively such that $$D = d + 3, d > 0$$. If $$\frac{p+q}{p-q} = \frac{19}{5}$$, then $$p - q$$ is equal to

If a curve $$y = y(x)$$ passes through the point $$\left(1, \frac{\pi}{2}\right)$$ and satisfies the differential equation $$(7x^4 \cot y - e^x \csc y) \frac{dx}{dy} = x^5, x \geq 1$$, then at $$x = 2$$, the value of $$\cos y$$ is :

The centre of a circle C is at the centre of the ellipse E : $$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$, $$a > b$$. Let C pass through the foci $$F_1$$ and $$F_2$$ of E such that the circle C and the ellipse E intersect at four points. Let P be one of these four points. If the area of the triangle $$PF_1F_2$$ is 30 and the length of the major axis of E is 17, then the distance between the foci of E is :

Let $$f(x) + 2f\left(\frac{1}{x}\right) = x^2 + 5$$ and $$2g(x) - 3g\left(\frac{1}{2}\right) = x$$, $$x > 0$$. If $$\alpha = \int_1^2 f(x)\,dx$$, and $$\beta = \int_1^2 g(x)\,dx$$, then the value of $$9\alpha + \beta$$ is :

Let A be the point of intersection of the lines $$L_1 : \frac{x-7}{1} = \frac{y-5}{0} = \frac{z-3}{-1}$$ and $$L_2 : \frac{x-1}{3} = \frac{y+3}{4} = \frac{z+7}{5}$$. Let B and C be the points on the lines $$L_1$$ and $$L_2$$ respectively such that $$AB = AC = \sqrt{15}$$. Then the square of the area of the triangle ABC is :

Let the mean and the standard deviation of the observation 2, 3, 3, 4, 5, 7, a, b be 4 and $$\sqrt{2}$$ respectively. Then the mean deviation about the mode of these observations is :