NTA JEE Mains 04th April 2024 Shift 1

If 2 and 6 are the roots of the equation $$ax^2 + bx + 1 = 0$$, then the quadratic equation whose roots are $$\frac{1}{2a+b}$$ and $$\frac{1}{6a+b}$$ is:

Let α and β be the sum and the product of all the non-zero solutions of the equation $$(\bar{z})^2 + |z| = 0,\ z \in \mathbb{C}$$. Then $$4(\alpha^2 + \beta^2)$$ is equal to:

There are 5 points $$P_1, P_2, P_3, P_4, P_5$$ on the side AB, excluding A and B, of a triangle ABC. Similarly there are 6 points $$P_6, P_7,\ldots, P_{11}$$ on the side BC and 7 points $$P_{12}, P_{13},\ldots, P_{18}$$ on the side CA of the triangle. The number of triangles, that can be formed using the points $$P_1, P_2,\ldots, P_{18}$$ as vertices, is:

Let the first three terms 2, p and q, with q ≠ 2, of a G.P. be respectively the 7th, 8th and 13th terms of an A.P. If the $$5^{th}$$ term of the G.P. is the $$n^{th}$$ term of the A.P., then n is equal to:

The sum of all rational terms in the expansion of $$\left(2^{\frac{1}{5}} + 5^{\frac{1}{3}}\right)^{15}$$ is equal to:

The vertices of a triangle are A(−1, 3), B(−2, 2) and C(3, −1). A new triangle is formed by shifting the sides of the triangle by one unit inwards. Then the equation of the side of the new triangle nearest to origin is:

A square is inscribed in the circle $$x^2 + y^2 - 10x - 6y + 30 = 0$$. One side of this square is parallel to y = x + 3. If $$(x_i, y_i)$$ are the vertices of the square, then $$\sum(x_i^2 + y_i^2)$$ is equal to:

Let $$\alpha, \beta \in {R}$$. Let the mean and the variance of 6 observations −3, 4, 7, −6, α, β be 2 and 23, respectively. The mean deviation about the mean of these 6 observations is:

Let $$\alpha \in (0,\infty)$$ and $$A = \begin{bmatrix}1 & 2 & \alpha\\ 1 & 0 & 1\\ 0 & 1 & 2\end{bmatrix}$$. If $$\det(\text{adj}(2A-A^T)\cdot\text{adj}(A-2A^T)) = 2^8$$, then $$(\det(A))^2$$ is equal to:

If the system of equations $$x + (\sqrt{2}\sin\alpha)y + (\sqrt{2}\cos\alpha)z = 0$$, $$x + (\cos\alpha)y + (\sin\alpha)z = 0$$, $$x + (\sin\alpha)y - (\cos\alpha)z = 0$$ has a non-trivial solution, then $$\alpha \in \left(0,\frac{\pi}{2}\right)$$ is equal to: