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NTA JEE Mains 24th january 2023 Shift 1 - Mathematics

For the following questions answer them individually

Let a tangent to the curve $$y^2 = 24x$$ meet the curve $$xy = 2$$ at the points $$A$$ and $$B$$. Then the midpoints of such line segments $$AB$$ lie on a parabola with the

The compound statement $$(\sim(P \wedge Q)) \vee ((\sim P) \wedge Q) \Rightarrow ((\sim P) \wedge (\sim Q))$$ is equivalent to

Let $$\alpha$$ be a root of the equation $$(a-c)x^2 + (b-a)x + (c-b) = 0$$ where $$a, b, c$$ are distinct real numbers such that the matrix $$\begin{pmatrix} \alpha^2 & \alpha & 1 \\ 1 & 1 & 1 \\ a & b & c \end{pmatrix}$$ is singular. Then the value of $$\frac{(a-c)^2}{(b-a)(c-b)} + \frac{(b-a)^2}{(a-c)(c-b)} + \frac{(c-b)^2}{(a-c)(b-a)}$$ is

Let $$PQR$$ be a triangle. The points $$A, B$$ and $$C$$ are on the sides $$QR, RP$$ and $$PQ$$ respectively such that $$\frac{QA}{AR} = \frac{RB}{BP} = \frac{PC}{CQ} = \frac{1}{2}$$. Then $$\frac{\text{Area}(\triangle PQR)}{\text{Area}(\triangle ABC)}$$ is equal to