NTA JEE Main 9th January 2019 Shift 2

If the circles $$x^2 + y^2 - 16x - 20y + 164 = r^2$$ and $$(x-4)^2 + (y-7)^2 = 36$$ intersect at two distinct points, then:

Let $$A(4, -4)$$ and $$B(9, 6)$$ be points on the parabola, $$y^2 = 4x$$. Let $$C$$ be chosen on the arc AOB of the parabola, where $$O$$ is the origin, such that the area of $$\triangle ACB$$ is maximum. Then, the area (in sq. units) of $$\triangle ACB$$, is:

A hyperbola has its centre at the origin, passes through the point $$(4, 2)$$ and has transverse axis of length 4 along the $$x$$-axis. Then the eccentricity of the hyperbola is:

For each $$x \in R$$, let $$[x]$$ be the greatest integer less than or equal to $$x$$. Then $$\lim_{x \to 0^{-}} \frac{x([x] + |x|)\sin[x]}{|x|}$$ is equal to:

The logical statement $$[\sim(\sim p \vee q) \vee (p \wedge r)] \wedge (\sim q \wedge r)$$ is equivalent to:

A data consists of $$n$$ observations: $$x_1, x_2, \ldots, x_n$$. If $$\sum_{i=1}^{n}(x_i + 1)^2 = 9n$$ and $$\sum_{i=1}^{n}(x_i - 1)^2 = 5n$$, then the standard deviation of this data is:

If $$A = \begin{bmatrix} e^t & e^{-t}\cos t & e^{-t}\sin t \\ e^t & -e^{-t}\cos t - e^{-t}\sin t & -e^{-t}\sin t + e^{-t}\cos t \\ e^t & 2e^{-t}\sin t & -2e^{-t}\cos t \end{bmatrix}$$, then $$A$$ is:

If the system of linear equations $$x - 4y + 7z = g$$; $$3y - 5z = h$$; $$-2x + 5y - 9z = k$$ is consistent, then:

If $$x = \sin^{-1}(\sin 10)$$ and $$y = \cos^{-1}(\cos 10)$$, then $$y - x$$ is equal to:

Let $$f: [0,1] \to R$$ be such that $$f(xy) = f(x) \cdot f(y)$$, for all $$x, y \in [0,1]$$, and $$f(0) \neq 0$$. If $$y = y(x)$$ satisfies the differential equation, $$\frac{dy}{dx} = f(x)$$ with $$y(0) = 1$$, then $$y\left(\frac{1}{4}\right) + y\left(\frac{3}{4}\right)$$ is equal to: