NTA JEE Main 9th January 2019 Shift 1

Equation of a common tangent to the circle, $$x^2 + y^2 - 6x = 0$$ and the parabola, $$y^2 = 4x$$ is:

Let $$0 < \theta < \frac{\pi}{2}$$. If the eccentricity of the hyperbola $$\frac{x^2}{\cos^2\theta} - \frac{y^2}{\sin^2\theta} = 1$$ is greater than 2, then the length of its latus rectum lies in the interval:

The value of $$\lim_{y \to 0} \frac{\sqrt{1 + \sqrt{1 + y^4}} - \sqrt{2}}{y^4}$$ is:

If the Boolean expression $$p \oplus q \wedge \sim p \odot q$$ is equivalent to $$p \wedge q$$, where $$\oplus$$, $$\odot \in \{\wedge, \vee\}$$, then the ordered pair $$(\oplus, \odot)$$ is:

5 students of a class have an average height 150 cm and variance 18 cm$$^2$$. A new student, whose height is 156 cm, joined them. The variance in cm$$^2$$ of the height of these six students is:

If $$A = \begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix}$$, then the matrix $$A^{-50}$$ when $$\theta = \frac{\pi}{12}$$, is equal to:

The system of linear equations
$$x + y + z = 2$$
$$2x + 3y + 2z = 5$$
$$2x + 3y + (a^2 - 1)z = a + 1$$

If $$\cos^{-1}\left(\frac{2}{3x}\right) + \cos^{-1}\left(\frac{3}{4x}\right) = \frac{\pi}{2}$$, ($$x > \frac{3}{4}$$), then $$x$$ is equal to:

For $$x \in R - \{0, 1\}$$, let $$f_1(x) = \frac{1}{x}$$, $$f_2(x) = 1 - x$$ and $$f_3(x) = \frac{1}{1-x}$$ be three given functions. If a function, $$J(x)$$ satisfies $$(f_2 \circ J \circ f_1)(x) = f_3(x)$$ then $$J(x)$$ is equal to:

Let $$f: R \to R$$ be a function defined as
$$f(x) = \begin{cases} 5, & \text{if } x \leq 1 \\ a + bx, & \text{if } 1 < x < 3 \\ b + 5x, & \text{if } 3 \leq x < 5 \\ 30, & \text{if } x \geq 5 \end{cases}$$

Then $$f$$ is: