Let $[x]$ denote the greatest integer less than or equal to $x$ and $f(x) = [\tan^2 x]$. Which of the following is true?

If a function $f(x)$ defined by $f(x) = \begin{cases} ax^2 + bx + c, & x \leq -1 \\ 2x^2 + 4x + 1, & -1 < x < 1 \\ cx^2 + bx + a, & x \geq 1 \end{cases}$ is continuous on $\mathbb{R}$,and $\lim_{x \rightarrow \frac{3}{2}} f(x) = 14$,then find $\lim_{x \rightarrow -2} f(x)$.

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

The values of $a$ and $b$ for which the function $f(x) = \begin{cases} 1+|\sin x|^{a/|\sin x|}, & -\pi / 6 < x < 0 \\ b, & x = 0 \\ e^{\tan 2 x / \tan 3 x}, & 0 < x < \pi / 6 \end{cases}$ is continuous at $x = 0$ are

If the function $f(x) = \begin{cases} \frac{1}{x} \log_{e}\left(\frac{1+\frac{x}{a}}{1-\frac{x}{b}}\right), & x < 0 \\ k, & x = 0 \\ \frac{\cos^{2} x - \sin^{2} x - 1}{\sqrt{x^{2}+1}-1}, & x > 0 \end{cases}$ is continuous at $x = 0$,then $\frac{1}{a} + \frac{1}{b} + \frac{4}{k}$ is equal to:

The number of discontinuities in $\mathbb{R}$ for the function $f(x)=\frac{x-1}{x^3+6x^2+11x+6}$ is