In the interval $(-2 \pi, 0)$, the function $f(x) = \sin \left(\frac{1}{x^3}\right)$

Let the function $f(x)$ be defined as: $f(x) = \begin{cases} [\tan(\frac{\pi}{4} + x)]^{\frac{1}{x}}, & x \neq 0 \\ k, & x = 0 \end{cases}$. If $f(x)$ is continuous at $x = 0$, then the value of $k$ is...

If $f(x) = \begin{cases} \frac{a + 3\cos x}{x^2}, & x < 0 \\ b\tan \left( \frac{\pi}{[x + 3]} \right), & x \geqslant 0 \end{cases}$ is continuous at $x = 0$,then:

If $f(x) = \begin{cases} \frac{1-\cos Kx}{x \sin x}, & \text{if } x \neq 0 \\ \frac{1}{2}, & \text{if } x=0 \end{cases}$ is continuous at $x=0$,then the value of $K$ is

If $f(x) = \begin{cases} \frac{\cos(ax) - \cos(bx)}{x^2}, & x \neq 0 \\ \frac{1}{2}(b^2 - a^2), & x = 0 \end{cases}$ where $a$ and $b$ are real and distinct constants,then:

The values of $A$ and $B$ such that the function $f(x) = \begin{cases} -2\sin x, & x \le -\frac{\pi}{2} \\ A\sin x + B, & -\frac{\pi}{2} < x < \frac{\pi}{2} \\ \cos x, & x \ge \frac{\pi}{2} \end{cases}$ is continuous everywhere are