If $f(x) = \begin{cases} \frac{1-\cos 4x}{x^2} & \text{if } x < 0 \\ a & \text{if } x = 0 \\ \frac{\sqrt{16+\sqrt{x}}-4}{\sqrt{x}} & \text{if } x > 0 \end{cases}$ is continuous at $x = 0$,then find the value of $a$.

Let $f: R \rightarrow R$ be defined by $f(x)=\begin{cases} \alpha+\frac{\sin [x]}{x}, & x>0 \\ 2, & x=0 \\ \beta+\left[\frac{\sin x-x}{x^3}\right], & x < 0 \end{cases}$. If $f$ is continuous at $x=0$,find the value of $\alpha + \beta$.

Let $[x]$ denote the greatest integer less than or equal to $x$. Then, the value of $\alpha$ for which the function $f(x)=\begin{cases} \frac{\sin [-x^2]}{[-x^2]}, & x \neq 0 \\ \alpha, & x=0 \end{cases}$ is continuous at $x=0$ is:

For every pair of continuous functions $f, g: [0, 1] \rightarrow \mathbb{R}$ such that $\max \{f(x): x \in [0, 1] \} = \max \{g(x): x \in [0, 1] \} = \lambda$,the correct statement$(s)$ is (are):

Let $f(x) = \frac{1 - x(1 + |1 - x|)}{|1 - x|} \cos \left(\frac{1}{1 - x}\right)$ for $x \neq 1$. Which of the following statements is/are true?
$(A)$ $\lim_{x \to 1^{-}} f(x) = 0$
$(B)$ $\lim_{x \to 1^{-}} f(x) = 1$
$(C)$ $\lim_{x \to 1^{+}} f(x) = 0$
$(D)$ $\lim_{x \to 1^{+}} f(x)$ does not exist.

If the function $f(x) = \begin{cases} \frac{\cos ax - \cos 9x}{x^2}, & x \neq 0 \\ 16, & x = 0 \end{cases}$ is continuous at $x = 0$, then $a =$