Define $f: R \to R$ by $f(x) = \begin{cases} \frac{1-\cos 4x}{x^2}, & x < 0 \\ a, & x = 0 \\ \frac{\sqrt{x}}{\sqrt{16+\sqrt{x}}-4}, & x > 0 \end{cases}$ Then the value of $a$ so that $f$ is continuous at $x = 0$ is:

$f(x) = \begin{cases} \frac{e^{\alpha x} - e^{x} - x}{x^{2}}, & x \neq 0 \\ \frac{3}{2}, & x = 0 \end{cases}$ Find the value of $\alpha$ for which the function $f$ is continuous.

Show that the function defined by $f(x)=|\cos x|$ is a continuous function.

If the function $f(x)$ defined by $f(x) = \begin{cases} ax + 1 & \text{if } x \leq 3 \\ bx + 3 & \text{if } x > 3 \end{cases}$ is continuous at $x = 3$, then $(a - b) = ..........$

Define $f(x) = \begin{cases} \frac{1-\sin x}{(\pi-2x)^2} & \text{, if } x \neq \frac{\pi}{2} \\ k & \text{, if } x = \frac{\pi}{2} \end{cases}$. If $f(x)$ is continuous at $x = \frac{\pi}{2}$,then $k =$

If the function $f(x) = \begin{cases} x + a^2\sqrt{2} \sin x, & 0 \le x < \pi/4 \\ x \cot x + b, & \pi/4 \le x < \pi/2 \\ b \sin 2x - a \cos 2x, & \pi/2 \le x \le \pi \end{cases}$ is continuous in the interval $[0, \pi]$,then the values of $(a, b)$ are: