If $f(x) = \frac{(e^{2x} - 1) \sin x^{\circ}}{x^2}, x \neq 0$ is continuous at $x = 0$,then $f(0) =$

Let $f(x) = \begin{cases} \frac{1 + \cos 2\pi x}{1 - \sin \pi x}, & x < \frac{1}{2} \\ p, & x = \frac{1}{2} \\ \frac{\sqrt{2x - 1}}{\sqrt{4 + \sqrt{2x - 1}} - 2}, & x > \frac{1}{2} \end{cases}$. If $f(x)$ is discontinuous at $x = \frac{1}{2}$,then:

Let $a, b \in R, b \neq 0$. Define a function $f(x) = \begin{cases} a \sin \frac{\pi}{2}(x-1), & \text{for } x \leq 0 \\ \frac{\tan 2x - \sin 2x}{bx^3}, & \text{for } x > 0 \end{cases}$. If $f$ is continuous at $x = 0$,then $10 - ab$ is equal to ...... .

On the interval $I = [-2, 2]$,the function $f(x) = \begin{cases} (x + 1) e^{-\left[ \frac{1}{|x|} + \frac{1}{x} \right]} & x \neq 0 \\ 0 & x = 0 \end{cases}$ is given. Which one of the following does not hold good?

Let $f(x) = \begin{cases} \frac{(x - 1)(6x - 1)}{2x - 1}, & \text{if } x \neq \frac{1}{2} \\ 0, & \text{if } x = \frac{1}{2} \end{cases}$. Then at $x = \frac{1}{2}$,

If $f(x) = \begin{cases} x \frac{e^{(1/x)} - e^{(-1/x)}}{e^{(1/x)} + e^{(-1/x)}}, & x \ne 0 \\ 0, & x = 0 \end{cases}$ then which of the following is true?