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

If $f: [0, 2) \to R$ is defined by $f(x) = \begin{cases} 1 + 2x^k, & 0 \le x < 1 \\ kx, & 1 \le x < 2 \end{cases}$ where $k > 0$ and $f$ is such that $\lim_{x \to 1^-} f(x) = \lim_{x \to 1^+} f(x)$,then the value of $k^2$ is:

If $f(x) = \frac{e^{x^2} - \cos x}{x^2}$ for $x \neq 0$ is continuous at $x = 0$,then $f(0) = $.

The function $f(x) = \frac{x^2 - 1}{x^3 - 1}$ is not defined at $x = 1$. If the function is continuous at $x = 1$,then the value of $f(1)$ will be:

If the function $f(x) = \begin{cases} 5x - 4, & 0 < x \le 1 \\ 4x^2 + 3bx, & 1 < x < 2 \end{cases}$ is continuous at every point of its domain,then the value of $b$ is

If the function $f(x) = \left( \frac{5x - 8}{8 - 3x} \right)^{\frac{3}{2x-4}}$ for $x \neq 2$ is continuous at $x = 2$, then the value of $f(2)$ is...