Find the values of $k$ so that the function $f$ is continuous at the indicated point. $f(x) = \begin{cases} kx^2, & \text{if } x \le 2 \\ 3, & \text{if } x > 2 \end{cases}$ at $x=2$.

If $f(x) = \frac{\ln(e^{x^2} + 2\sqrt{x})}{\sqrt{x}}$ is continuous at $x = 0$,then $f(0)$ must be equal to:

If $f(x) = \begin{cases} ax^2 - b, & 0 \le x < 1 \\ 2, & x = 1 \\ x + 1, & 1 < x \le 2 \end{cases}$ is continuous at $x = 1$,then the most suitable values of $a$ and $b$ are:

If for $p \neq q \neq 0$,the function $f(x) = \frac{\sqrt[7]{p(729+x)}-3}{\sqrt[3]{729+qx}-9}$ is continuous at $x=0$,then:

If $f(x) = \begin{cases} x^2, & \text{when } x \le 1 \\ x + 5, & \text{when } x > 1 \end{cases}$,then

Let the function $f(x) = \begin{cases} \frac{\log_{e}(1+5x) - \log_{e}(1+\alpha x)}{x} & \text{if } x \neq 0 \\ 10 & \text{if } x = 0 \end{cases}$ be continuous at $x = 0$. The value of $\alpha$ is equal to: