Examine whether the function $f$ given by $f(x) = x^{2}$ is continuous at $x = 0$.

Let $a, b, c$ be three real numbers. If the function $f(x) = \begin{cases} \cos(2x + \pi) & \text{if } x \leq 0 \\ ax^2 + b & \text{if } 0 < x < 1 \\ cx + 4 & \text{if } 1 \leq x \leq 2 \\ 3a + 1 & \text{if } x \geq 2 \end{cases}$ is continuous everywhere, then $b^2 - bc + c^2 =$

If a function $f$ defined by $f(x) = \begin{cases} \frac{1-\cos 4x}{x^2}, & x < 0 \\ a, & x=0 \\ \frac{\sqrt{x}}{\sqrt{16+\sqrt{x}}-4}, & x>0 \end{cases}$ is continuous at $x=0$,then $a=$

If $f(x) = \begin{cases} -x^3 + 1, & \text{if } -\infty < x \leq 1 \\ |x - 1| + \lambda, & \text{if } x > 1 \end{cases}$,then:

If the function $f(x)$ is continuous on its domain $[-2, 2]$,where $f(x) = \begin{cases} \frac{\sin ax}{x} + 3, & -2 \leq x < 0 \\ x + 5, & 0 \leq x \leq 1 \\ \sqrt{x^2 + 8} - b, & 1 < x \leq 2 \end{cases}$,then $7a + b + 1$ is equal to:

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