If $f(x) = \begin{cases} (1 + 2x)^{1/x}, & x \ne 0 \\ e^2, & x = 0 \end{cases}$,then:

The function $f(x) = \frac{1 - \sin x + \cos x}{1 + \sin x + \cos x}$ is not defined at $x = \pi$. The value of $f(\pi)$,so that $f(x)$ is continuous at $x = \pi$,is

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 the function $f(x)$,defined below,is continuous on the interval $[0, 8]$,then
$f(x) = \begin{cases} x^{2} + ax + b, & 0 \le x < 2 \\ 3x + 2, & 2 \le x \le 4 \\ 2ax + 5b, & 4 < x \le 8 \end{cases}$

Let $f: R \rightarrow R$ be a continuous function satisfying $f(0)=1$ and $f(2x)-f(x)=x$ for all $x \in R$. If $\lim_{n \rightarrow \infty} \{f(x)-f(\frac{x}{2^n})\} = G(x)$,then $\sum_{r=1}^{10} G(r^2)$ is equal to

If $f(x) = \begin{cases} x, & x \le 0 \\ 0, & x > 0 \end{cases}$ then the function $f(x)$ at $x = 0$ is: