Consider the function $f : \mathbb{R} \rightarrow \mathbb{R}$ defined by $f(x) = \begin{cases} (2 - \sin(\frac{1}{x}))|x|, & x \neq 0 \\ 0, & x = 0 \end{cases}$. Then $f$ is

Let $f : R \to R$ be a differentiable function satisfying $f''(3) + f'(2) = 0$. Then $\mathop {\lim }\limits_{x \to 0} {\left( {\frac{{1 + f\left( {3 + x} \right) - f\left( 3 \right)}}{{1 + f\left( {2 - x} \right) - f\left( 2 \right)}}} \right)^{\frac{1}{x}}}$ is equal to

Let $f(x)$ be defined in $[-2, 2]$ by
$f(x) = \begin{cases} \max(4 - x^2, 1 + x^2), & -2 < x < 0 \\ \min(4 - x^2, 1 + x^2), & 0 < x < 2 \end{cases}$
Then $f(x)$:

Given the following properties of a function $f(x)$:
$(i)$ $f(x)$ is continuous and defined for all real numbers.
$(ii)$ $f'(-5) = 0$; $f'(2)$ is not defined and $f'(4) = 0$.
$(iii)$ $(-5, 12)$ is a point on the graph of $f(x)$.
$(iv)$ $f''(2)$ is undefined,but $f''(x)$ is negative everywhere else.
$(v)$ The signs of $f'(x)$ are given by the following number line:
$f'(x)$ is positive for $x < -5$,negative for $-5 < x < 2$,positive for $2 < x < 4$,and negative for $x > 4$.
On the possible graph of $y = f(x)$,we have:

If a function $f(x)$ defined by $f(x)=\begin{cases} a e^{x}+b e^{-x}, & -1 \leq x<1 \\ c x^{2}, & 1 \leq x \leq 3 \\ a x^{2}+2 c x, & 3 < x \leq 4 \end{cases}$ is continuous for some $a, b, c \in R$ and $f'(0)+f'(2)=e$,then the value of $a$ is:

$\mathop {Lim}\limits_{\lambda \to 0} \,{\left( {\int\limits_0^1 {{{(1 + x)}^\lambda }dx} } \right)^{\frac{1}{\lambda }}}$ is equal to