Let f:[0,1]\(\rightarrow\)R be a function. suppose the function f is twice differentiable,f(0)=0=f(1) and satisfies f''(x)-2f'(x)+f(x)\(\geq\)ex,x\(\in\)[0,1], which of the following is true?
0<f(x)<\(\infty\)
\(-\frac{1}{2}<f(x)<\frac{1}{2}\)
\(-\frac{1}{4}<f(x)<1\)
\(-\infty<f(x)<0\)
Let $\left\lfloor t \right\rfloor$ be the greatest integer less than or equal to $t$. Then the least value of $p \in \mathbb{N}$ for which
\[ \lim_{x \to 0^+} \left( x \left\lfloor \frac{1}{x} \right\rfloor + \left\lfloor \frac{2}{x} \right\rfloor + \dots + \left\lfloor \frac{p}{x} \right\rfloor \right) - x^2 \left( \left\lfloor \frac{1}{x^2} \right\rfloor + \left\lfloor \frac{2}{x^2} \right\rfloor + \dots + \left\lfloor \frac{9^2}{x^2} \right\rfloor \right) \geq 1 \]
is equal to __________.
If \( f(x) \) is defined as follows:
$$ f(x) = \begin{cases} 4, & \text{if } -\infty < x < -\sqrt{5}, \\ x^2 - 1, & \text{if } -\sqrt{5} \leq x \leq \sqrt{5}, \\ 4, & \text{if } \sqrt{5} \leq x < \infty. \end{cases} $$ If \( k \) is the number of points where \( f(x) \) is not differentiable, then \( k - 2 = \)
Let a line passing through the point $ (4,1,0) $ intersect the line $ L_1: \frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4} $ at the point $ A(\alpha, \beta, \gamma) $ and the line $ L_2: x - 6 = y = -z + 4 $ at the point $ B(a, b, c) $. Then $ \begin{vmatrix} 1 & 0 & 1 \\ \alpha & \beta & \gamma \\ a & b & c \end{vmatrix} \text{ is equal to} $