组卷台

函数 $f\left( x\right)$ 与 $g\left( x\right)$ 在 $\left[ {a,b}\right]$ 上存在二阶导数,且 ${g}^{\prime \prime }\left( x\right) \neq 0,f\left( a\right) = f\left( b\right) = g\left( a\right) = g\left( b\right) = 0$ ,证明在 $\left( {a,b}\right)$ 内至少存在一点 $\xi$ ,满足 $\frac{f\left( \xi \right) }{g\left( \xi \right) } = \frac{{f}^{\prime \prime }\left( \xi \right) }{{g}^{\prime \prime }\left( \xi \right) }$ .

设 $y = y\left( x\right)$ 有一阶连续导数, $y\left( 0\right) = 1$ ,且满足 ${y}^{\prime }\left( x\right) + 3{\int }_{0}^{x}{y}^{\prime }\left( t\right) {dt} + {2x}{\int }_{0}^{1}y\left( {xu}\right) \mathrm{d}u + {\mathrm{e}}^{-x} = 0$ ,求 $y = \; y\left( x\right)$ .

求曲线 ${4y} = {\int }_{0}^{2}x\sqrt{{12} - {x}^{2}{u}^{2}}\,\mathrm{d}u\left( {x \geq 0}\right)$ 的全长.

设正项数列 $\left\{ {a}_{n}\right\}$ 单调递减,且级数 $\mathop{\sum }\limits_{n = 1}^{\infty }{a}_{n}$ 发散,记 ${x}_{n} = \frac{{a}_{2} + {a}_{4} + \cdots + {a}_{2n}}{{a}_{1} + {a}_{3} + \cdots + {a}_{{2n} - 1}}\left( {n = 1,2,\cdots }\right)$ ,证明 $\mathop{\lim }\limits_{n \rightarrow \infty }{x}_{n} = 1$

解欧拉方程 ${x}^{2}\frac{{d}^{2}y}{\,\mathrm{d}{x}^{2}} + {2x}\frac{\mathrm{d}y}{\,\mathrm{d}x} + {2y} = 0$.

计算 ${\int }_{1}^{+\infty }\frac{\arctan x}{{x}^{2}}\,\mathrm{d}x$.

在球面 $z = \sqrt{1 - {x}^{2} - {y}^{2}},x \geq 0,y \geq 0$ 上，取以 ${M}_{1}\left( {1,0,0}\right)$ ， ${M}_{2}\left( {0,1,0}\right)$ ， ${M}_{3}\left( {\frac{\sqrt{2}}{2},0,\frac{\sqrt{2}}{2}}\right)$ 为顶点的球面三角形 $\sum ,\sum$ 的法向量的方向余弦 $\cos \gamma > 0$ ,计算 $I = {\iint }_{\sum }\frac{x\,\mathrm{d}y\,\mathrm{d}z + z\,\mathrm{d}x\,\mathrm{d}y}{{x}^{2} + {y}^{2} + {z}^{2}}$ .

设 $\left\{ {a}_{n}\right\}$ 为正项数列,且 $\mathop{\sum }\limits_{n = 1}^{\infty }{a}_{n}$ 发散,则以下级数:
(1) $\mathop{\sum }\limits_{n = 1}^{\infty }\frac{{a}_{n}}{1 + {a}_{n}}$;
(2) $\mathop{\sum }\limits_{n = 1}^{\infty }\frac{{a}_{n}}{1 - {a}_{n}}$;
(3) $\mathop{\sum }\limits_{n = 1}^{\infty }\frac{{a}_{n}}{1 + {a}_{n}^{2}}$；
(4) $\mathop{\sum }\limits_{n = 1}^{\infty }\frac{{a}_{n}}{1 + {n}^{2}{a}_{n}}$.
一定收敛的个数为$\underline{\qquad}$.

A. 1
B. 2
C. 3
D. 4

设 $y = y\left( x\right)$ 由参数方程 $\left\{\begin{array}{l}x = \frac{t}{1 + {t}^{3}} \\ y = \frac{{t}^{2}}{1 + {t}^{3}}\end{array}\right.$ 确定，则曲线 $y = y\left( x\right)$ 的斜渐近线方程为$\underline{\qquad}$.