Benchmarks

19. 解: (1) 取 $EF$ 的中点 $G$ , 连接 $DG$ , $CG$

$\backslash therefore$ 四边形 $ADE\_1F\_1$ 和 $BCE\_2F\_2$ 都是菱形, $\backslash angle\; DE\_1F\_1\; =\; \backslash angle\; CE\_2F\_2\; =\; 60^\backslash circ$

$\backslash therefore\; DG\; \backslash perp\; EF$ .................................................... 1

又 $BC\; =\; 2$

$\backslash therefore\; DG\; =\; CG\; =\; \backslash sqrt\{3\}$ .................................................... 2

又 $CD\; =\; AB\; =\; \backslash sqrt\{6\}$

即有 $DG^2\; +\; CG^2\; =\; CD^2$ , $DG\; \backslash perp\; CG$ .................................................... 3

又 $CG\; \backslash cap\; EF\; =\; G$ , $CG,\; EF\; \backslash subset$ 平面 $BCEF$

$\backslash therefore\; DG\; \backslash perp$ 平面 $BCEF$ .................................................... 5

又 $DG\; \backslash subset$ 平面 $AFED$

$\backslash therefore$ 平面 $AFED\; \backslash perp$ 平面 $BCEF$ .................................................... 6

(2) 连接 $DF$ , $DB$

由 (1) 知 $DG\; \backslash perp$ 平面 $BCEF$ , 同理 $CG\; \backslash perp$ 平面 $AFED$ .................................................... 7

且 $S\_\{\backslash text\{菱形\}BCEF\}\; =\; 2\; \backslash times\; \backslash sqrt\{3\}\; =\; 2\backslash sqrt\{3\}$ , $S\_\{\backslash triangle\; ADF\}\; =\; \backslash frac\{1\}\{2\}\; \backslash times\; 2\; \backslash times\; \backslash sqrt\{3\}\; =\; \backslash sqrt\{3\}$ .................................................... 9

$$20.\; \backslash text\{解:\; (1)\; \}\; f\text{'}(x)\; =\; \backslash frac\{-mx^2\; +\; 2(m-1)x\; +\; 4\}\{e^x\}\; =\; \backslash frac\{-(mx+2)(x-2)\}\{e^x\}\; \backslash quad\; (m>0)\; \backslash quad\; \backslash text\{....................................................\; 1\}$$

令 $f\text{'}(x)=0$ , 解得 $x=-\backslash frac\{2\}\{m\}$ 或 $x=2$ , 且 .................................................... 2

当 $x\; \backslash in\; (-\backslash infty,\; -\backslash frac\{2\}\{m\}]$ 时, $f\text{'}(x)\; \backslash le\; 0$ , 当 $x\; \backslash in\; (-\backslash frac\{2\}\{m\},\; 2)$ 时, $f\text{'}(x)\; >\; 0$ ,

当 $x\; \backslash in\; [2,\; +\backslash infty)$ 时, $f\text{'}(x)\; \backslash le\; 0$ .................................................... 4

即 $f(x)$ 的单调增区间为 $(-\backslash frac\{2\}\{m\},\; 2)$ , 单调减区间为 $(-\backslash infty,\; -\backslash frac\{2\}\{m\}]$ , $[2,\; +\backslash infty)$ .................................................... 5

(2) 由 (1) 知, 当 $m>0$ , $x\; \backslash in\; [1,\; 2]$ 时, $f\text{'}(x)\; >\; 0$ 恒成立 .................................................... 6

所以 $f(x)$ 在 $[1,\; 2]$ 上为增函数,

$$\backslash text\{即\; \}\; f(x)\_\{\backslash max\}\; =\; f(2)\; =\; \backslash frac\{4m+2\}\{e^2\},\; \backslash quad\; f(x)\_\{\backslash min\}\; =\; f(1)\; =\; \backslash frac\{m\}\{e\}\; \backslash quad\; \backslash text\{....................................................\; 8\}$$

$$f(x\_1)\; -\; f(x\_2)\; \backslash text\{\; 的最大值为\; \}\; g(m)\; =\; f(x)\_\{\backslash max\}\; -\; f(x)\_\{\backslash min\}\; =\; \backslash frac\{(4-e)m+2\}\{e^2\}\; \backslash quad\; \backslash text\{....................................................\; 9\}$$

$\backslash therefore\; \backslash frac\{4\}\{e^2\}\; \backslash ge\; [f(x\_1)\; -\; f(x\_2)]$ 恒成立 .................................................... 10

$$\backslash therefore\; \backslash frac\{4\}\{e^2\}\; \backslash ge\; \backslash frac\{(4-e)m+2\}\{e^2\},$$

$$\backslash text\{即\; \}\; m\; \backslash le\; \backslash frac\{2\}\{4-e\}\; \backslash quad\; \backslash text\{....................................................\; 11\}$$

$$\backslash text\{又\; \}\; m\; >\; 0\; \backslash quad\; \backslash therefore\; m\; \backslash in\; \backslash left(0,\; \backslash frac\{2\}\{4-e\}\backslash right]$$

$$\backslash text\{故\; \}\; m\; \backslash text\{\; 的取值范围\}\; \backslash left(0,\; \backslash frac\{2\}\{4-e\}\backslash right]\; \backslash quad\; \backslash text\{....................................................\; 12\}$$

$$21.\; \backslash text\{解:\; (1)\; \}\; \backslash because\; e\; =\; \backslash frac\{c\}\{a\}\; =\; \backslash frac\{\backslash sqrt\{2\}\}\{2\},\; \backslash quad\; \backslash therefore\; c^2\; =\; \backslash frac\{1\}\{2\}a^2,\; \backslash quad\; \backslash therefore\; b^2\; =\; a^2\; -\; c^2\; =\; \backslash frac\{1\}\{2\}a^2.\; \backslash quad\; \backslash text\{....................................................\; 2\}$$

$\backslash therefore$ 椭圆的中心 $O$ 到直线 $x+y-2b=0$ 的距离为 $5\backslash sqrt\{2\}$ ,

$$\backslash therefore\; \backslash frac\{|-2b|\}\{\backslash sqrt\{2\}\}\; =\; 5\backslash sqrt\{2\},\; \backslash quad\; \backslash therefore\; b\; =\; 5\; \backslash quad\; \backslash therefore\; b^2\; =\; 25,\; \backslash quad\; a^2\; =\; 2b^2\; =\; 50.$$

$$\backslash therefore\; \backslash text\{椭圆\; \}\; C\; \backslash text\{\; 的方程为\; \}\; \backslash frac\{x^2\}\{50\}\; +\; \backslash frac\{y^2\}\{25\}\; =\; 1.\; \backslash quad\; \backslash text\{....................................................\; 4\}$$

(2) 由 (1) 可知 $F(5,\; 0)$ , 由题可知直线 $AB$ 的方程为 $y\; =\; \backslash sqrt\{2\}(x-5)$ , 与椭圆 $C$

$$\backslash text\{的方程联立\}\; \backslash begin\{cases\}\; y\; =\; \backslash sqrt\{2\}(x-5)\; \backslash \backslash \; \backslash frac\{x^2\}\{50\}\; +\; \backslash frac\{y^2\}\{25\}\; =\; 1\; \backslash end\{cases\},\; \backslash quad\; \backslash text\{消去\; \}\; y\; \backslash text\{\; 得\}\; \backslash quad\; x^2\; -\; 8x\; +\; 10\; =\; 0$$

设 $A(x\_1,\; y\_1)$ , $B(x\_2,\; y\_2)$ , 则有 $x\_1\; +\; x\_2\; =\; 8$ , $x\_1\; x\_2\; =\; 10$ . .................................................... 6

设 $Q(x,\; y)$ , 由 $\backslash overrightarrow\{OQ\}\; =\; \backslash lambda\; \backslash overrightarrow\{OA\}\; +\; \backslash mu\; \backslash overrightarrow\{OB\}$ 得 $(x,\; y)\; =\; \backslash lambda(x\_1,\; y\_1)\; +\; \backslash mu(x\_2,\; y\_2)\; =\; (\backslash lambda\; x\_1\; +\; \backslash mu\; x\_2,\; \backslash lambda\; y\_1\; +\; \backslash mu\; y\_2)$ ,

$$\backslash therefore\; \backslash begin\{cases\}\; x\; =\; \backslash lambda\; x\_1\; +\; \backslash mu\; x\_2\; \backslash \backslash \; y\; =\; \backslash lambda\; y\_1\; +\; \backslash mu\; y\_2\; \backslash end\{cases\}\; \backslash quad\; \backslash text\{....................................................\; 7\}$$

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