急求 2005高考数学广东卷 20题答案.
參考答案:注: sqrt() 为开方, <> 为 不等于
解：（1）
(I) 当k = 0时, 此时A点与D点重合, 折痕所在的直线方程y = 1/2,
(II) 当k<>0时, 设A点落在线段DC上的点A’(x0, 1), (0< = x0< = 2), 则直线OA’的斜率kOA’ = 1/x0,
∵折痕所在直线垂直平分OA’
∴kOA’* k = -1 ∴x0 = -k
又∵折痕所在的直线与OA’的交点坐标（线段OA’的中点）为M(-k/2, 1/2),
∴折痕所在的直线方程y-1/2 = k ( x + k/2 ), 即y = kx + k^2/2 + ½ ,
由( I ) ( II )得折痕所在的直线方程为：y = kx + k^2/2 + 1/2 (-2 < = k < = 0 )
（2）折痕所在的直线与坐标轴的交点坐标为E( 0 , (k^2+1)/2 ), F( -(k^2+1)/2k , 0 )
由（1）知, k = -x0 ∵0 < = x0 < = 2 ∴ -2 < = k < = 0,
设折痕长度为d, 所在直线的倾斜角为 t,
(I) 当 k = 0时, 此时A点与D点重合, 折痕的长为2
(II) 当 -2 <= k < 0时,
设 a = -(k^2+1) / 2k , b = (k^2+1)/2,
0 < a <= |AB| =2时, l与线段AB相交, 此时-2 <= k <= -2 + sqrt (3),
a > |AB| =2时, l与线段BC相交, 此时-2 + sqrt (3) < k < 0,
0 < b <= 1时, l与线段AD相交, 此时 -1 <= k < 0,
b > 1时, l与线段DC相交, 此时 -2 <= k < -1,
∴将k所在的分为3个子区间：
1) 当 -2 <= k < -1时, 折痕所在的直线l与线段DC、AB相交,
折痕的长 d = 1 / (sin t) = 1 /(|k| / sqrt (1+k^2) ) = sqrt (1+k^2)/|k| = sqrt (1 / k^2 +1),
∴sqrt (5)/2 <= d < sqrt (2)
2) 当 -1 <= k <= -2 +sqrt (3)时, 折痕所在的直线l与线段AD、AB相交,
d = sqrt ( ( (-1- k^2 ) / 2k)^2 + ( (1+k^2) / 2)^2) = sqrt ( k^4/4 + 3k^2/4 + 1/4k^2 + 3/4)
令g’(x) >= 0, 即k^3 + 3k/2 – 1/2k^3 >= 0, 即2k^6 + 3k^4 -1 <=0, 即(k^2+1)^2(k^2–1/2)<=0,
∵-1 <= k <= -2 +sqrt (3), ∴解得-sqrt (2)/2 <= k <= -2+sqrt (3)
令g’(x) <=0, 解得 -1 <= k <= -sqrt (2)/2,
故当 -1 <= k <= -sqrt (2)/2时, g(x)是减函数, 当-sqrt (2)/2 <=k <= -2+sqrt (3)时,g(x)是增函数,
∵g(-1) =2, g(-2+sqrt (3)) = 4(8-4sqrt (3)),
∴g(-1) < g(-2+sqrt (3)),
∴当k = -2+sqrt (3)时, g(-2+sqrt (3)) = 4(8-4sqrt (3)),
d = sqrt (g(-2+sqrt (3)) = 2 (sqrt (6)- sqrt (2)),
∴当-1 <= k <= -2+ sqrt (3)时, d<=2(sqrt (6) – sqrt (2)),
3)当 -2 + sqrt (3) < k < 0时, 折痕所在的直线l与线段AD、BC相交,
折痕的长d = 2 / |cos t| = 2 sqrt (1+k^2),
∴2 < l < 2 sqrt (8-4sqrt (3) ), 即2 < l < 2(sqrt (6) – sqrt (2)),
综上所述得, 当 k = -2 + sqrt (3)时, 折痕的长有最大值, 为2 (sqrt (6) – sqrt (2))．
