已知函数. (1)求函数的单调区间; (2)设,求在上的最大值; (3)试证明:对任意,不等式恒成立. 答案:解:(1)∵ 令得 显然是上方程的解 令,,则 ∴函数在上单调递增 ∴是方程的唯一解 ∵当时,当时 ∴函数在上单调递增,在上单调递减………………5分 (2)由(1)知函数在上单调递增,在上单调递减 故①当即时在上单调递增 ∴= ②当时在上单调递减 ∴= ③当,即时 …………………………He wrote a ______ composition in English but there were quite a few spelling mistakes.
[ ]A. two hundred wordsB. two-hundred-wordsC. two-hundred-wordD. two hundred word