2020年高考数学导数压轴题每日一题
例8（分类讨论，区间划分）已知函数f(x)?的导函数.
(1)设函数f(x)的图象与x轴交点为A,曲线y=f(x)在A点处的切线方程是y?3x?3,求a,b的值;
(2)若函数g(x)?e例8解:(Ⅰ)∵f(x)??ax1312x?ax?x?b(a?0),f'(x)为函数f(x)32?f'(x),求函数g(x)的单调区间.
1312x?ax?x?b(a?0), 322∴f'(x)?x?ax?1
∵f(x)在(1,0)处切线方程为y?3x?3, ?f'(1)?3∴?, ?f(1)?011∴a?1,b??. (各1分)
6f'(x)x2?ax?1(Ⅱ)g(x)?ax?(x?R). axee(2x?a)eax?a(x2?ax?1)eax??x[ax?(a2?2)]e?ax g'(x)?ax2(e)①当a?0时,g'(x)?2x,
x (??,0) - ↘ 0 0 (0,??) + 极小值 ↗ g(x)的单调递增区间为(0,??),单调递减区间为(??,0) ②当a?0时,令g'(x)?0,得x?0或x?(ⅰ)当
g'(x) g(x) 2?a a2?a?0,即0?a?2时, ax (??,0) 0 2?a2(0,)a + ↗ 2?a a0 极大值 22?a2(,??)a - ↘ g'(x) - ↘ 0 极小值 g(x) 2?a22?a2),单调递减区间为(??,0),(,??); g(x)的单调递增区间为(0,aa22?2x(ⅱ)当?a?0,即a?2时,g'(x)???2xe?0,
a
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故g(x)在(??,??)单调递减; (ⅲ)当
2?a?0,即a?2时, a2(??,?a)x a 2?a a0 极小值 2(?a,0)a + ↗ 0 (0,??) - ↘ g'(x) - ↘ 0 极大值 g(x) 2?a22?a2,0)上单调递增,在(0,??),(??,)上单调递减 g(x)在(aa综上所述,当a?0时,g(x)的单调递增区间为(0,??),单调递减区间为(??,0);
2?a2),单调递减区间为(??,0) 当0?a?2时,g(x)的单调递增区间为(0,a当a?当a?2,g(x)的单调递减区间为(??,??)
22时,g(x)的单调递增区间为(?a,0),单调递减区间为(0,、??)a2(??,?a)
a
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2020年高考数学导数压轴题每日一题 (8)
