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(1,4]
分析:(1)由三視圖可知:該幾何體是由上下兩部分組成,上面是一個(gè)圓錐,其高為2,底面半徑為1;下面是一個(gè)與圓錐底面同底的半球,半徑為1.據(jù)此即可計(jì)算出答案;
(2)利用導(dǎo)數(shù)和分類討論方法即可求出.
解答:(1)由三視圖可知:該幾何體是由上下兩部分組成,上面是一個(gè)圓錐,其高為2,底面半徑為1;下面是一個(gè)與圓錐底面同底的半球,半徑為1.
∴V=
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=
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;
(2)利用分類討論方法:
函數(shù)f(x)=ax
3-3x+1對(duì)于x∈[-1,1]總有f(x)≥0 成立?[f(x)]
min≥0,x∈[-1,1].
由已知可得:f
′(x)=3ax
2-3,
①當(dāng)a≤0時(shí),f
′(x)<0,∴函數(shù)f(x)在[-1,1]上單調(diào)遞減,∴[f(x)]
min=f(1)=a-2≥0,解得a≥2,與a≤0矛盾,故舍去;
②當(dāng)0<a≤1時(shí),
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,由x∈[-1,1]可得
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≤0,即函數(shù)f(x)在[-1,1]上單調(diào)遞減,∴[f(x)]
min=f(1)=a-2≥0,解得a≥2,無解;
③當(dāng)a>1時(shí),
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,由x∈[-1,1]可得
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≥0,即函數(shù)f(x)在[-1,1]上單調(diào)遞增,∴[f(x)]
min=f(-1)=4-a≥0,解得a≤4,∴1<a≤4;
綜上可知:1<a≤4.
點(diǎn)評(píng):由三視圖正確恢復(fù)原幾何體和熟練掌握利用導(dǎo)數(shù)研究函數(shù)的單調(diào)性及分類討論的思想方法是解題的關(guān)鍵.