【答案】
分析:A.當(dāng)x<0時(shí),利用基本不等式的性質(zhì),y=-
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≤-4,可知無(wú)最小值;
B.變形為
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,利用基本不等式的性質(zhì)可知:最小值大于4;
C.利用基本不等式的性質(zhì)即可判斷出滿(mǎn)足條件;
D.利用基本不等式的性質(zhì)可知:最小值大于4.
解答:解:A.當(dāng)x<0時(shí),
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=-4,當(dāng)且僅當(dāng)x=-2時(shí)取等號(hào).因此此時(shí)A無(wú)最小值;
B.
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=
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=4,當(dāng)且僅當(dāng)x
2+2=1時(shí)取等號(hào),但是此時(shí)x的值不存在,故不能取等號(hào),即y>4,因此B的最小值不是4;
C.
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=4,當(dāng)且僅當(dāng)
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,解得e
x=2,即x=ln4時(shí)取等號(hào),即y的最小值為4,因此C滿(mǎn)足條件;
D.當(dāng)0<x<π時(shí),sinx>0,∴
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=4,當(dāng)且僅當(dāng)
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,即sinx=2時(shí)取等號(hào),但是sinx不可能取等號(hào),故y>4,因此不滿(mǎn)足條件.
綜上可知:只有C滿(mǎn)足條件.
故選C.
點(diǎn)評(píng):熟練掌握基本不等式的性質(zhì)是解題的關(guān)鍵,特別注意“=”是否取到.