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2005年全国高考数学卷第22题为(Ⅰ)设函数f(x)=xlog2x+(1-x)log2(1-x)(0x1),求f(x)的最小值;(Ⅱ)设正数p1,p2,p3,…,p2n满足p1+p2+p3+…+p2n=1,证明p1log2p1+p2log2p2+p3log2p3+…+p2nlog2p2n≥-n.下面我们从试题背景、解法分析来剖析这一道题.1试题背景Jensen不等式f(x)是开
In 2005, the 22th national college entrance examination mathematics volume (I) set the function f (x) = xlog2x + (1-x) log2 (1-x) Let p1 + p2 + p3 + ... + p2n = 1 prove that p1log2p1 + p2log2p2 + p3log2p3 + ... + p2nlog2p2n≥-n. Here we analyze the problem from the test background and solution analysis.1 Questions background Jensen inequality f (x) is on