本刊2011年4月(上)期刊登了《例析函数的奇偶性》一文(下称文[1]),文中例4的解法存在瑕疵,为行文方便,将文[1]中的例4摘抄如下.例1(2010年高考江苏卷第5题)设函数f(x)=x(ex+ae-x)(x∈R)是偶函数,则实数a=_____.分析如果直接利用f(-x)=f(x)会浪费很多时间,我们可以看出g(x)=ex+ae-x为奇函数,由g(0)=0,从而直接得出答案a=-1.上述答案是正确的,但解法存在瑕疵,同学们都喜欢用这种方法.但是如果没有正确理 The article published in April 2011 (last issue) “parsing functions of parity” article (hereinafter referred to as the article [1]), the text of Example 4 there is a flaw in the solution for the convenience of the text, the text [1] 4 Excerpts are as follows: Example 1 (2010 Jiangsu entrance examination Question 5) Let the function f (x) = x (ex + ae-x) (x∈R) is an even function, the real number a = _____. Since f (x) = f (x) istes a lot of time, we can see that g (x) = ex + ae-x is an odd function and g The answer is correct, but there are flaws in the solution, students like to use this method, but if not correct