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技巧1:利用函数的奇偶性解答例1(2014年西安高级中学月考卷)设曲线f(x)=x2+1,其图像上任意一点(x,f(x))处的切线的斜率记为g(x),则函数y=g(x)cos x的部分图像可以为难度系数0.65分析观察选项可知,选项A、C为奇函数的图像,选项B、D为偶函数的图像,由此启迪我们:本题可借助函数的奇偶性加以灵活分析.解根据导数的几何意义,可知g(x)=2x,所以y=g(x)cos x=2xcos x.再结合奇偶性的定义,可知函数y=g(x)cos x是奇函数,所以排除选项B、D.又当
Example 1: Using the Function’s Parity Solution Example 1 (2014 Xi’an High School Monthly Test Paper) Let the curve f (x) be x2 + 1, the slope of the tangent at any point (x, f (x) Is g (x), the partial image of the function y = g (x) cos x can be a difficulty coefficient of 0.65. Analysis of the observation options shows that the options A and C are odd function images and the options B and D are even function images. This enlighten us: the problem can be flexibly analyzed by the function’s parity. According to the geometric meaning of the derivative, we know that g (x) = 2x, so y = g (x) cos x = 2xcos x. Combined with the definition of parity, We can see that the function y = g (x) cos x is an odd function, so exclude options B, D. And when