高考数学《函数与导数》课后练习 一、选择题1.()263,034,0x x x x f x x ⎧---≤=⎨->⎩,则函数()y f f x =⎡⎤⎣⎦的零点个数为( ) A .3B .5C .6D .7 【答案】D【解析】【分析】作出()f x 的图像,将()y f f x =⎡⎤⎣⎦的零点个数即()0f f x =⎡⎤⎣⎦的实数根个数,令()t f x =,解()0f t =有三个实数根,再结合图像即可得到答案.【详解】由题意,()y f f x =⎡⎤⎣⎦的零点个数即()0f f x =⎡⎤⎣⎦的实数根个数,作()f x 的图像如图所示, 设()t f x =,则()0f t =,当0t ≤时,即2630t t ---=,解得,1236,36t t =-=-当0t >时,即340t -=,解得33log 4t =; 结合图像知,()36f x =-()36f x =-+3()log 4f x =时有三个根,所以()0f f x =⎡⎤⎣⎦有7个根,即()y f f x =⎡⎤⎣⎦的零点个数为7. 故选:D【点睛】本题主要考查函数的零点问题、解函数值以及一元二次函数和指数函数的图像,考查学生数形结合的思想,属于中档题.