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江西省乐平市2023-2024学年度八年级下学期期中学业评价数学.考卷答案试卷答案
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分析求出函数的导数，由函数的零点存在定理可得f′（1）f′（3）＜0，进而验证a=4与a=$\frac{16}{9}$时是否符合题意，即可求答案．
解答解：f（x）的导数为f′（x）=$\frac{1}{x}$-$\frac{a}{（x+1）^{2}}$，
当f′（1）f′（3）＜0时，函数f（x）在区间（1，3）上只有一个极值点，
即为（1-$\frac{1}{4}$a）（$\frac{1}{3}$-$\frac{1}{16}$a）＜0，
解得4＜a＜$\frac{16}{3}$；
当a=4时，f′（x）=$\frac{1}{x}$-$\frac{4}{（x+1）^{2}}$=0，解得x=1∉（1，3），
当a=$\frac{16}{3}$时，f′（x）=$\frac{1}{x}$-$\frac{16}{3（x+1）^{2}}$=0在（1，3）上无实根，
则a的取值范围是4＜a＜$\frac{16}{3}$，且a∈N，即为a=5．
故选：A．
点评本题考查利用导数研究函数的极值问题，体现了转化的思想方法的运用，考查运算能力，属于中档题．
