每日一题_191014

已知函数\(f(x)=(2x+a)\left(|x-a|+|x+2a|\right)\) \((a<0)\),若\(f(1)+f(2)+\cdots+f(672)=0\),则满足\(f(x)=2019\)的\(x\)的值为\(\underline{\qquad\qquad}.\)
解析: 注意到\(y=2x+a\)的图象关于\(\left(-\dfrac{a}{2},0\right)\)中心对称,函数\[y=|x-a|+|x+2a|\]的图象关于\(x=-\dfrac{a}{2}\)轴对称,因此若记\(F(x)=f\left(x-\dfrac{a}{2}\right)\),则\[F(x)=2x\left(\left |x-\dfrac{3a}{2} \right |+\left|x+\dfrac{3a}{2}\right|\right).\]显然\(F(x)\)是定义在\(\mathbb{R}\)上的单调递增的奇函数.又\[ F\left(1+\dfrac{a}{2}\right)+F\left(2+\dfrac{a}{2}\right)+\cdots+F\left(672+\dfrac{a}{2}\right)=0.\qquad (\ast)\]若\[\left(k+\dfrac{a}{2}\right)+\left(673-k+\dfrac{a}{2}\right)>0,k=1,2,\cdots 336.\]则\[ F\left(k+\dfrac{a}{2}\right)>F\left[-\left(673-k+\dfrac{a}{2}\right)\right].\]从而\[ F\left(k+\dfrac{a}{2}\right)+F\left(673-k+\dfrac{a}{2}\right)>0,k=1,2,\cdots 336.\]与\((\ast)\)相矛盾,不符题设,舍去.若\[\left(k+\dfrac{a}{2}\right)+\left(673-k+\dfrac{a}{2}\right)<0,k=1,2,\cdots 336.\]同理可证明该种情形不符题设.因此综上可知必然有\[\left(k+\dfrac{a}{2}\right)+\left(673-k+\dfrac{a}{2}\right)=0,k=1,2,\cdots,336.\]
所以\(a=-673\),从而\[ F(x)=2x\left(\left| x+1009.5 \right |+|x-1009.5|\right).\]注意到\(F(0.5)=2019\),因此方程\(f(x)=2019\)的解为\(x=337\).