25~26高三上·安徽淮南一模·第14题

$4$

【题目】已知$a,b \in R$，若函数$f(x) = (x – a – 4)^2\ln\left(\dfrac{x – 2a}{2b – x}\right)$的图象存在对称中心，则$b=$_________.

【答案】$4$

【解析】由题可知，$\dfrac{x – 2a}{2b – x} > 0 \implies (x – 2a)(2b – x) > 0$，
令$(x – 2a)(2b – x) = 0$，解得$x = 2a$或$x = 2b$，
由题可知，$a \neq b$，
所以$f(x)$定义域为$(2a,2b)$或$(2b,2a)$.

因为函数$f(x)$的图象存在对称中心，
所以对称中心横坐标为区间中点$\dfrac{2a + 2b}{2} = a + b$，
因为$\ln\left(\dfrac{a + b – 2a}{2b – a – b}\right) = \ln\left(\dfrac{b – a}{b – a}\right) = 0$，即$f(a + b) = 0$，
所以函数$f(x)$图象的对称中心为$(a + b,0)$，
则$f(a + b + x) = -f(a + b – x)$.

$
\begin{align} f(a + b + x) &= (a + b + x – a – 4)^2\ln\left(\dfrac{a + b + x – 2a}{2b – a – b – x}\right) \ &= (x + b – 4)^2\ln\left(\dfrac{x + b – a}{b – a – x}\right), \ \ f(a + b – x) &= (a + b – x – a – 4)^2\ln\left(\dfrac{a + b – x – 2a}{2b – a – b + x}\right) \ &= (-x + b – 4)^2\ln\left(\dfrac{-x + b – a}{b – a + x}\right), \end{align}
$

因为$f(a + b + x) = -f(a + b – x)$，
所以$(x + b – 4)^2\ln\left(\dfrac{x + b – a}{b – a – x}\right) $$= -(-x + b – 4)^2\ln\left(\dfrac{-x + b – a}{b – a + x}\right)$，

因为$\ln\left(\dfrac{x + b – a}{b – a – x}\right) = -\ln\left(\dfrac{-x + b – a}{b – a + x}\right)$，
所以$(x + b – 4)^2 = (-x + b – 4)^2$对于定义域内任意$x$都成立，
所以$b – 4 = 0$，即$b = 4$.

故答案为：$4$.