每日导数79
放缩找点、多次隐零点代换
已知函数\(f(x)=\dfrac{1}{2}x^2\left(\ln x-\dfrac{1}{2}\right)+ax(\ln x-1),a\neq 0\)
(1)若\(a>0\),证明:\(f(x)\)有唯一的零点
(2)若\(f(x)>0\),求实数\(a\)的取值范围
解
(1)\(f(x)=0\)即\(\dfrac{x}{2}\left(\ln x-\dfrac{1}{2}\right)+a(\ln x-1)=0\)
分析:\(\lim\limits_{x\to 0}\varphi(x)=-\infty\),\(\lim\limits_{x\to +\infty}\varphi(x)=+\infty\),则要使得其只有一个零点,可能,大概,要么是增,要么一开始摆动一下,但是到后面就是恒定的大于0
记\(\varphi(x)=\dfrac{x}{2}\left(\ln x-\dfrac{1}{2}\right)+a(\ln x-1)\)
,\(\varphi^{\prime}(x)=\dfrac{1}{2}\ln x+\dfrac{1}{4}+\dfrac{a}{x}\)
则当\(x>1\)时,\(\varphi^{\prime}(x)>0\)
则\(\varphi(x)\)在\((1,+\infty)\)上增,并且\(\varphi(1)=-\dfrac{1}{4}-a<0\)
现在说明在\(x>1\)上有唯一的零点
因\(\ln x>1-\dfrac{1}{x}\)
则\(\varphi(x)>\dfrac{x}{2}\left(1-\dfrac{1}{x}-\dfrac{1}{2}\right)+a\left(1-\dfrac{1}{x}-1\right)=\dfrac{x}{4}-\dfrac{a}{x}-\dfrac{1}{2}=\dfrac{x^2-2x-4a}{4x}\)
二次函数\(x^2-2x-4a\)的零点为\(x_{1,2}=\dfrac{2\pm\sqrt{4+8a}}{2}=1\pm2\sqrt{1+2a}\)
则取\(x=1+2\sqrt{1+2a}>1\)有, \(\varphi\left(1+\sqrt{1+2a}\right)>0\)
由零点存在定理,\(\varphi(x)\)在\((1,1+2\sqrt{1+2a})\)存在至少一个零点
又因\(\varphi(x)\)在\((1,+\infty)\)上单调递增
则\(\varphi(x)\)在\((1,1+2\sqrt{1+2a})\)上存在唯一零点
现在说明\(x\in(0,1)\)上没有零点
因\(\ln x<x-1\),
则有
\(\varphi(x)=\dfrac{x}{2}\left(\ln x-\dfrac{1}{2}\right)+a(\ln x-1) <\dfrac{x}{2}\left(x-1-\dfrac{1}{2}\right)+a(x-1-1) =\dfrac{1}{2}\left[x^2+(2a-3)x-4a\right]\)
考虑二次函数\(g(x)=x^2+(2a-3)x-4a\)在\(0<x<1\)上正负
因其开口向上\(g(0)=-a<0,g(1)=-2-2a<0\)
则\(g(x)\)在\((0,1)\)上为负
从而\(\varphi(x)<g(x)<0\)
则\(\varphi(x)\)在\((0,1)\)上没有零点
综上\(\varphi(x)\)在\(x>0\)上有且仅有一个零点.
(2)由(1)得\(a\leq 0\),要使得\(f(x)>0\),要有\(f(1)=-\dfrac{1}{4}-a>0\),即\(a<-\dfrac{1}{4}\)
因\(a<0\),则\(\varphi^{\prime}(x)=\dfrac{1}{2}\ln x+\dfrac{1}{4}+\dfrac{a}{x}\)单调递增
不难得到\(\varphi^{\prime}(x)\)有唯一零点\(x_0\),即\(\dfrac{1}{2}\ln x_0+\dfrac{1}{4}+\dfrac{a}{x_0}=0\),
即\(2x_0\ln x_0+x_0+4a=0\)
并且\(x=x_0\)是\(\varphi(x)\)的极小值点
\(\varphi(x_0)=\dfrac{x_0}{2}\left(\ln x_0-\dfrac{1}{2}\right)+a(\ln x_0-1) =-\dfrac{1}{4}x_0-\dfrac{1}{4}x_0-2a+a\ln x_0\)
\(=-\dfrac{1}{2}x_0+a\ln x_0+x_0+\dfrac{1}{2}x_0 =(x_0+a)\ln x_0>0\)
因\(x_0>1\),则\(x_0>-a\)或\(a>-x_0\),则有
\(0=2x_0\ln x_0+x_0+4a>2x_0\ln x_0+x_0-4x_0=2x_0\ln x_0-3x_0\)
得\(x_0\in \left(1,e^{\frac{3}{2}}\right)\)
则\(a>-e^{\frac{3}{2}}\)
综上\(a\in\left(-e^{\frac{3}{2}},-\dfrac{1}{4}\right)\)