每日导数93

常规比值代换，极值点偏移

已知函数\(f(x)=x\ln x-a(2x^2+1)\)

(1)若\(a=-1\)，求\(f(x)\)在\(x=-1\)处的切线

(2)若函数\(f(x)\)有两个极值点\(x_1,x_2(x_1<x_2)\)

(i)求\(a\)的范围

(ii)证明:\(3x_1-x_2>\dfrac{1}{a}-2\)

解
(1)\(y=5x-2\)

(2)\(f^{\prime}(x)=\ln x+1-4ax\)

(i)记\(\varphi(x)=\ln x-4ax+1,\varphi^{\prime}(x)=\dfrac{1}{x}-4a\)

Case1当\(a\leq 0\)时，\(\varphi^{\prime}(x)\geq 0\)，此时\(\varphi(x)\)单调递增

则\(\varphi(x)\)至多一个零点，舍

Case2 当\(a>0\)时，\(\varphi^{\prime}\left(\dfrac{1}{4a}\right)=0\)

则\(\varphi(x)\)在\(\left(0,\dfrac{1}{4a}\right)\)上增，\(\left(\dfrac{1}{4a},+\infty\right)\)减

而\(x\to 0,\varphi(x)\to -\infty,x\to +\infty,\varphi(x)\to -\infty\)

则要使得\(\varphi(x)\)有两个零点，则有\(\varphi\left(\dfrac{1}{4a}\right)>0\)

即\(\ln 4a<0\)，得\(0<a<\dfrac{1}{4}\)

(ii)由(i),\(0<x_1<\dfrac{1}{4a}<x_2\)，设\(\dfrac{x_1}{x_2}=t,0<t<1\)，即\(x_1=tx_2\)

则有\(\begin{cases} \ln x_1-4ax_1+1=0\\ \ln x_2-4ax-2+1=0 \end{cases}\)，两式做差有\(\ln\dfrac{x_1}{x_2}+4a(x_2-x_1)=0\)

即有\(\ln t+4ax_2(1-t)=0\)，得\(x_2=\dfrac{\ln t}{4a(t-1)},x_2=\dfrac{t\ln t}{4a(t-1)}\)

则\(3_2-x_1=\dfrac{3\ln t}{4a(t-1)}-\dfrac{t\ ln t}{4a(t-1)}\)

则要证明:\(3x_2-x_1>\dfrac{1}{a}-2\)

即证明:\(\dfrac{3\ln t}{4a(t-1)}-\dfrac{t\ln t}{4a(t-1)}>\dfrac{1}{a}-2\)

即\(3\ln t-t\ln t-4(t-1)+8a(t-1)<0\)

记\(\gamma(t)=3\ln t-t\ln t-4(t-1)+8a(t-1),0<t<1\)

\(\gamma^{\prime}(t)=\dfrac{3}{t}-\ln t+8a-5<3-0+8a-5=-2+8a<-2+8\cdot\dfrac{1}{4}=-2=2=0\)

从而\(\gamma(t)\)单调递减，则\(\gamma(t)<\gamma(1)=0-0-0+0=0\)

得证.