难难难的双参问题
已知函数\(f(x)=\ln(1+x)+\dfrac{x^2}{2}\)
(1)当\(x>0\),比较\(f(x)\)与\(x\)的大小
(2)若函数\(g(x)=\cos x+\dfrac{x^2}{2}\),且\(f\left(e^{\frac{a}{2}}\right)=g(b)-1(a>0,b>0)\),证明:\(f(b^2)+1>g(a+1)\)
解
(1)做差,记\(\varphi(x)=f(x)-x=\ln(1+x)+\dfrac{x^2}{2}-x\)
\(\varphi^{\prime}(x)=\dfrac{1}{x+1}+x-1=\dfrac{1}{x+1}+x+1-2\geq 2-2=0\)
从而\(\varphi(x)\)单调递增,则\(\varphi(x)\geq \varphi(0)=0\)
从而\(f(x)\geq x\)
(2)设\(h(x)=f(x)+1-g(x)=\ln(1+x)+1-\cos x\),
当\(x>0\)时,\(1-\cos x\geq 0,\ln(1+x)>0\),则\(h(x)>0\)恒成立
由\(h\left(e^{\frac{a}{2}}\right)>0\),得\(f\left(e^{\frac{a}{2}}\right)+1>g\left(e^{\frac{a}{2}}\right)\)
从而\(g(b)>g\left(e^{\frac{a}{2}}\right)\)
不难证明\(g(x)\)单调递增
则\(g(b)>g\left(e^{\frac{a}{2}}\right)\)得到\(b>e^{\frac{a}{2}}\)
同理\(h(b)=f(b)+1-g(b)>0\)
得\(f(b^2)+1>g(b^2)\)
则要证\(f(b^2)+1>g(a+1)\)
即证:\(g(b^2)>g(a+1)\)
即证:\(b^2>a+1\)
因\(b>e^{\frac{a}{2}}\),则\(b^2>e^a\)
因\(e^a>a+1\)
则\(b^2>e^a>a+1\)
标签:right,frac,导数,ln,dfrac,每日,varphi,76,left From: https://www.cnblogs.com/manxinwu/p/18056803