伯努利不等式应用
已知函数\(f(x)=(1+x)^m-mx-1,x>-1,m>0\)且\(m\neq 1\)
\((1)\) 讨论\(f(x)\)单调性
\((2)\) 若\(\forall x\in\left(0,\dfrac{\pi}{2}\right)\cup \left(\dfrac{\pi}{2},\pi\right),a\sin x<(1+\cos^2x)^{\sin x}\)求\(a\)取值范围.
\((3)\) 证明:当\(x\in\left(0,\dfrac{\pi}{2}\right)\cup \left(\dfrac{\pi}{2},\pi\right)\),且\(n\in\mathbb{N},n\geq 2\)时,\(\dfrac{(1-\sqrt{\sin x})(1-\sqrt[3]{\sin x})\cdots(1-\sqrt[n]{\sin x})}{(1-\sin x)^{n-1}}>\dfrac{1}{n!}\)恒成立.
解
\((1)\) \(f^{\prime}(x)=m(1+x)^{m-1}-m=m\left[(1+x)^{m-1}-1\right]\)
当\(m>1\)时,\(f^{\prime}(x)=0\)得\(x_0=0\),此时\(f(x)\)在\((-1,0)\)上单调递减,\((0,+\infty)\)上单调递增.
当\(0<m<1\)时,,此时\(f(x)\)在\((-1,0)\)上单调递增,\((0,+\infty)\)
上单调递减
\((2)\) 由\((1)\)知当\(0<m<1\)时,\(f(x)<f(0)=0\)取\(m=\sin x,x=\cos^2x\)有
\((1+\cos^2x)^{\sin x}<\cos^2 x\sin x+1\)
则要使\(a\sin x<(1+\cos^2x)^{\sin x}\)成立则有
\(a\sin x<\cos^2x\sin x+1\)
当\(a\leq 1\)时,不等式一定成立
当\(a>1\)时,原不等式为\(a<\dfrac{\cos^2x\sin x+1}{\sin x}=\dfrac{\sin x-\sin^3x+1}{\sin x}=\dfrac{1}{\sin x}-\sin^2x+1\)
记\(\sin x=t,g(t)=\dfrac{1}{t}-t^2+1,t\in(0,1)\)
从而不难知道\(g(t)\)单调递减,从而\(a<1\)矛盾!
综上\(a\leq 1\)
\((3)\) \(\dfrac{(1-\sqrt{\sin x})(1-\sqrt[3]{\sin x})\cdots(1-\sqrt[n]{\sin x})}{(1-\sin x)^{n-1}}=\dfrac{1-\sqrt{\sin x}}{1-\sin x}\cdot\dfrac{1-\sqrt[3]{\sin x}}{1-\sin x}\cdots\dfrac{1-\sqrt[n]{\sin x}}{1-\sin x}\)
由\((1)\)得\((1+x)^m<mx+1\)
即\(1-(1+x)^m>-mx\xlongequal{1+x=\sin x}1-\sin^mx>m(1-\sin x)\)
即\(\dfrac{1-\sin^mx}{1-\sin x}>m\)
,\(m\)分别取\(\dfrac{1}{2},\dfrac{1}{3},\cdots \dfrac{1}{n}\)累乘有
\(\dfrac{(1-\sqrt{\sin x})(1-\sqrt[3]{\sin x})\cdots(1-\sqrt[n]{\sin x})}{(1-\sin x)^{n-1}}>\dfrac{1}{n!}\)
标签:right,21,导数,dfrac,每日,sqrt,cdots,pi,sin From: https://www.cnblogs.com/manxinwu/p/17936214.html