已知函数\(f(x)=m\tan x+2\sin x\),\(x\in\left[ 0,\dfrac{\pi}{2}\right)\),\(m\in\mathbb{R}\).
\((1)\) 若函数\(y=f(x)\)在\(x\in\left[ 0,\dfrac{\pi}{2}\right)\)上是单调函数,求实数\(m\)的取值范围;
\((2)\) 当\(m=1\)时,若对任意\(x\in\left[0,\dfrac{\pi}{2}\right)\),不等式\(f(x)\geqslant a{\ln}(x+1)\)恒成立,求实数\(a\)的取值范围.
解析:
\((1)\) 对\(f(x)\)求导可得$$
f'(x)=\dfrac{m}{\cos^2x}+2\cos x,x\in\left[0,\dfrac{\pi}{2}\right).$$由题可知方程\(f'(x)=0\)在区间\(\left[0,\dfrac{\pi}{2}\right)\)无实根,从而易得\(m\)的取值范围为
\(\left(-\infty,-2\right)\cup[0,+\infty)\).
\((2)\) 由端点分析易得\(a\)的讨论分界点为\(3\), 题中所给条件不等式等价于$$
\forall x\in\left[0,\dfrac{\pi}{2} \right),\tan x+2\sin x-a{\ln}(x+1)\geqslant 0.$$记上述不等式左侧为\(h(x)\),注意到\(h(0)=0\),
情形一 证法一 当\(a\leqslant 3\),由于我们熟知$$
\forall x\in\left[0,\dfrac{\pi}{2}\right),\tan x\geqslant x+\dfrac{x^3}{3},\sin x\geqslant x-\dfrac{x^3}{6}.$$则$$
\forall x\in\left[0,\dfrac{\pi}{3}\right),h(x)\geqslant \left(x+\dfrac{x3}{3}\right)+2\left(x-\dfrac{x3}{6}\right)-3{\ln}(x+1)\geqslant 0.$$
情形二 证法二 当\(a\leqslant 3\),仅需证明$$
\forall x\in\left[0,\dfrac{\pi}{2} \right),\tan x+2\sin x-3{\ln}(x+1)\geqslant 0.$$记上述不等式左侧为\(g(x)\),求导可得$$
g'(x)=\dfrac{1}{\cos^2x}+2\cos x-\dfrac{3}{x+1},x\in\left[0,\dfrac{\pi}{2}\right).$$
其中\(y=-\dfrac{1}{x+1}\)在\(\left[0,\dfrac{\pi}{2}\right)\)显然单调递增,易证\(y=\dfrac{1}{t^2}+2t,t\in\left(0,1\right]\)上单调递减,根据单调性的复合运算可知$$y=\dfrac{1}{\cos^2x}+2\cos x,x\in\left[ 0,\dfrac{\pi}{2}\right).$$为单调递增函数,从而\(g'(x)\)在其定义域内单调递增,因此\(\forall x\in\left[0,\dfrac{\pi}{2}\right),g'(x)\geqslant g'(0)=0\).因此\(g(x)\)单调递增,从而$$
\forall x\in\left[0,\dfrac{\pi}{2}\right),g(x)\geqslant 0.$$证毕.
情形二 当\(a>3\),则$$
h'(0)=3-a<0.$$此时必然$$
\exists x_0\in\left(0,\dfrac{\pi}{2}\right),\forall x\in\left(0,x_0\right),h'(x)<0,h(x)\leqslant h(0)=0.$$不符题设,舍去.
综上可得\(a\)的取值范围为\((-\infty,3]\).
