每日一题_191126

Posted 2020-11-22 math521

已知函数(f(x)=m an x+2sin x),(xinleft[ 0,dfrac{pi}{2} ight)),(minmathbb{R}).
((1)) 若函数(y=f(x))在(xinleft[ 0,dfrac{pi}{2} ight))上是单调函数,求实数(m)的取值范围;
((2)) 当(m=1)时,若对任意(xinleft[0,dfrac{pi}{2} ight)),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}+2cos x,xinleft[0,dfrac{pi}{2} ight).]由题可知方程(f'(x)=0)在区间(left[0,dfrac{pi}{2} ight))无实根,从而易得(m)的取值范围为
(left(-infty,-2 ight)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 xinleft[0,dfrac{pi}{2} ight), an x+2sin 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(aleqslant 3),由于我们熟知[ forall xinleft[0,dfrac{pi}{2} ight), an xgeqslant x+dfrac{x^3}{3},sin xgeqslant x-dfrac{x^3}{6}.]则[ forall xinleft[0,dfrac{pi}{3} ight),h(x)geqslant left(x+dfrac{x^3}{3} ight)+2left(x-dfrac{x^3}{6} ight)-3{ln}(x+1)geqslant 0.]
情形二 证法二 当(aleqslant 3),仅需证明[ forall xinleft[0,dfrac{pi}{2} ight), an x+2sin 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}+2cos x-dfrac{3}{x+1},xinleft[0,dfrac{pi}{2} ight).]
其中(y=-dfrac{1}{x+1})在(left[0,dfrac{pi}{2} ight))显然单调递增,易证(y=dfrac{1}{t^2}+2t,tinleft(0,1 ight])上单调递减,根据单调性的复合运算可知[y=dfrac{1}{cos^2x}+2cos x,xinleft[ 0,dfrac{pi}{2} ight).]为单调递增函数,从而(g'(x))在其定义域内单调递增,因此(forall xinleft[0,dfrac{pi}{2} ight),g'(x)geqslant g'(0)=0).因此(g(x))单调递增,从而[ forall xinleft[0,dfrac{pi}{2} ight),g(x)geqslant 0.]证毕.
情形二 当(a>3),则[ h'(0)=3-a<0.]此时必然[ exists x_0inleft(0,dfrac{pi}{2} ight),forall xinleft(0,x_0 ight),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]).