(1)证明:当 0<x<10<x<10<x<1 时,x−x2<sinx<xx-x^2<\sin x<xx−x2<sinx<x;
(2)已知函数 f(x)=cosax−ln(1−x2)f(x)=\cos ax-\ln(1-x^2)f(x)=cosax−ln(1−x2),若 x=0x=0x=0 是 f(x)f(x)f(x) 的极大值点,求 aaa 的取值范围。
先证明 sinx<xsinx \lt xsinx<x
令 h1(x)=x−sinxh_1(x) = x-sinxh1(x)=x−sinx
h1(0)=0\colorbox{yellow}{\(h_1(0) = 0\)}h1(0)=0
h1′(x)=1−cosxh_1'(x)= 1-cosxh1′(x)=1−cosx,
当 0<x<10<x<10<x<1 时,h′(x)>0 ⟹ h1(x)>0h'(x) \gt 0 \implies h_1(x) \gt 0h′(x)>0⟹h1(x)>0
∴sinx<x\therefore \colorbox{yellow}{\(sinx\lt x\)}∴sinx<x
再证明 x−x2<sinxx-x^2 \lt sinxx−x2<sinx
令 h2(x)=sinx−x+x2h_2(x) = sinx - x + x^2h2(x)=sinx−x+x2
h2(0)=0\colorbox{yellow}{\(h_2(0) = 0\)}h2(0)=0
h2′(x)=cosx−1+2xh_2'(x)= cosx - 1 + 2xh2′(x)=cosx−1+2x,
h2′(0)=0\colorbox{yellow}{\(h_2'(0) = 0\)}h2′(0)=0
h2′′(x)=2−sinxh_2''(x)= 2-sinxh2′′(x)=2−sinx,
当 0<x<10<x<10<x<1 时,h2′′(x)>0 ⟹ h2′(x)严格单调增 ⟹ h2′(x)>0h_2''(x) \gt 0 \implies h_2'(x)严格单调增 \implies h_2'(x) \gt 0h2′′(x)>0⟹h2′(x)严格单调增⟹h2′(x)>0
h2′(x)>0 ⟹ h2(x)严格单调增 ⟹ h2(x)>0h_2'(x) \gt 0 \implies h_2(x)严格单调增 \implies h_2(x) \gt 0h2′(x)>0⟹h2(x)严格单调增⟹h2(x)>0
∴x−x2<sinx\therefore \colorbox{yellow}{\(x-x^2\lt sinx\)}∴x−x2<sinx