Bài 104917

Question

Chứng minh rằng:
a)

$\sin x>x-\frac{x^3}{6}, \forall x \in (0;\frac{\pi}{2})$
b)

$\sin x+\tan x>2x, \forall x \in (0;\frac{\pi}{2})$
c)

$2\sin x+\tan x>3x, \forall x \in (0;\frac{\pi}{2})$

score 0 · Answer 1

a) Xét

$f(x)=\sin x-x+\frac{x^3}{6}, \forall x \in R$

$f'(x)=\cos x-1+-2\sin^2 \frac{x}{2}+\frac{x^2}{2}$

$=\frac{1}{2}[\frac{x^2}{2}-(\sin \frac{x}{2})^2]>0, \forall x >0$ ( Vì

$|\sin t|<t, \forall x >0$ )

$\Rightarrow f(x)>f(0)=0, \forall x >0$

$\Rightarrow \sin x>x-\frac{x^3}{6}, \forall x >0$

b) Xét

$f(x)=\sin x+\tan x-2x, x\in (0;\frac{\pi}{2})$

$f'(x)=\cos x+\frac{1}{\cos^2 x}-2>\cos^2 x+\frac{1}{\cos^2 x}-2$

$\geq (\cos x-\frac{1}{\cos x})^2\geq 0, \forall x\in (0;\frac{\pi}{2})$

$\Rightarrow f(x)>f(0)=0, \forall x\in (0;\frac{\pi}{2}) \Rightarrow \sin x+\tan x>2x, \forall x\in (0;\frac{\pi}{2})$

c) Xét

$f(x)=2\sin x+\tan x-3x, \forall x\in (0;\frac{\pi}{2})$

$f'(x)=2\cos x+\frac{1}{\cos^2 x}-3=\cos x+\cos x+\frac{1}{\cos^2 x}-3$

$>3\sqrt[3]{\cos x.\cos x.\frac{1}{\cos^2 x}}-3=0, \forall x\in (0;\frac{\pi}{2})$

$\Rightarrow f(x)>f(0)=0, \forall x\in (0;\frac{\pi}{2})$

$\Rightarrow 2\sin x+\tan x>3x, \forall x\in (0;\frac{\pi}{2})$

1 Đáp án