Bài 107335

Question

Cho

$\begin{cases} n\in Z,n\geq 2\\ 0<x<1 \end{cases}$

Chứng minh rằng:

$x^{n}\sqrt{1-x}<\frac{1}{\sqrt{2ne}}$ (với

$e=\mathop {\lim }\limits_{k \to +\infty }(1+\frac{1}{k})^{k}$ )

huyenhana164 · Answer 1 · 6/8/2012 10:14:31 AM

Xét:

$f(x)=x^{2n}(1-x)=x^{2n}-x^{2n+1},x\in (0,1)$

$f'(x)=2n.x^{2n-1}-(2n+1)x^{2n}$ (vì

$x\in (0,1)$ )

$f'(x)=0\Leftrightarrow x=\frac{2n}{2n+1}$

Bảng biến thiên:

$\Rightarrow x^{2n}(1-x)\leq (\frac{2n}{2n+1})^{2n}.\frac{1}{2n+1},\forall x\in (0,1)$

Do đó để chứng minh:

$x^{n}\sqrt{1-x}<\frac{1}{\sqrt{2ne}} \Leftrightarrow x^{2n}(1-x)<\frac{1}{2ne}$

Ta sẽ chứng minh:

$(\frac{2n}{2n+1})^{2n}.\frac{1}{2n+1}<\frac{1}{2ne}$

$\Leftrightarrow (1+\frac{1}{2n})^{2n+1}>e$

$\Leftrightarrow (1+\frac{1}{k})^{k+1}>e (k=2n>0) (*)$

Xét:

$g(x)=\ln x,x>0$

$g'(x)=\frac{1}{x}$

Theo định lý Lagrange:

$\exists c\in (k,k+1)$ :

$g(k+1)-g(k)=g'(c).(k+1-k)=g'(c)=\frac{1}{c}>\frac{1}{k+1}$

$\Rightarrow \ln \frac{k+1}{k}>\frac{1}{k+1}$

$\Rightarrow \ln (1+\frac{1}{k})^{k+1}>1$

$\Rightarrow (1+\frac{1}{k})^{k+1}>e$

$\Rightarrow (*)$ đúng.

$\Rightarrow$ (ĐPCM)

1 Đáp án