# Ngô Quốc Anh

## September 23, 2007

### “Hàm số” vs. “Số phức”

Filed under: Các Bài Tập Nhỏ — Ngô Quốc Anh @ 18:27

1) Let $z_{1},z_{2}\in\mathbb{C}$ such as $z_{1}= f(a)+if(b)$, $z_{2}= f(b)-f(a)i$ and also $\parallel z_{1}|-|z_{2}\parallel = |z_{1}|+|z_{2}|$, where $f$ is a differentiable function at $[a,b]$. Prove that exists $\xi_{1},\xi_{2}\in(a,b)$ such as $f'(\xi_{1})+f'(\xi_{2}) = 0$.

2) Let $f$ continuous function to $[a,b]$ such as $f(x)\neq0$ $\forall x\in[a,b]$. Also let $z\in\mathbb{C}$ such as $z+\frac{1}{z}= f(a)$ and $z^{2}+\frac{1}{z^{2}}= f^{2}(b)$. Prove that $i)|f(a)| > |f(b)|$ and $ii)$ the equation $f(b)+x^{3}f(a) = 0$ has at least one real root to $(-1,1)$.

3) Let $f$ differentiable function to $[a,b]$. Also let $z,w\in\mathbb{C}$ such as $|z-iw|^{2}= |z|^{2}+|iw|^{2}$ and $f(b)\neq0$ . Prove that exists $\xi\in (a,b)$ such as $\xi f'(\xi) = f(\xi)$.

### Một bài thi OLP’ SV

Filed under: Giải Tích 2 — Ngô Quốc Anh @ 14:55

Chứng minh rằng $\forall a\in\mathbb{R}$, $\int_{0}^{\infty}\frac{dx}{(1+x^{2})(1+x^{a})}=\frac{\pi}{4}$.

Chứng minh.

Using $x=\tan (\theta )$ $\int_{0}^{\infty }\frac{1}{\left(x^{2}+1\right)\left(x^{\alpha }+1\right)}\, dx=\int_{0}^{\frac{\pi }{2}}\frac{1}{\tan^{\alpha }(\theta )+1}\, d\theta$ $P_{1}=\int_{0}^{\frac{\pi }{2}}\frac{1}{\tan^{\alpha }(\theta )+1}\, d\theta =\int_{0}^{\frac{\pi }{2}}\frac{\cos^{\alpha }(\theta )}{\cos^{\alpha }(\theta )+\sin^{\alpha }(\theta )}\, d\theta$

Putting $\theta\to\frac{\pi }{2}-\theta$ $P_{2}=\int_{0}^{\frac{\pi }{2}}\frac{1}{\cot^{\alpha }(\theta )+1}\, d\theta =\int_{0}^{\frac{\pi }{2}}\frac{\sin^{\alpha }(\theta )}{\cos^{\alpha }(\theta )+\sin^{\alpha }(\theta )}\, d\theta$

Of course $P_{1}=P_{2}$ $P_{1}+P_{2}=\int_{0}^{\frac{\pi }{2}}\frac{\cos^{\alpha }(\theta )}{\cos^{\alpha }(\theta )+\sin^{\alpha }(\theta )}\, d\theta+\int_{0}^{\frac{\pi }{2}}\frac{\sin^{\alpha }(\theta )}{\cos^{\alpha }(\theta )+\sin^{\alpha }(\theta )}\, d\theta =\int_{0}^{\frac{\pi }{2}}\frac{\cos^{\alpha }(\theta )+\sin^{\alpha }(\theta )}{\cos^{\alpha }(\theta )+\sin^{\alpha }(\theta )}\, d\theta$ $P_{1}+P_{2}=\int_{0}^{\frac{\pi }{2}}1\, d\theta =\frac{\pi }{2}=2 P_{1}=2 P_{2}$
And we are done.

### Tính 2 tích phân

Filed under: Giải Tích 2 — Ngô Quốc Anh @ 14:51

Tính các tích phân sau

1) ${\int_{0}^{\frac{\pi}{2}}\frac{x}{\sin x+\cos x+1}dx}$
2) ${\int_{0}^{\frac{\pi}{2}}\frac{x^{2}}{\sin x+\cos x+1}dx}$