MATHEMATICAL OLYMPIAD STUDENT

(third-2012)

Exercise 1. For $f(x)=2(x-1)-\arctan x,x\in\mathbb{R}$.
a) Prove that $f(x)=0$ have only a root $a\in(1,\sqrt{3})$.
b) Let $\{u_{n}\}_{n=1}^{\infty}$ be a sequence defined by$\left\{\begin{matrix} u_1=\dfrac{3}{2}& \\ u_n=1+\dfrac{1}{2}\arctan x,&n\geq1 \end{matrix}\right.$.
Prove that $\{u_{n}\}_{n=1}^{\infty}$ converges to $a$.
Exercise 2. Let $f:[0,+\infty)\longrightarrow \mathbb{R}$ be a differentiable function such that $f(0)=1$. Prove that if $f'(x)\geq f(x)$ for all $x\in[0,+\infty)$, then the function $g(x)=f(x)-e^x$ is a increasing function.
Exercise 3. Let $f:[0,1]\longrightarrow \mathbb{R}$ be a integrable function such that $\int_0^1xf(x)dx=0$. Prove that
$\int_0^1f^2(x)dx\geq 4(\int_0^1f(x)dx)^2$.
Exercise 4. Let $f:\mathbb{R}\longrightarrow \mathbb{R}$ be a twice differentiable, $g:\mathbb{R}\longrightarrow \mathbb{R}^+$ be a function such that $f(x)+f''(x)=-xg(x)f'(x)$ for all $x\in\mathbb{R}$. Prove that $f(x)$ is bounded.
Exercise 5. Let $P(x)=\sum_{i=0}^{n}a_ix^i$ with $a_n>0$ be a $n$ degrees polynomial and have distinct $n$ roots. Prove that the polynomial $Q(x)=(P(x))^2-P'(x)$ only have
a) distinct $n+1$ roots if $n$ is odd.
b) distinct $n$ roots if $n$ is even.
Exercise 6. Find al function $f:\mathbb{R}\longrightarrow \mathbb{R}$ satisfies
$f(x+y)\geq f(x).f(y)\geq e^{x+y}$ for all $x,y\in \mathbb{R}$.

Exercise 3. We have $\int_0^1f^2(x)dx\int_0^2(3x-2)^2dx\geq (\int_0^1f(x)(3x-2)dx)^2$.

Assumming $g(x)=ax+b$ such that $\int_0^1g(x)dx=\int_0^1xg(x)=1$. Hence, $g(x)=6x-2$.
We have $\int_0^1(f(x)-6x+2)^2dx\geq 0$, but $\int_0^1(f(x)-6x+2)^2dx=\int_0^1(f(x))^2dx-4$

Problem 2. We have $g(0)=0$ and $g(x)$ be a continuous. Integral both sides, we get
$\int_0^xg(y)dy=(\int_0^yf(t)dt)^{2012}\geq 0$.
Assign $h(x)=\int_0^xg(t)dt$. So, we have $h'(x)=g(x)$.
Because $g(x)$ be a nonincreasing, so $g(x)\leq g(0),\forall x\in[0,+\infty)$. Therefore
$h'(x)\leq 0,\forall x\in[0,+\infty)$ or $h(x)\leq h(0)=0,\forall x\in[0,+\infty)$
Implies
$h(x)=0,\forall x\in[0,+\infty)$.
Similarly for $x\in(-\infty,0]$. We have $h(x)=0$ or $g(x)=0,\forall x\in \mathbb{R}$.
When $\int_0^xf(t)dt=0,\forall x\in\mathbb{R}$. Easily, we prove that $f(x)=0$

problem 3. For all $\epsilon>0,\exists n_0$ such that $|2a_{n+1}-a_n-2012|<\epsilon,\forall n\geq n_0$. Hence,
$|2(a_{n+1}-2012)-(a_n-2012)|<\epsilon$
Implies
$|a_{n+1}-2012|<\dfrac{\epsilon}{2}+\dfrac{1}{2}|a_n-2012| <\dots<\epsilon(\dfrac{1}{2}+\dfrac{1}{4}+\dots+\dfrac{1}{2^{n-n_0}})=\epsilon(1-(\dfrac{1}{2} )^{n-n_0})<\epsilon$
We have $\lim a_n=2012$

Problem 2. Every $x\in[0,2]$, we have
$f(x)=f(x)-f(0)=f'(\theta_1)x,\theta_1\in(0,x)\geq-2x$
$f(x)-f(1)=f'(\theta_2)(x-1),\theta_2\in(x,1)\Rightarrow f(x)\geq 2x-1$.
So,
$\int_0^1f(x)dx=\int_0^\frac{1}{4}f(x)dx+\int_\frac{1}{4}^1f(x)dx
\geq \int_0^\frac{1}{4}-2xdx+\int_\frac{1}{4}^1 (2x-1)dx=\dfrac{1}{8}$
But $f(x)=\left\{\begin{matrix}
-2x &x\in[0,\frac{1}{4}] \\
2x-1&x\in[\frac{1}{4},1]
\end{matrix}\right.$
isn't continuous at $x=\dfrac{1}{4} $.

Exercise 5. We have $f(t)=\int_0^1f(x)dx+\int_0^txf'(x)dx+\int_t^1(x-1)f'(x)dx$, forall $t\in [0,1]$.
Therefore $|f(t)|\leq \int_0^1|f(x)|dx+t\int_0^t|f'(x)|dx+(1-t)\int_t^1|f'(x)|dx$.
Final, we choose $t=\dfrac{1}{2} $.

Exercise 4: Find all integrable function $f:[0,1]\rightarrow \mathbb{R}$ such that
$f(x)\leq \int_0^x t^{2012}f(t)dt , \forall x \in [0,1]$.

Problem 5. Assign $g(x)=e^x[f(x)-1]+1$. We have $g'(x)=e^x[f(x)-1+f'(x)]<0$. So,
$e[f(1)-1]+1=g(1)<g(0)=0$
If $f(x)+f'(x)<1$ then there don't exist $f(x)$.

Consider $F(x)=\dfrac{1}{x} \int_0^{x}f(t)dt$. We have $f(x)=(xF(x))'$ and
$\int_0^{1}f(x)dx=(xF(x))|_0^{1}=F(1)$
$2\int_{\dfrac{1}{4}}^{\dfrac{3}{4}}f(x)dx=\dfrac{3}{2}F(\dfrac{3}{4})-\dfrac{1}{2}F(\dfrac{1}{4}) $.
Implies
$3F(\dfrac{3}{4} )-F(\dfrac{1}{4} )=2F(1)$
$\Leftrightarrow F(\dfrac{3}{4})-F(\dfrac{1}{4})=2(F(1)-F(\dfrac{3}{4}))$
So, there exist $\theta_{1}\in (\dfrac{1}{4},\dfrac{3}{4})$ and $\theta_{2}\in (\dfrac{3}{4},1)$ such that
$F'(\theta_1)\dfrac{1}{2}=2.F'(\theta_2)\dfrac{1}{4}$
$\Rightarrow F'(\theta_1)= F'(\theta_2)$
we have $\theta\in (\theta_1,\theta_2)$ such that $F''(\theta)=0$ or
$\dfrac{2}{\theta^3} \int_0^{\theta}f(t)dt-\dfrac{2}{\theta^2}f(\theta)+\dfrac{1}{\theta}f'(\theta)=0 $
Next, assignment $H(x)=2\int_0^{x}f(t)dt-2xf(x)+x^2f'(x)$
$\Rightarrow H(0)=0, H(\theta)=0$
$\exists x_0\in (0,\theta):H'(x_0)=0$
But $H'(x)=x^2f''(x)$
Final $f''(x_0)=0$

Bạn phải hiểu đây là sân chơi để khẳng định mình, khẳng định trường mình. Còn được gì, thì ngoài vật chất ''những tấm bằng khen" (rất khó để có) và tiền, thì cái bạn có nữa là cơ hội giao lưu bạn bè về toán học. Đề đại số những năm gần không "khó" lắm. Với theo tôi cái khó của người này đôi khi là cái dễ của bản thân nữa. Thân chào bạn.

nếu hệ là nhừ thế này $\left\{\begin{array}{l}x^3+y=2\\y^3+x=2\end{array}\right. $ thì có thể giải như sau;
trừ hai vế pt ta có pt $(x-y)(x^2+y^2+xy-1)=0\Leftrightarrow$ $x=y$ hoặc $x^2+y^2+xy-1=0$
với x=y thì $x^3+x-2=0 \Leftrightarrow (x-1)(x^2+x+2)=0$
với $X^2+y^2+xy-1=0$ thì bó tay

đề đúng rồi