Sunday, August 29, 2021

Find all polynomials $P(x)$ such that: $P(x-1).P(x+1)=P(x^2-1)$

Find all polynomials $P(x)$ such that: $P(x-1).P(x+1)=P(x^2-1)$

Solution

Suppose that $\alpha$ is a root of $P(x)$ then, $P(\alpha)=0$

Therefore, $P((\alpha+1)-1)=0$

Find the last two digits of : $N=(1!+2!+3!+.......+101!)^101$

Note: This is equivalent to finding $N(mod11)$ .

ie: The remainder when dividing $N$ by $100$ .

Observation: $10!\equiv0(mod100)$ Because, $10!=10...5...2$

Therefore, $N\equiv(1!+2!+3!+......+9!)^101(mod100)$

$N\equiv(1+2+6+24+20+20+40+20+80)^101(mod100)$

$N\equiv13^101(mod100)$

Prove that: $A_n=3^(n+3)-4^(4n+2)$ is divided by $11$

Solution

We will prove by Mathematics Induction:

If $n=0$ then $A_n=3^3-4^2=27-16=11$ It is true that $A_n$ is divided with $11$

Assuming that: $A_n$ is divided with $11$ for $n=k$ $(1)$

We will prove it is true for $n=k+1$ then $A_(k+1)$ is divided with $11$

Prove that: $forall x\inR$ : $:|acosx+bsinx|\leqsqrt(a^2+b^2)$

a. Prove that: $forall x\inR$ : $:|acosx+bsinx|\leqsqrt(a^2+b^2)$

b. Find the maximum and minimum of $f(x)=20cosx+21sinx+27$

Solution

a. Prove $|acosx+bsinx|\leqsqrt(a^2+b^2)$ $forall x\inR$

Assuming we have:

$acosx+bsinx=sqrt(a^2+b^2)(a/sqrt(a^2+b^2)cosx+b/sqrt(a^2+b^2)sinx)$

$X_1;X_2$ are the root of : $X^2-(2cost+3sint)X-11sin^2t=0$ Find the minimum of $A=X_1^2+X_1X_2+X_2^2$ .

$X_1;X_2$ are the root of : $X^2-(2cost+3sint)X-11sin^2t=0$

Find the minimum of $A=X_1^2+X_1X_2+X_2^2$ .

Solution

We can see that: $A=X_1^2+X_1X_2+X_2^2$

$=(X_1+X_2)^2-X_1X_2$

Following Vieta's Formulas $X^2-SX+P=0$ $(1)$

Find all Polynomials $P(x)$ which is satisfied that: $(x-2010)P(x+67)=xP(x)$ (2010 Baltic Way)

Find all Polynomials $P(x)$ which is satisfied that:

$(x-2010)P(x+67)=xP(x)$

(2010 Baltic Way)

Prove that: $tan^3x/(1-3tan^2x)=1/8(tan3x-3tanx)$

a. Prove that: $tan^3x/(1-3tan^2x)=1/8(tan3x-3tanx)$

As we knew: $tan3x=(3tanx-tan^3x)/(1-3tan^2x)$

$=(3tanx-9tan^3x+8tan^3x)/(1-3tan^2x)$

$=(8tan^3x+3tanx(1-3tan^2x))/(1-3tan^2x)$

$=(8tan^3x)/(1-3tan^2x)+3tanx$

Then, $tan3x-3tanx=(8tan^3x)/(1-3tan^2x)$

Find the limit of $S_n=2/(1.3)+2/(3.5)+2/(5.3)+.......+2/((2n+1)(2n+3))$

Solution

We can see that general term of this sequence is:

$2/((2k+1)(2k+3))=1/(2k+1)-1/(2k+3)$

If $g(x)=f(x)+1-x$ , find the value of $g(2020)$

It is given function $f(x)$ determine on $R$ , satisfied that:

$f(1)=1$

$f(x+5)\geqf(x)+5$

$f(x+1)\leqf(x)+1$

If $g(x)=f(x)+1-x$ , find the value of $g(2020)$ $\forallx,y\inR$

If $(1+2x+3x^2)^10=a_0+a_1x+a_2x^2+.......+a_(20)x^(20)$ then find the coefficients of $a_1$ ; $a_2$ ; $a_3$ and $a_20$

Solution

From our hypothesis we already had:

$(1+2x+3x^2)^10=a_0+a_1x+a_2x^2+.......+a_(20)x^(20)$

We will find the coefficients of $a_1$ ; $a_2$ ; $a_3$ and $a_20$

Which meant they are the coefficients of $x$ ; $x^2$ ; $x^3$ and $x^(20)$ .

We can rewrite $(1+2x+3x^2)^10=[1+x(2+3x)]^10$

We will use the Newton's Formula

$(a+b)^n$ = $\sum_{i=0}^nC(n,i)$ $a^(n-i).b^i$

Then, $[1+x(2+3x)]^10=C(10,0)+C(10,1)x(2+3x)+C(10,2)x^2.(2+3x)^2$

$+C(10,3)x^3.(2+3x)^3+....+C(10,10)x^10.(2+3x)^10$

We observe that:

$Coef(x)=2C(10,1)=20$ then, $a_1=20$

$Coef(x^2)=2^2C(10,2)=4.45=180$ then, $a_2=180$

$Coef(x^3)=12C(10,2)+8C(10,3)=45.12+120.8=1500$ then $a_3=1500$

$Coef(x^20)=C(10,10).3^10=3^10$ then, $a_20=3^10$

Prove $(21n+4)/(14n+3)$ Is Irreducible For Every Natural Number $n$

Prove

$(21n+4)/(14n+3)$ Is Irreducible For Every Natural Number

$n$ .

Solution

Solution 01

Denoting the greatest common divisor (GCD) of $a$ and $b$ $(a,b)$ and we will use Euclidean Algorithm Theory.

$(21n+4,14n+3)=(7n+1,14n+3)=(7n+1,1)=1$

It follows that $(21n+4)/(14n+3)$ is irreducible. Q.E.D

Solution 02

Vietnam Math Out Standing Student 2012-13

Problem 01: Solve the equation of:

$x^(4n)$ +

$\sqrt{x^{2n}+2012}$

$=2012$ for all

$n\in N$ .

Problem 02: It is given

$(U_n)$ which is determined by:

$U_1=3$

$U_(n+1)=1/3(2U_n+3/U_n^2)$ ; for all

$n\in N$ .

Let's finding the limit of:

$\lim_{n\rightarrow\infty}(u_n)$ .

Problem 03: It is given three non-negative real numbers

$x,y,z$ , Prove that:

$1/x+1/y+1/z$

$>36/(9+x^2y^2+y^2^2+z^2x^2)$ .

Problem 04: Find roots positive numbers of equation:

$\sqrt{x+2\sqrt3}=\sqrt y+\sqrt z$

Solution

Find all functions $f:R\rightarrow R$ such that: $f(x)f(y)=f(xy-1)+xf(y)+yf(x)$ $\forall$ $x,y$ $\inR$

Find all functions $f: R \rightarrow R$ Such that:
$f(x)f(y)=f(xy-1)+xf(y)+yf(x)$ $\forall$ $x ,y$ $\inR$

Solution

Replace :

$y=0$ we will see that:

$f(x).f(0)=f(-1)+x.f(0)$

Notice that:

$f(0)$ not equal to

$0$ then,

$f(x)=x+c$ which

$c$ is a constant. We replaced to our hypothesis unsatisfactory. Then,

$f(0)=0$ from that we can see that:

$f(-1)=0$ .

1970 IMO Problems And Solutions

Problem 01

Let $M$ be a point on the side $AB$ of $\triangle ABC$ . Let $r_1, r_2$ , and $r$ be the inscribed circles of triangles $AMC, BMC$ , and $ABC$ . Let $q_1, q_2$ , and $q$ be the radii of the exscribed circles of the same triangles that lie in the angle $ACB$ . Prove that

$\frac{r_1}{q_1} \cdot \frac{r_2}{q_2} = \frac{r}{q}$ .

Solution

1979 IMO Problems And Solutions

Problem 01

If $p$ and $q$ are natural numbers so that $\frac{p}{q}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+ \ldots -\frac{1}{1318}+\frac{1}{1319},$ Prove that $p$ is divisible with $1979$ .

Solution

1963 IMO Problems And Solutions

Problem 01

Find all real roots of the equation

$\sqrt{x^2-p}+2\sqrt{x^2-1}=x$ ,

where $p$ is a real parameter.

Solution

Problem 01

Real numbers $a_1, a_2, \dots , a_n$ are given. For each $i$ ( $1\le i\le n$ ) define

$d_i=\max\{a_j:1\le j\le i\}-\min\{a_j:i\le j\le n\}$

and let

$d=\max\{d_i:1\le i\le n\}$ .

(a) Prove that, for any real numbers $x_1\le x_2\le \cdots\le x_n$ ,

$\max\{|x_i-a_i|:1\le i\le n\}\ge \dfrac{d}{2} (*)$

(b) Show that there are real numbers $x_1\le x_2\le x_n$ such that equality holds in (*)

Solution

Problem 01

Given any set $A = \{a_1, a_2, a_3, a_4\}$ of four distinct positive integers, we denote the sum $a_1 +a_2 +a_3 +a_4$ by $s_A$ . Let $n_A$ denote the number of pairs $(i, j)$ with $1 \leq i < j \leq 4$ for which $a_i +a_j$ divides $s_A$ . Find all sets $A$ of four distinct positive integers which achieve the largest possible value of $n_A$

Solution

Problem 01

Let $\mathbb{Z}$ be the set of integers. Determine all functions $f : \mathbb{Z} \to \mathbb{Z}$ such that, for all integers $a$ and $b$ , $f(2a) + 2f(b) = f(f(a + b)).$

Solutions

Solution 1

Let us substitute $0$ in for $a$ to get $f(0) + 2f(b) = f(f(b)).$

Now, since the domain and range of $f$ are the same, we can let $x = f(b)$ and $f(0)$ equal some constant $c$ to get $c + 2x = f(x).$

Pages

Sunday, August 29, 2021

Saturday, August 28, 2021

Thursday, July 15, 2021

Wednesday, July 14, 2021

Monday, July 12, 2021

Thursday, July 8, 2021

Wednesday, July 7, 2021

Tuesday, July 6, 2021

Vietnam Math Out Standing Student 2012-13

Solution

Monday, July 5, 2021

Find all functions f: R \rightarrow Rf: R \rightarrow R Such that:f(x)f(y)=f(xy-1)+xf(y)+yf(x)f(x)f(y)=f(xy-1)+xf(y)+yf(x) \forall\forall x ,y x ,y \inR\inR

Solution

Sunday, April 4, 2021

Problem 01

Solution

Wednesday, March 31, 2021

Problem 01

Solution

Monday, March 29, 2021

Problem 01

Solution

Problem 01

Solution

Problem 01

Solution

Problem 01

Solutions

Solution 1

Find all functions $f: R \rightarrow R$ Such that:
$f(x)f(y)=f(xy-1)+xf(y)+yf(x)$ $\forall$ $x ,y$ $\inR$