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1  Revista, Técnicas, Cursos, Problemas / De oposición y olimpíadas / probability : 22/11/2019, 04:26:36 am
[texx]6[/texx] pair dice are thrown independently. The probability that there are exactly two different pairs, is

(like one example of [texx]2[/texx] different  pair is an ordered combination like [texx]2,2,1,2,5,6[/texx])
2  Revista, Técnicas, Cursos, Problemas / De oposición y olimpíadas / Inequality : 14/11/2019, 11:19:18 am
Let [texx]x_{1},x_{2},x_{3},\cdots, x_{n}[/texx] be [texx]n[/texx] distinct real number and [texx]n\geq 2[/texx] and [texx]x_{i}\in [-1,1]\;\forall i = 1,2 ,3,\cdots ,n.[/texx]

Then prove that [texx]\displaystyle \sum^{n}_{i=1}\frac{1}{p_{i}}\geq 2^{n-2}[/texx], where [texx]p_{i}=\prod_{j\neq i}|x_{j}-x_{i}|[/texx]
3  Revista, Técnicas, Cursos, Problemas / De oposición y olimpíadas / Re: functional equation : 14/11/2019, 11:14:34 am
Thanks Admin.
4  Revista, Técnicas, Cursos, Problemas / De oposición y olimpíadas / Re: Quadrilateral problem : 11/11/2019, 11:24:49 am
Thanks martiniano and Admin
5  Revista, Técnicas, Cursos, Problemas / De oposición y olimpíadas / functional equation : 11/11/2019, 11:23:41 am
If a function [texx]f[/texx] satisfies the relation [texx]f(x)f''(x)-f(x)f'(x)=(f'(x))^2\,\forall x \in\mathbb{R}[/texx]

and [texx]f(0)=f'(0)=1[/texx], then value of [texx]f(x)[/texx] is
6  Revista, Técnicas, Cursos, Problemas / De oposición y olimpíadas / Re: probability : 11/11/2019, 11:22:02 am
Thanks Admin and Richard.
7  Revista, Técnicas, Cursos, Problemas / De oposición y olimpíadas / probability : 08/11/2019, 03:36:48 am
Q: The sum of [texx]3[/texx] positive integer is [texx]20.[/texx] Then the probability that they form a Triangle is

i am assuming [texx]a+b+c=20,a,b,c \in \mathbb{N}[/texx] using wlog [texx]a\leq b \leq c[/texx]

for triangle [texx]a+b\geq c\Rightarrow a+b+c \geq 2c\Rightarrow c\leq 10[/texx] and [texx]a+b+c\leq 3c\Rightarrow 20\geq 3c\Rightarrow c\geq 7[/texx]

so range of [texx]7\leq c\leq 10[/texx]

For [texx]c=7.[/texx] we have [texx]a+b=13,[/texx] Then ordered pairs [texx](6,7)[/texx]

For [texx]c=8.[/texx] we have [texx]a+b=12,[/texx] Then ordered pairs [texx](6,6),(5,7),(4,8)[/texx]

For [texx]c=9.[/texx] we have [texx]a+b=11,[/texx] Then ordered pairs [texx](5,6),(4,7),(3,8),(2,9)[/texx]

For [texx]c=10.[/texx] we have [texx]a+b=10,[/texx] Then ordered pairs [texx](5,5),(4,6),(3,7),(2,8),(1,9)[/texx]

ordered pairs to form a triangle is [texx]13[/texx](favorable ways)

How do i calculate Total ways. please see it Thanks
8  Revista, Técnicas, Cursos, Problemas / De oposición y olimpíadas / Re: Polynomials : 08/11/2019, 03:25:11 am
Thanks Moderator got it
9  Revista, Técnicas, Cursos, Problemas / De oposición y olimpíadas / Polynomials : 07/11/2019, 10:34:31 am
If [texx]f[/texx] be a non zero polynomial such that [texx]f(1-x)=f(1+x)[/texx] for all real [texx]x[/texx]

And [texx]f(1)=0[/texx]. Then largest positive integer [texx]m[/texx] such that

[texx](x-1)^{m}[/texx] divides polynomial [texx]f(x)[/texx] for all polynomial [texx]f(x),[/texx] is
10  Revista, Técnicas, Cursos, Problemas / De oposición y olimpíadas / Re: Quadrilateral problem : 01/11/2019, 06:46:44 am
Thanks Admin Got it.

please explain me how i find [texx]|PC-PD|[/texx]
11  Revista, Técnicas, Cursos, Problemas / De oposición y olimpíadas / Quadrilateral problem : 22/10/2019, 11:32:27 am
Let [texx]A(z_{1}),B(z_{2}),C(Z_{3})[/texx] and [texx]D(z_{4})[/texx] are the vertices of an Trepezium in an Argand plane

Let [texx]|z_{1}-z_{2}|=4,|z_{3}-z_{4}|=10[/texx] and Diagonal [texx]AC[/texx] and [texx]BD[/texx] intersect

at [texx]P.[/texx] It is given that [texx]\displaystyle \arg\bigg(\frac{z_{4}-z_{2}}{z_{3}-z_{1}}\bigg)=\frac{\pi}{2}[/texx] and [texx]\displaystyle \arg\bigg(\frac{z_{3}-z_{2}}{z_{4}-z_{1}}=\frac{\pi}{4}\bigg).[/texx]

Then area of Triangle [texx]\displaystyle PCB[/texx] and [texx]|CP-DP|[/texx] is
12  Revista, Técnicas, Cursos, Problemas / De oposición y olimpíadas / Re: summation : 22/10/2019, 11:25:33 am
Thanks Admin Got it.
13  Matemática / Cálculo 1 variable / Re: Indefinite Integration : 14/10/2019, 10:16:50 am
Thanks Admin.
14  Revista, Técnicas, Cursos, Problemas / De oposición y olimpíadas / summation : 14/10/2019, 10:15:36 am
[texx]\displaystyle \sum^{n-1}_{i=0}\bigg(\sum^{n}_{j=i+1}\bigg(j\binom{n}{i}+i\binom{n}{j}\bigg)\bigg)=[/texx]
15  Revista, Técnicas, Cursos, Problemas / De oposición y olimpíadas / Re: Constant C : 14/10/2019, 10:14:50 am
Thanks Admin.
16  Matemática / Cálculo 1 variable / Indefinite Integration : 28/09/2019, 02:37:46 pm
[texx]\displaystyle\int \frac{2x+1}{(x^2+4x+1)\sqrt{x^2+4x+1}}dx[/texx]
17  Revista, Técnicas, Cursos, Problemas / De oposición y olimpíadas / Constant C : 28/09/2019, 02:31:11 pm
Find the largest constant [texx]C[/texx] for which [texx]\displaystyle \sum^{4}_{i=1}\bigg(x_{i}+\frac{1}{x_{i}}\bigg)^{3}\geq C[/texx] for all [texx]x_{1},x_{2},x_{3},x_{4}>0[/texx] such that [texx]x^{3}_{1}+x^{3}_{3}+3x_{1}x_{3}=x_{2}+x_{4}=1[/texx]
18  Revista, Técnicas, Cursos, Problemas / De oposición y olimpíadas / Re: Trigonometric Sum : 28/09/2019, 11:17:55 am
Thanks Admin and Moderator.
19  Revista, Técnicas, Cursos, Problemas / De oposición y olimpíadas / Trigonometric Sum : 20/09/2019, 04:39:27 am
 [texx]\displaystyle 4\sin \frac{\pi}{7}-\tan \frac{3\pi}{7}=[/texx]
20  Revista, Técnicas, Cursos, Problemas / De oposición y olimpíadas / Re: equation quadratic : 20/09/2019, 04:38:43 am
Thanks Admin it is Given for real numbers and answer given as for all odd integers [texx]n.[/texx]
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