My Math Forum Iranian college post graduation course entrance exams

 Math Events Math Events, Competitions, Meetups - Local, Regional, State, National, International

November 16th, 2010, 05:00 AM   #1
Newbie

Joined: Oct 2010

Posts: 6
Thanks: 0

Iranian college post graduation course entrance exams

Quote:
 Originally Posted by skipjack Using high school calculus, the substitution t ? t/a gives $\int_1\,^x\frac{e^{at}}{t}dt\,=\int_a\,^{ax}\frac{ e^{t}}{t/a}\cdot\frac{1}{a}dt\,=\int_a\,^{ax}\frac{e^{t}}{t }dt\,=\,F(ax)\,-\,F(a).$

January 20th, 2011, 08:18 AM   #2
Global Moderator

Joined: Dec 2006

Posts: 20,924
Thanks: 2203

Quote:
 Originally Posted by Saeid 36- Which substitution will make the equation y' + p(x)y = q(x)y^4 a linear equation?
Multiplying the equation by -3y^(-4) gives -3y^(-4)y' + -3p(x)y^(-3) = -3q(x), so the substitution z = y^(-3) will produce a linear equation.

 January 24th, 2011, 04:44 AM #3 Newbie   Joined: Oct 2010 Posts: 6 Thanks: 0 Re: Iranian college post graduation course entrance exams If you can solve any question, feel free to post your solution. Be aware that these problems are tricky and wrong answers are among choices!! There might be some translation and or type mistakes, If you have a better translation for some questions, your suggestion is welcome. This page is available in pgexam.e-rasht.com/civil2010.html
 January 24th, 2011, 05:18 AM #4 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: Iranian college post graduation course entrance exams 44) $\sum_{n=1}^{\infty}\frac{1}{n(n+4)}$ Use of partial fraction decomposition gives: $\frac{1}{4}\sum_{n=1}^{\infty}$$\frac{1}{n}-\frac{1}{n+4}$$=\frac{1}{4}$$\sum_{n=1}^{4}\frac{1 }{n}+\sum_{n=1}^{\infty}\frac{1}{n+4}-\sum_{n=1}^{\infty}\frac{1}{n+4}$$=$ $\frac{1}{4}$$1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}$$=\frac{25}{48}$ 40) We want to find y' in terms of both x and y: $\frac{y}{x^2}=k$ Implicitly differentiate with respect to x: $\frac{x^2y'-y(2x)}{x^4}=0$ $\frac{y'}{x^2}-2\frac{y}{x^3}=0$ Multiply through by $x^2$ and solve for y': $y'=2\frac{y}{x}$ The family of orthogonal curves will then satisfy: $y'=-\frac{x}{2y}$ Separate variables and integrate: $2\int y\,dy=-\int x\,dx$ $y^2=-\frac{x^2}{2}+k$ Multiply through by 2 and rearrange: $x^2+2y^2=k$
 January 24th, 2011, 06:06 AM #5 Senior Member   Joined: Feb 2009 From: Adelaide, Australia Posts: 1,519 Thanks: 3 Re: Iranian college post graduation course entrance exams 41) L'Hôpital's rule 42) Rewrite $\lim_{x\to\infty}f(x)$ as $\lim_{x\to 0}f(\frac1x)$, simplify, and use L'Hôpital's rule.
 January 24th, 2011, 08:16 AM #6 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: Iranian college post graduation course entrance exams 45) $x^2+y^2+z^2=1$ $F(x,y,z)=z^2+y^2+z^2-1=0$ $x\frac{\delta z}{\delta x}=-x\frac{F_x(x,y,z)}{F_z(x,y,z)}=-x\frac{2x}{2z}=-\frac{x^2}{z}$ $y\frac{\delta z}{\delta y}=-y\frac{F_y(x,y,z)}{F_z(x,y,z)}=-y\frac{2y}{2z}=-\frac{y^2}{z}$ $x\frac{\delta z}{\delta x}+y\frac{\delta z}{\delta y}=-$$\frac{x^2}{z}+\frac{y^2}{z}$$=-$$\frac{x^2+y^2}{z}$$=$$\frac{z^2-1}{z}$$=z-\frac{1}{z}$ 43) $V=\pi\int_0\,^1 1^2-$$x^3$$^2\,dx=\pi$x-\frac{x^7}{7}$_0^1=\frac{6\pi}{7}$
 February 27th, 2011, 03:22 AM #7 Newbie   Joined: Oct 2010 Posts: 6 Thanks: 0 Re: Iranian college post graduation course entrance exams The web page containing the questions is removed; I have uploaded them again, sorry for the inconvenience! Here are the questions:

 Tags college, entrance, exams, graduation, iranian, post

 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post sara213 New Users 15 August 3rd, 2011 06:45 PM ZardoZ Math Events 1 May 22nd, 2011 07:09 PM jessmari86 Calculus 4 April 11th, 2011 04:55 PM Amir Kiani Number Theory 5 September 6th, 2009 11:56 PM eagles24548 Algebra 5 May 10th, 2009 02:28 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top