Answer to Question #348073 in Discrete Mathematics for Api

Question #348073

The set S consists of all integers from 1 to 2007 inclusive. For how many

elements n in S is f(n)=2n³+n²-n-2 over n²-2 an integer?


1
Expert's answer
2022-06-08T13:42:54-0400
2x3+x2x2x22\dfrac{2x^3+x^2-x-2}{x^2-2}

=2x(x22)+(x22)+3xx22=\dfrac{2x(x^2-2)+(x^2-2)+3x}{x^2-2}

=2x+1+3xx22=2x+1+\dfrac{3x}{x^2-2}

Given nZ+n\in \Z^+

n=1,2(1)+1+3(1)(1)22=0,0Zn=1, 2(1)+1+\dfrac{3(1)}{(1)^2-2}=0, 0\in \Z



n2,3xx221n\ge 2, \dfrac{3x}{x^2-2}\ge1

x23x2x220\dfrac{x^2-3x-2}{x^2-2}\le0

x23x2x220\dfrac{x^2-3x-2}{x^2-2}\le0

x23x2=0x^2-3x-2=0

x1=3172,x2=3+172x_1=\dfrac{3-\sqrt{17}}{2}, x_2=\dfrac{3+\sqrt{17}}{2}


x(2,3172](2,3+172]x\in(-\sqrt{2}, \dfrac{3-\sqrt{17}}{2}]\cup (\sqrt{2}, \dfrac{3+\sqrt{17}}{2}]

Since nn is positive integer, we consider n=2,3.n=2,3.


n=2,3nn22=3(2)(2)22=3,integern=2,\dfrac{3n}{n^2-2}=\dfrac{3(2)}{(2)^2-2}=3, integer

n=3,3nn22=3(3)(3)22=97,is not integern=3,\dfrac{3n}{n^2-2}=\dfrac{3(3)}{(3)^2-2}=\dfrac{9}{7}, is\ not \ integer

n=1,n=3.n=1,n=3.



Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!

Leave a comment