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1. Find constants $A$ and $B$ such that \[\frac{x + 7}{x^2 - x - 2} = \frac{A}{x - 2} + \frac{B}{x + 1}\] for all $x$ such that $x\neq -1$ and $x\neq 2$. Give your answer as the ordered pair $(A,B)$.

 

2. a) Suppose that \[|a - b| + |b - c| + |c - a| = 20.\] What is the maximum possible value of $|a - b|$?

b) Suppose that \[|a - b| + |b - c| + |c - d| + \dots + |m-n| + |n-o| + \cdots+ |x - y| + |y - z| + |z - a| = 20.\] What is the maximum possible value of $|a - n|$?

Guest Apr 23, 2018
 #1
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\( \frac{x + 7}{x^2 - x - 2} = \frac{A}{x - 2} + \frac{B}{x + 1}\)

 

We can use partial fraction decmposition to find A, B

Factoring the denominator on the left side  we have   (x - 2) ( x + 1)

Multiply through by this common denominator  and we have

 

x + 7  =  A(x + 1)  + B(x - 2)   simplify

 

x + 7  =  Ax + A  + BX  - 2B        equate coefficients  and we get this system

 

A + B  = 1

A - 2B  = 7       subtract the second equation from the first

 

3B  = -6     divide both sides  by 3

B  = -2

 

Using the first equation to  find A

A +  -2   = 1         add 2 to  both sides

A  = 3

 

So  (A, B )   =  ( 3, -2)

 

 

cool cool cool 

CPhill  Apr 23, 2018

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