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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=/=-1 and x=/=2. Give your answer as the ordered pair (A,B).

 

2.

The function f(x) is defined for 1≤x≤5 as follows:
\(\[f(x) = \left\{ \begin{array}{cl} 2x + 1 & \text{if }1 \le x \le 2, \\ 7 - x & \text{if }2 < x \le 3, \\ 10 - 2x & \text{if }3 < x \le 4, \\ 10 - x & \text{if }4 < x \le 5. \end{array}\right.\]\)
Find all real numbers x such that f(x)=x.

If you find more than one answer, list them all, separated by commas.

 

3.

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 \( |a - n|?\)

 

4.

The function \(h(x)\) is defined as:
\(\[h(x) = \left\{ \begin{array}{cl} \lfloor 4x \rfloor & \text{if } x \le \pi, \\ 3-x & \text{if }\pi < x \le 5.2, \\ x^2& \text{if }5.2< x. \end{array}\right.\]\)
Find \(h(h(\sqrt{2}))\)

Guest Mar 16, 2018
edited by Guest  Mar 16, 2018
 #1
avatar+26720 
+2

Here's #1 for you:

 

Collect the terms on right hand side over a common denominator:

 

\(\frac{A}{x-2}+\frac{B}{x+1}\rightarrow\frac{A(x+1)+B(x-2)}{(x-2)(x+1)}\rightarrow\frac{(A+B)x+A-2B}{x^2-x-2}\)

 

Compare this with \(\frac{x+7}{x^2-x-2}\)

 

To make them the same compare the numerators.  It is clear we must have:

 

A + B = 1   and

A - 2B = 7

 

Subtract the second from the first to get: 3B = -6  so B = -3

Put this back into the first to get A - 3 = 1  so  A = 4

 

Hence (A, B) = (4, -3)

 

For #2  

Set x = each term in turn and solve for x.  if the result lies in the range given for each term then that result is valid.  If not it isn't.

 

For #3

Can't see the whole expression!

 

For #4

First plug sqrt(2) into h to get, say m = h(sqrt(2))

Then plug m into h to get h(m) which is h(h(sqrt(2))) 

.

Alan  Mar 17, 2018
edited by Alan  Mar 17, 2018
edited by Alan  Mar 17, 2018

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