+249 1. Let \(f(x) = \left\lfloor\dfrac{2 - 3x}{x + 5}\right\rfloor\). Evaluate \(F (1) + f (2) + f (3) + ... + f (999) + f (1000)\)
(This sum has 1000 terms, one for the result when we input each integer from 1 to 1000 into \(f\).)
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. 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 \((A, B)\).
+23252 LEt me try to answer question #3:
(x + 7) / (x2 - x - 2) = A / (x - 2) + B / (x + 1)
Since x2 - x - 2 = (x - 2)(x + 1)
---> (x + 7) / (x2 - x - 2) = (x + 7) / [ (x - 2)(x + 1) ] =
(x + 7) / [ (x - 2)(x + 1) ] = A / (x - 2) + B / (x + 1)
Writing with the common denominator of (x - 2)(x + 1)
(x + 7) / [ (x - 2)(x + 1) ] = [ A(x + 1) ] / [ (x - 2)(x + 1) ] + [B(x - 2) ] / [ (x - 2)(x + 1) ]
Since the denominators are equal, they can be ignored:
x + 7 = A(x + 1) + B(x - 2)
x + 7 = Ax + A + Bx - 2B
x + 7 = Ax + Bx + A - 2B
Set the x-term on the left side equal to the x-terms on the right side: x = Ax + Bx
---> x = (A + B)x ---> 1x = (A + B)x ---> 1 = A + B
Set the number term on the left side equal to the number terms on the right side: 7 = A - 2B
Solving: A + B = 1 ---> x -1 ---> -A - B = -1
A - 2B = 7 ---> A - 2B = 7
Adding down the columns: -3B = 6
B = -2
Since B = -2: A + B = 1 ---> A + -2 = 1 ---> A = 3
+23252 Let me try to answer question #2:
Set each part of f(x) = x for the domain of that part.
1) 2x + 1 = x for 1 <= x <= 2
x + 1 = 0
x = -1 Since this value is not within 1 <= x <= 2, there is no answer for this section.
2) 7 - x = x for 2 < x <= 3
7 = 2x
x = 3.5 Since this value is not within 2 < x <= 3, there is no answer for this section.
3) 10 - 2x = x for 3 < x <= 4
10 = 3x
x = 3 1/3 This is within the domain, so this is part of the answer.
4) 10 - x = x for 4 < x <= 5
10 = 2x
x = 5 This is within the domain, so this is part of the answer.
Answer: 3 1/3, 5
+23252 LEt me try to answer question #3:
(x + 7) / (x2 - x - 2) = A / (x - 2) + B / (x + 1)
Since x2 - x - 2 = (x - 2)(x + 1)
---> (x + 7) / (x2 - x - 2) = (x + 7) / [ (x - 2)(x + 1) ] =
(x + 7) / [ (x - 2)(x + 1) ] = A / (x - 2) + B / (x + 1)
Writing with the common denominator of (x - 2)(x + 1)
(x + 7) / [ (x - 2)(x + 1) ] = [ A(x + 1) ] / [ (x - 2)(x + 1) ] + [B(x - 2) ] / [ (x - 2)(x + 1) ]
Since the denominators are equal, they can be ignored:
x + 7 = A(x + 1) + B(x - 2)
x + 7 = Ax + A + Bx - 2B
x + 7 = Ax + Bx + A - 2B
Set the x-term on the left side equal to the x-terms on the right side: x = Ax + Bx
---> x = (A + B)x ---> 1x = (A + B)x ---> 1 = A + B
Set the number term on the left side equal to the number terms on the right side: 7 = A - 2B
Solving: A + B = 1 ---> x -1 ---> -A - B = -1
A - 2B = 7 ---> A - 2B = 7
Adding down the columns: -3B = 6
B = -2
Since B = -2: A + B = 1 ---> A + -2 = 1 ---> A = 3
+129899 (1) floor [ (2 - 3x) / ( x + 5) ]
Note f(1) = -1 and f(1000) = -3
f(1) = -1
f(2) = -1
f(3) = -1
f(4) = - 2
Now, we need to find out where the function = exactly -2
Solving this
[ (2 - 3x) / ( x + 5) ] = -2, we have x = 12
So.....from f(4) to f(12), the function takes on a floor value of -2 for each successive integer value of x
And from f(13) to f(1000) the function takes on a floor value of -3 for each successive integer value of x
So
f(1) + f(2) + f(3) + ... + f(999) + f(1000) =
3(-1) + 9(-2) + 988(-3) =
-2985
