1. Find all values of t such that \(\lfloor t\rfloor = 2t + 3.\) If you find more than one value, then list the values you find in increasing order, separated by commas.

2. 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).

3.Let \(f(x) = \left\lfloor\dfrac{2 - 3x}{x + 5}\right\rfloor\). Evaluate \(f(1)+f(2) + f(3) + \cdots + f(999)+f(1000).\) (This sum has 1000 terms, one for the result when we input each integer from 1 to 1000 into f.)

MIRB16 Aug 26, 2017

#1**+1 **

3. This is a "floor" function.....it returns the greatest integer less than or equal to f(x)

f(1) = floor ( -1/6) = -1

f(1000) = (-2998/1005) ≈ -2.98 ...so the floor of this = -3

We need to find the "break" points

We need to first determine what integer value [approximately] makes (2 -3x) / ( x + 5) = -1

So we have

(2-3x) / (x + 5) = -1

(2 - 3x) = -1(x + 5)

2 -3x = -x -5

7 = 2x

3.5 = x

Note that f(3) produces floor (-7/ 8) = -1

And f(4) produces floor (-10/9) = -2

Next.....we need to determine what integer value [ approximately] makes (2- 3x) / ( x + 5) = -2

So we have

(2 - 3x) / ( x + 5) = -2

2 - 3x = -2(x + 5)

2 -3x = -2x -10

12 = x

Note that f(12) produces floor (-34/17) = -2

And f(13) produces floor (-37/ 18) = -3

So...from f(1) to f(3)...we will have the sum 3(-1) = -3

And from f(4) to f(12) we will have the sum 9(-2) = -18

And from f(13) to f(1000) we will have the sum 988(-3) = -2964

So....the sum of the series = (-3 ) + ( -18 ) + ( -2964 ) = -2985

This graph shows the break points :

https://www.desmos.com/calculator/q95fsnkc0t

CPhill Aug 26, 2017

#2**0 **

The equation \(\left \lfloor{t}\right \rfloor=2t+3\) can be a tad difficult to parse. Sometimes, it is better to think of an easier problem before attempting one a harder one. Let's think about a basic equation that contains the floor function.

\(\left \lfloor{t}\right \rfloor=5\)

Before we can solve this, though, we have to understand the floor function. The floor function truncates any decimal. The solution set for this simplified equation, represented in a compound inequality, is \(5\leq t<6\). Any value for *t *that satisfies this restriction is a solution. We can use this logic to generalize the floor function, actually to \(\left \lfloor{x}\right \rfloor=a\rightarrow a\leq x < a+1\) . However, you must have caution when doing this, however. Not every solution is actually a solution. I'll touch on this later. Knowing this will be key to solving this equation. Knowing this, the floor function transforms from \(\left \lfloor{t}\right \rfloor=2t+3\) to \(2t+3\leq t<2t+3+1\). Let's solve this compound inequality. I'll solve each one separately.

\(2t+3\leq t\) | Subtract 2t from both sides. |

\(3\leq-t\) | Divide by -1 on both sides. |

\(-3\geq t\) | Of course, the inequality flips when dividing by a negative number. |

\(t\leq -3\) | |

Now, let's solve for the other inequality \(t<2t+3+1\):

\(t<2t+3+1\) | Subtract 2t from both sides. |

\(-t<4\) | Divide by -1 on both sides. Yet again, remember to flip the inequality sign. |

\(t>-4 \) | |

Of course, all solutions must adhere to both restrictions. Both inequalities show that t must be greater than -4 and less than or equal to -3. -3 is automatically a solution because it is an integer.

We aren't done yet, though. Are there are other solution that adhere to our current restriction that makes \(2t+3\) an integer? We don't even have to worry about the +3 because the sum of any integer and 3 will always yield another integer. Let's set this equal to 0:

\(2t+3=0\) | Subtract 3 on both sides. |

\(2t=-3\) | Divide by 2 on both sides. |

\(t=-\frac{3}{2}\) | |

What I have demonstrate here is that a number divided by 2 will make 2t+3 an integer. The only one that meets our description and our current restriction is -(7/2).

Therefore, there are 2 solutions to this equation

\(t_1=-\frac{7}{2}\)

\(t_2=-3\)

.TheXSquaredFactor Aug 26, 2017

#3**+1 **

Sorry, X^{2}......I think you made a slight mistake.....your solution of -3 is correct, but not -3/2

Floor (-3/2) = -2

But

2(-3/2) + 3 = 0

The other solution is t = -7/2

Floor (-7/2) = Floor (-3.5) = -4

And

2(-3.5) + 3 =

-7 + 3 = -4

Here's a graph of the solution points :

CPhill Aug 26, 2017

#4**+1 **

Yes, Cphill, I must concede to you and your wolframalpha...

What I was trying to do was find a number in the expression \(\frac{a}{2}\) such that\(a\in \mathbb{Z}\hspace{1mm}\text{and}\hspace{1mm}-4<\frac{a}{2}\leq -3\). And of course, the only two numbers that work is -7 and -6.

TheXSquaredFactor
Aug 26, 2017