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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
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4+0 Answers

 #1
avatar+78618 
+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

 

 

 

cool cool cool

CPhill  Aug 26, 2017
edited by CPhill  Aug 26, 2017
edited by CPhill  Aug 26, 2017
edited by CPhill  Aug 26, 2017
 #2
avatar+1364 
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 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
edited by TheXSquaredFactor  Aug 26, 2017
edited by TheXSquaredFactor  Aug 26, 2017
 #3
avatar+78618 
+1

 

Sorry, X2......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 :

 

 

 

 

 

cool cool cool

CPhill  Aug 26, 2017
 #4
avatar+1364 
+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

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