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1. The greatest integer function, $$\lfloor x\rfloor$$, denotes the largest integer less than or equal to x. For example, $$\lfloor3.5\rfloor=3$$, $$\lfloor\pi\rfloor=3$$ and $$\lfloor -\pi\rfloor=-4$$. Find the sum of the three smallest positive solutions to $$x-\lfloor x\rfloor=\frac1{\lfloor x\rfloor}.$$ Express your answer as a mixed number.

2. Let g(x) be a function piecewise defined as

$$g(x) = \left\{ \begin{array}{cl} -x & x\le 0, \\ 2x-41 & x>0. \end{array} \right.$$

If a is negative, find a so that $$g(g(g(10.5)))=g(g(g(a))).$$

Thank you

Sep 16, 2019

#1
+23850
+1

1.
The greatest integer function, $$\lfloor x\rfloor$$r, denotes the largest integer less than or equal to $$x$$. For example, $$\lfloor3.5\rfloor=3$$, and $$\lfloor -\pi\rfloor=-4$$.

Find the sum of the three smallest positive solutions to $$x-\lfloor x\rfloor=\frac1{\lfloor x\rfloor}$$.

$$\text{Let  \text{frac}(x) = x - \lfloor x\rfloor  } \\ \text{Let  x = \lfloor x\rfloor + \text{frac}(x)  }$$

I assume:

$$\begin{array}{|rcll|} \hline x-\lfloor x\rfloor &=& \frac1{\lfloor x\rfloor} \quad | \quad x - \lfloor x\rfloor = \text{frac}(x) \\ \text{frac}(x) &=& \frac1{\lfloor x\rfloor} \\ \mathbf{\lfloor x\rfloor \cdot \text{frac}(x)} &=& \mathbf{1} \\ \hline \end{array}$$

$$\begin{array}{|rclcl|c|} \hline \mathbf{\lfloor x\rfloor } &\cdot& \mathbf{\text{frac}(x)} &=& \mathbf{1}& x = \lfloor x\rfloor + \text{frac}(x) \\ \hline 2 &\cdot& \dfrac{1}{2} &=& 1 & 2\dfrac{1}{2} \\ \hline 3 &\cdot& \dfrac{1}{3} &=& 1 & 3\dfrac{1}{3} \\ \hline 4 &\cdot& \dfrac{1}{4} &=& 1 & 4\dfrac{1}{4} \\ \hline 5 &\cdot& \dfrac{1}{5} &=& 1 & 5\dfrac{1}{5} \\ \hline 6 &\cdot& \dfrac{1}{6} &=& 1 & 6\dfrac{1}{6} \\ \hline &\ldots &&&&\ldots \\ \hline \end{array}$$

The sum of the three smallest positive solutions is $$2\dfrac{1}{2}+3\dfrac{1}{3}+4\dfrac{1}{4} = 10\dfrac{1}{12} = 10.08\overline{3}$$

Sep 17, 2019
#2
+1

THX :)

Sep 17, 2019