1. Find the greatest integer x such that \(|x^2 - 98x+1| = 98x-x^2-1.\)
2. Find the greatest possible value for \( \frac{\sqrt{\lfloor x \rfloor}}{\lfloor \sqrt{x} \rfloor},\)where \(x \ge 1. \)
3. Let \\(f(x) = \left\{ \begin{array}{cl} ax+3, &\text{ if }x>2, \\ x-5 &\text{ if } -2 \le x \le 2, \\ 2x-b &\text{ if } x <-2. \end{array} \right.\)Find a+b if the piecewise function is continuous (which means that its graph can be drawn without lifting your pencil from the paper).
Thank you for the help!
3.
We have that
2(-2)- b = -2 - 5 and a(2) + 3 = 2 - 5
-4- b = -7 2a + 3 = -3
b = 3 2a = -6
a = -3
Here's the graph : https://www.desmos.com/calculator/avrddz9mrw
And a + b = 0
1.
Find the greatest integer x such that \(|x^2 - 98x+1| = 98x-x^2-1\).
\(\begin{array}{|rclrll|} \hline \mathbf{|x^2 - 98x+1|} &=& \mathbf{98x-x^2-1} \\\\ |x^2 - 98x+1| &=& -\underbrace{(x^2 - 98x+1)}_{<0!} \\\\ && x^2 - 98x+1 &<& 0 \\ && \left(x-\dfrac{98}{2}\right)^2 -\dfrac{98^2}{4} + 1 &<& 0 \\ && \left(x-\dfrac{98}{2}\right)^2 &<& \dfrac{98^2}{4} - 1 \\ && \left(x-\dfrac{98}{2}\right)^2 &<& \dfrac{98^2-4}{4} \\ && x-\dfrac{98}{2} &<& \dfrac{\sqrt{98^2-4}}{2} \\ && x &<& \dfrac{98}{2} +\dfrac{\sqrt{98^2-4}}{2} \\ && x &<& \dfrac{98+\sqrt{98^2-4}}{2} \\ && x &<& 97.9897948557 \\ &&\mathbf{ x_{\text{max}}} &=& \mathbf{97} \\ \hline \end{array} \)