scc404 wrote:
For each value of \(y\) greater than \(2 \sqrt{3}\), the function \(f(x)\) is such that the equation \(f(x) = y\) has a solution of the form \(x=\frac{y^2+12}{y}\)
Select one value for \(a\) and one value for \(b\) such that the given information implies \(f(a) = b\). Make only two selections, one in each column.
Some questions of DI have odd wording but you can figure out what to do by reading what is given and what is asked.
I struggled to understand what this means:
the function \(f(x)\) is such that the equation \(f(x) = y\) has a solution of the form \(x=\frac{y^2+12}{y}\)but I ignored it. I could see that there is a relation between x and y if f(x) = y and that's all I cared about because I was given f(a) = b. This meant that the relation will exist for a and b
\(a=\frac{b^2+12}{b} = b + \frac{12}{b}\)
y needs to be greater than 2*sqrt3 which means that b needs to be greater than 2*1.7 = 3.4 approx.
If b = 4, a = 4 + 3 = 7 (Not available in options)
If b = 6, a = 6 + 2 = 8 - Available
ANSWER: 8, 6