Grades 11-12 Video Solutions 2013-Levels 11&12 Video Solutions 2013 problem6

Video Transcription

Question number six, let F be a linear function for which the value at 2013 minus the value at 2001 is equal to 100. What is the value of F at 2031 minus the value of F at 2013?  We will need to recall the linearity property of a function.  If we have a linear function, then we can apply that function to a combination AX plus BY where A and B are constants, X and Y are variables,  and the function behaves as follows.  This is equal to A times the value at X plus B times the value at Y. So supposing for a moment that A and B are 1 and negative 1 respectively,  we can simplify the information that's given to us considerably.  First, we need to compute F evaluated at 2031 minus F evaluated at 2013, which by linearity is F evaluated  at the difference of the inputs. And so that simplifies to F evaluated at 18. Similarly, we make a computation with 2013 and 2001  and obtain the value 2013 minus 2001, which is F of 12, but that is given to us as 100.  And so we use linearity again. Now we let B be 0 and A be 3 halves so that 3 halves multiplied by F of 12 is equal to F evaluated at 3 halves times 12, and that's F evaluated at 18.  But we know that at the same time, 3 halves of the value of F at 12 is 3 halves times 100, and that's given to us. And so the answer here is equal to 150 or D.

Video Summary

The problem involves a linear function \( F \) where the difference \( F(2013) - F(2001) = 100 \). To find \( F(2031) - F(2013) \), we use the linearity property. By examining increments, \( F(18) \) equates to \( \frac{3}{2} \times F(12) \), therefore \( F(12) = 100 \) and \( F(18) = 150 \). Hence, the desired value is 150.