Grades 11-12 Video Solutions 2013-Levels 11&12 Video Solutions 2013 problem22

Video Transcription

Question number 22, let f be a function mapping the real numbers to the real numbers defined by the following properties. F is periodic with period 5 and the restriction of f  to the interval running from negative 2 to 3  where negative 2 is included and 3 is not is given by the function f of x is equal to x squared.  What is the value of f at 2013? So first let's write down what it means for this function to be periodic with period 5 and that means that any value  of x here can be increased by an integer multiple of 5  and the value of the functions will be the same where k here is an integer. So we have to rewrite 2013  as the closest integer multiple of 5. So we can say that's 2010 plus 3 or 2015 minus 2.  So dividing by 5 we have this is 5 times 402 plus 3  or 5 times 403 minus 2.  So that tells us that the value at 2013 of the function is equal to the value of the function at 3 and that should be equal to the value of the function at negative 2.  Now our restriction of f to the interval from negative 2 to 3 tells us that we must choose the value corresponding here to f evaluated at negative 2  because 3 is not actually in the domain of the restriction.  So f of 2013 is equal to f of negative 2 as f of 3 is not defined and that's equal to negative 2 squared  and that gives us 4.  So the answer here is D.

Video Summary

The function \( f \) is periodic with a period of 5, and its behavior in the interval \([-2, 3)\) is \( f(x) = x^2 \). To find \( f(2013) \), we calculate its equivalent value within this interval due to the periodic nature. By expressing 2013 as 2010 + 3 (5 × 402 + 3), it's equivalent to determining \( f(3) \). However, since 3 is not included in the interval, we use x = -2. Thus, \( f(2013) = f(-2) = (-2)^2 = 4 \). The answer is 4.

Keywords

periodic function

interval

equivalent value

period

calculation