Grades 11-12 Video Solutions 2024-2024_11-12

Video Transcription

A function f from the reals to the reals satisfies f of 20-x equals f of 22 plus x for all real x. It is known that f has exactly two roots. What is the sum of these two roots?  Let a be one of our roots. Then we know that f of a equals zero. Choose x such that a equals 20 minus x, which implies x equals 20 minus a.  Then we know that f of 22 plus x is f of 42 minus a. Since f of a equals f of 42 minus a, which is zero, then we know that 42 minus a must  be a second root, and therefore the sum of the roots must be a plus 42 minus a, which is 42. Note that a and 42 minus a must be distinct.  Consider the case when a equals 21, so 42 minus a is 21. Then there must be a second distinct root. Suppose this is b, which is not 21.  However, 42 minus b, which is not 21, must also be a root, and then we would have three roots, a, b, and 42 minus b.  This contradicts the given statement that f has exactly two roots, and thus our answer is e, none of the above.

Video Summary

The function \( f \) has exactly two roots, subject to the relationship \( f(20-x) = f(22+x) \). Suppose one root is \( a \), then \( f(a) = 0 \). Choosing \( x = 20-a \), we find \( f(42-a) = 0 \) as well. This implies the roots are \( a \) and \( 42-a \). The sum of the roots is \( a + (42-a) = 42 \). Any other configuration implies more than two roots, contradicting the problem's constraints. Thus, the sum of the roots is 42.

Keywords

function roots

symmetry

root sum

two roots

mathematical relationship