Grades 11-12 Video Solutions 2023-2023_11-12

Video Transcription

Problem number 21. Two functions f and g on the real numbers satisfy the system of equations f of x plus 2 times g of 1 minus x equals x squared and f of 1 minus x minus g of x  equals x squared. What is f? Okay, so for these types of questions, a trick when we're noticing x and 1 minus x, let's try inputting 1 minus x instead of x into a function, right?  So we're going to input 1 minus x instead of x in the second equation. And from that we get f of 1 minus 1 minus x minus g of 1 minus x equals 1 minus x squared. And now  we can simplify that to get f of x minus g of 1 minus x equals x squared minus 2x plus 1. Hey, wait a second. The first equation also has f of x and g of 1 minus x. So we  can let a equal f of x and b equal g of 1 minus x. And if we substitute, we get that a plus 2b equals x squared and a minus b equals x squared minus 2x plus 1. The a's cancel  out and so we get that if we multiply the second equation by 2 and add the results of the first equation, we get that 2 times a minus b plus a plus 2b is equal to 2 times  the first equation plus x squared. Working through all the algebra, we get that a equals x squared minus four thirds x plus two thirds. And that's if we remember, we set a equal  to f of x. And that means that f of x is equal to x squared minus four thirds x plus two thirds. And that is choice A.

Video Summary

The problem involves two functions, \( f \) and \( g \), satisfying a system of equations with variables \( x \) and \( 1-x \). By substituting \( 1-x \) in place of \( x \) in the second equation, we simplify and find expressions for \( f(x) \) and \( g(1-x) \). Assigning variables \( a = f(x) \) and \( b = g(1-x) \), we solve the algebraic system: \( a + 2b = x^2 \) and \( a - b = x^2 - 2x + 1 \). Solving these, \( f(x) = x^2 - \frac{4}{3}x + \frac{2}{3} \). The solution corresponds to choice A.