HMMT 二月 2011 · ALGGEO 赛 · 第 11 题
HMMT February 2011 — ALGGEO Round — Problem 11
题目详情
- Let f ( x ) = x + 6 x + c for all real numbers x , where c is some real number. For what values of c does f ( f ( x )) have exactly 3 distinct real roots?
解析
- Let f ( x ) = x + 6 x + c for all real numbers x , where c is some real number. For what values of c does f ( f ( x )) have exactly 3 distinct real roots? √ 11 − 13 Answer: Suppose f has only one distinct root r . Then, if x is a root of f ( f ( x )), it must 1 1 2 be the case that f ( x ) = r . As a result, f ( f ( x )) would have at most two roots, thus not satisfying 1 1 the problem condition. Hence f has two distinct roots. Let them be r 6 = r . 1 2 Since f ( f ( x )) has just three distinct roots, either f ( x ) = r or f ( x ) = r has one distinct root. Assume 1 2 2 without loss of generality that r has one distinct root. Then f ( x ) = x + 6 x + c = r has one root, so 1 1 2 that x + 6 x + c − r is a square polynomial. Therefore, c − r = 9, so that r = c − 9. So c − 9 is a 1 1 1 11 13 2 2 2 root of f . So ( c − 9) + 6( c − 9) + c = 0, yielding c − 11 c + 27 = 0, or ( c − ) = . This results to 2 2 √ 11 ± 13 c = . 2 √ √ √ √ √ 11 − 13 2 11 − 13 7+ 13 5 − 13 − 7 − 13 If c = , f ( x ) = x + 6 x + = ( x + )( x + ). We know f ( x ) = has a 2 2 2 2 2 √ √ − 5+ 13 − 7 − 13 double root, -3. Now > so the second root is above the vertex of the parabola, and is 2 2 hit twice. √ √ √ √ √ 11+ 13 11+ 13 7 − 13 5+ 13 − 7+ 13 2 If c = , f ( x ) = x + 6 x + = ( x + )( x + ). We know f ( x ) = has a 2 2 2 2 2 double root, -3, and this is the value of f at the vertex of the parabola, so it is its minimum value. √ √ √ − 5 − 13 − 7+ 13 − 5 − 13 Since < , f ( x ) = has no solutions. So in this case, f has only one real root. 2 2 2 √ 11 − 13 So the answer is c = . 2 Note: In the solutions packet we had both roots listed as the correct answer. We noticed this oversight on the day of the test and awarded points only for the correct answer.