HMMT 二月 2011 · ALGCALC 赛 · 第 12 题

英文解析

12. Let f ( x ) = x + 6 x + c for all real numbers x , where c is some real number. For what values of c doesf ( 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
    be the case that f ( x ) = r . As a result, f ( f ( x )) would have at most two roots, thus not satisfying 2
    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
    without loss of generality that r has one distinct root. Then f ( x ) = x + 6 x + c = r has one root, so 2
    1 1
    that x + 6 x + c − r is a square polynomial. Therefore, c − r = 9, so that r = c − 9. So c − 9 is a 2
    1 1 1
    2 2 11 2 13
    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 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
    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 = .
    Note: In the solutions packet we had both roots listed as the correct answer. We noticed this oversight 2
    on the day of the test and awarded points only for the correct answer.