HMMT 二月 2011 · ALGCALC 赛 · 第 16 题

英文解析

16. Let f ( x ) = x − r x + r for all real numbers x , where r and r are some real numbers. Define a
    2 3 2 3
    sequence { g } for all nonnegative integers n by g = 0 and g = f ( g ). Assume that { g } satisfiesn 0 n +1 n nthe following three conditions: (i) g < g and g > g for all 0 ≤ i ≤ 2011; (ii) there exists
    2 i 2 i +1 2 i +1 2 i +2
    a positive integer j such that g > g for all i > j , and (iii) { g } is unbounded. If A is the greatesti +1 i nnumber such that A ≤ | r | for any function f satisfying these properties, find A .
    Answer: 2 Consider the function f ( x ) − x . By the constraints of the problem, f ( x ) − x must be 2
    negative for some x , namely, for x = g , 0 ≤ i ≤ 2011. Since f ( x ) − x is positive for x of large
    2 i +1
    absolute value, the graph of f ( x ) − x crosses the x -axis twice and f ( x ) − x has two real roots, saya < b . Factoring gives f ( x ) − x = ( x − a )( x − b ), or f ( x ) = ( x − a )( x − b ) + x .
    Now, for x < a , f ( x ) > x > a , while for x > b , f ( x ) > x > b . Let c 6 = b be the number such thatf ( c ) = f ( b ) = b . Note that b is not the vertex as f ( a ) = a < b , so by the symmetry of quadratics, cr r +1
    b + c b + a
    2 2
    exists and = as the vertex of the parabola. By the same token, = is the vertex of
    2 2 2 2
    f ( x ) − x . Hence c = a − 1. If f ( x ) > b then x < c or x > b . Consider the smallest j such that g > b .
    Then by the above observation, g < c . (If g ≥ b then f ( g ) ≥ g ≥ b so by induction, g ≥ g forjj − 1 i i i i +1 iall i ≥ j . Hence j > 1; in fact j ≥ 4025.) Since g = f ( g ), the minimum value of f is less than c .
    j − 1 j − 2
    b + a − 1
    The minimum value is the value of f evaluated at its vertex, , so
    ( )2
    b + a − 1
    f < c
    ( ) ( )2
    b + a − 1 b + a − 1 b + a − 1 − a − b + < a − 1
    2 2 2
    1 − ( b − a ) b − a + 12 + < 0
    4 2
    3 ( b − a ) b − a 2 < −
    4 4 2
    4 < ( b − a − 1) .2
    Then either b − a − 1 < − 2 or b − a − 1 > 2, but b > a , so the latter must hold and ( b − a ) > 9.2
    Now, the discriminant of f ( x ) − x equals ( b − a ) (the square of the difference of the two roots) and 2
    2 2
    ( r + 1) − 4 r (from the coefficients), so ( r + 1) > 9 + 4 r . But r = g > g = 0 so | r | > 2.
    2 3 2 3 3 1 0 2
    We claim that we can make | r | arbitrarily close to 2, so that the answer is 2. First define G , i ≥ 0
    2 i
    √
    as follows. Let N ≥ 2012 be an integer. For ε > 0 let h ( x ) = x − 2 − ε, g ( x ) = − x + 2 + ε and 2
    G = 2 + ε , and define G recursively by G = g ( G ) , G = h ( G ). (These two equations areε
    2 N +1 i i ε i +1 i +1 iconsistent.) Note the following.
    √
    √
    (i) G < G and G > G for 0 ≤ i ≤ N − 1. First note G = − 4 + 2 ε > − 4 + 2 ε + ε =2
    2 i 2 i +1 2 i +1 2 i +2 2 N − 2 − ε . Let l be the negative solution to h ( x ) = x . Note that − 2 − ε < G < l < 0 since
    2 Nh ( G ) > 0 > G . Now g ( x ) is defined as long as x ≥ − 2 − ε , and it sends ( − 2 − ε, l ) into ( l, 0)
    2 N 2 N εand ( l, 0) into ( − 2 − ε, l ). It follows that the G , 0 ≤ i ≤ 2 N are well-defined; moreover, G < li 2 iand G > l for 0 ≤ i ≤ N − 1 by backwards induction on i , so the desired inequalities follow.
    2 i +1
    (ii) G is increasing for i ≥ 2 N + 1. Indeed, if x ≥ 2 + ε , then x − x = x ( x − 1) > 2 + ε so h ( x ) > x .2
    Hence 2 + ε = G < G < · · · .i
    2 N +1 2 N +2
    (iii) G is unbounded. This follows since h ( x ) − x = x ( x − 2) − 2 − ε is increasing for x > 2 + ε , so Gi iincreases faster and faster for i ≥ 2 N + 1.
    2 2
    Now define f ( x ) = h ( x + G ) − G = x + 2 G x + G − G − 2 − ε . Note G = h ( G ) while
    0 0 0 0 i +1 i
    ∞0
    g = f ( g ) = h ( g + G ) − G , so by induction g = G − G . Since { G } satisfies (i), (ii), andi +1 i i 0 0 i i 0 ii =0
    (iii), so does g .
    We claim that we can make G arbitrarily close to − 1 by choosing N large enough and ε small enough;ithis will make r = − 2 G arbitrarily close to 2. Choosing N large corresponds to taking G to be a 0
    2 0 0
    larger iterate of 2 + ε under g ( x ). By continuity of this function with respect to x and ε , it suffices toε
    Algebra & Calculus Individual Testtake ε = 0 and show that (letting g = g )
    ( n )0
    g (2) = g ( · · · g (2) · · · ) → − 1 as n → ∞ .
    ︸ ︷︷ ︸
    But note that for 0 ≤ θ ≤ ,πn
    ( ) ( )2
    √
    θ π θg ( − 2 cos θ ) = − 2 − 2 cos θ = − 2 sin = 2 cos − .
    2 2 2
    ( ( ))
    ( n ) π π n π ( n ) ( n − 1)
    Hence by induction, g ( − 2 cos θ ) = − 2 cos − + · · · + ( − 1) θ − . Hence g (2) = g ( − 2 cos 0)
    2 4 2 nπ π πconverges to − 2 cos( − + · · · ) = − 2 cos( ) = − 1, as needed.
    2 4 3