HMMT 二月 2007 · CALC 赛 · 第 9 题

HMMT February 2007 — CALC Round — Problem 9

[ 7 ] g is a twice differentiable function over the positive reals such that 3 ′ 4 ′′ g ( x ) + 2 x g ( x ) + x g ( x ) = 0 for all positive reals x. (1) lim xg ( x ) = 1 (2) x →∞ Find the real number α > 1 such that g ( α ) = 1 / 2.

解析

[ 7 ] g is a twice differentiable function over the positive reals such that 3 ′ 4 ′′ g ( x ) + 2 x g ( x ) + x g ( x ) = 0 for all positive reals x. (1) lim xg ( x ) = 1 (2) x →∞ 2 Find the real number α > 1 such that g ( α ) = 1 / 2. 6 3 ′ 4 ′′ Answer: . In the first equation, we can convert the expression 2 x g ( x )+ x g ( x ) into the derivative π of a product, and in fact a second derivative, by writing y = 1 /x . Specifically, ( ) ( ) ( ) 1 1 1 3 ′ 4 ′′ − 3 ′ − 4 ′′ 0 = g ( x ) + 2 x g ( x ) + x g ( x ) = g + 2 y g + y g y y y ( ) [ ( )] 1 d 1 − 2 ′ = g + − y g y d y y ( ) [ ( )] 2 1 d 1 = g + g 2 y d y y ( ) 1 Thus g = c cos( y ) + c sin( y ) or g ( x ) = c cos(1 /x ) + c sin(1 /x ). Now the second condition gives 1 2 1 2 y sin(1 /x ) 1 = lim c x + c · = c + lim c x 1 2 2 1 x →∞ x →∞ 1 /x It must be that c = 0 , c = 1. Now since 0 < 1 /α < 1, the value of α such that g ( α ) = sin(1 /α ) = 1 / 2 1 2 is given by 1 /α = π/ 6 and so α = 6 /π .