HMMT 二月 2006 · CALC 赛 · 第 10 题
HMMT February 2006 — CALC Round — Problem 10
题目详情
英文原题
- Suppose f and g are differentiable functions such that
′ ′ ′ ′
xg ( f ( x )) f ( g ( x )) g ( x ) = f ( g ( x )) g ( f ( x )) f ( x )
for all real x . Moreover, f is nonnegative and g is positive. Furthermore,
∫
− 2 aaef ( g ( x )) dx = 1 −
for all reals a . Given that g ( f (0)) = 1, compute the value of g ( f (4)).02
解析
英文解析
- Suppose f and g are differentiable functions such that
′ ′ ′ ′
xg ( f ( x )) f ( g ( x )) g ( x ) = f ( g ( x )) g ( f ( x )) f ( x )
for all real x . Moreover, f is nonnegative and g is positive. Furthermore,
∫
− 2 aaef ( g ( x )) dx = 1 −
for all reals a . Given that g ( f (0)) = 1, compute the value of g ( f (4)).02 − 161
Answer: e ore 16 − 2 a
Solution: Differentiating the given integral with respect to a gives f ( g ( a )) = e .
′ ′ ′ ′Nowd [ln ( f ( g ( x )))] f ( g ( x )) g ( x ) g ( f ( x )) f ( x ) d [ln ( g ( f ( x )))]
x = x = =
dx f ( g ( x )) g ( f ( x )) dxwhere the second equals sign follows from the given. Since ln ( f ( g ( x ))) = − 2 x , we have
2 − x − 162 − x + C = ln ( g ( f ( x ))), so g ( f ( x )) = Ke . It follows that K = 1 and g ( f (4)) = e . 4