题目内容
已知f(x)=2012lgx+
+
,若f(2012)=3,则f(
)=
| 2012 |
| log2x |
| x |
| x-1 |
| 1 |
| 2012 |
-2
-2
.分析:由f(2012)=3求得 2012×lg2012+
=2-
.再由 f(
)=-(2012×lg2012+
)-
,运算求得结果.
| 2012 |
| log22012 |
| 1 |
| 2011 |
| 1 |
| 2012 |
| 2012 |
| log22012 |
| 1 |
| 2011 |
解答:解:由题意可得f(2012)=2012×lg2012+
+
=3,∴2012×lg2012+
=2-
.
∴f(
)=2012×lg
+
+
=-2012×lg2012-
-
=-(2012×lg2012+
)-
=-(2-
)-
=-2,
故答案为-2.
| 2012 |
| log22012 |
| 2012 |
| 2011 |
| 2012 |
| log22012 |
| 1 |
| 2011 |
∴f(
| 1 |
| 2012 |
| 1 |
| 2012 |
| 2012 | ||
log2
|
| 1 |
| -2011 |
| 2012 |
| log22012 |
| 1 |
| 2011 |
| 2012 |
| log22012 |
| 1 |
| 2011 |
| 1 |
| 2011 |
| 1 |
| 2011 |
故答案为-2.
点评:本题主要考查求函数的值,体现了整体代换的思想,求得2012×lg2012+
=2-
,是解题的关键,属于基础题.
| 2012 |
| log22012 |
| 1 |
| 2011 |
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