1999 AHSME Problems/Problem 18
Contents
[hide]Problem
How many zeros does have on the interval ?
Solution
For we have , and the logarithm is a strictly increasing function on this interval.
is zero for all of the form , where . There are such in .
Here's the graph of the function on :
As we go closer to , the function will more and more wildly oscilate between and . This is how it looks like at .
And one more zoom, at .
Solution
If cos(log(x)) = zero, then log(x) = π/2 + nπ.
If we consider the limiting case as x approaches zero, log(x) approaches negative infinity.
If we consider the other boundary, x equals 1 where log(x) equals zero.
Due to the intermediate value theorem log(x) must contain all negative real numbers, and thus an infinite number of solutions to log(x) = π/2 + nπ. This means that cos(x) is zero an infinite number of times giving .
-PhysicsMan
See also
1999 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 17 |
Followed by Problem 19 | |
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