Difference between revisions of "1999 AHSME Problems/Problem 18"
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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. | 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 <math>\boxed{\text{( | + | This means that cos(x) is zero an infinite number of times giving <math>\boxed{\text{(E) infinitely\ many}}</math>. |
-PhysicsMan | -PhysicsMan |
Latest revision as of 00:08, 3 August 2024
Contents
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 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 | ||
All AHSME Problems and Solutions |
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