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1999 AHSME Problems/Problem 18: Difference between revisions

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== 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 <math>\boxed{\text{(D) infinitely\ many}}</math>.
-PhysicsMan


== See also ==
== See also ==
{{AHSME box|year=1999|num-b=17|num-a=19}}
{{AHSME box|year=1999|num-b=17|num-a=19}}
{{MAA Notice}}
{{MAA Notice}}

Revision as of 17:55, 26 June 2024

Problem

How many zeros does $f(x) = \cos(\log x)$ have on the interval $0 < x < 1$?

$\mathrm{(A) \ } 0 \qquad \mathrm{(B) \ } 1 \qquad \mathrm{(C) \ } 2 \qquad \mathrm{(D) \ } 10 \qquad \mathrm{(E) \ } \text{infinitely\ many}$

Solution

For $0 < x < 1$ we have $-\infty < \log x < 0$, and the logarithm is a strictly increasing function on this interval.

$\cos(t)$ is zero for all $t$ of the form $\frac{\pi}2 + k\pi$, where $k\in\mathbb{Z}$. There are $\boxed{\text{infinitely\ many}}$ such $t$ in $(-\infty,0)$.

Here's the graph of the function on $(0,1)$:

[asy] import graph; size(250,200,IgnoreAspect); real f(real t) {return cos(log(t));} draw(graph(f,0.01,1),red,"$\cos(\log(x))$"); xaxis("$x$",BottomTop,LeftTicks); yaxis("$y$",LeftRight,RightTicks(trailingzero)); attach(legend(),truepoint(E),20E,UnFill); [/asy]

As we go closer to $0$, the function will more and more wildly oscilate between $-1$ and $1$. This is how it looks like at $(0.0001,0.02)$.

[asy] import graph; size(250,200,IgnoreAspect); real f(real t) {return cos(log(t));} draw(graph(f,0.0001,0.02),red,"$\cos(\log(x))$"); xaxis("$x$",BottomTop,LeftTicks); yaxis("$y$",LeftRight,RightTicks(trailingzero)); attach(legend(),truepoint(E),20E,UnFill); [/asy]

And one more zoom, at $(0.000001,0.0005)$.

[asy] import graph; size(250,200,IgnoreAspect); real f(real t) {return cos(log(t));} draw(graph(f,0.000001,0.0005),red,"$\cos(\log(x))$"); xaxis("$x$",BottomTop,LeftTicks); yaxis("$y$",LeftRight,RightTicks(trailingzero)); attach(legend(),truepoint(E),20E,UnFill); [/asy]

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 $\boxed{\text{(D) infinitely\ many}}$.

-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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing

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