1965 AHSME Problems/Problem 23

Problem

If we write $|x^2 - 4| < N$ for all $x$ such that $|x - 2| < 0.01$, the smallest value we can use for $N$ is:

$\textbf{(A)}\ .0301 \qquad  \textbf{(B) }\ .0349 \qquad  \textbf{(C) }\ .0399 \qquad  \textbf{(D) }\ .0401 \qquad  \textbf{(E) }\ .0499\qquad$

Solution

There are two cases: (i) $x-2>0$ and (ii) $x-2<0$. In case (i), $0<x-2<0.01$, so $4<x+2<4.01$. In case (ii), $-0.01<x-2<0$, so $3.99<x+2<4$. Because $(x+2)$ in case (i) can take a larger value, we use it to determine the upper bound for $|x^2-4|$ under the given restrictions. Now, we can see that $|x^2-4|=|x-2||x+2|<0.01*4.01=\boxed{0.0401}$, which is answer choice $\fbox{\textbf{(D)}}$.

See Also

1965 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 22
Followed by
Problem 24
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 31 32 33 34 35 36 37 38 39 40
All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png