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 $\fbox{\textbf{(D)}}$ .

1965 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 22	Followed by Problem 24
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1965 AHSME Problems/Problem 23

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