Difference between revisions of "1952 AHSME Problems/Problem 36"

Revision as of 22:41, 12 April 2020

To be continuous at $x = - 1$ , the value of $\frac {x^3 + 1}{x^2 - 1}$ is taken to be:

$\textbf{(A)}\ - 2 \qquad \textbf{(B)}\ 0 \qquad \textbf{(C)}\ \frac {3}{2} \qquad \textbf{(D)}\ \infty \qquad \textbf{(E)}\ -\frac{3}{2}$

Factoring the numerator and denominator gives $\dfrac{(x+1)(x^{2}-x+1)}{(x+1)(x-1)}=1$ $\fbox{E}$

1952 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 35	Followed by Problem 37
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All AHSME Problems and Solutions

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 == Solution ==
+Factoring the numerator and denominator gives <cmath>\dfrac{(x+1)(x^{2}-x+1)}{(x+1)(x-1)}=1</cmath>
 <math>\fbox{E}</math>