1976 AHSME Problems/Problem 7

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If $x$ is a real number, then the quantity $(1-|x|)(1+x)$ is positive if and only if

$\textbf{(A) }|x|<1\qquad \textbf{(B) }|x|>1\qquad \textbf{(C) }x<-1\text{ or }-1<x<1\qquad\\ \textbf{(D) }x<1\qquad  \textbf{(E) }x<-1$

Solution

We divide our solution into three cases: that of $x < -1$, that of $-1 < x < 1$, and that of $x > 1$. (When $x = -1$ or $x = 1$, the expression is zero, therefore not positive.)

If $x < -1$, then the first factor is negative, and the second factor is also negative.

If $-1 < x < 1$, then the first factor is positive, and the second factor is also positive.

If $x > 1$, the first factor is negative, but the second factor is positive.

Combining this with the rules for signs and multiplication, we find that the expression is positive when $x < -1$ or when $-1 < x < 1$, so our answer is $\boxed{\textbf{(C)}}$ and we are done.

See also

1976 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 6
Followed by
Problem 8
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