Difference between revisions of "1976 AHSME Problems/Problem 7"

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If <math>-1 < x < 1</math>, then the first factor is positive, and the second factor is also positive.
 
If <math>-1 < x < 1</math>, then the first factor is positive, and the second factor is also positive.
  
If <math>x > 1</math>, the first factor is positive, but the second factor is negative.
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If <math>x > 1</math>, 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 <math>x < -1</math> or when <math>-1 < x < 1</math>, so our answer is <math>\boxed{\textbf{(C)}}</math> and we are done.
 
Combining this with the rules for signs and multiplication, we find that the expression is positive when <math>x < -1</math> or when <math>-1 < x < 1</math>, so our answer is <math>\boxed{\textbf{(C)}}</math> and we are done.

Latest revision as of 22:02, 24 September 2020

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
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

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