1974 AHSME Problems/Problem 15

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Problem

If $x<-2$, then $|1-|1+x||$ equals

$\mathrm{(A)\ } 2+x \qquad \mathrm{(B) \ }-2-x \qquad \mathrm{(C) \  } x \qquad \mathrm{(D) \  } -x \qquad \mathrm{(E) \  }-2$

Solution

Notice that, for $a<0$, $|a|=-a$, and for $a>0$, $|a|=a$.

Since $x<-2$, $1+x<1-2<0$, so $|1+x|=-x-1$. Therefore, $1-|1+x|=1+x+1=x+2<-2+2=0$, and so $|1-|1+x||=|x+2|=-2-x, \boxed{\text{B}}$.

See Also

1974 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 14
Followed by
Problem 16
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