1974 AHSME Problems/Problem 15

Problem

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

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

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 $|1-|1+x||=|x+2|=-2-x, \boxed{\text{B}}$ .

1974 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 14	Followed by Problem 16
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