Paper 1 · modulus / inequalities

Find the set of values of $x$ which satisfy $|3x + 2| < -x + 1$ .

-\tfrac{3}{2} < x < -\tfrac{1}{4}

-\tfrac{2}{3} < x < \tfrac{1}{4}

-\tfrac{1}{4} < x < 1

x < -\tfrac{3}{2}

x > -\tfrac{1}{4}

Explanation: For

3x+2\ge0

3x+2<-x+1

gives

x<-\tfrac14

, so

-\tfrac23\le x<-\tfrac14

. For

3x+2<0

-(3x+2)<-x+1

gives

x>-\tfrac32

, so

-\tfrac32<x<-\tfrac23

. Combining,

-\tfrac32<x<-\tfrac14

Derived from ZIMSEC Mathematics Paper 1, November 2011, Q1

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