Question

Let $x$ be a real number such that $5<|x - 1|<15$. Then

• A. $-18<x<-3$ or $3<x<19$
• B. $-14<x<-3$ or $6<x<17$
• C. $-16<x<-2$ or $6<x<20$
• D. $-14<x<-4$ or $6<x<16$
• E. $-10<x<-1$ or $3<x<18$

Hint:

For double inequalities with modulus, solve separately and take intersection.

Answer 1

The Correct Option is D

Concept:
• Solve compound inequality involving modulus

Step 1: Break inequality
\[ 5<|x-1|<15 \] \[ |x-1|>5 \quad \text{and} \quad |x-1|<15 \]

Step 2: Solve individually
\[ |x-1|>5 \Rightarrow x-1>5 \text{ or } x-1 6 \text{ or } x<-4 \] \[ |x-1|<15 \Rightarrow -15<x-1<15 \Rightarrow -14<x<16 \]

Step 3: Take intersection
\[ (-14 6 \text{ or } x<-4) \] \[ = (-14<x<-4) \cup (6<x<16) \] Final Conclusion:
Option (D)