A) \[(-3,5)\]
B) \[(5,9)\]
C) \[\left( -\frac{2}{3},8 \right)\]
D) \[\left( -8,\frac{2}{3} \right)\]
E) \[\left( -5,\frac{2}{3} \right)\]
Correct Answer: C
Solution :
\[|2x-3|<|x+5|\] \[\Rightarrow \]\[\pm (2x-3)<\pm (x+5)\] ?.. (i) First we take positive sign and negative sign on both sides, \[\Rightarrow \] \[2x-3<x+5\] \[\Rightarrow \] \[x<8\] ?.(ii) \[-(2x-3)<-(x+5)\] \[\Rightarrow \] \[2x-3>x+5\] \[\Rightarrow \] \[x>8\] ?.(iii) Now, we take negative sign in RHS and positive in LHS, \[\Rightarrow \] \[2x-3<-x-5\] \[\Rightarrow \] \[3x<-2\] \[\Rightarrow \] \[x<-\frac{2}{3}\] ?.. (iv) and we take negative sign LHS and positive in RHS, \[\Rightarrow \] \[-(2x-3)<x+5\] \[\Rightarrow \] \[-2x+3<x+5\] \[\Rightarrow \] \[-2<3x\] \[\Rightarrow \] \[-2/3<x\] ?.(v) From Eqs. (ii) and (v), \[-2/3<x<8\] \[\Rightarrow \] \[x\in (-2/3,8)\] From Eqs. (iii) and (iv), \[x\in (-\infty ,-2/3)\cup (8,\infty )\]
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