A) \[x\le -1\,\,or\,\,x\ge 3\]
B) \[x<-1\,\,or\,\,x>3\]
C) \[x\le -1\,\,or\,\,x>3\]
D) \[x<-3\,\,or\,\,x>2\]
Correct Answer: B
Solution :
[b] Let \[b=ar\] and \[c=a{{r}^{2}},\] so that a, b, c are in GP. \[\therefore \,\,\,\,a+b+c=xb\] \[\Rightarrow \,\,\,a+ar+a{{r}^{2}}=x.\,\,ar\Rightarrow \,{{r}^{2}}+(1-x)r+1=0\] ? (1) If r is real, then discriminant of \[(1)\ge 0\] \[\Rightarrow \,\,\,{{(1-x)}^{2}}-4.1.1\ge 0\Rightarrow {{x}^{2}}-2x-3\ge 0\] \[\Rightarrow \,\,\,\,\,(x+1)(x-3)\ge 0\Rightarrow x\le -1\] or \[x\ge 3.\] Now for \[x=3\] we get \[r=1,\] which will make \[a=b=c\] Also for \[x=-1,\] we get \[r=-1,\] for which \[a=c,\] thus \[x<-1\] or \[x>3\]
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