A) \[\left| \,\frac{r}{p}-7\ \right|\ \ge 4\sqrt{3}\]
B) \[\left| \ \frac{p}{r}-7\ \right|\ <4\sqrt{3}\]
C) All \[p\]and \[r\]
D) No \[p\] and \[r\]
Correct Answer: A
Solution :
\[p,\ q,\ r\] are positive and are in A.P. \[\therefore \ q=\frac{p+r}{2}\] ......(i) \[\because \] The roots of \[p{{x}^{2}}+qx+r=0\] are real \[\Rightarrow \] \[{{q}^{2}}\ge 4pr\]\[\Rightarrow \]\[{{\left[ \frac{p+r}{2} \right]}^{2}}\ge 4pr\] [using (i)] \[\Rightarrow \] \[{{p}^{2}}+{{r}^{2}}-14pr\ge 0\] \[\Rightarrow \] \[{{\left( \frac{r}{p} \right)}^{2}}-14\left( \frac{r}{p} \right)+1\ge 0\] \[(\because \ p>0\ \text{and}\ p\ne 0)\] \[\Rightarrow \] \[{{\left( \frac{r}{p}-7 \right)}^{2}}-48\ge 0\]\[\Rightarrow \]\[{{\left( \frac{r}{p}-7 \right)}^{2}}-{{(4\sqrt{3})}^{2}}\ge 0\] \[\Rightarrow \] \[\left| \ \frac{r}{p}-7\ \right|\ \ge 4\sqrt{3}\].
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