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 :
| \[\because p,q,r\]are in A.P. \[\therefore \]\[2q=p+r...(1)\] |
| \[\because \] roots of \[p{{x}^{2}}+qx+r=0\] are all real then \[{{q}^{2}}-4pr\ge 0\Rightarrow {{\left( \frac{p+r}{2} \right)}^{2}}-4pr\ge 0\] |
| [From (1)] |
| \[\Rightarrow \]\[{{(p+r)}^{2}}-16pr\ge 0\] \[\Rightarrow \]\[{{p}^{2}}+{{r}^{2}}-14pr\ge 0\] |
| \[\Rightarrow \]\[{{\left( \frac{r}{p} \right)}^{2}}-14\left( \frac{r}{p} \right)+1\ge 0\]\[\Rightarrow \]\[{{\left( \frac{r}{p}-7 \right)}^{2}}\ge 48\Rightarrow \left| \frac{r}{p}-7 \right|\ge 4\sqrt{3}\] |
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