A) \[(4/5,\ 2)\]
B) \[(2,\,\,\infty )\]
C) \[(-1,\,-4/5)\]
D) \[(-\infty ,\,-1)\]
Correct Answer: D
Solution :
Let \[f(x)=4{{x}^{2}}-20px+(25{{p}^{2}}+15p-66)=0\] .....(i) The roots of (i) are real if \[{{b}^{2}}-4ac=400{{p}^{2}}-16(25{{p}^{2}}+15p-66)\] \[=16(66-15p)\ge 0\] Þ \[p\le 22/5\] .....(ii) Both roots of (i) are less than 2. Therefore \[f(2)>0\] and sum of roots < 4. Þ \[{{4.2}^{2}}-20p.2+(25{{p}^{2}}+15p-66)>0\]and \[\frac{20p}{4}\]<4 Þ \[{{p}^{2}}-p-2>0\]and \[p<4/5\] Þ \[(p+1)(p-2)>0\]and \[p<4/5\] Þ \[p<-1\]or \[p>2\]and \[p<4/5\]Þ \[p<-1\] .....(iii) From (ii) and (iii), we get \[p<-1\] i.e. \[p\in (-\infty ,-1)\].
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