A) \[\frac{7}{20}\]
B) \[\frac{13}{20}\]
C) \[\frac{17}{20}\]
D) \[\frac{3}{20}\]
Correct Answer: B
Solution :
(b)The graph of \[y=16{{x}^{2}}+8\left( a+5 \right)x-7a-5\] is strictly above the x-axis |
\[\Rightarrow y>0\forall x\in R\] \[\Rightarrow 16{{x}^{2}}+8\left( a+5 \right)x+7a-5>0\forall x\in R\] |
The above inequality holds |
If discriminant |
\[\left[ \because \operatorname{coefficient}\,of\,{{x}^{2}}>0 \right]\] |
\[\Rightarrow 64{{\left( a+5 \right)}^{2}}-4.16\left( -7a-5 \right)<0\] |
\[\Rightarrow {{a}^{2}}+17a+30<0\] \[\Rightarrow \left( a+2 \right)\left( a+15 \right)<0\] \[\Rightarrow -15<a<-2\] |
Given \[-20\le a\le 0\] and favourable cases \[-15<a<-2\] |
\[\therefore \]Required probability \[=\frac{length\,of\,interval\left( -15,-2 \right)}{length\,ofinterval\,\left( -20,0 \right)}\]\[=\frac{-2-\left( -15 \right)}{0-\left( -20 \right)}=\frac{13}{20}\]. |
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