question_answer

If  \[a\in \left[ -20,0 \right],\]then probability that the graph of the function\[y=16{{x}^{2}}+8\left( a+5 \right)x-7a-5\] is strictly above the \[\operatorname{x}-axis\,is\]

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 \[=\frac{length\,of\,interval\left( -15,-2 \right)}{length\,ofinterval\,\left( -20,0 \right)}\]\[=\frac{-2-\left( -15 \right)}{0-\left( -20 \right)}=\frac{13}{20}\].