question_answer

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

A) \[\frac{7}{20}\]

B) \[\frac{13}{20}\]

C) \[\frac{17}{20}\]

D) \[\frac{3}{20}\]

Correct Answer: B

Solution :

the graph of \[y=16{{x}^{2}}+8(a+5)x-7a-5\]is strictly above the x-axis

\[\Rightarrow \]\[y>0\,\forall \,x\in \mathbf{R}\]\[\Rightarrow \]\[16{{x}^{2}}+8(a+5)x-7a-5>0\,\forall \,x\in R\]

The above inequality holds if discriminant <0

[\[\because \]coefficient \[{{x}^{2}}>0\]]

\[\Rightarrow \]\[64{{\left( a+5 \right)}^{2}}-4.16\left( -7a-50 \right)<0\]\[\Rightarrow \]\[{{a}^{2}}+17a+30<0\]\[\Rightarrow \]\[\left( a+2 \right)\left( a+15 \right)<0\]\[\Rightarrow \]\[-15<a<-2\]

Given \[-20\le a\le 0\] and favourable cases \[-15<a<-2\]

\[\therefore \] Required probability \[=\frac{length\,of\,interval\left( -15,-2 \right)}{length\,of\,interval\left( -20,0 \right)}\]

\[=\frac{-2-\left( -15 \right)}{0-\left( -20 \right)}=\frac{13}{20}\]