Question

If $a\in \left[ -20,0 \right]$ , then find the probability that the graph of the function $16{{x}^{2}}+8\left( a+5 \right)x-7a-5$ is strictly above the x-axis.
A. Required probability $=\dfrac{13}{20}$
B. Required probability $=\dfrac{12}{20}$
C. Required probability $=\dfrac{7}{20}$
D. Required probability $=\dfrac{8}{20}$

Answer 1

Hint: We need to find the probability of the graph of the function $16{{x}^{2}}+8\left( a+5 \right)x-7a-5$will be strictly above the x-axis. $16{{x}^{2}}+8\left( a+5 \right)x-7a-5$ is a quadratic function and for the graph of a quadratic function to be strictly above the x-axis, its discriminant should be less than 0. So, we will find out the value of the discriminant and keep it less than 0. Hence, we will find the values of ‘a’ for which the graph would be above x-axis. We also have been given a domain for ‘a’. Thus, we will know the length of both the intervals and hence we can find our required probability.

Complete step by step answer:
Let the given function be y.
Thus, y is given as:
$y=16{{x}^{2}}+8\left( a+5 \right)x-7a-5$
For the graph of y to be above the x-axis, discriminant of y should be less than 0.
 Discriminant of $y=a{{x}^{2}}+bx+c$ is given as $D={{b}^{2}}-4ac$
Here, $a=16,b=8\left( a+5 \right)$ and $c=-7a-5$
Thus, our discriminant will be given as:
$\begin{align}
  & D={{\left( 8\left( a+5 \right) \right)}^{2}}-4\left( 16 \right)\left( -7a-5 \right) \\
 & \Rightarrow D=64\left( {{a}^{2}}+10a+25 \right)+64\left( 7a+5 \right) \\
 & \Rightarrow D=64\left( {{a}^{2}}+10a+25+7a+5 \right) \\
 & \Rightarrow D=64\left( {{a}^{2}}+17a+30 \right) \\
 & \Rightarrow D=64\left( {{a}^{2}}+15a+2a+30 \right) \\
 & \Rightarrow D=64\left( a\left( a+15 \right)+2\left( a+15 \right) \right) \\
 & \Rightarrow D=64\left( a+2 \right)\left( a+15 \right) \\
\end{align}$
Now, for the graph of y to be above the x-axis, $D\text{ } < \text{ }0$
Putting the value of D, we will get:
$\begin{align}
  & 64\left( a+2 \right)\left( a+15 \right)\text{ } < \text{ }0 \\
 & \Rightarrow \left( a+2 \right)\left( a+15 \right)\text{ } < \text{ }0 \\
 & \Rightarrow -15\text{ } < a\text{ } < \text{ }-2 \\
\end{align}$
Thus, the interval of the required values of ‘a’ will be given as:
$a\in \left( -15,-2 \right)$
In the question, we have been given that a lies in the interval $\left[ -20,0 \right]$
Thus, the interval for possible values of ‘a’ is given as:
$a\in \left[ -20,0 \right]$
Thus, the length of the required interval of ‘a’ $=-2-\left( -15 \right)=13$
The length of the interval of the possible values of ‘a’ $=0-\left( -20 \right)=20$
Now, probability of any event is given as:
$P=\dfrac{\text{Favourable outcomes}}{\text{Total outcomes}}$
Here,
Favourable outcomes=length of required interval of ‘a’=13
Total outcomes=length of interval of possible values of ‘a’=20
Therefore, required probability is given as:
$P=\dfrac{13}{20}$

So, the correct answer is “Option A”.

Note: Keep in mind that here, the coefficient of ${{x}^{2}}$ in here is ‘16’ which is a positive real number, hence our graph will be an upward parabola and for the graph to be strictly above the axis, the quadratic equation should not have any solutions. Thus, we have kept the discriminant less than 0.