Question

If $a\in I$ and $a\in [-5,30]$ then the probability that the graph of $y=x^2+2(a+4)x-5a+64$ is strictly above the $x$-axis is

• A. $\frac{1}{6}$
• B. $\frac{7}{36}$
• C. $\frac{1}{2}$
• D. $\frac{3}{5}$

Answer 1

The correct option is C $\frac{1}{2}$

Given $a\in I$ and $a\in [-5,30]$

$\Rightarrow y=x^2+2(a+4)x-5a+64$

$a$ can assume values from $-5$ to $30.$

$\Rightarrow y=x^2+2(a+4)x-5a+64>0$ for all $a\in I$ and $x\in R$

disc$<0$ i.e, $b^2-4c<0$

$={\left[2\right(a+4\left)\right]}^2-4[-(5a+64\left)\right]$

$=a^2+18a+144.$

$\therefore$ The roots are $(-a,a),$ this point denotes that the graph is below $x-$axis.

$\therefore$ The probability of rest of points being strictly above the $x-$axis $=\frac{1}{2}.$

Hence, the answer is $\frac{1}{2}.$