Question

If $aϵ[-10,0]$, then the probability that the graph of the function $y=x^2+2(a+3)x-(2a+6)$ is strictly above x-axis is

• A. $\frac{3}{5}$
• B. $\frac{2}{5}$
• C. $\frac{1}{5}$
• D. None of these

Answer 1

The correct option is C $\frac{1}{5}$

(i)As $f\left(x\right)=y=x^2+2(a+3)x-2a-6>\forall xϵR$
(ii) $f\left(x\right)>0,$ conefficient of $x^2$ is $1>0$
(iii)$\therefore D<0$ $\therefore 4{(a+3)}^2+2a+6<0$ ${(a+3)}^2+2a+6<0$
$a^2+8a+15<0$ $(a+3)(a+5)<0$ $-5<a<-3$
Now $n\left(S\right)=$ length of interval, as $nϵ[-10,0]$ $=0-(-10)=10\therefore n\left(S\right)=10$
$n\left(A\right)=$ length of interval, when $-5<a<-3=-3-\left(5\right)=2$
$\therefore$ Required probability $=\frac{n\left(A\right)}{n\left(S\right)}=\frac{2}{10}=\frac{1}{5}$