Question

If $a \in [-10, 0]$, 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

Answer: (C)

(i) As $f(x) = y = x^2 + 2(a + 3) x - 2a - 6 > 0 $
$\,\forall \, x \in R$
(ii) $f(x) > 0$, coefficient of $x^2$ is $1 > 0$
(iii) $D < 0$
$\therefore 4(a + 3)^2 - 4(- 2a - 6) < 0$
$\Rightarrow (a + 3)^2 + 2a + 6 < 0$
$\Rightarrow a^2 + 8a + 15 < 0$
$\Rightarrow (a + 3) (a + 5) < 0$
$\Rightarrow - 5 < a < -3$
Now, $n(S) =$ length of interval, as $n \in [-10, 0]$
$= 0 - (- 10) = 10$
$\therefore n(S) = 10$
$n(A) =$ length of interval, when $- 5 < a < - 3$
$= - 3 - (- 5) = 2$
$\therefore $ Required probability $ = \frac{n(A)}{n(S)} $
$= \frac{2}{10} = \frac{1}{5}$
Alternative Solution :
As $f(x) = y > 0$ ,br> $\Rightarrow a > 0$ and $D < 0$
$\Rightarrow a^2 + 8a + 15 < 0$
$\Rightarrow -5 < a < -3$ (To be use for $n(A))$
$\therefore $ Required probability $ = \frac{n(A)}{n(S)}$
$ = \frac{\int\limits_{-5}^{-3} dx}{\int\limits_{-10}^{0} dx} = \frac{2}{10} = \frac{1}{5}$