Question

If $x^2 + 6x - 27 > 0$ and $x^2 - 3x - 4 < 0 $ then

• A. $x > 3$
• B. $x < 4$
• C. $3 < x < 4$
• D. $x = 3 \, \frac {1}{2}$

Answer 1

Answer: (C)

$x^2 + 6x - 27 \geq 0$ and $x^2 - 3x + 4 < 0$
$ \Rightarrow $ $ (x + 9) (x - 3) > 0$ and $(x - 4) (x + 1) < 0$.
$\Rightarrow $ $ x > - 9, x > 3 x - 4 > 0$
or $x < - 9, x < 3x + 1 < 0 $
$\Rightarrow $ $x > 3$ or $x < - 9 $ or $x - 4 < 0, x + 1 > 0 $
$i.e., \, x > 4, x < -1$ or $x < 4, x > - 1 $
$\therefore \, - 1 < x < 4$, other case is not possible.
Thus $x > 3 $ or $x < - 9$ and $- 1 < x < 4$.
Thus $3 < x < 4$.