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)

$3 < x < 4$

Answer 2

$x^2 + 6x - 27 \geq 0$ 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 $3 < x < 4$.