Question

The least integral value $\alpha$ of $x$ such that $\frac{x-5}{x^2+5x-14}>0$, satisfies the relation:

• A. ${\alpha }^2+3\alpha -4=0$
• B. ${\alpha }^2-5\alpha +4=0$
• C. ${\alpha }^2-7\alpha +6=0$
• D. ${\alpha }^2+5\alpha -6=0$

Answer 1

Answer: (D)

Solution: $0<\frac{x-5}{x^2+5x-14}=\frac{x-5}{(x+7)(x-2)}=E$ (say)
Sign of $E$ in different intervals are shown below

Thus, $\frac{x-5}{(x+7)(x-2)}>0$
$\Rightarrow -7<x<2$ or $x>5$.
Therefore the least integral value $\alpha$ of $x$ is -6 .
This value of $\alpha$ satisfies the relation ${\alpha }^2+5\alpha -6=0$.