Question

The least integral value $\alpha$ of $x$ such that $\frac{x-5}{x^2+5 x-14}>0$, 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: $0<\frac{x-5}{x^2+5 x-14}=\frac{x-5}{(x+7)(x-2)}= E$ (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 $\alpha^2+5 \alpha-6=0$.