Question

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

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

Answer 1

The correct option is B ${\alpha }^2+3\alpha -4=0$

REF.Image

Given $\frac{x-5}{x^2+5x-14}>0$

$=\frac{x-5}{x^2-2x+7x-14}>0$

$=\frac{x-5}{x(x-2)+7(x-2)}>0$

$=\frac{x-5}{(x-2)(x+7)}>0$ ___ (I)

Equation hold when $x>5$ and $xϵ(-\infty ,-7)\vee (2,\infty )$

$\therefore xϵ(5,\infty )$ ___(II)

Equation (I) also hold when $xϵ(-7,2)$ ___ (III)

From (II) & (III) least value in (III)

$\therefore xϵ(-7,2)$

clearly ${\alpha }^2+3\alpha -4=0$ so its lies the condition

$\alpha =1,\alpha =-4$