Question

If $\alpha ,\beta$ are the roots of equation $(k+1)x^2-(20k+14)x+91k+40=0;(\alpha <\beta )$, $k>0$, then

The larger root $\beta$ lies in the interval

• A. $(4,7)$
• B. $(7,10)$
• C. $(10,13)$
• D. None of these

Answer 1

Answer: (C)

$(10,13)$

$(k+1)x^2-(20k+14)x+91k+40=0$
$\Rightarrow k(x^2-20x+91)+x^2-14x+40=0$
$\Rightarrow k=-\frac{x^2-14x+40}{x^2-20x+91}$
$\Rightarrow k=-\frac{(x-4)(x-10)}{(x-13)(x-7)}>0(\because k>0)$
$\Rightarrow \frac{(x-4)(x-10)}{(x-13)(x-7)}<0$
$\Rightarrow x\in (4,7)$ or $(10,13)$
i.e. $\alpha ,\beta \in (4,7)$ or $(10,13)$
But, sum of the roots$=\frac{20k+14}{k+1}=\alpha +\beta$
$\frac{14k+14}{k+1}<\frac{20k+14}{k+1}<\frac{20k+20}{k+1}$
$\Rightarrow 14<\alpha +\beta <20$
If both $4<\alpha ,\beta <7$, then $\alpha +\beta <14$
If both $10<\alpha ,\beta <13$, then $\alpha +\beta >20$
$\Rightarrow$ One root should lie between$(4,7)$

and the other root should lie between $(10,13)$
So, smaller root $\alpha \in (4,7)$ and larger root $\beta \in (10,13)$.