Question

Let $k$ be the non-zero real number such that the quadratic equation $k{x}^2+2x+k=0$ has two distinct real roots $\alpha \text{and}\beta (\alpha <\beta ).$

If $\alpha <2$ and $\beta >5$, then

• A. $kϵ(-\frac{4}{5},-\frac{5}{13})$
• B. $kϵ(-1,-\frac{4}{5})\cup (-\frac{5}{13},1)$
• C. $kϵ(-\frac{5}{13},0)$
• D. $kϵ(-1,1)$

Answer 1

The correct option is C

$kϵ(-\frac{5}{13},0)$

Let $f\left(x\right)=k{x}^2+2x+k$
$kf\left(2\right)<0k(5k+4)<0$
$-\frac{4}{5}<k<\mathrm{0...}\left(i\right)$
Also,
$kf\left(5\right)<0\Rightarrow k(26k+10)<0\Rightarrow -\frac{5}{13}<k<\mathrm{0...}\left(ii\right)$
$\therefore$ from $\left(i\right)\text{and}\left(ii\right)$
$kϵ(-\frac{5}{13},0)$