Question

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

If $\beta <\sqrt{2}+\alpha$, then

• A. $kϵ(-\sqrt{\frac{2}{3}},\sqrt{\frac{2}{3}})$
• B. $kϵ(-\infty ,-\sqrt{\frac{2}{3}})\cup \left(\sqrt{\frac{2}{3},}\infty \right)$
• C. $kϵ(-1,-\sqrt{\frac{2}{3}})\cup \left(\sqrt{\frac{2}{3},}1\right)$
• D. $kϵ(-1,1)$

Answer 1

The correct option is C

$kϵ(-1,-\sqrt{\frac{2}{3}})\cup \left(\sqrt{\frac{2}{3},}1\right)$

$D=4-4k^2$
As D > 0
$kϵ(-1,1)....\left(i\right)$
Now $(\beta -\alpha {)}^2<2$
$\Rightarrow (\beta +\alpha {)}^2-4\alpha \beta <2\Rightarrow \frac{4}{k^2}-4<2\frac{4}{k^2}<6\Rightarrow 6k^2>4$
$\Rightarrow k^2>\frac{2}{3}....\left(ii\right)$ $\Rightarrow kϵ(-1,-\sqrt{\frac{2}{3}})\cup (\sqrt{\frac{2}{3}},1)$