Question

The set of values of $k$ for which the equation $x^4+(k-1)x^3+x^2+(k-1)x+1=0$ has $2$ positive and $2$ negative roots is

• A. $\varphi$
• B. $(-\infty ,-\frac{1}{2})$
• C. $(\frac{5}{2},\infty )$
• D. $(-\frac{1}{2},\frac{5}{2})$

Answer 1

The correct option is A $\varphi$

$x^4+(k-1)x^3+x^2+(k-1)x+1=0\cdots \left(1\right)$
Dividing the equation by $x^2$,
$\Rightarrow (x^2+\frac{1}{x^2})+(k-1)(x+\frac{1}{x})+1=0$
Assuming $x+\frac{1}{x}=t$
We know that $t\in (-\infty ,-2]\cup [2,\infty )$
${(x+\frac{1}{x})}^2=t^2\Rightarrow (x^2+\frac{1}{x^2})=t^2-2$
Now,
$t^2-2+(k-1)t+1=0\Rightarrow t^2+(k-1)t-1=0\cdots \left(2\right)$
Let the roots of the equation $\left(2\right)$ is $t_1,t_2$
For every $t\ge 2$, there exist $2$ positive values of $x.$
For every $t\le -2$, there exist $2$ negative values of $x.$
For every $t\in (-2,2)$, there exist no real roots for $x.$

So, for equation $\left(1\right)$ to have $2$ negative and $2$ positive roots, equation $\left(2\right)$ should have one root $t_1\le -2$ and another root $t_2\ge 2$, so the required conditions are,
Therefore $[-2,2]$ lies in between the roots,
$\left(i\right)f(-2)\le 0\Rightarrow 4-2(k-1)-1\le 0\Rightarrow k\ge \frac{5}{2}\left(ii\right)f\left(2\right)\le 0\Rightarrow 4+2(k-1)-1\le 0\Rightarrow k\le -\frac{1}{2}\therefore k\in \varphi$