Question

Let $f\left(x\right)=x^4+a{x}^3+b{x}^2+ax+1$ be a polynomial, where $a,b\in R$. If $b=-1$, then the range of $a$ for which $f\left(x\right)=0$ does not have real roots is

• A. $(-\frac{1}{2},\frac{1}{2})$
• B. $(-1,1)$
• C. $(-\frac{5}{2},\frac{5}{2})$
• D. $(-3,3)$

Answer 1

The correct option is A $(-\frac{1}{2},\frac{1}{2})$

$x^4+a{x}^3+b{x}^2+ax+1=0\cdots \left(1\right)x^4+a{x}^3-x^2+ax+1=0$
Dividing the equation by $x^2$,
$(x^2+\frac{1}{x^2})+a(x+\frac{1}{x})-1=0$
Let $x+\frac{1}{x}=t,t\in (-\infty ,-2]\cup [2,\infty )$
Now,
${(x+\frac{1}{x})}^2=t^2\Rightarrow x^2+\frac{1}{x^2}=t^2-2$
Now,
$t^2-2+at-1=0\Rightarrow t^2+at-3=0\cdots \left(2\right)$
let the roots of equation $2$ be $t_1,t_2(t_1<t_2)$
For every $t\ge 2$, there exist $2$ positive values of $x.$
For every $t\le -2$, there exist $2$ negative values of $x.$

Equation $\left(1\right)$ will have no real roots iff equation $\left(2\right)$ will have
(A) Non real roots or
(B) $t_1,t_2\in (-2,2)$
Case(A) : Non real roots,
$D<0\Rightarrow a^2+12<0\Rightarrow a\in \varphi \cdots \left(3\right)$

Case (B)$:t_1,t_2\in (-2,2)$
$-2<t_1,t_2<2\left(i\right)D\ge 0\Rightarrow a^2+12\ge 0\Rightarrow a\in R\left(ii\right)f(-2)>0\Rightarrow 1-2a>0\Rightarrow a<\frac{1}{2}\left(iii\right)f\left(2\right)>0\Rightarrow 2a+1>0\Rightarrow a>-\frac{1}{2}\left(iv\right)-2<-\frac{B}{2A}<2\Rightarrow -2<-\frac{a}{2}<2\Rightarrow 4>a>-4\therefore a\in (-\frac{1}{2},\frac{1}{2})\cdots \left(4\right)$

From equation $\left(3\right)$ and $\left(4\right)$,
$a\in (-\frac{1}{2},\frac{1}{2})$