Question

Let $f\left(x\right)=x^4+b{x}^3+1.$ A possible value of $b\in R$ for which the equation $f\left(x\right)=0$ has no real root is

• A. $\frac{-11}{5}$
• B. $\frac{-3}{2}$
• C. $2$
• D. $\frac{5}{2}$

Answer 1

The correct option is B $\frac{-3}{2}$

$f\left(x\right)=x^4+b{x}^3+1$
$f\left(x\right)=0$ has no real root. This means $f\left(x\right)>0$ for all $x\in R$
$f^{\prime }\left(x\right)=4x^3+3b{x}^2$
$f^{\prime }\left(x\right)=0\Rightarrow x^2(4x+3b)=0$
We can check by first derivative test that $x=\frac{-3b}{4}$ is the point of minima.

$f\left(\frac{-3b}{4}\right)>0$
$\Rightarrow {\left(\frac{-3b}{4}\right)}^4+b{\left(\frac{-3b}{4}\right)}^3+1>0$
$\Rightarrow b^4<\frac{256}{27}$
$\Rightarrow |b|<\frac{4}{\sqrt[4]{427}}=1.754$
$\therefore$ Option(b) is correct.