Question

Consider the equation $a{z}^2+z+1=0$ having purely imaginary root where $a=\cos \theta +i\sin \theta ,i=\sqrt{-1}$ and function $f\left(x\right)=x^3-3x^2+3(1+\cos \theta )x+5$, then answer the following questions

$i)$ Which of the following is true about $f\left(x\right)$?

• A. $f\left(x\right)$ decreases for $x\in [2n\pi ,(2n+1\left)\pi \right],n\in Z$
• B. $f\left(x\right)$ is non-monotonic function
• C. $f\left(x\right)$ increases for $x\in R$
• D. $f\left(x\right)$ decreases for $x\in \left[\right(2n-1)\frac{\pi }{2},(2n+1\left)\frac{\pi }{2}\right],n\in Z$

Answer 1

The correct option is C $f\left(x\right)$ increases for $x\in R$

Given:
$a{z}^2+z+1=0$ has purely imaginary root

$a{z}^2+z+1=0\dots \left(i\right)$
Taking conjugate on both sides
$\overline{a}{z}^2+\overline{z}+1=0$
(Since $z$ is purely imaginary, $\overline{z}=-z$)
$\overline{a}{z}^2-z+1=0\dots \left(ii\right)$
Eliminating $z$ from $\left(i\right)$ and $\left(ii\right)$ by cross-multiplication rule, we get
$(\overline{a}-a{)}^2+2(a+\overline{a})=0\dots (iii)$

Given: $a=\cos \theta +i\sin \theta ,\overline{a}=\cos \theta -i\sin \theta$
Substituting $a$ and $\overline{a}$ in $\left(iii\right)$,
$(-2i\sin \theta {)}^2+2(2\cos \theta )=0$
$\Rightarrow -4{\sin }^2\theta +4\cos \theta =0$
$\Rightarrow \cos \theta ={\sin }^2\theta \dots \left(iv\right)$

$f\left(x\right)=x^3-3x^2+3(1+\cos \theta )x+5$ (Given)
$f^{\prime }\left(x\right)=3x^2-6x+3(1+\cos \theta )$
Discriminant of $f^{\prime }\left(x\right)$ is
$36-36(1+\cos \theta )=-36\cos \theta =-36{\sin }^2\theta <0\Rightarrow f^{\prime }\left(x\right)>0\forall x\in R$

Hence, $f\left(x\right)$ is increasing $\forall x\in R$