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.

Number of roots of the equation $cos2\theta =cos\theta$ in $[0,4\pi ]$ are

• A. 2
• B. 3
• C. 4
• D. 6

Answer 1

The correct option is C 4

We have
$\underset{–}{a{z}^2+z+1}=0$ (1)
$\Rightarrow a{z}^2+z+1=0$ (taking conjugate of both sides)
$\Rightarrow \overline{a}{z}^2-z+1=0$ (2)
[Since $z$ is purely imaginary $\overline{z}=-z]$
Eliminating z from (1) and (2) by cross-multiplication rule,
$(\overline{a}-a{)}^2+2(a+\overline{a})=0\Rightarrow {\left(\frac{\overline{a}-a}{2}\right)}^2+\frac{a+\overline{a}}{2}=0$
$\Rightarrow -{\left(\frac{a-\overline{a}}{.2i}\right)}^2+\left(\frac{a+\overline{a}}{2}\right)=0\Rightarrow -si{n}^2\theta +cos\theta =0$
$\Rightarrow cos\theta =si{n}^2\theta$ (3)
Now,
$f\left(x\right)=x^3-3x^2+3(1+cos\theta )x+5$
$f^{\prime }\left(x\right)=3x^2-6x+3(1+cos\theta )$
Its discriminant is
$36-36(1+cos\theta )=-36cos\theta =-36si{n}^2\theta <0$
$\Rightarrow f\left(x\right)>0\forall xϵR$
Hence, $f\left(x\right)$ is increasing $\forall xϵR$, Also, $f\left(0\right)=5$, then $f\left(x\right)=0$ has one negative root. Now,
$cos2\theta =cos\theta \Rightarrow 1-si{n}^2\theta =cos\theta$
$\Rightarrow 1-2cos\theta =cos\theta$
$\Rightarrow cos\theta =1/3$
which has four roots for $\theta ϵ[0,4\pi ]$.