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 cos $2\theta$ = 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,

$a{z}^2+z+1=0$ (1)

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$\Rightarrow$ $a{z}^2+z+1=0$ (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,

$(\overline{a}-a{)}^2+2(a+\overline{a})=0\Rightarrow {\left(\frac{\overline{a}-a}{2}\right)}^2$ + $\frac{a+\overline{a}}{2}=0$

$\Rightarrow$ -${\left(\frac{a-\overline{a}}{2i}\right)}^2+\left(\frac{a+\overline{a}}{2i}\right)=0$ $\Rightarrow$ - $si{n}^2\theta +cos\theta =0$

$\Rightarrow$ $cos\theta =si{n}^2\theta$ (3)

Now,

$f\left(x\right)=x^3-3x^2+3(1$ + cos $\theta )x+5$

$f^{\prime }\left(x\right)=3x^3-6x+3(1$ + cos $\theta )$

Its discriminant is

$36-36(1+cos\theta )=-36cos\theta =-36si{n}^2\theta$ < 0

$\Rightarrow$ $f\left(x\right)$ > 0 $\forall$ $x$ $ϵ$ $R$

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

Also, $f\left(0\right)$ = 5, then $f\left(x\right)$ = 0 has one negative root. Now,

$cos2\theta =cos\theta \Rightarrow 1-2si{n}^2\theta =cos\theta$

$\Rightarrow$ $1-2cos\theta =cos\theta$

$\Rightarrow$ $cos\theta =\frac{1}{3}$

which has four roots for $\theta$ $ϵ$ $[0,4\pi ]$.