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

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

Answer 1

Answer: (A), (D)

$f\left(x\right)$ decreases for $x$ $ϵ$ $[2n\pi ,(2n+1\left)\pi \right]$, $n$ $ϵ$ $Z$
$f\left(x\right)$ increases for $x$ $ϵ$ $R$.

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,

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

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

$\Rightarrow$ $cos\theta =1/3$

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