Q50 / 160 · TS EAMCET 2024 (10 May Shift 2) · Mathematics
If $\omega$ is the complex cube root of unity and
$\left(\frac{a+b \omega+c \omega^2}{c+a \omega+b \omega^2}\right)^k+\left(\frac{a+b \omega+c \omega^2}{b+a \omega^2+c \omega}\right)^l=2$
then $2 k+l$ is always
then $2 k+l$ is always
- Adivisible by 2
- Bdivisible by 6
- Correct answerCdivisible by 3
- Ddivisible by 5
Solution
$\begin{array}{r}\left(\frac{a+b \omega+c \omega^2}{c+a \omega+b \omega^2}\right)^k+\left(\frac{a+b \omega+c \omega^2}{b+a \omega^2+c \omega}\right)^l=2 \\ \left(\frac{\omega\left(a+b \omega+c \omega^2\right)}{\omega\left(c+a \omega+b \omega^2\right)}\right)^k+\left(\frac{a+b \omega+c \omega^2}{b+a \omega^2+c \omega}\right)^l=2\end{array}$
$\begin{aligned} & =\left(a+b \omega+c \omega^2\right)^k\left(\mathrm{~b}+a \omega^2+c \omega\right)^l+\left(a+b \omega+c \omega^2\right)^l \\ & \left(b+a \omega^2+c \omega\right)^{\mathrm{k}}=2 \\ & \left(a+b \omega+c \omega^2\right)^{k+l} \frac{1}{\omega^{2 k+l}}\left(b+a \omega^2+c \omega\right)^{k+l} \\ & +\frac{1}{\omega^{2 k+l}}\left(b+a \omega^2+c \omega\right)^{k+l}=2\left(b+a \omega^2+c \omega\right)^{k+l}\end{aligned}$
$\Rightarrow \frac{1}{\omega^{2 k+l}}+\frac{1}{\omega^{2 k+l}}=2$
$\Rightarrow \omega^{2 k+l}=1 \Rightarrow 2 k+l$ is divisible by 3
Hence, the correct answer is option: 'C'.