MCQ
$f$ is a real valued function given by $\text{f(x)}=27\text{x}^3+\frac{1}{\text{x}^3}$ and $\alpha,\beta$ are roots of $3\text{x}+\frac{1}{\text{x}}=12$ Then,
  • A
    $\text{f}(\alpha)\neq\text{f}(\beta)$
  • B
    $\text{f}(\alpha)=10$
  • C
    $\text{f}(\beta)=-10$
  • None of these.

Answer

Correct option: D.
None of these.
$\text{f(x)}=27\text{x}^3+\frac{1}{\text{x}^3}$
$\Rightarrow\text{f(x)}=\Big(3\text{x}+\frac{1}{\text{x}}\Big)\Big(9\text{x}^2+\frac{1}{\text{x}^2}-3\Big)$
$\Rightarrow\text{f(x)}=\Big(3\text{x}+\frac{1}{\text{x}}\Big)\Big(\Big(3\text{x}+\frac{1}{\text{x}}\Big)^2-9\Big)$
$\Rightarrow\text{f}(\alpha)=\Big(3\alpha+\frac{1}{\alpha}\Big)\Big(\Big(3\alpha+\frac{1}{\alpha}\Big)^2-9\Big)$
Since $\alpha$ and $\beta$ are the roots of $3\text{x}+\frac{1}{\text{x}}=12,$
$3\alpha+\frac{1}{\alpha}=12$ and $3\beta+\frac{1}{\beta}=12$
$\Rightarrow\text{f}(\alpha)=12\big((12)^2-9\big)$ and $\text{f}(\beta)=12\big((12)^2-9\big)$
$\Rightarrow\text{f}(\alpha)=\text{f}(\beta)=\big((12)^2-9\big)$

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