Question

f is a real valued function given by $f\left(x\right)=27x^3+\frac{1}{x^3}$and α, β are roots of $3x+\frac{1}{x}=12$. Then,
(a) f(α) ≠ f(β)
(b) f(α) = 10
(c) f(β) = −10
(d) None of these

Answer 1

(d) None of these

Given:
$f\left(x\right)=27x^3+\frac{1}{x^3}$
$\Rightarrow f\left(x\right)=\left(3x+\frac{1}{x}\right)\left(9x^2+\frac{1}{x^2}-3\right)$
$\Rightarrow f\left(x\right)=\left(3x+\frac{1}{x}\right)\left({\left(3x+\frac{1}{x}\right)}^2-9\right)$
$\Rightarrow f\left(\alpha \right)=\left(3\alpha +\frac{1}{\alpha }\right)\left({\left(3\alpha +\frac{1}{\alpha }\right)}^2-9\right)$
Since α and β are the roots of $3x+\frac{1}{x}=12$,
$3\alpha +\frac{1}{\alpha }=12\mathrm{and}3\beta +\frac{1}{\beta }=12$
$\Rightarrow f\left(\alpha \right)=12\left({\left(12\right)}^2-9\right)$ and $f\left(\beta \right)=12\left({\left(12\right)}^2-9\right)$
$\Rightarrow f\left(\alpha \right)=f\left(\beta \right)=12\left({\left(12\right)}^2-9\right)$

Disclaimer: The question in the book has some error, so none of the options are matching with the solution. The solution is created according to the question given in the book.