Question

If f is a real valued function given by $f\left(x\right)=27x^3+\frac{1}{x^3}$ and $\alpha ,\beta$ are roots of $3x+\frac{1}{x}=2$. Then,

• A. $f\left(\alpha \right)\ne f\left(\beta \right)$
• B. $f\left(\alpha \right)=10$
• C. $f\left(\beta \right)=-10$
• D. None of these

Answer 1

The correct option is C

$f\left(\beta \right)=-10$

Given:
$f\left(x\right)=27x^3+\frac{1}{x^3}\Rightarrow f\left(x\right)=(3x+\frac{1}{x})(9x^2+\frac{1}{x^2}-3)\Rightarrow f\left(x\right)=(3x+\frac{1}{x})({(3x+\frac{1}{x})}^2-9)\Rightarrow f\left(\alpha \right)=(3\alpha +\frac{1}{\alpha })({(3\alpha +\frac{1}{\alpha })}^2-9)$
$Sin\alpha$ and $\beta$ are the roots of $3x+\frac{1}{x}=2$
$3\alpha +\frac{1}{\alpha }=2$ and $3\beta +\frac{1}{\beta }=2$
$\Rightarrow f\left(\alpha \right)=2\left(\right(2{)}^2-9)$ and
$f\left(\beta \right)=2\left(\right(2{)}^2-9)$
$\Rightarrow f\left(\alpha \right)=f\left(\beta \right)=2\left[\right(2{)}^2-9]=2[4-9]=-10\therefore (c\left)f\right(\beta )=-10$