Question

If $f\left(x\right)=64x^3+\frac{1}{x^3}$ and $\alpha ,\beta$ are the roots of $4x+\frac{1}{x}=3$. Then,

• A. $f\left(\alpha \right)=f\left(\beta \right)=-9$
• B. $f\left(\alpha \right)=f\left(\beta \right)=63$
• C. $f\left(\alpha \right)\ne f\left(\beta \right)$
• D. None of these

Answer 1

The correct option is A

$f\left(\alpha \right)=f\left(\beta \right)=-9$

Given:
$f\left(x\right)=64x^3+\frac{1}{x^3}\Rightarrow f\left(x\right)=(4x+\frac{1}{x})(16x^2+\frac{1}{x^2}-4)\Rightarrow f\left(x\right)=(4x+\frac{1}{x})({(4x+\frac{1}{x})}^2-12)\Rightarrow f\left(\alpha \right)=(4\alpha +\frac{1}{\alpha })({(4\alpha +\frac{1}{\alpha })}^2-12)$ and
$f\left(\beta \right)=(4\beta +\frac{1}{\beta })({(4\beta +\frac{1}{\beta })}^2-12)$
Since $\alpha$ and $\beta$ are the roots of $4x+\frac{1}{x}=3$
$4\alpha +\frac{1}{\alpha }=3$ and $4\beta +\frac{1}{\beta }=3\Rightarrow f\left(\alpha \right)=3\left(\right(3{)}^2-12)=-9$ and $f\left(\beta \right)=3$
$\left(\right(3{)}^2-12)=-9\Rightarrow f(\alpha )=f(\beta )=-9$