Question

If $f\left(x\right)=64x^3+\frac{1}{x^3}$ and $\alpha .\beta$ are the roots of $4x+\frac{1}{x}=2$, then

• A. $f\left(\alpha \right)=-64$
• B. $f\left(\beta \right)=-8$
• C. $f\left(\beta \right)=-16$
• D. $f\left(\alpha \right)=-24$

Answer 1

The correct option is C $f\left(\beta \right)=-16$

We have,

$f\left(x\right)=64x^3+\frac{1}{x^3}$

$\alpha$ and $\beta$ be the roots

$4x+\frac{1}{x}=2$ ………$\left(1\right)$

On cube both side and we get,

${(4x+\frac{1}{x})}^3=2^3$

$(4x{)}^3+{\left(\frac{1}{x}\right)}^3+3\times 4x\times \frac{1}{x}(4x+\frac{1}{x})=8$

$64x^3+\frac{1}{x^3}+12(4x+\frac{1}{x})=8$

$f\left(x\right)+12\times 2=8$

$f\left(x\right)+24=8$

$f\left(x\right)=8-24$

$f\left(x\right)=-16$

Now,

$\alpha$ and $\beta$ are the roots

$f\left(\alpha \right)=f\left(\beta \right)=-16$

Hence, this is the answer.