Question

If $f\left(x\right)=27x^3+\frac{1}{x^3}$ and $\alpha ,\beta$ are the roots of $3x+\frac{1}{x}=2$ is

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

Answer 1

Answer: (A), (C)

The correct options are
A $f\left(\beta \right)=-10$
C $f\left(\alpha \right)=f\left(\beta \right)$

$3x+\frac{1}{x}=2$

Cubing the above equation, we get

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

$\therefore 27x^3+\frac{1}{x^3}+3\left(3x\right)\left(\frac{1}{x}\right)(3x+\frac{1}{x})=8$

$\therefore 27x^3+\frac{1}{x^3}+9(3x+\frac{1}{x})=8$

$\therefore 27x^3+\frac{1}{x^3}+9\left(2\right)=8$

$\therefore 27x^3+\frac{1}{x^3}=-10$

$\alpha$ and $\beta$ are roots of above equation.

$\therefore 27{\alpha }^3+\frac{1}{{\alpha }^3}=-10$ ...(1)

and $27{\beta }^3+\frac{1}{{\beta }^3}=-10$ ...(2)

$f\left(x\right)=27x^3+\frac{1}{x^3}$

$f\left(\alpha \right)=27{\alpha }^3+\frac{1}{{\alpha }^3}$

$⟹f\left(\alpha \right)=-10$ ...(from 1)

Similarly, $f\left(\beta \right)=27{\beta }^3+\frac{1}{{\beta }^3}$

$⟹f\left(\beta \right)=-10$ ...(from 2)

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

So, the correct options are option (A) and (C)