# f is a real valued function given by f(x)

Question:

$f$ is a real valued function given by $f(x)=27 x^{3}+\frac{1}{x^{3}}$ and $\alpha, \beta$ are roots of $3 x+\frac{1}{x}=12$. Then,

(a) f(α) ≠ f(β)

(b) f(α) = 10

(c) f(β) = −10

(d) None of these

Solution:

(d) None of these

Given:

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

$\Rightarrow f(x)=\left(3 x+\frac{1}{x}\right)\left(9 x^{2}+\frac{1}{x^{2}}-3\right)$

$\Rightarrow f(x)=\left(3 x+\frac{1}{x}\right)\left(\left(3 x+\frac{1}{x}\right)^{2}-9\right)$

$\Rightarrow f(\alpha)=\left(3 \alpha+\frac{1}{\alpha}\right)\left(\left(3 \alpha+\frac{1}{\alpha}\right)^{2}-9\right)$

Since $\alpha$ and $\beta$ are the roots of $3 x+\frac{1}{x}=12$,

$3 \alpha+\frac{1}{\alpha}=12$ and $3 \beta+\frac{1}{\beta}=12$

$\Rightarrow f(\alpha)=12\left((12)^{2}-9\right)$ and $f(\beta)=12\left((12)^{2}-9\right)$

$\Rightarrow f(\alpha)=f(\beta)=12\left((12)^{2}-9\right)$

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.