Question

If  \[f\left( x \right) = 64 x^3 + \frac{1}{x^3}\] and α, β are the roots of \[4x + \frac{1}{x} = 3\] . Then,

 

Options

• (a) f(α) = f(β) = −9

• (b) f(α) = f(β) = 63

• (c) f(α) ≠ f(β)

• (d) none of these

   

Answer 1

(a) f(α) = f(β) = −9  

Given:\[f\left( x \right) = 64 x^3 + \frac{1}{x^3}\]

\[\Rightarrow f\left( x \right) = \left( 4x + \frac{1}{x} \right)\left( 16 x^2 + \frac{1}{x^2} - 4 \right)\]

\[\Rightarrow f\left( x \right) = \left( 4x + \frac{1}{x} \right)\left( \left( 4x + \frac{1}{x} \right)^2 - 12 \right)\]

\[\Rightarrow f\left( \alpha \right) = \left( 4\alpha + \frac{1}{\alpha} \right)\left( \left( 4\alpha + \frac{1}{\alpha} \right)^2 - 12 \right)\text{ and } f\left( \beta \right) = \left( 4\beta + \frac{1}{\beta} \right)\left( \left( 4\beta + \frac{1}{\beta} \right)^2 - 12 \right)\] Since α and β are the roots of \[4x + \frac{1}{x} = 3\] \[4\alpha + \frac{1}{\alpha} = 3 \text{ and } 4\beta + \frac{1}{\beta} = 3\] \[\Rightarrow f\left( \alpha \right) = 3\left( \left( 3 \right)^2 - 12 \right) = - 9\]  and \[f\left( \beta \right) = 3\left( \left( 3 \right)^2 - 12 \right) = - 9\] \[\Rightarrow f\left( \alpha \right) = f\left( \beta \right) = - 9\]