Question

f is a real valued function given by \[f\left( x \right) = 27 x^3 + \frac{1}{x^3}\] and α, β are roots of \[3x + \frac{1}{x} = 12\] . Then,

 

 

Options

• (a) f(α) ≠ f(β)

• (b) f(α) = 10

• (c) f(β) = −10

• (d) None of these

   

Answer 1

(d) None of these

Given: \[f\left( x \right) = 27 x^3 + \frac{1}{x^3}\] \[\Rightarrow f\left( x \right) = \left( 3x + \frac{1}{x} \right)\left( 9 x^2 + \frac{1}{x^2} - 3 \right)\]

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

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

Since α and β are the roots of

 

\[3x + \frac{1}{x} = 12\]

\[3\alpha + \frac{1}{\alpha} = 12 \text{ and  } 3\beta + \frac{1}{\beta} = 12\]\[\Rightarrow f\left( \alpha \right) = 12\left( \left( 12 \right)^2 - 9 \right)\] and \[f\left( \beta \right) = 12\left( \left( 12 \right)^2 - 9 \right)\] \[\Rightarrow f\left( \alpha \right) = f\left( \beta \right) = 12\left( \left( 12 \right)^2 - 9 \right)\]