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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
Solution
(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 \[f\left( \beta \right) = 3\left( \left( 3 \right)^2 - 12 \right) = - 9\] \[\Rightarrow f\left( \alpha \right) = f\left( \beta \right) = - 9\]
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