If the function f ( x ) = 2 x 3 - 9 a x 2 + 12 a 2 x + 1 , a > 0 has a local maximum at x = and a local minimum at x = 2 , then and 2 are the roots of the equation : [2024]

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Solution

Q.
If the function $f (x) = 2 x^{3} - 9 a x^{2} + 12 a^{2} x + 1, a > 0$ has a local maximum at $x = α$ and a local minimum at $x = α^{2},$ then $α$ and $α^{2}$ are the roots of the equation : [2024]

1 $x^{2} + 6 x + 8 = 0$

2 $8 x^{2} - 6 x + 1 = 0$

3 $8 x^{2} + 6 x - 1 = 0$

4 $x^{2} - 6 x + 8 = 0$

Ans.
(4)

$We have, f (x) = 2 x^{3} - 9 a x^{2} + 12 a^{2} x + 1$

$\Rightarrow f' (x) = 6 x^{2} - 18 a x + 12 a^{2} = 6 (x^{2} - 3 a x + 2 a^{2})$

$\Rightarrow 6 (x - a) (x - 2 a) = 0$

$f' (x) = 0 \Rightarrow x = a, 2 a$

$Now, f'' (x) = 12 x - 18 a$

$f'' (a) = 12 a - 18 a = - 6 a < 0 (maximum)$

$f'' (2 a) = 24 a - 18 a = 6 a > 0 (minimum)$

$∴ f (x) has a local maximum at x = a and minimum at x = 2 a .$

$∴ α = a and α^{2} = 2 a$

$\Rightarrow a^{2} = 2 a \Rightarrow a = 0, 2 \Rightarrow a = 2 = α (∵ a > 0)$

$\Rightarrow α^{2} = 4$

$∴ Equation having roots α and α^{2} is$

$x^{2} - (α + α^{2}) x + α \cdot α^{2} = 0$

$\Rightarrow x^{2} - (2 + 4) x + 2 \times 4 = 0 \Rightarrow x^{2} - 6 x + 8 = 0$

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