If the function $\mathrm{f}(\mathrm{x})=2 \mathrm{x}^{3}-9 \mathrm{ax}^{2}+12 \mathrm{a}^{2} \mathrm{x}+1, \mathrm{a}>0$ has a local maximum at $\mathrm{x}=\alpha$ and a local minimum $\mathrm{x}=\alpha^{2}$, then $\alpha$ and $\alpha^{2}$ are the roots of the equation :

JEE Main · 2024 · 08 Apr 2024 · Shift 2 · Application of Derivatives

If the function

f (x) = 2 x^{3} - 9 ax^{2} + 12 a^{2} x + 1, a > 0

has a local maximum at

x = α

and a local minimum

x = α^{2}

, then

α

and

α^{2}

are the roots of the equation :

Answer: (A)

f^{'} (x) = 6 x^{2} - 18 a x + 12 a^{2} = 0 ⟹ x^{2} - 3 a x + 2 a^{2} = 0 ⟹ (x - α) (x - α^{2}) = 0

with

α + α^{2} = 3 a

α \cdot α^{2} = 2 a^{2}

.
From the product:

α^{3} = 2 a^{2}

. Cubing the sum:

(α + α^{2})^{3} = 27 a^{3} ⟹ α^{3} + α^{6} + 3 α^{3} \cdot α^{2} (α + α^{2}) = 27 a^{3}

.
Substituting

α^{3} = 2 a^{2}

and

α + α^{2} = 3 a

2 a^{2} + 4 a^{4} + 3 (2 a^{2}) (3 a) = 27 a^{3} ⟹ 4 a^{4} - 9 a^{3} + 2 a^{2} = 0 ⟹ 4 a^{2} - 9 a + 2 = 0

(for

a \neq = 0

)

⟹ (4 a - 1) (a - 2) = 0

a = 1/4

makes

α^{2} < α

, inconsistent with a local max at

α

and local min at

α^{2} > α

for

a > 0

; so

a = 2

.
With

a = 2

α + α^{2} = 6

α α^{2} = 8

, giving the quadratic

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

JEE Main · 2024 · 08 Apr 2024 · Shift 2 · Single correct option