If the function f(x)=2 x^3-9 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 :

Correct answer: A. x^2-6 x+8=0 aligned & α+α^2=3 a & α × α^2=2 a^2 & & (α+α^2)^3=27 a^3 & 2 a^2+4 a^4+3(3 a)(2 a^2)=27 a^3 & 2+4 a^2+18 a=27 a & 4 a^2-9 a+2=0 & 4 a^2-8 a-a+2=0 & (4 a-1)(a-2)=0 a=2 & so 6 x^2-36 x+48=0 aligned x^2-6 x+8=0 ...(1) If we take a=(1)/(4) then α=(1)/(2) which is not possible

JEE Main 20248 AprilEvening Shift

Exam: JEE Main
Year: 2024
Date: 8 April
Shift: Evening Shift
Subject: Mathematics
Chapter: Application of Derivatives

JEE Main 2024 · 8 April · Evening Shift · Mathematics · Application of Derivatives

If the function f(x)=2 x^3-9 x^2+12 a^2 x+1, a>0 has a local...

If the function

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

has a local maximum at

x=\alpha

and a local minimum at

x=\alpha^2

, then

\alpha

and

\alpha^2

are the roots of the equation :

Options

A $x^2-6 x+8=0$
B $x^2+6 x+8=0$
C $8 x^2+6 x-1=0$
D $8 x^2-6 x+1=0$

Correct answer

A. $x^2-6 x+8=0$

Step-by-step solution

\begin{aligned} & \alpha+\alpha^2=3 \mathrm{a} & \alpha \times \alpha^2=2 \mathrm{a}^2 \\ & \quad \downarrow \\ & \left(\alpha+\alpha^2\right)^3=27 \mathrm{a}^3 \\ & \Rightarrow 2 \mathrm{a}^2+4 \mathrm{a}^4+3(3 \mathrm{a})\left(2 \mathrm{a}^2\right)=27 \mathrm{a}^3 \\ & \Rightarrow 2+4 \mathrm{a}^2+18 \mathrm{a}=27 \mathrm{a} \\ & \Rightarrow 4 \mathrm{a}^2-9 \mathrm{a}+2=0 \\ & \Rightarrow 4 \mathrm{a}^2-8 \mathrm{a}-\mathrm{a}+2=0 \\ & \Rightarrow(4 \mathrm{a}-1)(\mathrm{a}-2)=0 \Rightarrow \mathrm{a}=2 \\ & \text{ so } 6 \mathrm{x}^2-36 \mathrm{x}+48=0\end{aligned}

\Rightarrow x^2-6 x+8=0

...(1) If we take

\mathrm{a}=\frac{1}{4}

then

\alpha=\frac{1}{2}$ which is not possible

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