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 x=α 2 , then α and α 2 are the roots of the equation :

Mathematics · Application of Derivatives

JEE Main 2024 — 8 April, Shift 2 — Question 14

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

$x=\alpha^{2}$ , then $\alpha$ and $\alpha^{2}$ are the roots of the equation :

$\alpha+\alpha^{2}=3 \mathrm{a} \& \alpha \times \alpha^{2}=2 \mathrm{a}^{2}$

$\left(\alpha+\alpha^{2}\right)^{3}=27 a^{3}$

$\Rightarrow 2 \mathrm{a}^{2}+4 \mathrm{a}^{4}+3(3 \mathrm{a})\left(2 \mathrm{a}^{2}\right)=27 \mathrm{a}^{3}$

$\Rightarrow 2+4 \mathrm{a}^{2}+18 \mathrm{a}=27 \mathrm{a}$

$\Rightarrow 4 \mathrm{a}^{2}-9 \mathrm{a}+2=0$ $\Rightarrow 4 \mathrm{a}^{2}-8 \mathrm{a}-\mathrm{a}+2=0$

$\Rightarrow(4 \mathrm{a}-1)(\mathrm{a}-2)=0 \Rightarrow \mathrm{a}=2$

so $6 x^{2}-36 x+48=0$

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

If we take $\mathrm{a}=\frac{1}{4}$

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

Answer key and solution verified before publishing.

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