Mathematics · Application of Derivatives

JEE Main 2024 — 8 April, Shift 1 — Question 5

The number of critical points of the function $f(x)=(x-2)^{2 / 3}(2 x+1)$ is :

Answer: A

$f(x)=(x-2)^{2 / 3}(2 x+1)$

$f^{\prime}(x)=\frac{2}{3}(x-2)^{-1 / 3}(2 x+1)+(x-2)^{2 / 3}$

$f^{\prime}(x)=2 \times \frac{(2 x+1)+(x-2)}{3(x-2)^{1 / 3}}$

$\frac{3 x-1}{(x-2)^{1 / 3}}=0$

Critical points $\mathrm{x}=\frac{1}{3}$ and $\mathrm{x}=2$

Answer key and solution verified before publishing.

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