JEE Main · 2024 · 29 Jan 2024 · Shift 2 · Application of Derivatives

1. (A)exactly one point of local minima and no point of local maxima
2. (B)exactly one point of local maxima and no point of local minima
3. (C)exactly one point of local maxima and exactly one point of local minima✓ Correct
4. (D)exactly two points of local maxima and exactly one point of local minima

Answer: (C)

Solution

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

Sign of $f^{\prime }(x)$:

• For $x<-1$: $x^{1/3}<-1$, so numerator and denominator are both negative $\Rightarrow f^{\prime }(x)>0$.
• For $-1<x<0$: $x^{1/3}\in (-1,0)$, numerator positive, denominator negative $\Rightarrow f^{\prime }(x)<0$.
• For $x>0$: both positive $\Rightarrow f^{\prime }(x)>0$.

So $f^{\prime }$ changes from $+$ to $-$ at $x=-1$ (local maximum) and from $-$ to $+$ at $x=0$ (local minimum, where $f^{\prime }$ is undefined but changes sign).

So $f$ has exactly one local maximum and exactly one local minimum.

JEE Main · 2024 · 29 Jan 2024 · Shift 2 · Single correct option