JEE Advanced · Mathematics · 25. AOD
The total number of local maxima and local minima of the function \(f(x)=\left\{\begin{array}{lc}(2+x)^3 ; & -3 < x \leq-1 \\ x^{\frac{2}{3}} ; & -1 < x < 2\end{array}\right.\) is
- A
0
- B
1
- C
2
- D
3
Answer & Solution
Correct Answer
(C)
2
Step-by-step Solution
Detailed explanation
Given that, \(f(x)=\left\{\begin{array}{l}(2+x)^3 ;-3 < x \leq-1 \\ x^{2 / 3} ; \quad-1 < x < 2\end{array}\right.\)
\[
\Rightarrow \quad f^{\prime}(x)=\left\{\begin{array}{l}
3(2+x)^2 ;-3 < x \leq 1 \\
\frac{2}{3} x^{-\frac{1}{3}} ; \quad-1 < x < 2
\end{array}\right.
\]
Clearly, \(f^{\prime}(x)\) changes its sign at \(x=-1\) from \(+\) ve to -ve and so \(f(x)\) has local maxima at \(x=-1\).
Also, \(f^{\prime}(0)\) does not exist but \(f^{\prime}\left(0^{-}\right) < 0\) and \(f^{\prime}\left(0^{+}\right) < 0\).
It can only be inferred that \(f(x)\) has a possibility of a minima at \(x=0\).
Hence, the given function has one local maxima at \(x=-1\) and one local minima at \(x=0\).
\[
\Rightarrow \quad f^{\prime}(x)=\left\{\begin{array}{l}
3(2+x)^2 ;-3 < x \leq 1 \\
\frac{2}{3} x^{-\frac{1}{3}} ; \quad-1 < x < 2
\end{array}\right.
\]

Clearly, \(f^{\prime}(x)\) changes its sign at \(x=-1\) from \(+\) ve to -ve and so \(f(x)\) has local maxima at \(x=-1\).
Also, \(f^{\prime}(0)\) does not exist but \(f^{\prime}\left(0^{-}\right) < 0\) and \(f^{\prime}\left(0^{+}\right) < 0\).
It can only be inferred that \(f(x)\) has a possibility of a minima at \(x=0\).
Hence, the given function has one local maxima at \(x=-1\) and one local minima at \(x=0\).
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