JEE Advanced · Mathematics · 25. AOD
Let \(f: I R \rightarrow I R\) be defined as \(f(x)=|x|+\left|x^{2}-1\right|\). The total number of points at which \(f\) attains either a local maximum or a local minimum is
- A 9
- B 4
- C 5
- D 3
Answer & Solution
Correct Answer
(C) 5
Step-by-step Solution
Detailed explanation
\(f(x)=|x|+\left|x^{2}-1\right|=\left\{\begin{array}{c}-x+x^{2}-1, x < -1 \\ -x-x^{2}+1,-1 \leq x \leq 0 \\ x-x^{2}+1,0 < x < 1 \\ x^{2}+x-1, \quad x \geq 1\end{array}\right.\)
\(\therefore \quad f^{\prime}(x)=\left[\begin{array}{ccc}
2 x-1 & , & x < -1 \\
-2 x-1 & , & -1 \leq x \leq 0 \\
-2 x+1 & , & 0 < x < 1 \\
2 x+1 & , & x>1
\end{array}\right.\)
Critical points are \(\frac{1}{2}, \frac{-1}{2},-1,0\) and 1 .
We observe at five points \(f^{\prime}(x)\) changes its sign \(\therefore\) There are 5 points at which either local maximum or local minimum.
\(\therefore \quad f^{\prime}(x)=\left[\begin{array}{ccc}
2 x-1 & , & x < -1 \\
-2 x-1 & , & -1 \leq x \leq 0 \\
-2 x+1 & , & 0 < x < 1 \\
2 x+1 & , & x>1
\end{array}\right.\)
Critical points are \(\frac{1}{2}, \frac{-1}{2},-1,0\) and 1 .
We observe at five points \(f^{\prime}(x)\) changes its sign \(\therefore\) There are 5 points at which either local maximum or local minimum.
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