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JEE Advanced 2012Paper 1
Exam
JEE Advanced
Year
2012
Paper
Paper 1
Subject
Mathematics
Chapter
Application of Derivatives

JEE Advanced 2012 · Paper 1 · Mathematics · Application of Derivatives

Let f: I R arrow I R be defined as f(x)=|x|+|x^ 2 -1| . The total...

Let f:IR→IRf: I R \rightarrow I R be defined as f(x)=∣x∣+∣x2−1∣f(x)=|x|+\left|x^{2}-1\right|. The total number of points at which ff attains either a local maximum or a local minimum is

Correct answer

5

Step-by-step solution

f(x)=|x|+\left|x^{2}-1\right|=\left\{−x+x2−1,x<−1−x−x2+1,−1≤x≤0x−x2+1,0<x<1x2+x−1,x≥1\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[2x−1,x<−1−2x−1,−1≤x≤0−2x+1,0<x<12x+1,x>1\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 12,−12,−1,0\frac{1}{2}, \frac{-1}{2},-1,0 and 1 . We observe at five points f′(x)f^{\prime}(x) changes its sign ∴\therefore There are 5 points at which either local maximum or local minimum.
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