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JEE Advanced 2012 · Application of Derivatives

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

Answer: 99

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Written solution

let f: IR → IR be defined as f(x) = |x| + |x² - 1|. The total number of points at which f attains either a local maximum or a local minimum is

f(x)=∣x∣+∣x2−1∣f(x)=|x|+|x^2-1|

To find: total number of local maxima/minima of f\text{To find: total number of local maxima/minima of } f

∣x∣=0⇒x=0;x2−1=0⇒x=±1|x|=0 \Rightarrow x=0;\quad x^2-1=0 \Rightarrow x=\pm 1

breakpoints: −1, 0, 1\text{breakpoints: } -1,\ 0,\ 1

f(x)= x²-x-1, x<-1 -x²-x+1, -1≤ x<0 -x²+x+1, 0≤ x<1 x²+x-1, x≥ 1

Plan: f′ on each piece; check sign change at −1,0,1\text{Plan: } f' \text{ on each piece; check sign change at } -1,0,1

x<−1: f′=2x−1;−1<x<0: f′=−2x−1x<-1:\ f'=2x-1;\qquad -1<x<0:\ f'=-2x-1

0<x<1: f′=−2x+1;x>1: f′=2x+10<x<1:\ f'=-2x+1;\qquad x>1:\ f'=2x+1

x<−1: f′=2x−1<0;x>1: f′=2x+1>0x<-1:\ f'=2x-1<0;\qquad x>1:\ f'=2x+1>0

−1<x<0: f′=−2x−1=0⇒x=−12,f′′<0⇒local max-1<x<0:\ f'=-2x-1=0 \Rightarrow x=-\dfrac{1}{2},\quad f''<0 \Rightarrow \text{local max}

0<x<1: f′=−2x+1=0⇒x=12,f′′<0⇒local max0<x<1:\ f'=-2x+1=0 \Rightarrow x=\dfrac{1}{2},\quad f''<0 \Rightarrow \text{local max}

x=−1: f′ changes −→+⇒local minx=-1:\ f'\ \text{changes } -\to+ \Rightarrow \text{local min}

x=0: f′ changes −→+⇒local minx=0:\ f'\ \text{changes } -\to+ \Rightarrow \text{local min}

x=1: f′ changes −→+⇒local minx=1:\ f'\ \text{changes } -\to+ \Rightarrow \text{local min}

maxima=2,minima=3  ⇒  total=5\text{maxima}=2,\qquad \text{minima}=3 \;\Rightarrow\; \text{total}=5

Total number of local extrema=5\text{Total number of local extrema} = 5

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