JEE Advanced 2013 Paper 2Question 1
The function $f(x)=2|x|+|x+2|-||x+2|-2|x||$ has a local minimum or a local maximum at $x=$
Differentiation and its Application
Solution
Concept. $a+b-|a-b|=2\min(a,b)$. With $a=|x+2|$, $b=2|x|$ the function is a simple broken line. Formulas used. (1) $a+b-|a-b|=2\min(a,b)$Step 1 - Rewrite. By (1), $f(x)=2\min\big(|x+2|,\,2|x|\big)$. Step 2 - Switch points. $|x+2|=2|x|$ gives $x=2$ or $x=-\frac23$. Step 3 - The graph.Step 4 - Check each option. $x=-2$: $f=0$ and $f\ge0$ everywhere, so a local minimum. $x=-\frac23$: $f=2(x+2)$ on the left (slope $2$), $f=-4x$ on the right (slope $-4$): local maximum. $x=2$: slopes $4$ then $2$, both positive: no extremum. $x=\frac23$: $f=4x$ on both sides: no extremum. Answer: (A), (B)