Application Of Derivatives Ques 84
The function $f(x)=2|x|+|x+2|-||x+2|-2|x||$ has a local minimum or a local maximum at $x$ is equal to
(2013 Adv.)
-2
(b) $\frac{-2}{3}$
2
(d) $2 / 3$
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Answer:
Correct Answer: 84.$(a, b)$
Solution:
Formula:
Maxima and Minima of functions of one variable :
- PLAN
$ \begin{aligned} & \text { We know that, }|x|=\begin{array}{c} x, \quad \text { if } x \geq 0 \text{ and } x \in \mathbb{R} \\ -x, \quad \text { if } x<0 \end{array} \\ & \Rightarrow \quad|x-a|=\begin{array}{cc} x - a, & \text{ if } x \geq a \ -(x-a), & \text { if } x<a \end{array} \end{aligned} $
and for non-differentiable continuous function, the maximum or minimum can be checked with a graph
Here, $f(x)=2|x|+|x+2|-||x+2|-2|x||$ $-2x-(x+2)+(x-2)$, if when $x \leq-2$ $-2x+x+2+3x+2, \quad$ if $\quad$ when $-2<x \leq-2/3$
$ \begin{aligned} & =\quad-4 x, \quad \text { if } \quad -\frac{2}{3}<x \leq 0 \\ & 4x, \quad \text{if } 0 < x \leq 2 \\ & 2 x+4, \quad \text { if } \quad x>2 \\ & -2 x-4, \text { if } \quad x \leq-2 \\ & 2x+4, \quad \text { if } \quad-2<x \leq-2/3 \\ & =-4x, \text { if } \quad-\frac{2}{3}<x \leq 0 \\ & 4x, \quad \text { if } \quad 0<x \leq 2 \\ & 2 x+4, \quad \text { if } \quad x>2 \end{aligned} $
Graph of $y=f(x)$ is shown as
BETA
