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\mathrm{f}(\mathrm{x})=|| x|-1| is not differentiable at

Option: 1

0


Option: 2

\pm 1,0


Option: 3

1


Option: 4

\pm 1


Answers (1)

best_answer

There are two moduli in the problem.

\mathrm{y}=|| x|-1|. Hence we have to consider the cases when \mathrm{x} \geq 0, \mathrm{x}<0,|\mathrm{x}| \geq 1 or |\mathrm{x}| \leq 1.

Now x \geq 1 \Rightarrow x^2-1 \geq 0 \Rightarrow(x+1)(x-1) \geq 0

 \Rightarrow \quad x \geq 1 \text { or } x \leq-1 \\                           .......(i)

|x| \leq 1 \Rightarrow x^2-1 \leq 0 \Rightarrow(x+1)(x-1) \leq 0 \\\Rightarrow-1 \leq \mathrm{x} \leq 1 \\                                            ...........(ii)

 \therefore \quad y=|x|-1, x \geq 1 \text { or } x \leq-1 \\   .......(iii)

\quad=-(|x|-1)=1-|x|,-1 \leq x \leq 1   ......(iv)   Now consider |\mathrm{x}|-\mathrm{x}, \mathrm{x} \geq 0,|\mathrm{x}|=-\mathrm{x}, \mathrm{x}<0

Now we redefine the given function

\begin{array}{ll} y=-x-1 \, \, \, \, \, \, \, \, \, \, \, x \leq-1 \\ y=x+1 \, \, \, \, \, \, \, \, \, \, \, -1 \leq x<0 \\ y=-x+1 \, \, \, \, \, \, \, \, \, \, \, 0 \leq x<1 \\ y=x-1 \, \, \, \, \, \, \, \, \, \, \, \, \,\, \, x \leq1 \end{array}

From the graph it is clear that f(x) is not differentiable at x = -1, 0, 1.

 

 

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Pankaj

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