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  • Number of points where <img alt="\mathrm{f(x)=x^{2}-\left|x^{2}-1\right|+2|| x|-1|+2|x|-7}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7Bf%28x%29%3Dx%5E%7B2%7D-%5Cleft%7Cx%5E%7B

Number of points where \mathrm{f(x)=x^{2}-\left|x^{2}-1\right|+2|| x|-1|+2|x|-7}  is non-differentiable is

Option: 1

0


Option: 2

1


Option: 3

2


Option: 4

3


Answers (1)

best_answer

Clearly \mathrm{f(x)} is continuous for all real \mathrm{x}.

\mathrm{f(x)=x^{2}-\left|x^{2}-1\right|+2|| x|-1|+2|x|-7}

\mathrm{ = \begin{cases}x^{2}-\left(x^{2}-1\right)+2(-x-1)-2 x-7, & x<-1 \\ x^{2}-\left(1-x^{2}\right)+2(x+1)-2 x-7, & -1 \leq x \leq 0 \\ x^{2}-\left(1-x^{2}\right)+2(1-x)+2 x-7, & 0 \leq x<1 \\ x^{2}-\left(x^{2}-1\right)+2(x-1)+2 x-7, & x>1\end{cases}}
\mathrm{ = \begin{cases}-4 x-8, & x<-1 \\ 2 x^{2}-6, & -1 \leq x<0 \\ 2 x^{2}-6, & 0 \leq x<1 \\ 4 x-8, & x>1\end{cases}}

\mathrm{ f^{\prime}(x)= \begin{cases}-4, & x<-1 \\ 4 x, & -1<x<0 \\ 4 x, & 0<x<1 \\ 4, & x>1\end{cases} }
Clearly, \mathrm{ f(x) } is differentiable for all real \mathrm{ x}.
 

Posted by

Ritika Harsh

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