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  • Number of points where <img alt="f(x)=x^2-\left|x^2-1\right|+2|| x|-1|+2|x|-7" src="https://entrancecorner.oncodecogs.com/gif.latex?f%28x%29%3Dx%5E2-%5Cleft%7Cx%5E2-1%5Cright%7C&plus;2%7C%

Number of points where 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 f(x) is continuous for all real x.

\mathrm{\begin{aligned} & f(x)=x^2-\left|x^2-11+2\right| \operatorname{tx}-1|+2| x \mid-7 \\ & = \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} \\ & = \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} \end{aligned}}

\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 differential for all real x.

 

Posted by

himanshu.meshram

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