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  • Let <img alt="\mathrm{f(x)=\left(x^2-4\right)\left|\left(x^3-6 x^2+11 x-6\right)\right|+\frac{x}{1+|x|}.}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7Bf%28x%29%3D%5Cleft%28x%5E

Let \mathrm{f(x)=\left(x^2-4\right)\left|\left(x^3-6 x^2+11 x-6\right)\right|+\frac{x}{1+|x|}.} The set of points at which the function f(x) is not differentiable is

Option: 1

\{-2,2,1,3\}


Option: 2

\{-2,0,3\}


Option: 3

\{-2,2,0\}


Option: 4

\{1,3\}


Answers (1)

\mathrm{ f(x)=\left(x^2-4\right)\left|\left(x^3-6 x^2+11 x-6\right)\right|+\frac{x}{1+|x|} }
\mathrm{\frac{x}{1+|x|}} is always differentiable.

\mathrm{(x-2)(x+2)|(x-1)(x-2)(x-3)|} is not differentiable at \mathrm{x=1,3.}

So, f(x) is not differentiable at \mathrm{x=1,3.}

Posted by

Sumit Saini

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