Let f:(−∞,∞)→(−∞,∞) be given by f(x)=x|x|+|x−1|−|x−2|2+|x−3|3 If A denotes the set of points wheref (x) is not differentiable, then A =

# Let $f:\left(-\mathrm{\infty },\mathrm{\infty }\right)\to \left(-\mathrm{\infty },\mathrm{\infty }\right)$ be given by $f\left(x\right)=x|x|+|x-1|-|x-2{|}^{2}+|x-3{|}^{3}$ If A denotes the set of points where is not differentiable, then A =

1. A

{3}

2. B

{1}

3. C

{0}

4. D

{2}

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### Solution:

$h\left(x\right)=x|x|=\left\{\begin{array}{cc}{x}^{2},& x\ge 0\\ -{x}^{2},& x<0\end{array}\right\$is a differentiable function

$u\left(x\right)=|x-2{|}^{2}=\left(x-2{\right)}^{2}$ is a differentiable function

$v\left(x\right)=|x-3{|}^{3}=\left\{\begin{array}{cc}\left(x-3{\right)}^{3},& x\ge 3\\ -\left(x-3{\right)}^{3},& x<3\end{array}\right\$is a differentiable
function

$w\left(x\right)=|x-1|$ is not differentiable at $x=1$.

$f$ is differentiable everywhere except at $x=1$

Hence $A\mid =\left\{1\right\}$

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