Number of points where $f (x) = (1 - x) ∣ ∣ x - x^{2} ∣ ∣ + x$ is not differentiable is

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Solution

The correct option is B 0
$f | x | = (1 - x) | x (1 - x) | + x$
$| x (1 - x) | = {\begin{matrix} x (1 - x) w h e n x (1 - x) \geq 0 x (x - 1) w h e n x (1 - x) < 0 \end{matrix} | x (1 - x) | = {\begin{matrix} x (1 - x) w h e n x \in [0, 1] x (x - 1) w h e n x \in (- \infty, \infty) \cup (1, \infty) \end{matrix}$
$f (x) = ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ \begin{matrix} x {(1 - x)}^{2} + x x \in [0, 1] = x (1 + x^{2} - 2 x + 1) = x (x^{2} - 2 x + 2) x (x - 1) (1 - x) + x x \in (- \infty, 0) \cup (1, \infty) = x (x^{2} + 2 x) = - x^{3} + 2 x^{2} \end{matrix}$
draw the graph of function $f (x)$ .
$f^{'} (1^{-}) = {\frac{d}{d x} (x^{3} - 2 x^{2} + 2 x) ∣ ∣ ∣}_{x = 1}$
$= {(3 x^{2} - 4 x + 2) ∣ ∣}_{x = 1}$
$f^{'} (1^{+}) = {\frac{d}{d x} (- x^{3} + 2 x^{2}) ∣ ∣}_{x = 1} = {(- 3 x^{2} + 4 x) ∣ ∣}_{x = 1} = 1$
$f^{'} (0^{+}) = {(3 x^{2} - 4 x + 2) ∣ ∣}_{x = 0} = 2$
$f^{'} (0^{-}) = {(- 3 x^{2} + 4 x) ∣ ∣}_{x = 0} = 0$
$f^{'} (0^{+}) \neq f^{'} (0^{-})$ hence not differentiable at $x = 0$
so $(B) 1$ nor differentiability of point.

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