Let $f : R \to R$ is a function which is defined by f(x)=max{x, $x^{3}$ }. The set of all points on which f(x) is not differentiable is

{-1, 1}

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{-1, 0}

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{0, 1}

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{-1, 0, 1}

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Solution

The correct option is C {-1, 0, 1}
The graphs of $y = x$ and $y = x^{3}$ are given in the adjoining figure.
Thus,
$f (x) = ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ \begin{matrix} x & if x \leq - 1 x^{3} & if - 1 \leq x < 0 x & if 0 \leq x \leq 1 x^{3} & if x \geq 1 \end{matrix}$
Clearly f is not differentiable at $x = - 1, 0, 1$ as there are corner points at $x = - 1, 0, 1.$

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