Let $f : (-1,1) \to R$ be a function defined by $f(x) = max \{- | x |,- \sqrt{ 1- x^2} \}$. If $K$ be the set of all points at which $f$ is not differentiable, then $K$ has exactly :
To determine the points at which the function $f(x) = \max \{-|x|, -\sqrt{1-x^2}\}$ is not differentiable, we need to analyze each component of this maximum function within its domain $(-1, 1)$.
First, consider the function $h(x) = -|x|$. The absolute value function $|x|$ is not differentiable at $x = 0$ because it has a cusp there. Therefore, $h(x)$ is not differentiable at $x = 0$.
Next, consider the function $g(x) = -\sqrt{1-x^2}$. This function is differentiable everywhere in $(-1, 1)$ except potentially at the endpoints $x = -1$ and $x = 1$. However, since the domain of our function $f(x)$ is (-1, 1)$ (excluding -1 and $1$), differentiability at these endpoints does not need to be considered.
Now analyze the points where $h(x)$ and $g(x)$ switch dominance, i.e., where $- | x | = - \sqrt{1 - x^2}$. Equating these, we get:
$|x| = \sqrt{1 - x^2}$.
Squaring both sides yields:
$x^2 = 1 - x^2$. This simplifies to:
$2x^2 = 1$, hence $x^2 = \frac{1}{2}$.
Therefore, $x = \pm \frac{1}{\sqrt{2}}$.
Based on the analysis, the set $K$ of points where $f(x)$ is not differentiable includes $x = 0$, $x = \frac{1}{\sqrt{2}}$, and $x = -\frac{1}{\sqrt{2}}$.
Thus, $K$ has exactly three elements, which confirms the correct answer: Three elements.
