Question

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 :

• A. Three elements
• B. One element
• C. Five elements
• D. Two elements

Answer 1

The Correct Option is A

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 $(-1, 1)$.

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 $x = 0$.
2. 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.
3. 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.