Let `f: (-1,1)toR` be a function defind by f(x) =max. `{-absx,-sqrt(1-x^2)}`. If K is the set of all points at which f is not differentiable, then K has set of all points at which f is not differentible, then K has exactly

Options

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

Explanation

To solve the problem, we need to analyze the function \( f(x) = \max\{-|x|, -\sqrt{1-x^2}\} \) defined on the interval \((-1, 1)\) and determine the points where it is not differentiable.

### Step 1: Understand the Components of the Function
The function is defined as the maximum of two functions:
1. \( g_1(x) = -|x| \)
2. \( g_2(x) = -\sqrt{1-x^2} \)

### Step 2: Analyze Each Function
- The function \( g_1(x) = -|x| \) is a V-shaped graph that opens downwards, with a vertex at the origin (0,0) and slopes of -1 on both sides.
- The function \( g_2(x) = -\sqrt{1-x^2} \) represents the lower half of a circle with radius 1, centered at the origin. This function is defined for \( x \in (-1, 1) \).

### Step 3: Find Points of Intersection
To find where \( f(x) \) changes from one function to another, we need to set \( g_1(x) = g_2(x) \):
\[ -|x| = -\sqrt{1-x^2} \]
Squaring both sides (noting that both sides are negative):
\[ x^2 = 1 - x^2 \]
\[ 2x^2 = 1 \implies x^2 = \frac{1}{2} \implies x = \pm\frac{1}{\sqrt{2}} \approx \pm 0.707 \]

### Step 4: Identify Points of Non-Differentiability
Next, we need to check the points where the function \( f(x) \) may not be differentiable. The potential points of non-differentiability occur where:
1. The two functions intersect (which we found to be \( x = \pm\frac{1}{\sqrt{2}} \)).
2. The corners or sharp points of the piecewise functions, which occur at \( x = 0 \) for \( g_1(x) \).

### Step 5: List the Points
Thus, the points where \( f(x) \) is not differentiable are:
1. \( x = -\frac{1}{\sqrt{2}} \)
2. \( x = 0 \)
3. \( x = \frac{1}{\sqrt{2}} \)

### Conclusion
The set \( K \) of points where \( f \) is not differentiable contains exactly 3 points:
\[ K = \left\{-\frac{1}{\sqrt{2}}, 0, \frac{1}{\sqrt{2}}\right\} \]

Thus, the answer is that \( K \) has exactly 3 points.