Question

If \( f(x) \) is defined as follows: 
$$ f(x) = \begin{cases} 4, & \text{if } -\infty < x < -\sqrt{5}, \\ x^2 - 1, & \text{if } -\sqrt{5} \leq x \leq \sqrt{5}, \\ 4, & \text{if } \sqrt{5} \leq x < \infty. \end{cases} $$ If \( k \) is the number of points where \( f(x) \) is not differentiable, then \( k - 2 = \) 
 

• A. 2
• B. 1
• C. 0
• D. 3

Hint:

For piecewise functions, check the left-hand and right-hand derivatives at the boundary points to determine non-differentiability.

Answer 1

The Correct Option is C

Step 1: {Check the points of non-differentiability}
We know that for a function to be differentiable at a point, both the left-hand derivative (LHD) and right-hand derivative (RHD) must be equal at that point. At \( x = -\sqrt{5} \), the left-hand derivative is 0, but the right-hand derivative is \( 2x = -2\sqrt{5} \). Therefore, \( f(x) \) is not differentiable at \( x = -\sqrt{5} \). Similarly, at \( x = \sqrt{5} \), the left-hand derivative is \( 2x = 2\sqrt{5} \), and the right-hand derivative is 0, meaning \( f(x) \) is not differentiable at \( x = \sqrt{5} \). Thus, \( k = 2 \). 
Step 2: {Calculate \( k - 2 \)}
Since \( k = 2 \), we find: \[ k - 2 = 2 - 2 = 0 \]

Answer 2

Step 1: Analyze the function \( f(x) \)
The function \( f(x) \) is piecewise defined as:
\[ f(x) = \begin{cases} 4, & \text{if } -\infty < x < -\sqrt{5}, \\ x^2 - 1, & \text{if } -\sqrt{5} \leq x \leq \sqrt{5}, \\ 4, & \text{if } \sqrt{5} \leq x < \infty. \end{cases} \]
We need to determine where the function is not differentiable. The function is potentially not differentiable at the points where the definition of \( f(x) \) changes, i.e., at \( x = -\sqrt{5} \) and \( x = \sqrt{5} \).

Step 2: Check differentiability at \( x = -\sqrt{5} \)
At \( x = -\sqrt{5} \), the function changes from a constant \( f(x) = 4 \) to a quadratic function \( f(x) = x^2 - 1 \).
For differentiability at \( x = -\sqrt{5} \), we need the following:
1. The function should be continuous at \( x = -\sqrt{5} \), i.e., the left-hand limit (from the constant function) must equal the right-hand limit (from the quadratic function). We check continuity:
\[ \lim_{x \to -\sqrt{5}^-} f(x) = 4, \quad \lim_{x \to -\sqrt{5}^+} f(x) = (-\sqrt{5})^2 - 1 = 4. \]
So, \( f(x) \) is continuous at \( x = -\sqrt{5} \).
2. The derivatives from both sides must also be equal:
- The derivative of the constant function is 0.
- The derivative of \( x^2 - 1 \) is \( 2x \), and at \( x = -\sqrt{5} \), we get \( 2(-\sqrt{5}) = -2\sqrt{5} \).
Since the derivatives from the left and right are not equal, \( f(x) \) is not differentiable at \( x = -\sqrt{5} \).

Step 3: Check differentiability at \( x = \sqrt{5} \)
At \( x = \sqrt{5} \), the function changes from a quadratic function \( f(x) = x^2 - 1 \) to a constant function \( f(x) = 4 \).
For differentiability at \( x = \sqrt{5} \), we need:
1. Continuity at \( x = \sqrt{5} \):
\[ \lim_{x \to \sqrt{5}^-} f(x) = (\sqrt{5})^2 - 1 = 4, \quad \lim_{x \to \sqrt{5}^+} f(x) = 4. \]
So, \( f(x) \) is continuous at \( x = \sqrt{5} \).
2. The derivatives from both sides must also be equal:
- The derivative of \( x^2 - 1 \) is \( 2x \), and at \( x = \sqrt{5} \), we get \( 2(\sqrt{5}) = 2\sqrt{5} \).
- The derivative of the constant function is 0.
Since the derivatives from the left and right are not equal, \( f(x) \) is not differentiable at \( x = \sqrt{5} \).

Step 4: Conclusion
The function is not differentiable at two points: \( x = -\sqrt{5} \) and \( x = \sqrt{5} \). Therefore, the number of points where \( f(x) \) is not differentiable is 2.

The given question asks for \( k - 2 \), where \( k \) is the number of points where \( f(x) \) is not differentiable. Since \( k = 2 \), we have:
\[ k - 2 = 0. \]

Thus, the correct answer is:

0