`"If "f(x)={{:(,-x-(pi)/(2),x le -(pi)/(2)),(,-cos x,-(pi)/(2) lt x le 0),(,x-1,0 lt x le 1),(,"In "x,x gt 1):}` then which one of the following is not correct?

Options

A. f(x) is continuous at `x=-pi//2`
B. f(x) is not differentiable at x=0
C. f(x) is differtiable at x=1
D. f(x) is differentiable at x=-3/2

Answer

To solve the problem step by step, we need to analyze the function \( f(x) \) defined piecewise and check the continuity and differentiability at the specified points.

### Step 1: Define the function
The function \( f(x) \) is defined as:
\[ f(x) = \begin{cases} -x - \frac{\pi}{2} & \text{if } x \leq -\frac{\pi}{2} \\ -\cos x & \text{if } -\frac{\pi}{2} < x \leq 0 \\ x - 1 & \text{if } 0 < x \leq 1 \\ \ln x & \text{if } x > 1 \end{cases} \]

### Step 2: Check continuity at \( x = -\frac{\pi}{2} \)
To check continuity at \( x = -\frac{\pi}{2} \):
- Calculate \( f(-\frac{\pi}{2}) \):
\[ f(-\frac{\pi}{2}) = -(-\frac{\pi}{2}) - \frac{\pi}{2} = 0 \]
- Calculate the left-hand limit as \( x \) approaches \( -\frac{\pi}{2} \):
\[ \lim_{x \to -\frac{\pi}{2}^-} f(x) = 0 \]
- Calculate the right-hand limit as \( x \) approaches \( -\frac{\pi}{2} \):
\[ \lim_{x \to -\frac{\pi}{2}^+} f(x) = -\cos(-\frac{\pi}{2}) = -0 = 0 \]
Since both limits and \( f(-\frac{\pi}{2}) \) are equal, \( f(x) \) is continuous at \( x = -\frac{\pi}{2} \).

### Step 3: Check differentiability at \( x = 0 \)
To check differentiability at \( x = 0 \):
- Calculate the left-hand derivative:
\[ f'(0^-) = \lim_{h \to 0^-} \frac{f(h) - f(0)}{h} = \lim_{h \to 0^-} \frac{-\cos(h) - (-1)}{h} = \lim_{h \to 0^-} \frac{-\cos(h) + 1}{h} \]
The limit does not exist because \( -\cos(h) + 1 \) approaches \( 0 \) but does not have a defined slope.

- Calculate the right-hand derivative:
\[ f'(0^+) = \lim_{h \to 0^+} \frac{f(h) - f(0)}{h} = \lim_{h \to 0^+} \frac{(h - 1) - (-1)}{h} = \lim_{h \to 0^+} \frac{h}{h} = 1 \]
Since the left-hand derivative does not exist and the right-hand derivative is \( 1 \), \( f(x) \) is not differentiable at \( x = 0 \).

### Step 4: Check differentiability at \( x = 1 \)
To check differentiability at \( x = 1 \):
- Calculate the left-hand derivative:
\[ f'(1^-) = \lim_{h \to 0^-} \frac{f(1 + h) - f(1)}{h} = \lim_{h \to 0^-} \frac{(1 + h - 1) - 0}{h} = \lim_{h \to 0^-} \frac{h}{h} = 1 \]
- Calculate the right-hand derivative:
\[ f'(1^+) = \lim_{h \to 0^+} \frac{f(1 + h) - f(1)}{h} = \lim_{h \to 0^+} \frac{\ln(1 + h) - 0}{h} = \lim_{h \to 0^+} \frac{h}{h} = 1 \]
Since both derivatives are equal, \( f(x) \) is differentiable at \( x = 1 \).

### Step 5: Check differentiability at \( x = -\frac{3\pi}{2} \)
Since \( -\frac{3\pi}{2} < -\frac{\pi}{2} \), we use the first piece of the function:
\[ f(x) = -x - \frac{\pi}{2} \]
This is a linear function, and thus differentiable everywhere, including at \( x = -\frac{3\pi}{2} \).

### Conclusion
Now, we summarize our findings:
1. \( f(x) \) is continuous at \( x = -\frac{\pi}{2} \) (Correct)
2. \( f(x) \) is not differentiable at \( x = 0 \) (Correct)
3. \( f(x) \) is differentiable at \( x = 1 \) (Incorrect, it is not differentiable)
4. \( f(x) \) is differentiable at \( -\frac{3\pi}{2} \) (Correct)

Thus, the statement that is **not correct** is that \( f(x) \) is differentiable at \( x = 1 \).