Consider the function `f(x)={{:(sin(x-4).tan^(-1)((1)/(x-4)),x ne 4),(0,x=4):}`, then

Options

A. f(x) is contunuous and differentiable at `x=4`
B. `f(x)` is continuous but non differentiable at x= 4
C. `f(x)` is discontinuous but differentiable at `x=4`
D. f(x) is dicountinuous and non differentiable at x = 4

Answer

To determine the continuity and differentiability of the function

\[ f(x) = \begin{cases} \sin(x-4) \tan^{-1}\left(\frac{1}{x-4}\right), & x \neq 4 \\ 0, & x = 4 \end{cases} \]

we will follow these steps:

### Step 1: Check for Continuity at \( x = 4 \)

To check for continuity at \( x = 4 \), we need to verify that:

\[ \lim_{x \to 4} f(x) = f(4) \]

We know that \( f(4) = 0 \).

#### Finding the Limit:

We will calculate the limit as \( x \) approaches 4 from both sides (left-hand limit and right-hand limit).

1. **Left-hand limit (LHL)**:
\[ \lim_{x \to 4^-} f(x) = \lim_{x \to 4^-} \sin(x-4) \tan^{-1}\left(\frac{1}{x-4}\right) \]
As \( x \to 4^- \), \( x - 4 \to 0^- \) (a small negative number). Thus:
- \( \sin(x-4) \to \sin(0) = 0 \)
- \( \tan^{-1}\left(\frac{1}{x-4}\right) \to \tan^{-1}(-\infty) = -\frac{\pi}{2} \)

Therefore:
\[ \lim_{x \to 4^-} f(x) = 0 \cdot \left(-\frac{\pi}{2}\right) = 0 \]

2. **Right-hand limit (RHL)**:
\[ \lim_{x \to 4^+} f(x) = \lim_{x \to 4^+} \sin(x-4) \tan^{-1}\left(\frac{1}{x-4}\right) \]
As \( x \to 4^+ \), \( x - 4 \to 0^+ \) (a small positive number). Thus:
- \( \sin(x-4) \to \sin(0) = 0 \)
- \( \tan^{-1}\left(\frac{1}{x-4}\right) \to \tan^{-1}(\infty) = \frac{\pi}{2} \)

Therefore:
\[ \lim_{x \to 4^+} f(x) = 0 \cdot \left(\frac{\pi}{2}\right) = 0 \]

Since both limits equal \( f(4) \):
\[ \lim_{x \to 4} f(x) = 0 = f(4) \]
Thus, \( f(x) \) is continuous at \( x = 4 \).

### Step 2: Check for Differentiability at \( x = 4 \)

To check for differentiability at \( x = 4 \), we need to find the left-hand derivative (LHD) and right-hand derivative (RHD) at \( x = 4 \).

1. **Left-hand derivative (LHD)**:
\[ f'(4^-) = \lim_{h \to 0^-} \frac{f(4+h) - f(4)}{h} = \lim_{h \to 0^-} \frac{\sin(h) \tan^{-1}\left(\frac{1}{h}\right) - 0}{h} \]
As \( h \to 0^- \):
- \( \sin(h) \approx h \)
- \( \tan^{-1}\left(\frac{1}{h}\right) \to -\frac{\pi}{2} \)

Therefore:
\[ f'(4^-) = \lim_{h \to 0^-} \frac{h \cdot \left(-\frac{\pi}{2}\right)}{h} = -\frac{\pi}{2} \]

2. **Right-hand derivative (RHD)**:
\[ f'(4^+) = \lim_{h \to 0^+} \frac{f(4+h) - f(4)}{h} = \lim_{h \to 0^+} \frac{\sin(h) \tan^{-1}\left(\frac{1}{h}\right) - 0}{h} \]
As \( h \to 0^+ \):
- \( \sin(h) \approx h \)
- \( \tan^{-1}\left(\frac{1}{h}\right) \to \frac{\pi}{2} \)

Therefore:
\[ f'(4^+) = \lim_{h \to 0^+} \frac{h \cdot \left(\frac{\pi}{2}\right)}{h} = \frac{\pi}{2} \]

Since \( f'(4^-) = -\frac{\pi}{2} \) and \( f'(4^+) = \frac{\pi}{2} \), we conclude that:

\[ f'(4^-) \neq f'(4^+) \]

Thus, \( f(x) \) is not differentiable at \( x = 4 \).

### Final Conclusion

The function \( f(x) \) is continuous at \( x = 4 \) but not differentiable at \( x = 4 \).