Let y = f(x) be defined in [a, b], then
(i) Test of continuity at `x = c, a lt c lt b`
(ii) Test of continuity at x = a
(iii) Test of continuity at x = b
Case I Test of continuity at `x = c, a lt c lt b`
If y = f(x) be defined at x = c and its value f(c) be equal to limit of f(x) as `x rarr c` i.e. f(c) = `lim_(x rarr c) f(x)`
or `lim_(x rarr c^(-))f(x) = f(c) = lim_(x rarr c^(+)) f(x)`
or LHL = f(c) = RHL
then, y = f(x) is continuous at x = c.
Case II Test of continuity at x = a
If RHL = f(a)
Then, f(x) is said to be continuous at the end point x = a
Case III Test of continuity at x = b, if LHL = f(b)
Then, f(x) is continuous at right end x = b.
If `f(x) = {{:(sin x",",x le 0),(tan x",",0 lt x lt 2pi),(cos x",",2pi le x lt 3pi),(3pi",",x = 3pi):}`, then f(x) is discontinuous at

Options

A. `(pi)/(2),(3pi)/(2),2pi,3pi`
B. `0, (pi)/(2),pi,(3pi)/(2), 3pi`
C. `(pi)/(2), 2pi`
D. None of these

Answer

To determine the points of discontinuity for the function \( f(x) \) defined as:

\[ f(x) = \begin{cases} \sin x & \text{for } x \leq 0 \\ \tan x & \text{for } 0 < x < 2\pi \\ \cos x & \text{for } 2\pi \leq x < 3\pi \\ 3\pi & \text{for } x = 3\pi \end{cases} \]

we will analyze the continuity at specific points: \( x = 0 \), \( x = \frac{\pi}{2} \), \( x = \frac{3\pi}{2} \), \( x = 2\pi \), and \( x = 3\pi \).

### Step 1: Test of Continuity at \( x = 0 \)

1. **Calculate Left Hand Limit (LHL)**:
\[ \text{LHL} = \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \sin x = \sin(0) = 0 \]

2. **Calculate Right Hand Limit (RHL)**:
\[ \text{RHL} = \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} \tan x = \tan(0) = 0 \]

3. **Value of the function at \( x = 0 \)**:
\[ f(0) = \sin(0) = 0 \]

4. **Conclusion**:
Since LHL = RHL = \( f(0) \), the function is continuous at \( x = 0 \).

### Step 2: Test of Continuity at \( x = \frac{\pi}{2} \)

1. **Calculate LHL**:
\[ \text{LHL} = \lim_{x \to \frac{\pi}{2}^-} f(x) = \lim_{x \to \frac{\pi}{2}^-} \tan x = \tan\left(\frac{\pi}{2}\right) \text{ (undefined)} \]

2. **Calculate RHL**:
\[ \text{RHL} = \lim_{x \to \frac{\pi}{2}^+} f(x) = \lim_{x \to \frac{\pi}{2}^+} \tan x = \tan\left(\frac{\pi}{2}\right) \text{ (undefined)} \]

3. **Conclusion**:
Since LHL and RHL do not exist, the function is discontinuous at \( x = \frac{\pi}{2} \).

### Step 3: Test of Continuity at \( x = \frac{3\pi}{2} \)

1. **Calculate LHL**:
\[ \text{LHL} = \lim_{x \to \frac{3\pi}{2}^-} f(x) = \lim_{x \to \frac{3\pi}{2}^-} \tan x = \tan\left(\frac{3\pi}{2}\right) \text{ (undefined)} \]

2. **Calculate RHL**:
\[ \text{RHL} = \lim_{x \to \frac{3\pi}{2}^+} f(x) = \lim_{x \to \frac{3\pi}{2}^+} \tan x = \tan\left(\frac{3\pi}{2}\right) \text{ (undefined)} \]

3. **Conclusion**:
Since LHL and RHL do not exist, the function is discontinuous at \( x = \frac{3\pi}{2} \).

### Step 4: Test of Continuity at \( x = 2\pi \)

1. **Calculate LHL**:
\[ \text{LHL} = \lim_{x \to 2\pi^-} f(x) = \lim_{x \to 2\pi^-} \tan x = \tan(2\pi) = 0 \]

2. **Calculate RHL**:
\[ \text{RHL} = \lim_{x \to 2\pi^+} f(x) = \lim_{x \to 2\pi^+} \cos x = \cos(2\pi) = 1 \]

3. **Conclusion**:
Since LHL \( \neq \) RHL, the function is discontinuous at \( x = 2\pi \).

### Step 5: Test of Continuity at \( x = 3\pi \)

1. **Calculate LHL**:
\[ \text{LHL} = \lim_{x \to 3\pi^-} f(x) = \lim_{x \to 3\pi^-} \cos x = \cos(3\pi) = -1 \]

2. **Calculate RHL**:
\[ \text{RHL} = \lim_{x \to 3\pi^+} f(x) = 3\pi \]

3. **Conclusion**:
Since LHL \( \neq \) RHL, the function is discontinuous at \( x = 3\pi \).

### Final Conclusion

The function \( f(x) \) is discontinuous at the following points:
- \( x = \frac{\pi}{2} \)
- \( x = \frac{3\pi}{2} \)
- \( x = 2\pi \)
- \( x = 3\pi \)