If `f(x)={(1,xlt0),(1+sinx,0 le x le(pi)/2),(2+(x-(pi)/2),(pi)/2lex):}` then which of the following is true for f(x)?

Options

A. continuous at x=0
B. continuous at `x=(pi)/2`
C. Non differentiable at x=0
D. Differentiable at `x=(pi)/2`

Answer

To determine the properties of the function \( f(x) \) given by:

\[ f(x) = \begin{cases} 1 & \text{if } x < 0 \\ 1 + \sin x & \text{if } 0 \leq x \leq \frac{\pi}{2} \\ 2 + (x - \frac{\pi}{2}) & \text{if } x > \frac{\pi}{2} \end{cases} \]

we will check for continuity and differentiability at the points \( x = 0 \) and \( x = \frac{\pi}{2} \).

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

To check continuity at \( x = 0 \), we need to verify that the left-hand limit (LHL) equals the right-hand limit (RHL) and both equal \( f(0) \).

- **Left-hand limit (LHL)** as \( x \) approaches \( 0 \) from the left:
\[ \text{LHL} = \lim_{x \to 0^-} f(x) = 1 \]

- **Right-hand limit (RHL)** as \( x \) approaches \( 0 \) from the right:
\[ \text{RHL} = \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} (1 + \sin x) = 1 + \sin(0) = 1 \]

- **Value at \( x = 0 \)**:
\[ f(0) = 1 + \sin(0) = 1 \]

Since LHL = RHL = \( f(0) = 1 \), the function is continuous at \( x = 0 \).

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

Now, we check continuity at \( x = \frac{\pi}{2} \).

- **Left-hand limit (LHL)** as \( x \) approaches \( \frac{\pi}{2} \) from the left:
\[ \text{LHL} = \lim_{x \to \frac{\pi}{2}^-} f(x) = \lim_{x \to \frac{\pi}{2}^-} (1 + \sin x) = 1 + \sin\left(\frac{\pi}{2}\right) = 1 + 1 = 2 \]

- **Right-hand limit (RHL)** as \( x \) approaches \( \frac{\pi}{2} \) from the right:
\[ \text{RHL} = \lim_{x \to \frac{\pi}{2}^+} f(x) = \lim_{x \to \frac{\pi}{2}^+} \left(2 + (x - \frac{\pi}{2})\right) = 2 + \left(\frac{\pi}{2} - \frac{\pi}{2}\right) = 2 \]

- **Value at \( x = \frac{\pi}{2} \)**:
\[ f\left(\frac{\pi}{2}\right) = 1 + \sin\left(\frac{\pi}{2}\right) = 2 \]

Since LHL = RHL = \( f\left(\frac{\pi}{2}\right) = 2 \), the function is continuous at \( x = \frac{\pi}{2} \).

### Step 3: Check Differentiability at \( x = 0 \)

To check differentiability at \( x = 0 \), we need to find the left-hand derivative (LHD) and the right-hand derivative (RHD).

- **Left-hand derivative (LHD)**:
\[ \text{LHD} = \lim_{h \to 0^-} \frac{f(0 + h) - f(0)}{h} = \lim_{h \to 0^-} \frac{(1 + \sin h) - 1}{h} = \lim_{h \to 0^-} \frac{\sin h}{h} = 1 \]

- **Right-hand derivative (RHD)**:
\[ \text{RHD} = \lim_{h \to 0^+} \frac{f(0 + h) - f(0)}{h} = \lim_{h \to 0^+} \frac{(1 + \sin h) - 1}{h} = \lim_{h \to 0^+} \frac{\sin h}{h} = 1 \]

Since LHD = RHD = 1, the function is differentiable at \( x = 0 \).

### Step 4: Check Differentiability at \( x = \frac{\pi}{2} \)

Now, we check differentiability at \( x = \frac{\pi}{2} \).

- **Left-hand derivative (LHD)**:
\[ \text{LHD} = \lim_{h \to 0^-} \frac{f\left(\frac{\pi}{2} + h\right) - f\left(\frac{\pi}{2}\right)}{h} = \lim_{h \to 0^-} \frac{(1 + \sin(\frac{\pi}{2} - h)) - 2}{h} = \lim_{h \to 0^-} \frac{(1 + \cos h) - 2}{h} = \lim_{h \to 0^-} \frac{\cos h - 1}{h} \]
The limit evaluates to \( -\infty \).

- **Right-hand derivative (RHD)**:
\[ \text{RHD} = \lim_{h \to 0^+} \frac{f\left(\frac{\pi}{2} + h\right) - f\left(\frac{\pi}{2}\right)}{h} = \lim_{h \to 0^+} \frac{(2 + h) - 2}{h} = \lim_{h \to 0^+} \frac{h}{h} = 1 \]

Since LHD \( \neq \) RHD, the function is not differentiable at \( x = \frac{\pi}{2} \).

### Conclusion

The function \( f(x) \) is continuous at both \( x = 0 \) and \( x = \frac{\pi}{2} \), but it is not differentiable at \( x = \frac{\pi}{2} \).