Let `f:[a,b]to[1,oo)` be a continuous function and let `g:RtoR` be defined as
`g(x)={(0,"if",xlta),(int_(a)^(x)f(t)dt,"if",alexleb),(int_(a)^(b)f(t)dt,"if",xgtb):}` Then

Options

A. (a) `g(x)` is continuous but not differentiable at a
B. (b) `g(x)` is differentiable on `R`
C. (c) `g(x)` is continuous but not differentiable at `b`
D. (d) `g(x)` is continuous and differentiable at either a or b but not both

Explanation

To solve the given problem step by step, we will analyze the function \( g(x) \) defined in the question and check its continuity and differentiability at the endpoints \( a \) and \( b \).

### Step 1: Define the function \( g(x) \)

The function \( g(x) \) is defined as follows:

\[ g(x) = \begin{cases} 0 & \text{if } x < a \\ \int_a^x f(t) \, dt & \text{if } a \leq x \leq b \\ \int_a^b f(t) \, dt & \text{if } x > b \end{cases} \]

### Step 2: Check continuity at \( x = a \)

To check continuity at \( x = a \), we need to find:

1. \( g(a) \)
2. \( \lim_{x \to a^-} g(x) \)
3. \( \lim_{x \to a^+} g(x) \)

- **Finding \( g(a) \)**:
\[ g(a) = \int_a^a f(t) \, dt = 0 \]

- **Finding \( \lim_{x \to a^-} g(x) \)**:
\[ \lim_{x \to a^-} g(x) = 0 \]

- **Finding \( \lim_{x \to a^+} g(x) \)**:
\[ \lim_{x \to a^+} g(x) = \int_a^a f(t) \, dt = 0 \]

Since \( g(a) = \lim_{x \to a^-} g(x) = \lim_{x \to a^+} g(x) = 0 \), \( g(x) \) is continuous at \( x = a \).

### Step 3: Check continuity at \( x = b \)

Now, we check continuity at \( x = b \):

1. \( g(b) \)
2. \( \lim_{x \to b^-} g(x) \)
3. \( \lim_{x \to b^+} g(x) \)

- **Finding \( g(b) \)**:
\[ g(b) = \int_a^b f(t) \, dt \]

- **Finding \( \lim_{x \to b^-} g(x) \)**:
\[ \lim_{x \to b^-} g(x) = \int_a^b f(t) \, dt \]

- **Finding \( \lim_{x \to b^+} g(x) \)**:
\[ \lim_{x \to b^+} g(x) = \int_a^b f(t) \, dt \]

Since \( g(b) = \lim_{x \to b^-} g(x) = \lim_{x \to b^+} g(x) \), \( g(x) \) is continuous at \( x = b \).

### Step 4: Check differentiability at \( x = a \) and \( x = b \)

- **At \( x < a \)**:
\[ g'(x) = 0 \]

- **At \( a < x < b \)**:
Using Leibniz's rule, we differentiate \( g(x) \):
\[ g'(x) = f(x) \]

- **At \( x > b \)**:
\[ g'(x) = 0 \]

Now, we check differentiability at the endpoints:

- At \( x = a \):
\[ \lim_{x \to a^-} g'(x) = 0 \quad \text{and} \quad \lim_{x \to a^+} g'(x) = f(a) \]
Since \( f(a) \geq 1 \), \( g'(x) \) is not equal from both sides, so \( g(x) \) is not differentiable at \( x = a \).

- At \( x = b \):
\[ \lim_{x \to b^-} g'(x) = f(b) \quad \text{and} \quad \lim_{x \to b^+} g'(x) = 0 \]
Again, since \( f(b) \geq 1 \), \( g(x) \) is not differentiable at \( x = b \).

### Conclusion

The function \( g(x) \) is continuous at both \( x = a \) and \( x = b \), but it is not differentiable at these points.