Let f(x) be a function defined in `1lexleoo` by
`f(x)={{:(2-x" for "1lexle2),(3x-x^(2)" for "xgt2):}`
Consider the following statements :
1. The function is continuous at every point in the interval `(1,oo)`.
2. The function is differentiable at x=1.5.
Which of the above statements is /are correct?

Options

A. 1 only
B. 2 only
C. Both 1 and 2
D. Neither 1 nor 2

Explanation

To solve the problem, we need to analyze the function \( f(x) \) defined in the interval \( [1, \infty) \) as follows:

\[ f(x) = \begin{cases} 2 - x & \text{for } 1 \leq x \leq 2 \\ 3x - x^2 & \text{for } x > 2 \end{cases} \]

We need to evaluate the two statements provided:

1. The function is continuous at every point in the interval \( (1, \infty) \).
2. The function is differentiable at \( x = 1.5 \).

### Step 1: Check Continuity of \( f(x) \)

To check if \( f(x) \) is continuous at every point in \( (1, \infty) \), we need to check the continuity at the point where the definition of the function changes, which is at \( x = 2 \).

**Left-hand limit as \( x \) approaches 2:**
\[ \lim_{x \to 2^-} f(x) = f(2) = 2 - 2 = 0 \]

**Right-hand limit as \( x \) approaches 2:**
\[ \lim_{x \to 2^+} f(x) = 3(2) - (2)^2 = 6 - 4 = 2 \]

**Value of the function at \( x = 2 \):**
\[ f(2) = 0 \]

Since the left-hand limit (0) does not equal the right-hand limit (2), we conclude that \( f(x) \) is not continuous at \( x = 2 \).

### Conclusion for Statement 1:
The function is not continuous at \( x = 2 \), hence the statement "The function is continuous at every point in the interval \( (1, \infty) \)" is **false**.

### Step 2: Check Differentiability at \( x = 1.5 \)

Since \( 1.5 \) is in the interval \( [1, 2] \), we will use the definition of the derivative to check differentiability.

**Right-hand derivative at \( x = 1.5 \):**
\[ f'(1.5) = \lim_{h \to 0} \frac{f(1.5 + h) - f(1.5)}{h} \]
For \( 1.5 + h \) (where \( h \) is small and positive), we have:
\[ f(1.5) = 2 - 1.5 = 0 \]
Thus,
\[ f'(1.5) = \lim_{h \to 0} \frac{(2 - (1.5 + h)) - 0}{h} = \lim_{h \to 0} \frac{0.5 - h}{h} = \lim_{h \to 0} \left(\frac{0.5}{h} - 1\right) = -1 \]

**Left-hand derivative at \( x = 1.5 \):**
\[ f'(1.5) = \lim_{h \to 0} \frac{f(1.5 - h) - f(1.5)}{-h} \]
For \( 1.5 - h \) (where \( h \) is small and positive), we have:
\[ f(1.5 - h) = 2 - (1.5 - h) = 0.5 + h \]
Thus,
\[ f'(1.5) = \lim_{h \to 0} \frac{(0.5 + h) - 0}{-h} = \lim_{h \to 0} \frac{0.5 + h}{-h} = -1 \]

Since both the left-hand derivative and right-hand derivative are equal, we conclude that \( f(x) \) is differentiable at \( x = 1.5 \).

### Conclusion for Statement 2:
The function is differentiable at \( x = 1.5 \), hence the statement "The function is differentiable at \( x = 1.5 \)" is **true**.

### Final Answer:
- Statement 1: False
- Statement 2: True

Thus, the correct option is that only the second statement is true.