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. f'(2+0) does not exist.
2. f'(2-0) does not exist.
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 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 check the existence of the derivatives \( f'(2+0) \) and \( f'(2-0) \).

### Step 1: Calculate \( f(2) \)

First, we find the value of the function at \( x = 2 \):
\[ f(2) = 2 - 2 = 0 \]

### Step 2: Calculate the right-hand derivative \( f'(2+0) \)

To find \( f'(2+0) \), we need to use the definition of the derivative:
\[ f'(2+0) = \lim_{h \to 0^+} \frac{f(2+h) - f(2)}{h} \]

For \( x > 2 \), the function is defined as \( f(x) = 3x - x^2 \). Thus, we calculate:
\[ f(2+h) = 3(2+h) - (2+h)^2 = 6 + 3h - (4 + 4h + h^2) = 2 - h - h^2 \]

Now substituting into the derivative limit:
\[ f'(2+0) = \lim_{h \to 0^+} \frac{(2 - h - h^2) - 0}{h} = \lim_{h \to 0^+} \frac{2 - h - h^2}{h} \]
\[ = \lim_{h \to 0^+} \left( \frac{2}{h} - 1 - h \right) \]

As \( h \to 0^+ \), \( \frac{2}{h} \to \infty \). Therefore, \( f'(2+0) \) does not exist.

### Step 3: Calculate the left-hand derivative \( f'(2-0) \)

Now we calculate \( f'(2-0) \):
\[ f'(2-0) = \lim_{h \to 0^+} \frac{f(2-h) - f(2)}{-h} \]

For \( x < 2 \), the function is defined as \( f(x) = 2 - x \). Thus, we calculate:
\[ f(2-h) = 2 - (2-h) = h \]

Now substituting into the derivative limit:
\[ f'(2-0) = \lim_{h \to 0^+} \frac{h - 0}{-h} = \lim_{h \to 0^+} -1 = -1 \]

### Conclusion

We have found:
- \( f'(2+0) \) does not exist (it approaches infinity).
- \( f'(2-0) = -1 \).

Thus, the statements are:
1. \( f'(2+0) \) does not exist. (True)
2. \( f'(2-0) \) does exist. (False)

Therefore, only the first statement is correct.

### Final Answer:
Only statement 1 is correct.

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