lf is a differentiable function satisfying `f(1/n)=0,AA n>=1,n in I`, then

Options

A. (a)`f(x) = 0, x in (0, 1]`
B. (b)`f'(0) = 0 = f(0)`
C. (c)f(0) = 0 but f'(0) not necessarily zero
D. (d)`|f(x)| le 1, x in (0, 1]`

Answer

To solve the problem, we need to analyze the given function \( f \) which is differentiable and satisfies the condition \( f\left(\frac{1}{n}\right) = 0 \) for all integers \( n \geq 1 \).

### Step-by-step Solution:

1. **Understanding the Given Condition**:
We have \( f\left(\frac{1}{n}\right) = 0 \) for \( n = 1, 2, 3, \ldots \). This means:
- \( f(1) = 0 \)
- \( f\left(\frac{1}{2}\right) = 0 \)
- \( f\left(\frac{1}{3}\right) = 0 \)
- \( f\left(\frac{1}{4}\right) = 0 \)
- And so on.

2. **Taking the Limit**:
As \( n \) approaches infinity, \( \frac{1}{n} \) approaches 0. Therefore, we can write:
\[ \lim_{n \to \infty} f\left(\frac{1}{n}\right) = f(0) \]
Since \( f\left(\frac{1}{n}\right) = 0 \) for all \( n \geq 1 \), we have:
\[ \lim_{n \to \infty} f\left(\frac{1}{n}\right) = 0 \]
Thus, \( f(0) = 0 \).

3. **Using the Differentiability of \( f \)**:
Since \( f \) is differentiable at \( x = 0 \), we can express the derivative at that point as:
\[ f'(0) = \lim_{h \to 0} \frac{f(h) - f(0)}{h} \]
Given that \( f(0) = 0 \), this simplifies to:
\[ f'(0) = \lim_{h \to 0} \frac{f(h)}{h} \]

4. **Finding \( f(h) \)**:
We know that \( f\left(\frac{1}{n}\right) = 0 \) for infinitely many points approaching 0. This suggests that \( f(x) \) must be 0 in a neighborhood around \( x = 0 \). Hence, we can conclude:
\[ f(x) = 0 \quad \text{for all } x \]

5. **Conclusion**:
Since \( f(x) = 0 \) for all \( x \), the derivative \( f'(x) \) is also:
\[ f'(x) = 0 \quad \text{for all } x \]
Thus, \( f'(0) = 0 \).

### Final Answer:
The correct option is \( f'(0) = 0 \).