A function `f:R to R` is differernitable and satisfies the equation `f((1)/(n))`=0 for all integers `n ge 1`, then

Options

A. `f(x)=0"for all x"in (0,1]`
B. `f(0)=f'(0)`
C. `f(0)=0` but f'(0) need not be equal to 0
D. `|f(x)| le 1"for all"x in[0,1]`

Explanation

To solve the problem, we need to analyze the function \( f: \mathbb{R} \to \mathbb{R} \) given that it 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 know that \( f\left(\frac{1}{n}\right) = 0 \) for \( n = 1, 2, 3, \ldots \). This means that the function takes the value 0 at the points \( \frac{1}{1}, \frac{1}{2}, \frac{1}{3}, \ldots \).

2. **Convergence of the Sequence**:
The sequence \( x_n = \frac{1}{n} \) converges to 0 as \( n \) approaches infinity. Since \( f \) is continuous (as it is differentiable), we can use the property of continuity.

3. **Applying Continuity**:
By the continuity of \( f \) at \( x = 0 \):
\[ \lim_{n \to \infty} f\left(\frac{1}{n}\right) = f(0) \]
Since \( f\left(\frac{1}{n}\right) = 0 \) for all \( n \), we have:
\[ \lim_{n \to \infty} 0 = f(0) \]
Thus, \( f(0) = 0 \).

4. **Finding the Derivative at Zero**:
Now, we will find \( f'(0) \). The derivative at a point is defined as:
\[ f'(0) = \lim_{x \to 0} \frac{f(x) - f(0)}{x - 0} = \lim_{x \to 0} \frac{f(x)}{x} \]
Since \( f(0) = 0 \), this simplifies to:
\[ f'(0) = \lim_{x \to 0} \frac{f(x)}{x} \]

5. **Using the Condition Again**:
We can also evaluate this limit using the points \( x = \frac{1}{n} \) as \( n \) approaches infinity. For \( x = \frac{1}{n} \):
\[ f\left(\frac{1}{n}\right) = 0 \implies \frac{f\left(\frac{1}{n}\right)}{\frac{1}{n}} = \frac{0}{\frac{1}{n}} = 0 \]
Therefore, as \( n \to \infty \):
\[ \lim_{n \to \infty} \frac{f\left(\frac{1}{n}\right)}{\frac{1}{n}} = 0 \]
This implies:
\[ f'(0) = 0 \]

### Conclusion:
From the above steps, we conclude that:
- \( f(0) = 0 \)
- \( f'(0) = 0 \)