If f(x) is a differentiable function such that f `R to R` and `f( (1)/(n)) =0 AA n ge 1,n in l,` then

Options

A. `f(x) =0 AA x in (0,1]`
B. `f(0) =0= f'(0)`
C. `f(0) =0 ` but ` f(0)` may or may not be 0
D. `| f (x) | le 1, AA x in [0,1]`

Answer

To solve the problem, we need to analyze the given function \( f(x) \) and its behavior at specific points. The problem states that \( f\left(\frac{1}{n}\right) = 0 \) for all integers \( n \geq 1 \). We will follow these steps to find the required information.

### Step-by-step Solution:

1. **Understanding the Given Condition:**
The function \( f \) is defined such that \( f\left(\frac{1}{n}\right) = 0 \) for every integer \( n \geq 1 \). 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:**
As \( n \) increases, \( \frac{1}{n} \) approaches \( 0 \). Therefore, we can conclude:
\[ \lim_{n \to \infty} f\left(\frac{1}{n}\right) = f(0) = 0 \]
This implies that \( f(0) = 0 \).

3. **Differentiating the Function:**
Now, we differentiate the function with respect to \( n \):
\[ \frac{d}{dn} f\left(\frac{1}{n}\right) = f'\left(\frac{1}{n}\right) \cdot \left(-\frac{1}{n^2}\right) \]
Since \( f\left(\frac{1}{n}\right) = 0 \), differentiating both sides gives:
\[ f'\left(\frac{1}{n}\right) \cdot \left(-\frac{1}{n^2}\right) = 0 \]

4. **Analyzing the Derivative:**
From the equation above, since \( -\frac{1}{n^2} \) is never zero for \( n \geq 1 \), we conclude that:
\[ f'\left(\frac{1}{n}\right) = 0 \]
for all integers \( n \geq 1 \).

5. **Taking the Limit of the Derivative:**
Again, as \( n \) approaches infinity:
\[ \lim_{n \to \infty} f'\left(\frac{1}{n}\right) = f'(0) = 0 \]

6. **Final Conclusions:**
We have established that:
- \( f(0) = 0 \)
- \( f'(0) = 0 \)

### Final Result:
Thus, we conclude that both \( f(0) \) and \( f'(0) \) are equal to \( 0 \).