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(B) Given that $f\left( \frac{1}{n} \right) = 0$ for all $n \in \{1, 2, 3, \dots\}$.Since $f(x)$ is continuous (as it is differentiable),we have $f(0)

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$f(0) = 0$ and $f'(0) = 0$

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(B) Given that $f\left( \frac{1}{n} ight) = 0$ for all $n \in \{1, 2, 3, \dots\}$.Since $f(x)$ is continuous (as it is differentiable),we have $f(0) = \lim_{n 	o \infty} f\left( \frac{1}{n} ight) = 0$.Now,consider the sequence $x_n = \frac{1}{n}$. We have $f(x_n) = 0$ and $f(0) = 0$.By the definition of the derivative at $x = 0$:$f'(0) = \lim_{x 	o 0} \frac{f(x) - f(0)}{x - 0} = \lim_{n 	o \infty} \frac{f(1/n) - f(0)}{1/n - 0} = \lim_{n 	o \infty} \frac{0 - 0}{1/n} = 0$.Thus,$f(0) = 0$ and $f'(0) = 0$.

If $f(x)$ is a differentiable function such that $f: R \to R$ and $f\left( \frac{1}{n} \right) = 0$ for all $n \ge 1, n \in I$,then:

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