If f is differentiable function satisfying f(1n)=0, for all n≥1,n∈I, then

f'(0)=limx→0f(x)−f(0)x−0=limx→0f(x)x=limn→∞f(1/n)1/n=0f(0)=f(0+)=limn→∞f(1n)=0f(1)=f12=f13=…=limn→∞ f1n=0 as f1n=0,n∈ integers and n≥1. ⇒ limn→∞ f1n =0⇒ f(0) =0∴ f(x)=0⇒ f′(x)=0⇒f′(0)=0 Hence, f(0)=f′(0)=0

If f is differentiable function satisfying f(1n)=0, for all n≥1,n∈I, then

f'(0)=limx→0f(x)−f(0)x−0=limx→0f(x)x=limn→∞f(1/n)1/n=0f(0)=f(0+)=limn→∞f(1n)=0f(1)=f12=f13=…=limn→∞ f1n=0 as f1n=0,n∈ integers and n≥1. ⇒ limn→∞ f1n =0⇒ f(0) =0∴ f(x)=0⇒ f′(x)=0⇒f′(0)=0 Hence, f(0)=f′(0)=0

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If $f$ is differentiable function satisfying $f (\frac{1}{n}) = 0,$ for all $n \geq 1, n \in I$ , then

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$| f (x) | \leq 1, x \in (0, 1]$

$f (0) = 0$ , but $f' (0)$ not necessarily zero

$f (x) = f^{'} (x) = 0, x \in (0, 1]$

$f^{'} (0) = 0 = f (0)$

answer is B.

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Detailed Solution

$f' (0) = \lim_{x \to 0} \frac{f (x) - f (0)}{x - 0} = \lim_{x \to 0} \frac{f (x)}{x} = \lim_{n \to \infty} \frac{f (1 / n)}{1 / n} = 0$ $f (0) = f (0^{+}) = \lim_{n \to \infty} f (\frac{1}{n}) = 0$

$f (1) = f (\frac{1}{2}) = f (\frac{1}{3}) = \dots = lim_{n \to \infty} f (\frac{1}{n}) = 0 as f (\frac{1}{n}) = 0, n \in integers and n \geq 1 . \begin{aligned} \Rightarrow lim_{n \to \infty} f (\frac{1}{n}) = 0 \\ \Rightarrow f (0) = 0 \end{aligned}$