Question:
If f (x) is continuous and differentiable function and f (1/n) = 0 $\forall$ n $\ge$ 1and n $\in$ I, then
If f (x) is continuous and differentiable function and f (1/n) = 0 $\forall$ n $\ge$ 1and n $\in$ I, then
Updated On: Jun 14, 2022
- f(x) = 0, x $\in$ (0, 1]
- f(0) = 0, f '(0) = 0
- f(0) = 0 = f '(0), x $\in$ (0, 1]
- f(0) = 0 and f '(0) need not to be zero
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The Correct Option is B
Solution and Explanation
Given that f (x) is a continuous and differentiable function and $f \left( \frac{1}{x} \right) = 0 , x = n \in I $
$\therefore \, f(0^{+} ) = f \left( \frac{ 1}{\infty}\right) = 0 $
Since R.H.L. = 0,
$\therefore \, f(0) = 0 $ for $f(x)$ to be continuous.
Also $f'(0) = \displaystyle \lim_{h \to 0} \frac{f(h) - f(0)}{h - 0} = \displaystyle \lim_{h \to 0} \frac{f(h)}{h} = 0 $
$ = 0 $ [Using f (0) = 0 and $f (0^+) = 0$]
Hence $f (0) = 0, f ' (0) = 0$
$\therefore \, f(0^{+} ) = f \left( \frac{ 1}{\infty}\right) = 0 $
Since R.H.L. = 0,
$\therefore \, f(0) = 0 $ for $f(x)$ to be continuous.
Also $f'(0) = \displaystyle \lim_{h \to 0} \frac{f(h) - f(0)}{h - 0} = \displaystyle \lim_{h \to 0} \frac{f(h)}{h} = 0 $
$ = 0 $ [Using f (0) = 0 and $f (0^+) = 0$]
Hence $f (0) = 0, f ' (0) = 0$
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Concepts Used:
Continuity
A function is said to be continuous at a point x = a, if
limx→a
f(x) Exists, and
limx→a
f(x) = f(a)
It implies that if the left hand limit (L.H.L), right hand limit (R.H.L) and the value of the function at x=a exists and these parameters are equal to each other, then the function f is said to be continuous at x=a.
If the function is undefined or does not exist, then we say that the function is discontinuous.
Conditions for continuity of a function: For any function to be continuous, it must meet the following conditions:
- The function f(x) specified at x = a, is continuous only if f(a) belongs to real number.
- The limit of the function as x approaches a, exists.
- The limit of the function as x approaches a, must be equal to the function value at x = a.