Let a function f(x) = begin{cases} -ln(3x - [ | vedclass

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Question

(D) The function is defined as $f(x) = -\ln({3x})$ when $3x 
eq n$,where ${3x}$ is the fractional part of $3x$. When $3x = n$,$f(x) = \ln(\operatorna

Vedclass · Accepted Answer

$14$

Vedclass · Answer

(D) The function is defined as $f(x) = -\ln({3x})$ when $3x 
eq n$,where ${3x}$ is the fractional part of $3x$. When $3x = n$,$f(x) = \ln(\operatorname{sgn}(n)) = \ln(1) = 0$.Thus,$f(x) = \begin{cases} -\ln({3x}) & ; 3x 
eq n \ 0 & ; 3x = n \end{cases}$.Since ${3x} \in [0, 1)$,$-\ln({3x})$ is defined for ${3x} \in (0, 1)$. As ${3x} 	o 1^-$,$-\ln({3x}) 	o 0^+$. At $3x = n$,$f(x) = 0$.This means the function $f(x)$ approaches $0$ at every point $x = \frac{n}{3}$ for $n \in \{1, 2, \dots, 14\}$ within the interval $(0, 5)$.Since the function values are always non-negative (as $-\ln({3x}) > 0$ for ${3x} \in (0, 1)$ and $f(n/3) = 0$),the minimum value of the function is $0$.This minimum occurs at all points $x = \frac{n}{3}$ for $n = 1, 2, \dots, 14$.Therefore,there are $14$ such points.

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