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  • Let f(x) be defined for all x > 0 and be continuous. Let f(x) satisfy <img alt="\mathrm{f\left(\frac{x}{y}\right)=f(x)-f(y)}" src="https://entrancecorner.oncodecogs.com/gif.latex?

Let f(x) be defined for all x > 0 and be continuous. Let f(x) satisfy \mathrm{f\left(\frac{x}{y}\right)=f(x)-f(y)} for all x, y and f(e)=1. Then

Option: 1

f(x) is bounded.


Option: 2

\mathrm{f\left(\frac{1}{x}\right) \rightarrow 0} as x \rightarrow 0.


Option: 3

x f(x) \rightarrow 1 as x \rightarrow 0


Option: 4

f(x) = ln x.


Answers (1)

best_answer

Since \mathrm{\log \left(\frac{x}{y}\right)=\log x-\log y \text { and } \log (e)=1 \text {. }}

Therefore f(x) = log x. Clearly f(x) is unbounded because \mathrm{f(x) \rightarrow-\infty ~as ~x \rightarrow 0 ~and ~f(x) \rightarrow+~\infty ~as ~x \rightarrow \infty}

We have, \mathrm{f\left(\frac{1}{x}\right)=\log \left(\frac{1}{x}\right)=-\log x. ~As ~x \rightarrow 0, f\left(\frac{1}{x}\right) \rightarrow \infty }.

Now, \mathrm{{x} f(x)=x \log x }.

\mathrm{ \Rightarrow \lim _{x \rightarrow 0} x f(x) =\lim _{x \rightarrow 0} x \log x=\lim _{x \rightarrow 0} \frac{\log x}{1 / x} \\ }
\mathrm{ =\lim _{x \rightarrow 0} \frac{1 / x}{-1 / x^2}=-\lim _{x \rightarrow 0} x=0 }
Hence, (d) is the correct alternative.

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Rishi

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