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If $f(x)=\frac{x}{\ln x}$ & $g(x)=\frac{\ln x}{x}$. Then identify the correct statement.

A) $\frac{1}{g(x)}$ and $f(x)$ are identical functions

B) $\frac{1}{f(x)}$ and $g(x)$ are identical functions

C) $f(x)\cdot g(x)=1 \forall x>0$

D) $\frac{1}{f(x)\cdot g(x)}=1 \forall x>0$

I don't have the solution but as per the answer key Only A is the correct statement , B,C,D are incorrect statement .

My Approach for B let $t(x)=\frac{1}{f(x)}$ , now the question is whether $t(x)$ & $g(x)$ are identical function, my thought would be that they are identical function because for identical function we need to check domain and range on $t(x)$ and not on its reciprocal.But on contrary in the ANSWER Key this is mentioned as INCORRECT.

Regarding C and D I don't know why it is incorrect.

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You're on the right track: the key to this question is to identify the domains of all the functions involved. Here are some of them.

  • Consider the function $g(x)=\dfrac{\ln x}{x}$. The domain of this function is $x>0$, i.e. $x\in(0,+\infty)$. (I hope you understand why.)

  • Consider the function $G(x)=\dfrac{1}{g(x)}$. This function, since it involves $g(x)$, is only defined when $g(x)$ is defined; and moreover, we can't have $g(x)=0$. So we take the domain of $g(x)$, i.e. $x\in(0,+\infty)$, and exclude all the points where $g(x)=\dfrac{\ln x}{x}=0$. There's only one such point, $x=1$. Therefore, the domain of $G(x)=\dfrac{1}{g(x)}$ is $x\in(0,1)\cup(1,+\infty)$.

Similarly, you can find the domains of $f(x)$, $\dfrac{1}{f(x)}$, etc. And you should see that the domains of the two functions in (A) are the same, but in (B) they are not.

As for (C) and (D), again it's about the domains. For example, in (C), the expression $f(x)\cdot g(x)$ is undefined at $x=1$ because $f(x)$ is undefined at $x=1$. So the statement is not true at $x=1$ (the left-hand side is undefined, and as such it's not equal to $1$), and thus we can't say that it holds for all $x>0$.

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The important point is that $\ln 1=0$, so you can't divide by it. In A you get a $\ln x$ in both denominators, so both sides of the equation are undefined at $x=1$. In B, $g(x)$ is nicely defined for all $x \gt 0$, but $f(1)$ is not so $\frac 1{f(1)}$ is not either. C and D both fail for $x=1$ for the same reason.

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