Question

$f\left(x\right)=\frac{x}{lnx}$ and $g\left(x\right)=\frac{lnx}{x}$. Then identify the CORRECT statement

• A. $\frac{1}{g\left(x\right)}$ and $f\left(x\right)$ are identical functions
• B. $\frac{1}{f\left(x\right)}$ and $g\left(x\right)$ are non-identical functions
• C. $f\left(x\right).g\left(x\right)<0;\forall x>0$
• D. $\frac{1}{f\left(x\right).g\left(x\right)}=-1\forall x>0$

Answer 1

Answer: (A), (B)

The correct options are
A $\frac{1}{g\left(x\right)}$ and $f\left(x\right)$ are identical functions
B $\frac{1}{f\left(x\right)}$ and $g\left(x\right)$ are non-identical functions

$f\left(x\right)=\frac{x}{\ln x}$

$\ln x$ is defined for $x>0$.

Also $\ln x$ is in denominator in $f\left(x\right)$, so cannot be equal to zero. Hence $x\ne 1$.

$\therefore$ Domain of $f\left(x\right),$ $x=(0,\infty )-\left\{1\right\}$

$\frac{1}{f\left(x\right)}=\frac{1}{\frac{x}{lnx}}$

$\frac{1}{f\left(x\right)}$ is defined when $\frac{x}{\ln x}\ne 0$

$x\ne 0$ and $\ln x\ne 0$. Also $\ln x$ is defined for $x>0$.

Domain of $\frac{1}{f\left(x\right)}$ is $(0,\infty )-\left\{1\right\}$

$g\left(x\right)=\frac{\ln x}{x}$

Domain of $g\left(x\right)$ is $(0,\infty )$

$\frac{1}{g\left(x\right)}=\frac{1}{\frac{\ln x}{x}}$

$\frac{\ln x}{x}\ne 0$

$\ln x\ne 0⟹x\ne 1$

For $\ln x$ to be defined $x>0$

Domain of $\frac{1}{g\left(x\right)}$ is $(0,\infty )-\left\{1\right\}$

Now $f\left(x\right)=\frac{x}{\ln x}$ and $\frac{1}{g\left(x\right)}=\frac{1}{\frac{\ln x}{x}}=\frac{x}{\ln x}$

Hence range of $f\left(x\right)$ and $\frac{1}{g\left(x\right)}$ are same.

Also their domains are same $(0,\infty )-\left\{1\right\}$

Hence $f\left(x\right)$ and $\frac{1}{g\left(x\right)}$ are identical.

And $\frac{1}{f\left(x\right)}$ and $g\left(x\right)$ are non -identical.

In options C and D there is $x=1$ in the domain which is not possible