Question

Let $f:(e,\infty )\to R$ be defined by $f\left(x\right)=\ell n\left[\ell n\right(\ell nx\left)\right]$, then.

• A. $f$ is one-one but not onto
• B. $f$ is onto but not one-one
• C. $f$ is both one-one and onto
• D. the range of $f$ is equal to its codomain

Answer 1

Answer: (C), (D)

The correct options are
C $f$ is both one-one and onto
D the range of $f$ is equal to its codomain

Given $f:(e,\infty )\to R$ is defined by

$f\left(x\right)=\log \left[\log \right(\log x\left)\right]$
Injective:

Let $x,y\in (e,\infty )$ such that

$f\left(x\right)=f\left(y\right)$

$\Rightarrow \log \left[\log \right(\log x\left)\right]=\log \left[\log \right(\log y\left)\right]$

$\Rightarrow \log \left(\log x\right)=\log \left(\log y\right)$

$\Rightarrow \log x=\log y$

$\Rightarrow x=y$

Hence, $f$ is one-one.
Surjective:

Let $y=\log \left[\log \right(\log x\left)\right]$

$\Rightarrow \log \left(\log x\right)=e^y$

$\Rightarrow \log x=e^{e^y}$

$\Rightarrow x=e^{e^{e^y}}$

Clearly $R\left(f\right)=R$

So, $R\left(f\right)=$Co-domain$\left(f\right)$

Hence, $f$ is onto.