Question

Let $f$ and $g$ be two functions such that $f:[e,e^2]\to [1,2],f\left(x\right)=\ln x$ and $g:[1,2]\to [e,e^2],g\left(x\right)=e^x.$ Then, $fog$ is

• A. neither one-one nor onto function.
• B. onto but not one-one function.
• C. one-one and onto function.
• D. one-one but not onto function.

Answer 1

The correct option is C one-one and onto function.

In the given domain, both $f\left(x\right)$ and $g\left(x\right)$ are one - one functions.
Also range of $f\left(x\right)$ is $[\ln e,\ln e^2]$ which can be simplified as $[1,2]$. Hence it is onto function.
Similarly, Range of $g\left(x\right)$ is $[e,e^2]$ so, it is also onto function.
Clearly, both $f$ and $g$ are bijective and Range of $g=$ Domain of $f$
So, $fog$ is also bijective function.

Alternate Solution:

Here range of $g\left(x\right)=$domain of $f\left(x\right)$
So, $f\left(g\right(x\left)\right)=\ln \left(g\right(x)=\ln (e^x)=x$

Also, domain of $fog\left(x\right)$ will be domain of $g\left(x\right)$ i.e. $[1,2]$
and range of $fog\left(x\right)$ will be $[1,2].$
Hence, for given domain and range $fog\left(x\right)$ will be both one - one and onto function.