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Question

Let f and g be two functions such that f:[e,e2][1,2],f(x)=lnx and g:[1,2][e,e2],g(x)=ex. Then, fog is

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

The correct option is C one-one and onto function.
In the given domain, both f(x) and g(x) are one - one functions.
Also range of f(x) is [lne,lne2] which can be simplified as [1,2]. Hence it is onto function.
Similarly, Range of g(x) is [e,e2] 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(x)=domain of f(x)
So, f(g(x))=ln(g(x)=ln(ex)=x

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

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