Periodic Function interval used of another non-periodic function From any non-periodic function we can manage functions which will be periodic by expanding the function with the condition f(x+T)=f(x) where ‘T’ is the length of interval for y=f(x) which we need to expand. Considering this answer the below question 
If g (x) is a periodic function with period ‘1’ defined from $f(x)=e^{2x},x\in[0,1) $ then the number of solutions to the equation g(x)=[x] is (where [.] represents greatest integer function)
(A) 6           
(B) 7           
(C) 8           
(D) none of these
I am not able to understand what the question is asking hence not able to approach ,one point I can understand that the length of interval 'T' is $[1,e^2]$ where $e^2$ is approximately equal to 7.38.
 A: What the question is saying is that $g(x)$ is periodic, with a period $1$, and with its values in any one period being given by $f(y)$ for $y \in [0,1)$, with $y$ being the change in value of $x$ from the start of the period (I used $y$ instead of $x$ as in the question to help avoid confusion). The question doesn't give any specific starting point for the periods, but there is some point (actually, an infinite # of them, but just choose one) $x_0$ where $g(x_0) = f(0)$, e.g., a value in $[0,1)$. Nonetheless, regardless of which $x_0$ you use, you then have for any point $x$, where $x - x_0 = n + r$ for some integer $n$ and real $0 \le r \lt 1$, that $g(x) = f(r)$.
Visually, as shown in the example graph below where $x_0 = 0.42$, $g(x)$ is a periodic, discontinuous function, with the values of $f(y)$ for $y \in [0,1)$ being used repeatedly over every section of length $1$, starting from some point and repeating both before & after that point. Also, this means the range of $g(x)$ would be the range of $f(y)$, i.e., $[1,e^2)$.
$g(x)$ meeting the question requirements">
To determine the answer, you need to consider how many solutions there are for $g(x) = \lfloor x \rfloor$ over an appropriate range of $x$ values. I'm leaving it to you to do this part, but an important thing to note is that for all periodic functions, the range of the function is always the same from any starting point over a set of $x$ values which are one period in length (i.e., the value of $x_0$ doesn't really matter), e.g., as in the diagram above, between $1$ and $2$, you have the latter part of $f(y)$ and then the missing, initial part of $f(y).
A: The are $x$s $\in [0, 1)$ where $e^{2x} = 1,2,3,4,5,6,7$ so there an $x \in [n,n+1)$ there $g(x) = e^{2(x-n)} = n = [x]$.  There are $7$ such $n$s.  That's all there is to it.

Okay, slightly more detail.
$g(x)$ is periodic.  If $x \in [0,1)$ then $g(x) = 2^{x}$.  If $x \in [n,n+1)$ then $g(x) = g(x-n) = e^{2(x-n)}$.
So if $x \in [n,n+1)$ then $[x] = n$.
When does $g(x)=e^{2(x-n)} = [x] = n$.
Well.... when $2(x-n) = 2\{x-n\} = \ln n$ or when $x-n = \frac {\ln n}2$ or $x = n+ \frac {\ln n}2$.
As $0 \le x-n < 1$ then we must have $0\le \ln n < 2$. or that $n=1,2,3,4,5,6,7$.
And if so we just need to have $x= n + \frac {\ln n}2$ where $n=1,2,3.....,7$
==== tl;dr below ======
$g(x)$ is the periodic function that if $x \in [n,n+1)$ then $g(x) = e^{2(x-n)}$.
Or in other words $g(x) = e^{2\{x\}}$
So the question is when does $g(x)= e^{2\{x\}} = [x]$.
Well $[x]$ is an integer. When is $g(x)$ an integer?
so the very lowest that $g(x)$ can be is $g(n) = e^0 = 1$.
And the limit of the highest $g(x)$ can approach but not reach is $\lim_{x\to (n+1)^-} g(x) = e^{2*1} = e^2\approx 7.389$
And $e^{2x}$ is continuous so between $[n,n+1)$ the function $g(x)$ goes from $e^{2*0} =1$ through $\lim_{x\to (n+1)^-} g(x) = e^{2*1} = e^2\approx 7.389$.
So for every interval $[n,n+1)$ the function $g(x)$ will hit the integers $1$ through $7$.
So somewhere in $[1,2)$ there is an $x$ where $g(x) = e^{2\{x\}} = [x]=1$.  In fact it is $g(1) = e^{2\{1\}} = e^{2*0} = 1= [1]$.
And somewhere in $[2,3)$ there ans $x$ where $g(x) = e^{2\{x\}} = [x]=2$.  We could even go as for as solving this. $0 < \ln 2 < 2$ and so if $x = 2+\frac{\ln 2}2$ then $g(x) = e^{2\{2+\frac {\ln 2}2\}} = e^{\ln 2} = 2 = [2+\ln 2]$.
And so on.  In $[3, 4)$ we have $g(3 + \frac {\ln 3}2) = 3=[3+ \frac {\ln 3}2]$ and so on.
In $[n, n+1)$ where $\ln n < 2$ we have $g(n+\frac {\ln n}2) = e^{2*\frac {\ln n}2}= n = [n+\frac {\ln n}2)$.
And $1,....,7$ are the only integers where $\ln n < 2$.
