f(x) and g(x) both defined R $\to$ R are two non-constant continuous functions then
Statement – 1: If g (x) is periodic then f(g(x)) is also periodic
Statement – 2: If g(x) and f(x) are both aperiodic then f(g (x)) is aperiodic.
Statement – 3: If g(x) is periodic with fundamental period T then f(g(x)) is also periodic with fundamental period T
Statement – 4: If g(x) is periodic with fundamental period T then f(g(x)) is also periodic with period T  

 Which of the following must be truth value of above statements in that order.

For the statement I

Suppose that the function $g\left(x\right)$ is periodic, and its period is $T$, it implies that $g\left(x+T\right)=g\left(x\right)$

Suppose that  $h\left(x\right)=f\left(g\left(x\right)\right)$

it implies that 

$\begin{array}{rcl}h\left(x+T\right)& =& f\left(g\left(x+T\right)\right)\\ & =& f\left(g\left(x\right)\right)\\ & =& h\left(x\right)\end{array}$

Therefore, the function $f\left(g\left(x\right)\right)$ is periodic function. 

Hence, the statement I is true. 

-For the statement 2 :  If g(x) and f(x) are both aperiodic then f(g (x)) is aperiodic.

This is false statement. 

Suppose that $g\left(x\right)=x^{\frac{1}{3}},f\left(x\right)=\sin \left(x^3\right)$ are not periodic functions, $f\left(g\left(x\right)\right)=\sin x$ is a periodic function with period $2\mathrm{\pi }$

For the statement 3 : -If g(x) is periodic with fundamental period T then f(g(x)) is also periodic with fundamental period T

This is false statement

Suppose that $g\left(x\right)=\cos x$ and $f\left(x\right)=\cos x$

Here $g\left(x\right)$ is periodic with fundamental period $2\mathrm{\pi }$

And 

$\begin{array}{rcl}h\left(x\right)& =& f\left(g\left(x\right)\right)\\ & =& \cos \left(\cos \left(x\right)\right)\\ h\left(\mathrm{\pi }+\mathrm{x}\right)& =& \cos \left(\cos \left(\mathrm{\pi }+\mathrm{x}\right)\right)\\ & =& \cos \left(-\cos x\right)\\ & =& \cos \left(\cos x\right)\end{array}$

Therefore, the function $h\left(x\right)$ is a periodic function whose fundamental period is $\mathrm{\pi }$ not $2\mathrm{\pi }$

Observing the above statement, the statement 4 is true.