If $f:\mathbb{R}\rightarrow \mathbb{{R}}$ and $f(x)$ satisfying $f(2x+3)+f(2x+5) = 2$ . The period of $f(x)$ If $f:\mathbb{R}\rightarrow \mathbb{{R}}$ and $f(x)$ be a function satisfying $f(2x+3)+f(2x+5) = 2$. Then period 
of function $f(x)$ is.
$\bf{My\; Solution::}$ Let $(2x+3) = t\;,$ Then equation is $f(t)+f(t+2) = 2$
Now Replace $t\rightarrow (t+2)\;,$ we get $f(t+2)+f(t+4) = 2$
So from these two equation, we get $f(t+4) = f(t)$
So period of function $f(t)$ is $=4$
But answer Given as $=2$  . But in Solution it is given as $ = 2$
I did not understand where I have done Wrong.
Help me
Thanks
 A: $\displaystyle f(2x+3) + f(2x+5) = 2$
Replace $x$ with $x+1$:
$\displaystyle f(2x+5) + f(2x+7) = 2$
Subtracting and rearranging, $\displaystyle f(2x+3) = f(2x+7)$
Hence $\displaystyle f(2x+3) = f((2x+3)+4)$
At this point, you can conclude that the period is $4$. If this is difficult to see, you can replace $(2x+3)$ with $t$, as the OP has done.
It might be tempting to do this:
$\displaystyle f(2(x+\frac{3}{2})) = f(2(x+\frac{7}{2}))$
and conclude that the period is $\displaystyle \frac{7}{2} - \frac{3}{2} = 2$.
But this would be a mistake. This ($2$) is the period of $f(2x)$. The period of $f(x) = 4$.
If you're still unconvinced, here's another example. If we define $f(x)=\sin 2x$, then the period of $f(x)$ is $\pi$. But if we define $f(2x)=\sin 2x$, then the period of $f(x)= \sin x$ is $2\pi$, even if the period of $\sin 2x=f(2x)$ remains $\pi$. This is what's happening in this case. The period of $f(2x)$ is $2$, but the period of $f(x)$ is actually $4$.
A: Period is usually defined as the smallest $p>0$ such that $f(x+p)=f(x)$.

$$\large\underline{\text{Counter-Example}}$$
Consider the function $$f(x)=1+\sin{\frac{3\pi x}{2}}$$
See for yourself that $$f(2x+3)+f(2x+5)=2$$
And that $$f(x+\frac{4}{3})=f(x).$$
By properties of $\sin$ function, it can be shown that $\frac{4}{3}$ is the period.

Conclusion
There is no point in arguing that this is the period or that is the period (unless you can show that it is the 'smallest'). None of the answers or comments have shown that either $2$ or $4$ is the smallest.

$$\underline{\text{Why 2 cannot be the period}}$$
By the way, we can prove that $2$ cannot be period of this function. That, if we assume $2$ is the period  then the given equation can be reduced to $$2f(2x+3)=2 \iff f(2x+3)=1$$ or simply $f$ is a constant function(which we don't consider periodic). 
A: The function is $t$-periodic with period 4. However, as $t = 2x + 3$, this means that $f$ is $x$-periodic with period 2, since an increment of 2 to $x$ will give an increment of 4 to $t$.
