Check whether the function defined by $f\left( x+\lambda \right)=1+\sqrt{2f\left( x \right)-{{f}^{2}}\left( x \right)},x\in \mathsf{\mathbb{R}},\lambda >0$ is periodic or not. If yes, then find its period. \[\]

Hint: We use the domain and range of the square root function to find the range of $f\left( x \right)$. We take $f\left( x+\lambda \right)-1=\sqrt{2f\left( x \right)-{{f}^{2}}\left( x \right)}$ and square both side, form a complete square ${{\left( f\left( x \right)-1 \right)}^{2}}$, replace $x$ with $x+\lambda $ and simplify to get squared expressions both side We take square root both side to get absolute values and use the range of $f\left( x \right)$ to get the period.

Complete step-by-step solution
We know that if the function $f$ is periodic then there exists nonzero constant $T$ such that
\[f\left( x+T \right)=f\left( x \right)\]
We are given the real valued function;
\[f\left( x+\lambda \right)=1+\sqrt{2f\left( x \right)-{{f}^{2}}\left( x \right)},x\in \mathsf{\mathbb{R}},\lambda >0\]
We know that the square root also returns non-negative values. So we have $\sqrt{2f\left( x \right)-{{f}^{2}}\left( x \right)}\ge 0$, then we have;
\[\begin{align}
  & f\left( x+\lambda \right)\ge 1+0 \\
 & \Rightarrow f\left( x+\lambda \right)\ge 1 \\
\end{align}\]
Since square root is a strictly increasing function with given condition $\lambda >0$ we have;
\[f\left( x \right)\ge 1....\left( 1 \right)\]
We know from the definition of square root function that the square root takes only non-negative values. So we have;
\[\begin{align}
  & 2f\left( x \right)-{{f}^{2}}\left( x \right)\ge 0 \\
 & \Rightarrow f\left( x \right)\left( 2-f\left( x \right) \right)\ge 0 \\
\end{align}\]
We multiply both side negative sign to have;
\[\Rightarrow f\left( x \right)\left( f\left( x \right)-2 \right)\le 0.\]
We see from the above inequality that either we have $f\left( x \right)\ge 0,f\left( x \right)-2\le 0$ or $f\left( x \right)\le 0,f\left( x \right)-2\ge 0$. The latter deduction $f\left( x \right)\le 0,f\left( x \right)-2\ge 0$ is not possible since we have already obtained $f\left( x \right)\ge 1$. So we have;
\[f\left( x \right)\ge 0,f\left( x \right)-2\le 0.....\left( 2 \right)\]
We take intersection of intervals or use wavy curve method to find the range of the functions $f\left( x \right)$ as
\[f\left( x \right)\in \left[ 1,2 \right]\]
Now let us consider
\[\begin{align}
  & f\left( x+\lambda \right)=1+\sqrt{2f\left( x \right)-{{f}^{2}}\left( x \right)} \\
 & \Rightarrow f\left( x+\lambda \right)-1=\sqrt{2f\left( x \right)-{{f}^{2}}\left( x \right)} \\
\end{align}\]
We take square both sides to have;
\[\Rightarrow {{\left( f\left( x+\lambda \right)-1 \right)}^{2}}=2f\left( x \right)-{{f}^{2}}\left( x \right)\]
Let us add $-1$ and $1$ in the right hand side of the above equation to get a complete square as;
\[\begin{align}
  & \Rightarrow {{\left( f\left( x+\lambda \right)-1 \right)}^{2}}=2f\left( x \right)-{{f}^{2}}\left( x \right)+1-1 \\
 & \Rightarrow {{\left( f\left( x+\lambda \right)-1 \right)}^{2}}=1-\left( {{f}^{2}}\left( x \right)-2f\left( x \right)+1 \right) \\
 & \Rightarrow {{\left( f\left( x+\lambda \right)-1 \right)}^{2}}=1-{{\left( f\left( x \right)-1 \right)}^{2}}.....\left( 3 \right) \\
\end{align}\]
Now let us replace $x$ with $x+\lambda $ in the above step and have;
\[\Rightarrow {{\left\{ f\left( x+2\lambda \right)-1 \right\}}^{2}}=1-{{\left( f\left( x+\lambda \right)-1 \right)}^{2}}\]
We put the value of ${{\left( f\left( x+\lambda \right)-1 \right)}^{2}}$ in the right hand side of the equation obtained from (3). We have;
\[\begin{align}
  & \Rightarrow {{\left\{ f\left( x+2\lambda \right)-1 \right\}}^{2}}=1-\left\{ 1-{{\left( f\left( x \right)-1 \right)}^{2}} \right\} \\
 & \Rightarrow {{\left\{ f\left( x+2\lambda \right)-1 \right\}}^{2}}=1-1+{{\left( f\left( x \right)-1 \right)}^{2}} \\
 & \Rightarrow {{\left\{ f\left( x+2\lambda \right)-1 \right\}}^{2}}={{\left( f\left( x \right)-1 \right)}^{2}} \\
\end{align}\]
Let us take square root both side and have;
\[\left| f\left( x+2\lambda \right)-1 \right|=\left| f\left( x \right)-1 \right|\]
Since the range of $f\left( x \right)$ is $\left[ 1,2 \right]$ $f\left( x+2\lambda \right)-1\ge 1-1=0$ and $f\left( x \right)-1\ge 1-1=0$. So we deduce the absolute values as;
\[\begin{align}
  & f\left( x+2\lambda \right)-1=f\left( x \right)-1 \\
 & \Rightarrow f\left( x+2\lambda \right)=f\left( x \right) \\
\end{align}\]
So the function is periodic and the period is $2\lambda$. \[\]

Note: We note that the least period if positive is called the fundamental period of function. Since We have used the identity $\sqrt{{{\left( f\left( x \right) \right)}^{2}}}=\left| f\left( x \right) \right|$ here with any real valued function $f\left( x \right)$. We must be careful when we are finding the range of $f\left( x \right)$ not $\left[ 0,2 \right]$. The square root function strictly increases in the domain $\left[ 0,\infty \right)$ because if $x > y $then $\sqrt{x} > \sqrt{y}$.