The function $f$ is defined by $f\left(x\right)=x^p(1-x{)}^q$ for all $x\in R$, where $p,q$ are positive integers, has a maximum value, for $x$ is equal to

The given function is  $f\left(x\right)=x^p(1-x{)}^q$

Differentiate the given function $f\left(x\right)=x^p(1-x{)}^q$

$$\begin{gathered} f\text{'}\left(x\right)=(1-x{)}^q\left(p{x}^{p-1}\right)-x^p\left(q(1-x{)}^{q-1}\right) \\ f\text{'}\left(x\right)=(1-x{)}^q{x}^p\left(\frac{p}{x}-\frac{q}{1-x}\right) \end{gathered}$$

For maximum value of $f\left(x\right)$ quate $f^{\text{'}}\left(x\right)=0$

$$\begin{gathered} f^{\text{'}}\left(x\right)=0 \\ (1-x{)}^q\left(p{x}^{p-1}\right)-x^p\left(q(1-x{)}^{q-1}\right)=0 \end{gathered}$$

Step $2$

Evaluate for $x$ by solving the above equation

$(1-x{)}^q\left(p{x}^{p-1}\right)=x^p\left(q(1-x{)}^{q-1}\right)$

$$\begin{gathered} p\frac{(1-x{)}^q}{(1-x{)}^{q-1}}=q\frac{x^p}{x^{p-1}} \\ p(1-x)=qx \\ x(p+q)=p \\ x=\frac{p}{p+q} \end{gathered}$$

Hence, if the equation is $f\left(x\right)=x^p(1-x{)}^q$ then the maximum value for $x$ is equal to $\frac{p}{p+q}·$

Therefore, the correct answer is option $\left(4\right)$.