The random  variable X can take only the values 1 and 2. Given that P(X=1)=p and that $E(X^2)=E(X)+1.$ The value of p is :

$\frac{1}{4}$ $\frac{1}{2}$ $\frac{2}{3}$ $\frac{1}{3}$

$\frac{1}{2}$