Decision boundary for quadratic discriminant analysis in 1D 1/2 puntos (calificado) Consider the general case where the two classes have different means and possibly different variances: XC=0N(0,02)XC=1N(1,12) and the prior distribution of C is P(C=1)=pP(C=0)=1p. To find the decision boundary, we need to solve the following equation for x : P(C=0)P(X=xC=0)=P(C=1)P(X=xC=1). It is easier to solve this by first taking the logarithm on both sides. Rearranging the terms, we get: ln[P(C=0)P(X=xC=0)]ln[P(C=1)P(X=xC=1)]=0. 1. What is the degree of the equation? Assume 02=12 and 0=1. quadratic equation in x. linear equation in x none of the above 2. Report the coefficient of x2 on the left hand side in the equation above. (If applicable, type mu_0, mu_1 for 0,1 respectively, sigma_0 2 for 02, and sigma_1 \2 for 12 respectively.) Coefficient of x2