Let f(x) = vmatrix ^2 x & ^2 x & 1 ^2 x & ^2 x & 1 -1 & 2 & 2 vmatrix . If the equation f(x) = 0 has exactly 15 solutions in the interval [0, n 3 ] , where n N , then the sum of all possible values of n is :

Step-by-step solution

First, we simplify the determinant f(x) . Applying the column operation C₁ C₁ + C₂ : f(x) = vmatrix ^2 x + ^2 x & ^2 x & 1 ^2 x + ^2 x & ^2 x & 1 -1 + 2 & 2 & 2 vmatrix = vmatrix 1 & ^2 x & 1 1 & ^2 x & 1 1 & 2 & 2 vmatrix Applying row operations R₂ R₂ - R₁ and R₃ R₃ - R₁ : f(x) = vmatrix 1 & ^2 x & 1 0 & ^2 x - ^2 x & 0 0 & 2 - ^2 x & 1 vmatrix Expanding along the first column: f(x) = 1 (( ^2 x - ^2 x) 1 - 0) = - 2x The equation f(x) = 0 becomes 2x = 0 . The general solution is 2x = (2k-1) 2 x = (2k-1) 4 , where k