If ∣∣ ∣ ∣∣xnxx+2xx+4ynyn+2yn+4znzn+2zn+4∣∣ ∣ ∣∣=(1y2−1x2)(1z2−1y2)(1x2−1z2) then n=
Using factor theorem the factors of y3 − 2y2 − y + 2 is
∀y∈R, f(y) =∣∣ ∣ ∣∣1yy+12yy(y−1)y(y+1)3y(y−1)y(y−1)(y−2)y(y2−1)∣∣ ∣ ∣∣ ,then ∫π/2−π/2f(y2+2)dy equals