Let f(x)=x2−2x+3x2−2x−8,x∈R−{−2,4} The range of f is

Let y=x2−2x+3x2−2x−8,x∈R−{−2,4}Note that y π 1.Now,                               (y−1)x2−2x(y−1)−(8y+3)=0As x is real          ∆≥0                       4(y−1)2+4(y−1)(8y+3)≥0⇒              (y−1)[y−1+8y+3]≥0.⇒               (y−1)(y+2/9)≥0⇒               y≤−2/9 or y>1.[∵y≠1] Thus,  y∈     R−−29,1