Exams › JEE Advanced › Maths
Correct answer: x in (-1, 1] union {2} union [3, infinity)
Even-power factors only create zeros (which satisfy <= 0) and never change sign. The odd factors (x-3)³ and (1-x) plus the linear denominator (x+1) determine the sign intervals. x = -1 is excluded (denominator zero). x = 2 is an isolated solution (a squared factor vanishes there).