If set A=x∣x2(5−x)(1−2x)(5x+1)(x+2)<0 and set B=x∣3x+16x3+x2−x>0, then A∩B does not contain

We have x2(x−5)(2x−1)(5x+1)(x+2)<0From the sign scheme, solution is−2,−15∪12,5               (i)Now 3x+1x6x2+x−1>0⇒ 3x+1x(3x−1)(2x+1)>0From the sign scheme solution is−∞,−12∪−13,0∪13,∞               (ii)Clearly, from (i) and (ii), option(s) (1) and (3) are correct.