The point (a2,a+1)  lies in the angle between the lines 3x – y + 1 = 0 and  x + 2y – 5 = 0 containing the origin if

Since origin & the point (a2,a+1)  lie on the some side of both the lines, therefore we have  3a2−(a+1)+1>0  i.e.  a(3a−1)>0  gives    a∈(−∞,0)∪(13,∞)    and                                                                      a2+2(a+1)−5<0  i.e.  a2+2a+3<0  i.e.  (a−1)(a+3)<0  gives  a∈(−3,1) Intersection of the above inequalities gives a∈(−3,0)∪(13,1)