Let f(x)=3x3+bx2+bx+3 (A) All the roots are negative. So, the number of sign change in f(−x) should be 3 , which is possible only when b is positive. According to options (A) ⟶ IV. (B) Clearly for any value of b,f(x)=0 has root x=−1. ∴f(x)=0⇒(x+1)(3x2+(b−3)x+3)=0 ⇒3x2+(b−3)x+3≠0 [ ∵ two roots are complex] ⇒(b−3)2−4(3)(3)<0 ⇒b2−6b−27<0 ⇒(b+3)(b−9)<0 ⇒−3<b<9 Hence, B ⟶ II. (C) We know that one root is x=−1 (Negative). Two roots are complex if −3<b<9. If b=−3, then f(x)=0 ⇒(x+1)(x−1)2=0 In this case two roots are positive and equal.(D) All roots will be real and distinct if and only if 3x2+(b−3)x+3=0 have two distinct real solutions other, than -1 . (b−3)2−4(3)(3)>0 ⇒b2−6b−27>0 ⇒(b+3)(b−9)>0 ⇒b∈(−∞,−3)∪(9,∞) Thus, D ⟶ III Therefore, A⟶IV,B⟶II,C⟶V,D⟶III