The set of all values of ‘a’ for which the roots of the equation (a+1)x2−3ax+4a=0(a≠−1) are real and greater than 1 is

(a+1)x2−3ax+4a=0D=9a2−16a(a+1)≥0 ⇒ -7a2-16a≥0 ⇒a7a+16≤0      ⇒a∈-167,0.......1x1>1,x2>1∴(a+1)f(1)>0 and 3a2(a+1)>1⇒(a+1)(2a+1)>0 a<-1 or a >-12 and a−2a+1>0……...(2) from  1,2 ∴a∈−167,−1