If α, β are two distinct real roots of the equation ax3+x−1−a=0(a≠−1,0) none of which is equal to unity, then the value of limx→1/α (1+a)x3−x2−ae1−αx−1(x−1) is alα(kα−β) then the value of kl is

We have,ax3+x−1−a=0⇒ ax3−1+(x−1)=0⇒(x−1)ax2+ax+(a+1)=0It is given that a., pare two distinct roots of ax3+x−1−a=0 or(x−1)ax2+ax+(a+1)=0 Therefore, α, β are roots ofax2+ax+(a+1)=0 and hence, 1α,1β are roots of(a+1)x2+ax+a=0.∴ (a+1)x−1αx−1β=(a+1)x2+ax+aNow,limx→1/α (1+a)x3−x2−ae1−αx−1(x−1)=limx→1/α ax3−1+x2(x−1)e1−αx−1(x−1)=limx→1/α ax2+x+1+x2e1−αx−1(1−αx)=limx→1/α (a+1)x2+ax+a1−αx∴alα(kα−β)=(a+1)α2β(α−β)=−1αlimx→1/α (a+1)(x−1)x−1α(x)1−αx∴−(a+1)αlimx→1/αx−1β x−(a+1)α1α−1β=(a+1)α2β(α−β)⇒ alα(kα−β)=(a+1)aα(a+1)(α−β) ∵αβ=a+1a⇒ l(kα−β)=(α−β)⇒l=k=1⇒kl=1