Q.

For t > –1, let αt and βt be the roots of the equation ((t+2)1/71)x2+((t+2)1/61)x+((t+2)1/211)=0. If limt1+αt=a and limt1+βt=b, then 72(a+b)2 is equal to __________.          [2025]


Ans.

(98)

a+b=limt1+(αt+βt)

=limt1+[(t+2)161](t+2)171          [Sum of roots]

Let t + 2 = y, we get

a+b=limy1+y1/61y1/71=-76

So, 72(a+b)2=72(4936)=98.