The values of ‘a’ for which exactly one root of the equation eax2−e2ax+ea−1=0 lies between 1 and 2 are given by
< a <
(eα−e2α+eα−1)(4eα−2e2α+eα−1)<0 ⇒(e2α−2eα+1)(2e2α−5eα+1)<0 Let x=eα ⇒(x−1)2(2x2−5x+1)=0 ⇒(x−1)2(x−5−√174)(x−5+√174)<0 ⇒5−√174 < x < 5+√174 ⇒log(5−√174) < x < log(5+√174)