IIT JEE , AIEEE Forum - Mathematics , Algebra

waterdemon

My method is a bit different .

e^ax² - e^2ax + e^a - 1 = 0 lies between 1 and 2.

Let e^a= b

and let y = e^ax² - e^2ax + e^a - 1 = 0

Therefore ,

y = bx² - b²x + b - 1 = 0.

We know that 1<y<2

So we can write it as

(y-1)(y-2) < 0

(bx² - b²x + b - 2) (bx² - b²x + b - 3) < 0

So we now have two quadratics Q₁ and Q₂.

No we will have two cases.

Q₁ < 0 and Q₂> 0 : and : Q₁ > 0 and Q₂ < 0

Get the solution for both and take thier intersection to get the answer,

Hope it helped you.

Cheers !!!!!!!!!!!!!!!!!!!