IIT JEE , AIEEE Forum - Mathematics , Algebra

alisha_gupta_27

For what value of a, for which exactly one root of the equation e^ax² - e^2ax +e^a - 1=0 lies between 1 and 2 ?

lazycol

let f(x) = e^ax² - e^2ax +e^a - 1

f(1) = e^ax² - e^2ax +e^a - 1 = -(e^a - 1)²ie f(1) < 0

so 4 f 2 hav a root in (1,2) f(2) shud b >0

f(2) =4e^a - 2e^2a +e^a - 1 > 0

2e^2a-5e^a + 1 < 0 ie e^a ( (5-

17)/4 , (5+

17)/4 )

a

(log [ (5-

17)/4, log [(5+

17)/4] )

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 !!!!!!!!!!!!!!!!!!!

lazycol

if m is a root of f,it means at dat value f(m)=0

so it can b in 2 ways f(x) < 0, 4 x<m ; f(x) =0 at x=m & f(x) >0 4 x>m

OR f(x) < 0, 4 x<m ; f(x) =0 at x=m & f(x) <0 4 x>m

f(1) < 0 and f has a root in (1,2)

means f(2) shud b>0 {becoz 1< root & 2>root}

f(2) is another quadratic in e^a

i think u r aware of d fact dat if f(x) < 0;x lies bw d roots of f(x)

so v find e^a,hence a

hope u got it

rate me plz