The values of ‘a’ for which exactly one root of the equation $e^{a} x^{2} - e^{2 a} x + e^{a} - 1 = 0$ lies between 1 and 2 are given by

< a <

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

< <

No worries! We‘ve got your back. Try BYJU‘S free classes today!

< a <

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A
< a <

$(e^{α} - e^{2 α} + e^{α} - 1) (4 e^{α} - 2 e^{2 α} + e^{α} - 1)$ <0 $\Rightarrow (e^{2 α} - 2 e^{α} + 1) (2 e^{2 α} - 5 e^{α} + 1)$ <0 $L e t x = e^{α}$ $\Rightarrow (x - 1)^{2} (2 x^{2} - 5 x + 1) = 0$ $\Rightarrow (x - 1)^{2} (x - \frac{5 - \sqrt{17}}{4}) (x - \frac{5 + \sqrt{17}}{4})$ <0 $\Rightarrow \frac{5 - \sqrt{17}}{4}$ < x < $\frac{5 + \sqrt{17}}{4}$ $\Rightarrow l o g (\frac{5 - \sqrt{17}}{4})$ < x < $l o g (\frac{5 + \sqrt{17}}{4})$

Suggest Corrections

Similar questions

^{'} a^{'}

x^{2} - 2 a x + a^{2} - 1 = 0

^{'} a^{'}

For the equation $x^{2}$ - 2ax + $a^{2}$ - 1 = 0, the values of 'a' for which 3 lies in between the roots of given equation is

For the equation $x^{2}$ - (a - 3) x + a = 0 (a ∈ R), find the values of 'a' such that exactly one root lies in between 1 and 2.

x^{2} - (a + 1) x + 2 a = 0

(0, 3)

a

a

2^{a} x^{2} - 4^{a} x - 2^{a} - 1 = 0

1

2

Join BYJU'S Learning Program