Question

The values of ‘a’ for which exactly one root of the equation e^ax² – e^2ax + e^a – 1 = 0 lies between 1 and 2 are given by –

• A. ln$\frac{(5 - \sqrt{17})}{4}$< a <ln$\frac{5 + \sqrt{17}}{4}$
• B. 0 < a < 100
• C. ln$\frac{5}{4}$< a <ln$\frac{10}{3}$
• D. None of these

Answer 1

Answer: (A)

ln$\frac{(5 - \sqrt{17})}{4}$< a <ln$\frac{5 + \sqrt{17}}{4}$

Answer 2

(e^a – e^2a + e^a – 1) (4e^a – 2e^2a + e^a – 1) < 0

(e^2a – 2e^a + 1) (2e^2a – 5e^a + 1) < 0

Let, x = e^a

⇒ (x – 1)² (2x² – 5x + 1) < 0

⇒ (x – 1)² $\left( x - \frac { 5 - \sqrt { 17 } } { 4 } \right)$ $\left( x - \frac{5 + \sqrt{17}}{4} \right)$ < 0

⇒ $\frac{5 - \sqrt{17}}{4}$ < x < $\frac{5 + \sqrt{17}}{4}$

⇒ ln $\left( \frac{5 - \sqrt{17}}{4} \right)$ < a < ln $\left( \frac{5 + \sqrt{17}}{4} \right)$.