Let x denote the greatest integer less than or equal to x . Then, the values of x ∈ R satisfying the equation e x 2 + e x + 1 - 3 = 0 lie in the interval:

Step-by-step solution

Given equation is e x 2 + e x + 1 - 3 = 0 We know that x + I = x + I ,   I ∈ Z , where · , represents the greatest integer function. ⇒ e x 2 + e x + 1 - 3 = 0 Let, e x = t ⇒ t 2 + t - 2 = 0 ⇒ t + 2 t - 1 = 0 ⇒ t = - 2 ,   1 But e x = - 2 , ( Not possible because we know that e x > 0 ) Hence, e x = 1 and also, we know that if x = 1 ,   ⇒ 1 ≤ x < 2 ∴   1 ≤ e x < 2 And, now by using the definition of logarithm, we have, if e x =