Math miscellaneousHard
Question
The number of real solution of equation 1 + |ex - 1| = ex(ex - 2) is:-
Options
A.1
B.2
C.3
D.No solution
Solution
1 + |ex - 1| = e2x - 2x + 1 - 1
1 + |ex - 1| = |ex - 1|2 - 1
|ex - 1|2 - |ex - 1| - 2 = 0
let |ex - 1| = t
t2 - t - 2 = 0
(t - 2) (t + 1) = 0
t = 2, -1
|ex - 1| = 2
ex - 1 = ± 2 ∵ ex = - 1 is not possible
ex = 3
x = ln3
Only one solution.
1 + |ex - 1| = |ex - 1|2 - 1
|ex - 1|2 - |ex - 1| - 2 = 0
let |ex - 1| = t
t2 - t - 2 = 0
(t - 2) (t + 1) = 0
t = 2, -1
|ex - 1| = 2
ex - 1 = ± 2 ∵ ex = - 1 is not possible
ex = 3
x = ln3
Only one solution.
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