Question
The number of real solutions of $$1 + \left| {{e^x} - 1} \right| = {e^x}\left( {{e^x} - 2} \right)$$ is
A.
0
B.
1
C.
2
D.
4
Answer :
1
Solution :
$$\eqalign{
& 2 + \left| {{e^x} - 1} \right| = {\left( {{e^x}} \right)^2} - 2{e^x} + 1 = {\left| {{e^x} - 1} \right|^2} \cr
& \therefore \,\,{\left| {{e^x} - 1} \right|^2} - \left| {{e^x} - 1} \right| - 2 = 0 \cr
& {\text{or, }}\left| {{e^x} - 1} \right| = 2, - 1 \cr
& \Rightarrow \,\,\left| {{e^x} - 1} \right| = 2 \cr
& \Rightarrow \,\,{e^x} - 1 = 2, - 2 \cr
& \Rightarrow \,\,{e^x} = 3, - 1 \cr
& \Rightarrow \,\,{e^x} = 3. \cr} $$