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The number of real solutions of 1 + |e^x – 1| = e^x(e^x – 2) , is.. a)0 b)1 c)2 d)4

2 years ago

Answers : (1)

Ram Janam
29 Points
							
CASE 1...
for ex greater than equal to 1...bcoz mod value is +ve or -ve depending on value of ex 
1+ex-1=ex(ex-2)
ex=ex(ex-2)
ex=3.....therefore  ex has one solution which is also greater than 1
CASE 2
for ex less than 1 and greater than zero..[ex is always +ve number]
1-(ex-1)=ex(ex-2)
2-ex=e2x-2ex
e2x-ex-2=0
put ex =t and solve using quadratic formula
t2-t-2=0
it gives t=2,-1
ex=2,-1
2 is more than 1 but we have taken condition that ex is less than 1 so it cant be solution.
-1 is -ve so it cant be solution bcoz ex is always +ve
so number of real solution of this equation is 1..ANS
 
2 years ago
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