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The number of real solutions of 1 + |e^x – 1| = e^x(e^x – 2) , is..a)0b)1c)2d)4
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 1CASE 2for ex less than 1 and greater than zero..[ex is always +ve number]1-(ex-1)=ex(ex-2)2-ex=e2x-2exe2x-ex-2=0put ex =t and solve using quadratic formulat2-t-2=0it gives t=2,-1ex=2,-12 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 +veso number of real solution of this equation is 1..ANS
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