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  •   Forum  >> Algebra  >> theory of equation  

    Tapasranjan Das  
    Total Posted Questions : 53
       

    Q: The value of 'a' for which the equation (x-1)2=|x-a| has exactly three real solutions is/are

    A. 1/4

    B. 3/4

    C. 1

    D. 5/4

      Post Your Reply




    Total Replies: 1
    Profile Message

    askIITians expert badiuddin.ismu
    Total Posted Answers: 1758
     

    Dear Tapasranjan

    (x-1)2=|x-a|

    case: x≥a

     (x-1)2=(x-a)

    x2 +1 -2x =x-a

    x2  -3x+a+1 =0

     x = [3 ±√9-4(a+1)]/2

     for real value of x

    9-4(a+1)≥0

    5/4 ≥a

     

     case: x<a

    (x-1)2=-(x-a)

    x2 +1 -2x =-x+a

    x2  -x+1-a =0

     x = [1 ±√1-4(1-a)]/2

     for real value of x

    1-4(1-a)≥0

    a≥3/4

    for exactly 3 value  in any one case discrimnent must be zero

    so either a =3/4 or a =5/4

     but for a=3/4 both case give only single value

    so solution is 5/4

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    Regards,
    Askiitians Experts
    Badiuddin

    2/8/2010 3:16:18 PM