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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
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
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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