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

 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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