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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```
7 years ago

147 Points
```										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
Please feel free to post as many doubts on our discussion forum as you can.If you find any question Difficult to understand - post it here and we will get you the answer and detailed  solution very  quickly. We are all IITians and here to help you in your IIT JEE preparation.All the best. Regards,Askiitians ExpertsBadiuddin
```
7 years ago
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