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# Q: The value of 'a' for which the equation (x-1)2=|x-a| has exactly three real solutions is/areA. 1/4B. 3/4C. 1D. 5/4

147 Points
11 years ago

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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