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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

Hi badiuddin the above question is from AIST FIITJEE there they gave the answer of this question as 3/4,1 and 5/4.....but they don't give the detail solution.....and u gave the ans only 5/4....
```

6 years ago

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```										Hi,
This Q can be solved by opening the modulus sign with + and - signs, one at a time. In this case you get a max. of 4 roots, two for each + and - signs. For getting exactly 3 roots, following may be the cases,
a) One of the equations (modulus with + or - sign) has both equal roots, and the other 1
has two distinct roots.
b) Both + and - give you distinct roots, but 1 root of + and 1 root of - become same.
Now when opened with '+' : a1= 1, b1=3 and c1=(1+a). Solving b1^2-4a1c1=0, we get a =5/4.
Similarly for opening with '-' sign: a2=1, b2= -1, c2=(1-a). Solving this you get a=3/4.

Now there can be the last possibility: Put both b1^2 - 4a1c1 > 0 and b2^2 - 4a2c2 > 0.
You get: a< 5/4 and a > 3/4, which gives you the last value, that is 1.
Please always specify, whether the Q has single or multiple correct option, as you are specified in the exam, it is incomplete otherwise, as a Q is left after we get a value and nothing is mentioned to proceed after that.
Thank you
```
6 years ago
```										Dear Tapasranjan
I have given the range of a for which given  equation has  real solution
3/4 ≤ a ≤ 5/4
for exactly 3 solution ,
1)  any equation has 1 root and the other equation has 2 root then 3 solution are possible .
2)  Or both the equation has 2 roots , but one root in both the equation is common then also exactly   3 solution is possible .

1st case is possible at end point of the calculated range of a  ie  a=3/4  and  5/4 (because at this point discrimnent of one equation is zero  already derived in earlier post )
and for the second case
formed two equation x2   -3x+a+1 =0   for x≥a
x2   -x+1-a =0     for x<a
let β is a common root
then it is satisfied by both the equation

β2   -3β+a+1 =0
β2   -β+1-a =0
solve for β
β =a
put this value of in any one of the equation
a2   -a+1-a =0
(a-1)2 =0
a =1
so for a=1   also exactly 3 solution exicts

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

```
6 years ago

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