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Tapasranjan Das Grade: 12
        

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....

7 years ago

Answers : (2)

Avijit Arya
33 Points
										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.
So you get your final answer as:a = 3/4, 1, 5/4.
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
7 years ago
Badiuddin askIITians.ismu Expert
147 Points
										

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.

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All the best.
 
Regards,
Askiitians Experts
Badiuddin




7 years ago
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