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....

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

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

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