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f(x)=x^2-2a x+a(a+1),f:[a ,oo)vec[a ,oo)...

`f(x)=x^2-2a x+a(a+1),f:[a ,oo)vec[a ,oo)dot` If one of the solution of the equation `f(x)=f^(-1)(x)i s 5049 ,` then the other may be (a)5051 (b) 5048 (c) 5052 (d) 5050

A

5051

B

5048

C

5052

D

5050

Text Solution

Verified by Experts

The correct Answer is:
B, D
`f(x)=x^(2)-2ax+a(a+1)`
` f(x)=(x-a)^(2)+a, x in [a,oo)`
Let `y=(x-a)^(2)+a. " Clearly " y ge a.` Thus,
`(x-a)^(2)=y-a`
`or x=a+sqrt(y-a)`
` :. f^(-1)(x)=a+sqrt(x-a)`
Now, ` f(x) =f^(-1)(x)`
`or (x-a)^(2)+a=a+sqrt(x-a)`
`(x-a)^(2) =sqrt(x-a)`
` or (x-a)^(4)=(x-a)`
i.e., `x= a or (x-a)^(3) =1`
i.e., ` x=a or a+1 `
If ` a=5049, " then " a+1=5050.`
If `a+1=5049, " then " a=5048.`
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