If a function f : R → R is given by the functional equation,
f ( x − f ( y ) ) = f ( f ( y ) ) + x f ( y ) + f ( x ) − 1 .
Now let's take the function f ( x ) where x is obviously a real number as per our definition of the function.
If the value of
∫ 0 ∣ z ∣ f − 1 ( t ) . d t = b a m / n
Here, a is a prime number. And gcd(a,b)=1 . ||Find the value of f ( 1 ) + f ( 2 ) + a + b + m + n + ∣ z ∣ .
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Thanks bro.
But there is one thing. While making the question I had to think more than solvers .... As if I would have kept z=2 for into... Then the integral would have been undefined as it's not invertible till that
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Haan voh bhi hai.....but wait a sec..... original question hai kya??!!
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Toh aur Kya. Koi book mein aisa qs Thori na rahta Hain.
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@Md Zuhair – Sahi hai broooo!! Functional equation to main solve hi karta hun....kuch Naya idea nahi aata ....😂lol
very nyc problem @Md Zuhair
How can we take f(y) = 0 without proving the function surjective??? There is clearly a flaw in the argument .......or am i missing something?
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U r correct. Let me see....
@rayyan shahid Oh Man!! Yes you are right......I don't know why I did not think of it......sorry
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But I have seen this in one of the IMO problems. I guess... I have missed certain statements!
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@Md Zuhair – @Md Zuhair Bro.....statements mili?? How to solve it now?? I guess Rayyan Shahid is right.......surjective waala constraint missing hai ismein......
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Nice question @Md Zuhair
After taking f(y) = 0 , we can observe that f(0) = 1........ Next, putting y as 0 we obtain a recurrence relation......after determining some values i.e. f(1) = 1/2 and f(-1) = 1/2....... We find out that the function is
f(x) = 1 - (x^2)/2...........
The rest is then easy......!!