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Bunuel
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Danielpgcanadasb
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Bunuel
Let \(f(x + 2) + f(5x + 6) = 2x – 1\) for all real x. Find the value of f(1).

(A) –2
(B) –1
(C) -5/2
(D) -3/2
(E) None of these

Since, it is valid for all real values of x.
Let's try putting x = -1
\(f(x + 2) + f(5x + 6) = 2x – 1\)
\(f(1) + f(1) = -2 – 1\)
\(2f(1) = -3\)
\(f(1) = -3/2\) - (D),IMO!



why is x =-1 and not positive 1 ? Please respnond.

Hi Danielpgcanadasb,
The question stem tells us that it is valid for all real values of x. So, x can be any real number.
The reason why I picked x to be 1 is because we want to find the value of f(1).
If I put x = 1, I'll get:
\(f(x + 2) + f(5x + 6) = 2x – 1\)
\(f(1 + 2) + f(5*1 + 6) = 2*1 – 1\)
\(f(3) + f(11) = 1\)
I'm not getting the value of f(1) by putting x = 1. I hope this helps :)
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