Let f be a function whose domain is all real numbers- If fx-2fx-2001x-1 - 4013-x for all x not equal to 1- then the value of f2003 is

F(x)+2f(x+2001x 1)=4013 x Put x=2, we have nbsp; f(2)+2f(2+20012 1)=4013 2 ⇒ f(2)+2f(2003)=4011 ⋯(1) Again put x=2003, we have f(2003)+2f(2003+20012003 1)=4 ...

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Standard XII

Mathematics

Fractional Part Function

Let f be a fu...

Question

Let $f$ be a function whose domain is all real numbers. If $f (x) + 2 f (\frac{x + 2001}{x - 1}) = 4013 - x$ for all $x$ not equal to $1,$ then the value of $f (2003)$ is

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Solution

$f (x) + 2 f (\frac{x + 2001}{x - 1}) = 4013 - x$
Put $x = 2,$ we have
$f (2) + 2 f (\frac{2 + 2001}{2 - 1}) = 4013 - 2$
$\Rightarrow f (2) + 2 f (2003) = 4011 \dots (1)$
Again put $x = 2003,$ we have
$f (2003) + 2 f (\frac{2003 + 2001}{2003 - 1}) = 4013 - 2003$
$\Rightarrow f (2003) + 2 f (2) = 2010 \dots (2)$
From $(1)$ and $(2),$ we have
$3 f (2003) = 2 \times 4011 - 2010 = 6012$
$∴ f (2003) = 2004$

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Let f be a function whose domain is all real numbers. If f(x)+2f(x+2001x−1)=4013−x for all x not equal to 1, then the value of f(2003) is

f(x)+2f(x+2001x−1)=4013−x Put x=2, we have f(2)+2f(2+20012−1)=4013−2 ⇒f(2)+2f(2003)=4011 ⋯(1) Again put x=2003, we have f(2003)+2f(2003+20012003−1)=4013−2003 ⇒f(2003)+2f(2)=2010 ⋯(2) From (1) and (2), we have 3f(2003)=2×4011−2010=6012 ∴f(2003)=2004

Let $f$ be a function whose domain is all real numbers. If $f (x) + 2 f (\frac{x + 2001}{x - 1}) = 4013 - x$ for all $x$ not equal to $1,$ then the value of $f (2003)$ is