JEE MAIN - Mathematics (2005 - No. 2)
A real valued function f(x) satisfies the functional equation
f(x - y) = f(x)f(y) - f(a - x)f(a + y)
where a is given constant and f(0) = 1, f(2a - x) is equal to
f(x - y) = f(x)f(y) - f(a - x)f(a + y)
where a is given constant and f(0) = 1, f(2a - x) is equal to
- f(x)
f(x)
f(a) + f(a - x)
f(- x)
Explanation
$$f\left( {2a - x} \right) = f\left( {a - \left( {x - a} \right)} \right)$$
$$ = f\left( a \right)f\left( {x - a} \right) - f\left( 0 \right)f\left( x \right)$$
$$ = f\left( a \right)f\left( {x - a} \right) - f\left( x \right)$$
$$ = - f\left( x \right)$$
$$\left[ {} \right.$$ as $$x = 0,y = 0,f\left( 0 \right) = {f^2}\left( 0 \right) - {f^2}\left( a \right)$$
$$\,\,\,\,\,\,\,\, \Rightarrow {f^2}\left( a \right) = 0 \Rightarrow f\left( a \right) = \left. 0 \right]$$
$$ \Rightarrow f\left( {2a - x} \right) = - f\left( x \right)$$
$$ = f\left( a \right)f\left( {x - a} \right) - f\left( 0 \right)f\left( x \right)$$
$$ = f\left( a \right)f\left( {x - a} \right) - f\left( x \right)$$
$$ = - f\left( x \right)$$
$$\left[ {} \right.$$ as $$x = 0,y = 0,f\left( 0 \right) = {f^2}\left( 0 \right) - {f^2}\left( a \right)$$
$$\,\,\,\,\,\,\,\, \Rightarrow {f^2}\left( a \right) = 0 \Rightarrow f\left( a \right) = \left. 0 \right]$$
$$ \Rightarrow f\left( {2a - x} \right) = - f\left( x \right)$$
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