If $f (0) = 1, f (1) = 2$ and $f (x + 2) = 2 f (x + 2) + f (x + 1)$ , then.

f (3) = - 1

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f (4) = 3

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f (5) = 0

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f (6) = - 2

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Solution

The correct option is C $f (6) = - 2$
$f (x + 2) = 2 f (x + 2) + f (x + 1)$
$f (x + 2) + f (x + 1) = 0$
$f (x + 2) = - f (x + 1)$
Now let $x = 0$
Then
$f (2) = - f (1) = - 2$

$x = 1$
$f (3) = - f (2) = 2$

$x = 2$
$f (4) = - f (3) = - 2$

$x = 3$
$f (5) = - f (4) = 2$

And at $x = 4$
$f (6) = - f (5) = - 2$

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