Question

If $f\left(0\right)=0,f\left(1\right)=1,f\left(2\right)=2$ and $f\left(x\right)=f\left(x-2\right)+f\left(x-3\right)$ for $x=3,4,\mathrm{5.....}$ then $f\left(9\right)=?$

• A. $12$
• B. $13$
• C. $14$
• D. $10$

Answer 1

The correct option is D $10$

Solution -

$f\left(0\right)=0,f\left(1\right)=1,f\left(2\right)=2$

$f\left(x\right)=f(x-2)+f(x-3)$

$f\left(3\right)=f\left(0\right)+f\left(1\right)=1$

$f\left(4\right)=f\left(2\right)+f\left(1\right)=3$ $f\left(5\right)=f\left(3\right)+f\left(2\right)=3$

$f\left(9\right)=f\left(7\right)+f\left(6\right)$

$f\left(9\right)=f\left(5\right)+f\left(4\right)+f\left(3\right)+f\left(4\right)$

$=3+3+1+3$

$=10$

D is correct