Question

If f is a function such that $f\left(0\right)=2,f\left(1\right)=3$ $f(x+2)=2f\left(x\right)-f(x+1)$ for every real x, then $f\left(5\right)=$

• A. 2
• B. 13
• C. 3
• D. none

Answer 1

The correct option is A 2

Given,

$f(x+2)=2f\left(x\right)-f(x+1)$

when $x=0$

$f(0+2)=2f\left(0\right)-f(0+1)$

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

$=2\left(2\right)-3$

$\therefore f\left(2\right)=1$

when $x=1$

$f(1+2)=2f\left(1\right)-f(1+1)$

$f\left(3\right)=2f\left(1\right)-f\left(2\right)$

$=2\left(3\right)-1$

$\therefore f\left(3\right)=5$

when $x=2$

$f(2+2)=2f\left(2\right)-f(2+1)$

$f\left(4\right)=2f\left(2\right)-f\left(3\right)$

$=2\left(1\right)-5$

$\therefore f\left(4\right)=-3$

when $x=3$

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

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

$=2\left(5\right)-(-3)$

$\therefore f\left(5\right)=13$