Question

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

• A. $7$
• B. $13$
• C. $1$
• D. $5$

Answer 1

The correct option is B $13$

Given: $f(x+2)=2f\left(x\right)-f(x+1)\forall x\in R$
$f\left(0\right)=2,f\left(1\right)=3$

Putting $x=0,$ we get
$f\left(2\right)=2f\left(0\right)-f\left(1\right)\Rightarrow f\left(2\right)=2\left(2\right)-3=1$

Putting $x=1,$ we get
$f\left(3\right)=2f\left(1\right)-f\left(2\right)\Rightarrow f\left(3\right)=2\left(3\right)-1=5$

Putting $x=2,$ we get
$f\left(4\right)=2f\left(2\right)-f\left(3\right)\Rightarrow f\left(4\right)=2\left(1\right)-5=-3$

Putting $x=3,$ we get
$f\left(5\right)=2f\left(3\right)-f\left(4\right)\Rightarrow f\left(5\right)=2\left(5\right)-(-3)=13$