Question

If $f\left(x\right)$ 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 $\forall x\in R$, then $f\left(5\right)$ is:

• A. $5$
• B. $13$
• C. $3$
• D. $-5$

Answer 1

The correct option is C $3$

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

Put,$x=0f\left(2\right)=2f\left(0\right)-f\left(1\right)f\left(2\right)=4-3=1f\left(2\right)=2f\left(0\right)-f(2+1)f\left(3\right)=2f\left(0\right)-f\left(2\right)f\left(3\right)=4-1f\left(3\right)=3$

Put,$x=2f\left(4\right)=2f\left(2\right)-3f\left(4\right)=1$

Put,$x=3f\left(5\right)=2f\left(3\right)-f(3+1)f\left(5\right)=4-1=3$