Question

If $f:Z\to Z$ be a linear function such that $f\left(2\right)+f\left(1\right)=4,f\left(2\right)-2f\left(1\right)=-1$, then $f\left(3\right)$ is

• A. $1$
• B. $3$
• C. $2$
• D. $\frac{2}{3}$

Answer 1

The correct option is B $3$

We know that linear function can be written as,
$f\left(x\right)=ax+b$
Now,
$f\left(2\right)=2a+bf\left(1\right)=a+b$
So,
$f\left(2\right)+f\left(1\right)=4\Rightarrow 2a+b+a+b=4\Rightarrow 3a+2b=4\cdots \left(1\right)$
$f\left(2\right)-2f\left(1\right)=-1\Rightarrow 2a+b-2a-2b=-1\Rightarrow -b=-1\Rightarrow b=1$
Using equation $\left(1\right)$,
$3a+2\times 1=4\Rightarrow a=\frac{2}{3}$
So,
$f\left(x\right)=\frac{2x}{3}+1\therefore f\left(3\right)=2+1=3$