Mathematics - Functions Question with Solution | TestHub

Given,

$f\left(\mathrm{n}\right)+\frac{1}{\mathrm{n}}f(\mathrm{n}+1)=1,\forall \mathrm{n}\in \{1,2,3\}$

$\Rightarrow nf\left(\mathrm{n}\right)+f(\mathrm{n}+1)=n$

At $n=1$,

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

At $n=2$,

$2f\left(2\right)+f\left(3\right)=2$    ....(2)

At $n=3$,

$3f\left(3\right)+f\left(4\right)=3$      ....(3)

Put the value of $f\left(2\right)$ from equation (1) in equation (2),

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

$\Rightarrow f\left(3\right)=2f\left(1\right)$       ....(4)

Put the value of $f\left(3\right)$ from equation (4) in equation (3),

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

$\Rightarrow f\left(4\right)=3-6f\left(1\right)$    

$\because f:\{1,2,3,4\}\to \{\mathrm{a}\in \mathrm{\mathbb{Z}}:|\mathrm{a}|\le 8\}$

$\Rightarrow -8\le f\left(4\right)\le 8$

$\Rightarrow -8\le 3-6f\left(1\right)\le 8$

$\Rightarrow -11\le -6f\left(1\right)\le 5$

$\Rightarrow \frac{-5}{6}\le f\left(1\right)\le \frac{11}{6}$

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

Case I: $f\left(1\right)=0$

$\Rightarrow f\left(2\right)=1, f\left(3\right)=0, f\left(4\right)=3$

Case II: $f\left(1\right)=1$

$\Rightarrow f\left(2\right)=0, f\left(3\right)=2, f\left(4\right)=-3$

Therefore, two such functions are possible.