### Functions Question 6

#### Question 6 - 25 January - Shift 2

The number of functions $f:{1,2,3,4} \to{a \in \mathbb{Z}:|a| \leq 8}$ satisfying $f(n)+$ $\frac{1}{n} f(n+1)=1, \forall n \in{1,2,3}$ is

(1) 3

(2) 4

(3) 1

(4) 2

## Show Answer

#### Answer: (4)

#### Solution:

#### Formula: Operations on functions

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

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

$f(n+1)$ must be divisible by $n$

$f(4) \Rightarrow-6,-3,0,3,6$

$f(3) \Rightarrow-8,-6,-4,-2,0,2,4,6,8$

$f(2) \Rightarrow-8$, 8

$f(1) \Rightarrow-8$, 8

$\frac{f(4)}{3}$ must be odd since $f(3)$ should be even therefore 2 solution possible.

$f(4)$ | $f(3)$ | $f(2)$ | $f(1)$ |
---|---|---|---|

-3 | 2 | 0 | 1 |

3 | 0 | 1 | 0 |