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 |
