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Mathematics Question on Functions

The number of functions f:1,2,3,4→a∈:Z∣a∣≤8f:\\{1,2,3,4\\} \rightarrow\\{ a \in: Z|a| \leq 8\\} satisfying f(n)+1nf(n+1)=1,∀n∈1,2,3f(n)+\frac{1}{n} f( n +1)=1, \forall n \in\\{1,2,3\\} is

A

2

B

1

C

4

D

3

Answer

2

Explanation

Solution

The correct answer is (A) : 2
f:1,2,3,4→a∈Z:∣a∣≤8f:{1,2,3,4}→{a∈Z:∣a∣≤8}
f(n)+12(n+1)=1,∀n∈1,2,3f(n)+\frac{1}{2}(n+1)=1,∀n∈{1,2,3}
f(n+1) must be divisible by n
f(4)⇒−6,−3,0,3,6
f(3)⇒−8,−6,−4,−2,0,2,4,6,8
f(2)⇒−8,……⋯⋯⋯,8
f(1)⇒−8,…………….8
3f(4)​ must be odd since f(3) should be even therefore 2 solution possible.

f(4)f(3)f(2)f(1)
-3201
3010