Question

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

• A. 2
• B. 1
• C. 4
• D. 3

Answer 1

The Correct Option is A

To satisfy \(f(n) + 1\) divisible by \(n\), the function values at each \(n\) must be such that:

\( f(1) + 1 \equiv 0 \pmod{1} \quad \text{(always true)} \)

\( f(2) + 1 \equiv 0 \pmod{2} \Rightarrow f(2) = 1, 3, 5, \text{ or } 7 \)

\( f(3) + 1 \equiv 0 \pmod{3} \Rightarrow f(3) = 2, 5, \text{ or } 8 \)

\(f(4)\) (no restriction for divisibility by 4)

Each selection for \(f(2)\) and \(f(3)\) determines a function. The absence of restrictions on \(f(4)\) means it can take any value from 1 to 8, resulting in several possible functions. Without specific overlaps or further restrictions, we compute:

\( 4 \times 3 \times 8 = 96 \)

However, ensuring each value remains distinct reduces the count significantly, often requiring case-by-case analysis or additional constraints provided in the problem. Assuming a simpler case without restrictions on distinctness or with overlap, the more likely number is fewer than 96.

Answer 2

The correct answer is (A) : 2
\(f:{1,2,3,4}→{a∈Z:∣a∣≤8}\)
 \(f(n)+\frac{1}{2}(n+1)=1,∀n∈{1,2,3}\)
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)
-3	2	0	1
3	0	1	0