JEE Main 2023 — Mathematics Algebra

Given,

$$f(n)+\frac{1}{n}f(n+1)=1,\forall n\in {1,2,3}$$f(n)+n1​f(n+1)=1,∀n∈1,2,3

$$\Rightarrow nf(n)+f(n+1)=n$$⇒nf(n)+f(n+1)=n

At $n=1$n=1,

$f(1)+f(2)=1$f(1)+f(2)=1 ....(1)

At $n=2$n=2,

$2f(2)+f(3)=2$2f(2)+f(3)=2 ....(2)

At $n=3$n=3,

$3f(3)+f(4)=3$3f(3)+f(4)=3 ....(3)

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

$$2(1-f(1))+f(3)=2$$2(1−f(1))+f(3)=2

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

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

$$3(2f(1))+f(4)=3$$3(2f(1))+f(4)=3

$$\Rightarrow f(4)=3-6f(1)$$⇒f(4)=3−6f(1)

$$\because f:{1,2,3,4}\rightarrow {a\in \mathbb{Z}:|a|\leq 8}$$∵f:1,2,3,4→a∈Z:∣a∣≤8

$$\Rightarrow -8\leq f(4)\leq 8$$⇒−8≤f(4)≤8

$$\Rightarrow -8\leq 3-6f(1)\leq 8$$⇒−8≤3−6f(1)≤8

$$\Rightarrow -11\leq -6f(1)\leq 5$$⇒−11≤−6f(1)≤5

$$\Rightarrow \frac{-5}{6}\leq f(1)\leq \frac{11}{6}$$⇒6−5​≤f(1)≤611​

$$\Rightarrow f(1)=0,1$$⇒f(1)=0,1

Case I: $f(1)=0$f(1)=0

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

Case II: $f(1)=1$f(1)=1

$$\Rightarrow f(2)=0,f(3)=2,f(4)=-3$$⇒f(2)=0,f(3)=2,f(4)=−3

Therefore, two such functions are possible.