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1.Relation and Function
hard
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
$3$
B
$4$
C
$1$
D
$2$
(JEE MAIN-2023)
Solution
$f:\{1,2,3,4\} \rightarrow\{ a \in Z 😐 a | \leq 8\}$
$f( n )+\frac{1}{ n } f ( n +1)=1, \forall n \in\{1,2,3\}$
$f( n +1)$ must be divisible by $n$
$f(4) \Rightarrow-6,-3,0,3,6$
$f(3) \Rightarrow-8,-6,-4,-2,0,2,4,6,8$
$f(2) \Rightarrow-8, \ldots \ldots \ldots \ldots \ldots, 8$
$f(1) \Rightarrow-8, \ldots \ldots \ldots \ldots \ldots, 8$
$\frac{f(4)}{3}$ must be odd since $f(3)$ should be even therefore $2$ solution possible.
| $f(4)$ | $f(3)$ | $f(2)$ | $f(1)$ |
| $-3$ | $2$ | $0$ | $1$ |
| $3$ | $0$ | $1$ | $0$ |
Standard 12
Mathematics
