- Home
- Standard 12
- Mathematics
1.Relation and Function
hard
$\mathrm{f}(\mathrm{n})+\frac{1}{\mathrm{n}} \mathrm{f}(\mathrm{n}+1)=1, \forall \mathrm{n} \in\{1,2,3\}$
को संतुष्ट करने वाले फलनों
$\mathrm{f}:\{1,2,3,4\} \rightarrow\{\mathrm{a} \in \mathbb{Z}|\mathrm{a}| \leq 8\}$
की संख्या है -
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