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1.Relation and Function
hard
$f(1)+f(2)+3 f(3)+\ldots+x f(x)=x(x+1) f(x) ; x \geq 2$ જ્યાં $f(1)=1$ નું સમાધાન કરતો વિધેય $f: N \rightarrow R$ ધ્યાને લો તો $\frac{1}{f(2022)}+\frac{1}{f(2028)}=............$
A
$8200$
B
$8000$
C
$8400$
D
$8100$
(JEE MAIN-2023)
Solution
Given for $x \geq 2$
$f(1)+2 f(2)+\ldots \ldots+x f(x)=x(x+1) f(x)$
$\text { replace } x \text { by } x +1$
$\Rightarrow \quad x(x+1) f(x)+(x+1) f(x+1)$
$=(x+1)(x+2) f(x+1)$
$\Rightarrow \quad \frac{x}{f(x+1)}+\frac{1}{f(x)}=\frac{(x+2)}{f(x)}$
$\Rightarrow \quad x f(x)=(x+1) f(x+1)=\frac{1}{2}, x \geq 2$
$f (2)=\frac{1}{4}, f (3)=\frac{1}{6}$
$\text { Now } f (2022)=\frac{1}{4044}$
$f(2028)=\frac{1}{4056}$
So, $\frac{1}{f(2022)}+\frac{1}{f(2028)}=4044+4056=8100$
Standard 12
Mathematics
