Functions Ans 11
Question 11 - 29 January - Shift 2
Consider a function $f: \mathbb{N} \to \mathbb{R}$, satisfying
$f(1)+2 f(2)+3 f(3)+\ldots+x f(x)=x(x+1) f(x) ; x \geq 2$
with $f(1)=1$. Then $\frac{1}{f(2022)}+\frac{1}{f(2028)}$ is equal to
(1) 8200
(2) 8000
(3) 8400
(4) 8100
Show Answer
Answer: (4)
Solution:
Formula: Operations on functions
Given for $x \geq 2$
$ f(1)+2 f(2)+\ldots \ldots+xf(x)=x(x+1) f(x) $
replace $x$ by $x+1$
$ \begin{aligned} & \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(x)=(x+1) f(x+1)=\frac{1}{2}, x \geq 2 \\ & f(2)=\frac{1}{4}, f(3)=\frac{1}{6} \text{ hathongonongon } \end{aligned} $
Now $f(2022)=\frac{1}{4044}$
$ f(2028)=\frac{1}{4056} $
So, $\frac{1}{f(2022)}+\frac{1}{f(2028)}=4044+4056=8100$
