Mathematics - Functions Question with Solution | TestHub

Given:

$f\left(1\right)+2f\left(2\right)+3f\left(3\right)+\dots +xf\left(x\right)=x\left(x+1\right)f\left(x\right)$

$\Rightarrow f\left(1\right)+2f\left(2\right)+3f\left(3\right)+\dots +\left(x-1\right)f\left(x-1\right)=x\left(x-1\right)f\left(x-1\right)$

Now when $x=2$, then

$f\left(1\right)+2f\left(2\right)=6f\left(2\right)$

$\Rightarrow 1=4f\left(2\right)$

$\Rightarrow f\left(2\right)=\frac{1}{4}$

When $x=3$, 

$f\left(1\right)+2f\left(2\right)=9f\left(3\right)\Rightarrow f\left(3\right)=\frac{1}{6}$

When $x=4$, $f\left(1\right)+2f\left(2\right)+3f\left(3\right)=16f\left(4\right)\Rightarrow f\left(4\right)=\frac{1}{8}$

So,

$f\left(x\right)=\frac{1}{2x}$

Hence, 

$\frac{1}{f\left(2022\right)}+\frac{1}{f\left(2028\right)}=4044+4056=8100$