$f(x)=\sum_{r=1}^{2021} \frac{r}{rx^2-1},\;F(x)=\int f(x)\;dx,\; g(x)=\sum_{r=1}^{2022} \tan\bigg(\frac{\pi r x}{2022}\bigg),\; G(x)=\int g(x)dx.$

Question

Let $$f(x)=\sum_{r=1}^{2021} \frac{r}{rx^2-1},\;F(x)=\int f(x)\;dx,\; g(x)=\sum_{r=1}^{2022} \tan\bigg(\frac{\pi r x}{2022}\bigg),\; G(x)=\int g(x)dx.$$

$A_r=\{x \;|\; f(x)=r, r\in \mathbb R,\; x\in (-1,1)\},\;$ $B_r=\{y \;| \;g(y)=r, r\in \mathbb R,\; y\in (-1,1)\}$

$1.$ Value of $n(A_1)+n(B_2)$ is

(A) $6060$

(B) $6062$

(C) $6064$

(D) N.O.T.

Note: Where $n(A_1)$ denotes number of elements in set $A_1$

I don't know how to evaluate $f(x).$ Any hint will be appreciated.

for $n(A_1)$ i tried to do $f(x)=1$ but i couldn't think further.

as suggested by @Jakobian, I am adding image of source question.

And I think $r$ in $A_r$ is different from $r$ used for $f(x)$

My main doubt is how we will find $f(x)$

Answer 1

This is a very interesting question I guess it is from FIITJEE AITS , and you are preparing for IIT just like me ..

Numerical approach is the only feasible way to solve this question .

We have to find the no. of solutions to $f(x)=1$ and $g(x)=2$.

let us take a function similar to $f(x)$ having only 1 term

$f_1(x)= \frac{1}{1-x^2} $ ; it has $0$ solution in $x \in (-1,1) $

$f_2(x)= \frac{1}{1-x^2}+\frac{2}{2-x^2} $ ; it has $2$ solution in $x \in (-1,1) $

$f_3(x)= \frac{1}{1-x^2}+\frac{2}{2-x^2}+\frac{3}{3-x^2} $ ; it has $4$ solution in $x \in (-1,1) $ ...

Similarly

$f_n(x)= \sum_{r=1}^n \frac{r}{r-x^2} $ ; it has $2(n-1)$ solution in $x \in (-1,1) $

$hence, n(A)=4040$

Same can be done for finding number of solutions of $g(x)$ but there is one key point which has to be taken in consideration while solving for $g(x)$

here

$g_1(x)= tan(\pi x \frac{1}{1})$ has $2$ solutions in $x\in (-1,1)$

$g_2(x)= tan(\pi x \frac{1}{2})+ tan(\pi x \frac{2}{2})$ has $3$ solutions in $x\in (-1,1)$

$g_3(x)= tan(\pi x \frac{1}{2})+ tan(\pi x \frac{2}{3})+ tan(\pi x \frac{3}{3})$ has $4$ solutions in $x\in (-1,1)$

and $g_n(x)= \sum_{r=1}^{n}tan(\pi x \frac{r}{n})$ has $n+1$ solutions in $x\in (-1,1)$

and hence $n(B)=2022+1=2023$

In order to find number of solutions use a rough plot of the graph here is some reference.

$f_4(x)$ ; $g_7(x)$ .