Let N be the set of all natural numbers and f | vedclass

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(B) Given the equation $x - f(x) = 19[\frac{x}{19}] - 90[\frac{f(x)}{90}]$.Rearranging the terms,we get $x - 19[\frac{x}{19}] = f(x) - 90[\frac{f(x)}{

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(B) Given the equation $x - f(x) = 19[\frac{x}{19}] - 90[\frac{f(x)}{90}]$.Rearranging the terms,we get $x - 19[\frac{x}{19}] = f(x) - 90[\frac{f(x)}{90}]$.This is equivalent to $x \pmod{19} = f(x) \pmod{90}$.Let $x = 1990$. Then $1990 = 19 	imes 104 + 14$,so $1990 \equiv 14 \pmod{19}$.Thus,$f(1990) \equiv 14 \pmod{90}$.This means $f(1990) = 90k + 14$ for some integer $k$.We are given $1990 Substituting $f(1990) = 90k + 14$,we get $1990 $1976 Dividing by $90$,we get $21.95 Since $k$ must be an integer,$k = 22$ or $k = 23$.If $k = 22$,$f(1990) = 90(22) + 14 = 1980 + 14 = 1994$.If $k = 23$,$f(1990) = 90(23) + 14 = 2070 + 14 = 2084$.Both values satisfy the condition $1990 Thus,there are $2$ possible values for $f(1990)$.

List-$I$	List-$II$
$A$. Range of $\sec ^{-1}\left[1+\cos ^2 x\right]$, where $[.]$ denotes the greatest integer function	$I$. Odd function
$B$. Domain of $f(x)$ where $f\left(x+\frac{1}{x}\right)=x^2+\frac{1}{x^2}$	$II$. $\left\{0, \frac{1}{2}\right\}$
$C$. $f(x+y)=f(x)+f(y) ; f(1)=5$	$III$. $\left\{\sec ^{-1} 5, \sec ^{-1} 4\right\}$
$D$. $\sin ^{-1} x-\cos ^{-1} x+\sin ^{-1}(1-x)=0 \Rightarrow x \in$	$IV$. $R$
	$V$. $\left\{\sec ^{-1} 1, \sec ^{-1} 2\right\}$

Let $N$ be the set of all natural numbers and $f: N \rightarrow N$ be such that $1990 < f(1990) < 2100$ and satisfies the equation $x-f(x)=19[\frac{x}{19}]-90[\frac{f(x)}{90}]$,where $[y]$ denotes the greatest integer less than or equal to $y$. Then the number of possible values of $f(1990)$ is

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