Composition of Functions Questions in English

(C) $f \circ g(x) = f(\sqrt{x^2 + 1}) = (x^2 + 1) - 1 = x^2$.
Since the codomain of $f \circ g(x)$ is $R$ and the range is $[0, \infty)$,$f \circ g$ is not onto,therefore it is not invertible.
Now,$(h \circ f \circ g)(x) = h(f \circ g(x)) = h(x^2)$.
Since $x^2 \geq 0$ for all $x \in R$,we have $h(x^2) = x^2$.
Thus,$(h \circ f \circ g)(x) = x^2$.
Also,$h(x)$ is not an identity function because for $x < 0$,$h(x) = 0 \neq x$.

(A) Given $f(x) = \frac{x-2}{2x+1}$.
We are given the equation $f(f(x)) = -x$.
Substituting $f(x)$ into the expression:
$\frac{f(x)-2}{2f(x)+1} = -x$
$\frac{\frac{x-2}{2x+1}-2}{2(\frac{x-2}{2x+1})+1} = -x$
$\frac{x-2-2(2x+1)}{2(x-2)+1(2x+1)} = -x$
$\frac{x-2-4x-2}{2x-4+2x+1} = -x$
$\frac{-3x-4}{4x-3} = -x$
$\frac{3x+4}{4x-3} = x$
$3x+4 = x(4x-3)$
$3x+4 = 4x^2-3x$
$4x^2-6x-4 = 0$
Dividing by $2$,we get $2x^2-3x-2 = 0$.
Since $\alpha$ and $\beta$ are roots of this quadratic equation,by Vieta's formulas:
$\alpha+\beta = \frac{3}{2}$ and $\alpha\beta = -1$.
We need to find $4(\alpha^2+\beta^2)$.
Using the identity $\alpha^2+\beta^2 = (\alpha+\beta)^2 - 2\alpha\beta$:
$\alpha^2+\beta^2 = (\frac{3}{2})^2 - 2(-1) = \frac{9}{4} + 2 = \frac{17}{4}$.
Therefore,$4(\alpha^2+\beta^2) = 4 \times \frac{17}{4} = 17$.

(B) Given the functions $f(x) = 2x + 1$ and $g(x) = x^2 - 2$.
To find the composite function $(g \circ f)(x)$,we use the definition $(g \circ f)(x) = g(f(x))$.
Substitute $f(x)$ into $g(x)$:
$g(f(x)) = g(2x + 1)$.
Since $g(x) = x^2 - 2$,we replace $x$ with $(2x + 1)$:
$g(2x + 1) = (2x + 1)^2 - 2$.
Expand the square using the identity $(a + b)^2 = a^2 + 2ab + b^2$:
$(2x + 1)^2 = (2x)^2 + 2(2x)(1) + 1^2 = 4x^2 + 4x + 1$.
Now,subtract $2$:
$4x^2 + 4x + 1 - 2 = 4x^2 + 4x - 1$.
Therefore,$(g \circ f)(x) = 4x^2 + 4x - 1$.

(A) Given $f(x) = \min \{x - [x], -x - [-x]\}$ for $x \in (0, 1)$.
Since $x \in (0, 1)$,$[x] = 0$.
Also,for $x \in (0, 1)$,$-x \in (-1, 0)$,so $[-x] = -1$.
Substituting these values into the function:
$f(x) = \min \{x - 0, -x - (-1)\} = \min \{x, 1 - x\}$.
Now,we evaluate the composition $(f \circ f \circ f \circ f)(x)$:
If $x \in (0, 1/2]$,then $x \le 1 - x$,so $f(x) = x$.
Then $f(f(x)) = f(x) = x$,and so on. Thus $(f \circ f \circ f \circ f)(x) = x$.
If $x \in (1/2, 1)$,then $1 - x < x$,so $f(x) = 1 - x$.
Then $f(f(x)) = f(1 - x)$. Since $1 - x \in (0, 1/2)$,$f(1 - x) = 1 - x$.
Wait,let us re-evaluate: $f(1-x) = \min \{1-x, 1-(1-x)\} = \min \{1-x, x\} = 1-x$ if $1-x < x$ (i.e.,$x > 1/2$).
Actually,for $x \in (0, 1)$,$f(x) = \min \{x, 1-x\}$.
$f(f(x)) = f(\min \{x, 1-x\})$.
If $x \in (0, 1/2]$,$f(x) = x$,so $f(f(x)) = f(x) = x$.
If $x \in (1/2, 1)$,$f(x) = 1-x$,so $f(f(x)) = f(1-x) = \min \{1-x, 1-(1-x)\} = \min \{1-x, x\} = 1-x$.
Thus,$(f \circ f)(x) = f(x)$.
Consequently,$(f \circ f \circ f \circ f)(x) = f(f(x)) = f(x)$.

(D) Given $f(x) = x^3$ and $g(x) = 3^x$.
We need to solve $(f \circ g)(x) = (g \circ f)(x)$ for $x \neq 0$.
$(f \circ g)(x) = f(g(x)) = f(3^x) = (3^x)^3 = 3^{3x}$.
$(g \circ f)(x) = g(f(x)) = g(x^3) = 3^{x^3}$.
Equating the two expressions: $3^{3x} = 3^{x^3}$.
Since the bases are equal,we equate the exponents: $3x = x^3$.
Rearranging the terms: $x^3 - 3x = 0$.
Factoring out $x$: $x(x^2 - 3) = 0$.
Since $x \neq 0$,we have $x^2 - 3 = 0$.
Thus,the required quadratic equation is $x^2 - 3 = 0$.

(B) Given,$f(x) = [x]$ and $g(x) = |x|$.
We need to find $(g \circ f)\left(\frac{-5}{3}\right)$.
By definition,$(g \circ f)\left(\frac{-5}{3}\right) = g\left(f\left(\frac{-5}{3}\right)\right)$.
First,calculate $f\left(\frac{-5}{3}\right) = \left[\frac{-5}{3}\right]$.
Since $\frac{-5}{3} = -1.666...$,the greatest integer less than or equal to $-1.666...$ is $-2$.
So,$f\left(\frac{-5}{3}\right) = -2$.
Now,substitute this into $g(x)$:
$(g \circ f)\left(\frac{-5}{3}\right) = g(-2) = |-2|$.
Since the modulus of $-2$ is $2$,we get $(g \circ f)\left(\frac{-5}{3}\right) = 2$.

(B) Given,$f(x)=x^2$ and $g(x)=\frac{1}{x^2}$.
We need to find the value of $x^4(f \circ g)(x)$.
First,calculate the composite function $(f \circ g)(x)$:
$(f \circ g)(x) = f(g(x)) = f\left(\frac{1}{x^2}\right) = \left(\frac{1}{x^2}\right)^2 = \frac{1}{x^4}$.
Now,multiply this by $x^4$:
$x^4(f \circ g)(x) = x^4 \times \frac{1}{x^4} = 1$.
Thus,the correct option is $B$.

(C) Given $g(x) = 1 + x - [x]$.
Since $x - [x] = \{x\}$,where $\{x\}$ is the fractional part function,we have $g(x) = 1 + \{x\}$.
We know that the range of the fractional part function $\{x\}$ is $[0, 1)$.
Therefore,$g(x) = 1 + \{x\} \in [1, 2)$.
Now,we evaluate $f(g(x))$. Since $g(x) \geq 1$ for all $x \in \mathbb{R}$,and the definition of $f(x)$ states that $f(x) = 5$ for all $x > 0$,it follows that $f(g(x)) = 5$ for all $x \in \mathbb{R}$.

(B) Given,$(g \circ f)^{-1}(t) = t-2$.
Taking the inverse on both sides,we get $(g \circ f)(t) = (t-2)^{-1}$.
Since the inverse of $h(t) = t-2$ is $h^{-1}(t) = t+2$,we have $(g \circ f)(t) = t+2$.
This implies $g(f(t)) = t+2$.
Substituting $f(t) = 3t-2$,we get $g(3t-2) = t+2$.
Let $u = 3t-2$. Then $3t = u+2$,which means $t = \frac{u+2}{3}$.
Substituting this into the expression for $g(u)$:
$g(u) = \frac{u+2}{3} + 2 = \frac{u+2+6}{3} = \frac{u+8}{3}$.
Replacing $u$ with $t$,we obtain $g(t) = \frac{t+8}{3}$.

(C) Given,$f(x) = (2020 - x^{2019})^{1 / 2019}$.
First,we find $f \circ f(x) = f(f(x))$.
$f(f(x)) = (2020 - (f(x))^{2019})^{1 / 2019}$.
Substituting $f(x)$,we get $f(f(x)) = (2020 - ((2020 - x^{2019})^{1 / 2019})^{2019})^{1 / 2019}$.
$f(f(x)) = (2020 - (2020 - x^{2019}))^{1 / 2019} = (x^{2019})^{1 / 2019} = x$.
Since $f \circ f(x) = x$,the function $f$ is its own inverse (an involution).
Therefore,$(f \circ f \circ f \circ f)(x) = (f \circ f)(f \circ f(x)) = f \circ f(x) = x$.
Thus,$(f \circ f \circ f \circ f) \left( \frac{2019}{2020} \right) = \frac{2019}{2020}$.

(C) Given the function $f(x) = \frac{3x+5}{7x-3}$.
To find $f^{-1}(x)$,let $y = \frac{3x+5}{7x-3}$.
$y(7x-3) = 3x+5 \Rightarrow 7xy - 3y = 3x+5$.
$x(7y-3) = 3y+5 \Rightarrow x = \frac{3y+5}{7y-3}$.
Thus,$f^{-1}(x) = \frac{3x+5}{7x-3} = f(x)$. So,option $A$ is true.
Now,$(f \circ f)(x) = f(f(x)) = \frac{3(\frac{3x+5}{7x-3})+5}{7(\frac{3x+5}{7x-3})-3} = \frac{9x+15+35x-15}{21x+35-21x+9} = \frac{44x}{44} = x$. So,option $B$ is true.
Since $(f \circ f)(x) = x$,then $(f \circ f \circ f)(x) = f((f \circ f)(x)) = f(x) \neq x$.
Also,$(f \circ f \circ f \circ f)(x) = (f \circ f)(f \circ f)(x) = x$. So,option $D$ is true.
Therefore,the statement that is not true is $(f \circ f \circ f)(x) = x$.

(B) Given that the composite function $g \circ f: A \rightarrow C$ is onto.
By definition,for every element $z \in C$,there exists an element $x \in A$ such that $(g \circ f)(x) = z$.
This can be written as $g(f(x)) = z$.
Since $f(x) = y$ for some $y \in B$,we have $g(y) = z$.
This implies that for every $z \in C$,there exists at least one $y \in B$ such that $g(y) = z$.
Therefore,$g$ must be an onto (surjective) function.
Thus,the necessary condition is that $g$ is onto.

(C) Given that $(g \circ j \circ f)(x) = f(x)$ for all $x \in D$.
Substituting the definitions of the functions:
$g(j(f(x))) = f(x)$
Since $g(x) = 1 - x$,we have:
$1 - j(f(x)) = f(x)$
Substituting $f(x) = \frac{1}{x}$:
$1 - j(\frac{1}{x}) = \frac{1}{x}$
Rearranging the equation to solve for $j(\frac{1}{x})$:
$j(\frac{1}{x}) = 1 - \frac{1}{x}$
Let $t = \frac{1}{x}$. Then $x = \frac{1}{t}$.
Substituting $t$ into the equation:
$j(t) = 1 - t$
Thus,$j(x) = 1 - x = g(x)$.

(D) Given $f(x) = \begin{cases} |[x-5]|, & x < 5 \\ [|x-5|], & x \geq 5 \end{cases}$
First,we calculate $f\left(-\frac{7}{2}\right)$.
Since $-\frac{7}{2} = -3.5 < 5$,we use the first case:
$f\left(-\frac{7}{2}\right) = |[-\frac{7}{2} - 5]| = |[-8.5]| = |-9| = 9$.
Now,we calculate $(f \circ f)\left(-\frac{7}{2}\right) = f(f(-\frac{7}{2})) = f(9)$.
Since $9 \geq 5$,we use the second case:
$f(9) = [|9-5|] = [|4|] = 4$.
Thus,$(f \circ f)\left(-\frac{7}{2}\right) = 4$.

(A) Given,$g(x)=x^2+x-2$ and $\frac{1}{2}(g \circ f)(x)=2 x^2-5 x+2$.
Multiplying by $2$,we get $g(f(x))=4 x^2-10 x+4$.
Substituting $f(x)$ into $g(x)$,we have $(f(x))^2+(f(x))-2=4 x^2-10 x+4$.
Assuming $f(x)$ is a linear polynomial $f(x)=ax+b$,we get $(ax+b)^2+(ax+b)-2=4 x^2-10 x+4$.
Expanding the left side: $a^2 x^2+(2ab+a)x+(b^2+b-2)=4 x^2-10 x+4$.
Comparing coefficients:
$1) a^2=4 \Rightarrow a=2$ or $a=-2$.
$2) 2ab+a=-10$.
$3) b^2+b-2=4 \Rightarrow b^2+b-6=0 \Rightarrow (b+3)(b-2)=0 \Rightarrow b=-3$ or $b=2$.
If $a=2$,then $2(2)b+2=-10 \Rightarrow 4b=-12 \Rightarrow b=-3$. Thus $f(x)=2x-3$.
If $a=-2$,then $2(-2)b-2=-10 \Rightarrow -4b=-8 \Rightarrow b=2$. Thus $f(x)=-2x+2$.
Comparing with the given options,$f(x)=2x-3$ is the correct choice.

(C) Let $g(x) = f(f(x))$.
Given $f(x) = \begin{cases} 1+x, & 0 \leq x \leq 2 \\ 3-x, & 2 < x \leq 3 \end{cases}$.
For $0 \leq x \leq 2$,$f(x) = 1+x$. Since $0 \leq x \leq 2$,$1 \leq 1+x \leq 3$.
If $1 \leq 1+x \leq 2$ (i.e.,$0 \leq x \leq 1$),then $f(f(x)) = f(1+x) = 1+(1+x) = 2+x$.
If $2 < 1+x \leq 3$ (i.e.,$1 < x \leq 2$),then $f(f(x)) = f(1+x) = 3-(1+x) = 2-x$.
For $2 < x \leq 3$,$f(x) = 3-x$. Since $2 < x \leq 3$,$0 \leq 3-x < 1$.
Thus,$f(f(x)) = f(3-x) = 1+(3-x) = 4-x$.
So,$g(x) = \begin{cases} 2+x, & 0 \leq x \leq 1 \\ 2-x, & 1 < x \leq 2 \\ 4-x, & 2 < x \leq 3 \end{cases}$.
Checking continuity at $x=1$:
$LHL = \lim_{x \to 1^-} (2+x) = 3$,$RHL = \lim_{x \to 1^+} (2-x) = 1$. Since $LHL \neq RHL$,it is discontinuous at $x=1$.
Checking continuity at $x=2$:
$LHL = \lim_{x \to 2^-} (2-x) = 0$,$RHL = \lim_{x \to 2^+} (4-x) = 2$. Since $LHL \neq RHL$,it is discontinuous at $x=2$.
Therefore,$f(f(x))$ is discontinuous at $x=1$ and $x=2$.

(B) Given the functions $f(x)$ and $g(x)$ defined on the set of real numbers $R$.
First,evaluate $(f \circ g)(\pi)$:
Since $\pi$ is an irrational number,$g(\pi) = 0$.
Since $0$ is a rational number,$f(g(\pi)) = f(0) = 0$.
Next,evaluate $(g \circ f)(e)$:
Since $e$ is an irrational number,$f(e) = 1$.
Since $1$ is a rational number,$g(f(e)) = g(1) = -1$.
Finally,calculate the sum:
$(f \circ g)(\pi) + (g \circ f)(e) = 0 + (-1) = -1$.

(D) Given $f(x)=e^x$ and $g(x)=\ln(x)$ for $x \in [1, \infty)$.
We define the composition $(f \circ g)(x) = f(g(x))$.
Substituting the functions,we get $(f \circ g)(x) = e^{\ln(x)}$.
Since $e^{\ln(x)} = x$ for all $x > 0$,we have $(f \circ g)(x) = x$ for $x \in [1, \infty)$.
The function $h(x) = x$ defined on the domain $[1, \infty)$ is an identity function.
An identity function is both one-one (injective) and onto (surjective) on its domain and range.
Therefore,the function is bijective.
Hence,option $D$ is correct.

(B) Given: $f: A \rightarrow B$ and $g: B \rightarrow C$.
Also,$g \circ f: A \rightarrow C$ is a bijection.
First,we prove $f$ is an injection:
Let $x_1, x_2 \in A$ such that $f(x_1) = f(x_2)$.
Then $g(f(x_1)) = g(f(x_2))$,which implies $(g \circ f)(x_1) = (g \circ f)(x_2)$.
Since $g \circ f$ is a bijection,it is injective,so $x_1 = x_2$.
Thus,$f$ is an injection.
Next,we prove $g$ is a surjection:
Let $z \in C$. Since $g \circ f: A \rightarrow C$ is a bijection,it is surjective.
Therefore,there exists $x \in A$ such that $(g \circ f)(x) = z$.
This means $g(f(x)) = z$.
Since $f(x) \in B$,let $y = f(x)$. Then $g(y) = z$ for some $y \in B$.
Thus,$g$ is a surjection.
Therefore,$f$ is an injection and $g$ is a surjection.

(B) Given that $g \circ f: A \rightarrow C$ is one-one.
By definition,if $g(f(x_1)) = g(f(x_2))$,then $x_1 = x_2$ for all $x_1, x_2 \in A$.
Suppose $f(x_1) = f(x_2)$. Then $g(f(x_1)) = g(f(x_2))$.
Since $g \circ f$ is one-one,this implies $x_1 = x_2$.
Thus,$f$ must be one-one.
However,$g$ does not necessarily have to be one-one. For example,if $A = \{1\}$,$B = \{2, 3\}$,$C = \{4\}$,$f(1) = 2$,$g(2) = 4$,$g(3) = 4$,then $g \circ f(1) = 4$ is one-one,but $g$ is not one-one.
Therefore,$f$ is one-one and $g$ need not be one-one.

(C) Given,$f(x) = ax + b$ and $g(x) = cx^3 + d$.
$(f \circ g)(x) = f(g(x))$.
$(f \circ g)(x) = a(cx^3 + d) + b = acx^3 + ad + b$.
Let $y = (f \circ g)(x) = acx^3 + ad + b$.
To find the inverse,solve for $x$ in terms of $y$:
$y - ad - b = acx^3$.
$x^3 = \frac{y - ad - b}{ac}$.
$x = \left( \frac{y - ad - b}{ac} \right)^{\frac{1}{3}}$.
Replacing $y$ with $x$,we get $(f \circ g)^{-1}(x) = \left( \frac{x - ad - b}{ac} \right)^{\frac{1}{3}}$.

(B) Given $f(x) = 1 - x$,$g(x) = \frac{1}{1 - x}$,and $h(x) = \frac{1}{x}$.
We are given the equation $f(F(h(x))) = g(x)$.
Substituting the expressions for $f$ and $g$,we get $1 - F(h(x)) = \frac{1}{1 - x}$.
Rearranging for $F(h(x))$,we have $F(h(x)) = 1 - \frac{1}{1 - x} = \frac{1 - x - 1}{1 - x} = \frac{-x}{1 - x} = \frac{x}{x - 1}$.
Let $t = h(x) = \frac{1}{x}$. Then $x = \frac{1}{t}$.
Substituting $x = \frac{1}{t}$ into the expression for $F(h(x))$,we get $F(t) = \frac{1/t}{1/t - 1} = \frac{1/t}{(1 - t)/t} = \frac{1}{1 - t}$.
Thus,$F(x) = \frac{1}{1 - x} = g(x)$.
Therefore,$F(2022) = g(2022)$.

(B) Given $f(x) = \frac{2x - 3}{3x - 2}$.
Calculate $f(f(x))$:
$f(f(x)) = \frac{2(\frac{2x - 3}{3x - 2}) - 3}{3(\frac{2x - 3}{3x - 2}) - 2}$
$= \frac{2(2x - 3) - 3(3x - 2)}{3(2x - 3) - 2(3x - 2)}$
$= \frac{4x - 6 - 9x + 6}{6x - 9 - 6x + 4}$
$= \frac{-5x}{-5} = x$.
Since $f(f(x)) = x$,the function is its own inverse.
For any even $n$,$f_n(x) = x$.
Since $32$ is an even number,$f_{32}(x) = x$.

(C) Given $f(x) = -|x|$.
First,we evaluate $(f \circ f \circ f)(x)$:
$f(f(f(x))) = f(f(-|x|)) = f(-|-|x||) = f(-|x|) = -|-|x|| = -|x| = f(x)$.
Wait,let us re-evaluate carefully:
$f(x) = -|x|$.
$f(f(x)) = f(-|x|) = -|-|x|| = -|x| = f(x)$.
Since $f(f(x)) = f(x)$,it follows that $(f \circ f \circ f)(x) = f(f(f(x))) = f(f(x)) = f(x)$.
Similarly,$(f \circ f \circ f)(-x) = f(f(f(-x))) = f(f(-x)) = f(-x)$.
Therefore,$(f \circ f \circ f)(x) + (f \circ f \circ f)(-x) = f(x) + f(-x)$.
Since $f(x) = -|x|$,we have $f(-x) = -|-x| = -|x| = f(x)$.
Thus,$f(x) + f(-x) = f(x) + f(x) = 2f(x)$.

(B) Given $f(x) = 3x-2$ and $g(x) = x^2+2$.
We need to find $(g \circ f)(x) + (f \circ g)(x)$.
First,$(g \circ f)(x) = g(f(x)) = (f(x))^2 + 2 = (3x-2)^2 + 2 = 9x^2 - 12x + 4 + 2 = 9x^2 - 12x + 6$.
Next,$(f \circ g)(x) = f(g(x)) = 3(g(x)) - 2 = 3(x^2+2) - 2 = 3x^2 + 6 - 2 = 3x^2 + 4$.
Adding these,$(g \circ f + f \circ g)(x) = (9x^2 - 12x + 6) + (3x^2 + 4) = 12x^2 - 12x + 10$.
Now,check the options by substituting $f(x)$ and $g(x)$:
$12g(x) - 4f(x) - 22 = 12(x^2+2) - 4(3x-2) - 22 = 12x^2 + 24 - 12x + 8 - 22 = 12x^2 - 12x + 10$.
Thus,the correct option is $B$.

(B) To find $\lim _{x \rightarrow 0} g(f(x))$,we evaluate the left-hand and right-hand limits.
For the left-hand limit $(x \rightarrow 0^-)$: $f(x) = 2-x$. As $x \rightarrow 0^-$,$f(x) \rightarrow 2^+$. Since $f(x) > 2$,we use the definition $g(x) = x-7$. Thus,$\lim _{x \rightarrow 0^-} g(f(x)) = \lim _{f(x) \rightarrow 2^+} (f(x)-7) = 2-7 = -5$.
For the right-hand limit $(x \rightarrow 0^+)$: $f(x) = x+2$. As $x \rightarrow 0^+$,$f(x) \rightarrow 2^+$. Again,since $f(x) > 2$,we use $g(x) = x-7$. Thus,$\lim _{x \rightarrow 0^+} g(f(x)) = \lim _{f(x) \rightarrow 2^+} (f(x)-7) = 2-7 = -5$.
Since both limits are equal,$\lim _{x \rightarrow 0} g(f(x)) = -5$.

(A) Given the function $f:[0,1] \rightarrow [0,1]$ defined as $f(x) = \begin{cases} x & \text{if } x \in Q \\ 1-x & \text{if } x \notin Q \end{cases}$.
We need to find the set $S = \{x \in [0,1] : (f \circ f)(x) = x\}$.
Case $1$: If $x \in Q$,then $f(x) = x$. Since $x \in [0,1]$,$x$ is rational,so $f(x) \in Q$. Thus,$(f \circ f)(x) = f(f(x)) = f(x) = x$. This holds for all $x \in Q \cap [0,1]$.
Case $2$: If $x \notin Q$,then $f(x) = 1-x$. Since $x$ is irrational and $x \in [0,1]$,$1-x$ is also irrational and $1-x \in [0,1]$. Thus,$(f \circ f)(x) = f(f(x)) = f(1-x) = 1-(1-x) = x$. This holds for all $x \in [0,1] \setminus Q$.
Since $(f \circ f)(x) = x$ for all $x \in [0,1]$,the set $S$ is the entire interval $[0,1]$.

(A) Given,$f(x) = (p - x^n)^{1/n}$,where $p > 0$.
To find $f[f(x)]$,we substitute $f(x)$ into the function $f$:
$f[f(x)] = f((p - x^n)^{1/n})$
$= (p - ((p - x^n)^{1/n})^n)^{1/n}$
$= (p - (p - x^n))^{1/n}$
$= (p - p + x^n)^{1/n}$
$= (x^n)^{1/n}$
$= x$.

(C) Given,$g\{f(x)\}=|\sin x|$ and $f\{g(x)\}=(\sin \sqrt{x})^2$.
Let us test the option $f(x)=\sin ^2 x$ and $g(x)=\sqrt{x}$.
First,calculate $f\{g(x)\}$:
$f\{g(x)\} = f(\sqrt{x}) = \sin ^2(\sqrt{x}) = (\sin \sqrt{x})^2$.
This matches the given condition.
Next,calculate $g\{f(x)\}$:
$g\{f(x)\} = g(\sin ^2 x) = \sqrt{\sin ^2 x} = |\sin x|$.
This also matches the given condition.
Therefore,the correct choice is $f(x)=\sin ^2 x$ and $g(x)=\sqrt{x}$.

(A) Given,$f(x)=x^2-3$.
First,we calculate $(f \circ f \circ f)(-1)$:
$f(-1) = (-1)^2 - 3 = -2$
$f(f(-1)) = f(-2) = (-2)^2 - 3 = 1$
$f(f(f(-1))) = f(1) = (1)^2 - 3 = -2$.
Next,we calculate $(f \circ f \circ f)(0)$:
$f(0) = (0)^2 - 3 = -3$
$f(f(0)) = f(-3) = (-3)^2 - 3 = 6$
$f(f(f(0))) = f(6) = (6)^2 - 3 = 33$.
Next,we calculate $(f \circ f \circ f)(1)$:
$f(1) = (1)^2 - 3 = -2$
$f(f(1)) = f(-2) = (-2)^2 - 3 = 1$
$f(f(f(1))) = f(1) = (1)^2 - 3 = -2$.
Summing these values:
$(f \circ f \circ f)(-1) + (f \circ f \circ f)(0) + (f \circ f \circ f)(1) = -2 + 33 - 2 = 29$.
Now,checking the options:
$f(4 \sqrt{2}) = (4 \sqrt{2})^2 - 3 = 32 - 3 = 29$.
Thus,the expression is equal to $f(4 \sqrt{2})$.

(C) Given that $f(x)=|x|$ and $g(x)=[x-3]$.
For $-\frac{8}{5} < x < \frac{8}{5}$,the range of $f(x)=|x|$ is $0 \leq f(x) < \frac{8}{5}$,which is $0 \leq f(x) < 1.6$.
We need to find the values of $g(f(x)) = [f(x)-3]$.
Case $1$: If $0 \leq f(x) < 1$,then $-3 \leq f(x)-3 < -2$. Thus,$[f(x)-3] = -3$.
Case $2$: If $1 \leq f(x) < 1.6$,then $-2 \leq f(x)-3 < -1.4$. Thus,$[f(x)-3] = -2$.
Combining these cases,the set of values is $\{-3, -2\}$.

(B) Given,$f(x)=x-[x]$ and $g(x)=[x]$ for $x \in R$.
We need to find $f(g(x))$.
Substituting $g(x)$ into $f(x)$,we get:
$f(g(x)) = f([x])$.
By the definition of $f(x)$,$f([x]) = [x] - [[x]]$.
Since $[x]$ is an integer,the greatest integer function of an integer is the integer itself,i.e.,$[[x]] = [x]$.
Therefore,$f(g(x)) = [x] - [x] = 0$.

(C) Given functions are $f(x) = 2x + 3$ and $g(x) = x^2 + 7$.
We need to find $x$ such that $g(f(x)) = 8$.
First,calculate the composite function $g(f(x))$:
$g(f(x)) = g(2x + 3) = (2x + 3)^2 + 7$.
Set this equal to $8$:
$(2x + 3)^2 + 7 = 8$.
Subtract $7$ from both sides:
$(2x + 3)^2 = 1$.
Taking the square root of both sides:
$2x + 3 = 1$ or $2x + 3 = -1$.
Case $1$: $2x + 3 = 1 \Rightarrow 2x = -2 \Rightarrow x = -1$.
Case $2$: $2x + 3 = -1 \Rightarrow 2x = -4 \Rightarrow x = -2$.
Thus,the values of $x$ are $-1$ and $-2$.

(B) Given $f(x) = \begin{cases} 0, & x \in \mathbb{Q} \\ 1, & x \notin \mathbb{Q} \end{cases}$ and $g(x) = \begin{cases} -1, & x \in \mathbb{Q} \\ 0, & x \notin \mathbb{Q} \end{cases}$.
Since $\pi$ is an irrational number,$g(\pi) = 0$. Since $0$ is a rational number,$f(g(\pi)) = f(0) = 0$.
Since $e$ is an irrational number,$f(e) = 1$. Since $1$ is a rational number,$g(f(e)) = g(1) = -1$.
Therefore,$(f \circ g)(\pi) + (g \circ f)(e) = 0 + (-1) = -1$.

(D) Given $f(x) = (20 - x^4)^{1/4}$.
First,calculate $f(1/2)$:
$f(1/2) = (20 - (1/2)^4)^{1/4} = (20 - 1/16)^{1/4} = (319/16)^{1/4}$.
Now,calculate $f(f(1/2)) = f((319/16)^{1/4})$:
$f((319/16)^{1/4}) = (20 - ((319/16)^{1/4})^4)^{1/4}$.
$= (20 - 319/16)^{1/4}$.
$= ((320 - 319)/16)^{1/4}$.
$= (1/16)^{1/4}$.
$= (1/2^4)^{1/4} = 1/2 = 2^{-1}$.

(D) We have,$f(x) = \begin{cases} 1 + x, & 0 \leq x \leq 2 \\ 3 - x, & 2 < x \leq 3 \end{cases}$.
$f(f(x))$ is defined as:
For $0 \leq f(x) \leq 2$,$f(f(x)) = 1 + f(x)$.
For $2 < f(x) \leq 3$,$f(f(x)) = 3 - f(x)$.
Evaluating for $x \in [0, 3]$:
If $0 \leq x \leq 1$,then $1 \leq f(x) \leq 2$,so $f(f(x)) = 1 + (1 + x) = 2 + x$.
If $1 < x \leq 2$,then $2 < f(x) \leq 3$,so $f(f(x)) = 3 - (1 + x) = 2 - x$.
If $2 < x \leq 3$,then $0 \leq f(x) < 1$,so $f(f(x)) = 1 + (3 - x) = 4 - x$.
Thus,$f(f(x)) = \begin{cases} 2 + x, & 0 \leq x \leq 1 \\ 2 - x, & 1 < x \leq 2 \\ 4 - x, & 2 < x \leq 3 \end{cases}$.
Checking continuity at $x = 1$: $\lim_{x \to 1^-} f(f(x)) = 3$,$\lim_{x \to 1^+} f(f(x)) = 1$. Since $3 \neq 1$,it is discontinuous at $x = 1$.
Checking continuity at $x = 2$: $\lim_{x \to 2^-} f(f(x)) = 0$,$\lim_{x \to 2^+} f(f(x)) = 2$. Since $0 \neq 2$,it is discontinuous at $x = 2$.
Therefore,$a = 1$ and $b = 2$.
Calculating $2a + 3b = 2(1) + 3(2) = 2 + 6 = 8$.

(C) Given $f(x)=e^x$.
$h(x)=(f \circ f)(x) = f(f(x)) = f(e^x) = e^{e^x}$.
Now,differentiate $h(x)$ with respect to $x$:
$h^{\prime}(x) = \frac{d}{dx}(e^{e^x}) = e^{e^x} \cdot \frac{d}{dx}(e^x) = e^{e^x} \cdot e^x$.
Since $h(x) = e^{e^x}$,we have $h^{\prime}(x) = h(x) \cdot e^x$.
Therefore,$\frac{h^{\prime}(x)}{h(x)} = e^x$.
Now,consider $\log h(x) = \log(e^{e^x}) = e^x \cdot \log e = e^x$.
Comparing the two results,we get $\frac{h^{\prime}(x)}{h(x)} = \log h(x)$.

(D) Given $f(x) = \sqrt{x}$ and $g(x) = 1 + x^2$.
The composite function $(f \circ g)(x) = f(g(x)) = \sqrt{1 + x^2}$.
To find the derivative $(f \circ g)^{\prime}(x)$,we use the chain rule:
$(f \circ g)^{\prime}(x) = \frac{d}{dx} (1 + x^2)^{1/2} = \frac{1}{2}(1 + x^2)^{-1/2} \cdot \frac{d}{dx}(1 + x^2)$.
$(f \circ g)^{\prime}(x) = \frac{1}{2\sqrt{1 + x^2}} \cdot (2x) = \frac{x}{\sqrt{1 + x^2}}$.
Now,substitute $x = 1$ into the derivative:
$(f \circ g)^{\prime}(1) = \frac{1}{\sqrt{1 + 1^2}} = \frac{1}{\sqrt{2}}$.

(A) For $f \circ g(x) = f(g(x)) = \sqrt{(\sqrt{x})^2 - 3\sqrt{x} + 2} = \sqrt{x - 3\sqrt{x} + 2}$.
For the domain $S$,we require $x \ge 0$ and $x - 3\sqrt{x} + 2 \ge 0$.
Let $u = \sqrt{x}$,then $u^2 - 3u + 2 \ge 0 \implies (u-1)(u-2) \ge 0$.
This gives $u \le 1$ or $u \ge 2$,so $\sqrt{x} \le 1$ or $\sqrt{x} \ge 2$.
Thus,$x \in [0, 1] \cup [4, \infty)$,so $S = [0, 1] \cup [4, \infty)$.
For $g \circ f(x) = g(f(x)) = \sqrt{\sqrt{x^2 - 3x + 2}}$.
For the domain $T$,we require $x^2 - 3x + 2 \ge 0 \implies (x-1)(x-2) \ge 0$.
This gives $x \le 1$ or $x \ge 2$,so $T = (-\infty, 1] \cup [2, \infty)$.
Now,$S \cap T = ([0, 1] \cup [4, \infty)) \cap ((-\infty, 1] \cup [2, \infty)) = [0, 1] \cup [4, \infty)$.
Since $S \cap T = S$,and $S$ is a union of two disjoint intervals,the intersection is not a single interval.

(A) Given $g(f(x)) = |\sin x|$ and $f(g(x)) = (\sin \sqrt{x})^2$.
Let us test option $A$: $f(x) = \sin ^2 x$ and $g(x) = \sqrt{x}$.
Then $g(f(x)) = g(\sin ^2 x) = \sqrt{\sin ^2 x} = |\sin x|$. This matches the first condition.
Next,$f(g(x)) = f(\sqrt{x}) = \sin ^2(\sqrt{x}) = (\sin \sqrt{x})^2$. This matches the second condition.
Therefore,the correct option is $A$.

(D) Given $f(x) = \frac{3x - 4}{2x - 3}$.
First,find $f(f(x))$:
$f(f(x)) = f\left(\frac{3x - 4}{2x - 3}\right) = \frac{3\left(\frac{3x - 4}{2x - 3}\right) - 4}{2\left(\frac{3x - 4}{2x - 3}\right) - 3}$
$= \frac{3(3x - 4) - 4(2x - 3)}{2(3x - 4) - 3(2x - 3)} = \frac{9x - 12 - 8x + 12}{6x - 8 - 6x + 9} = \frac{x}{1} = x$.
Now,find $f(f(f(x)))$:
$f(f(f(x))) = f(f(f(x))) = f(x) = \frac{3x - 4}{2x - 3}$.

(B) Given,$f(x)=2^{100} x+1$ and $g(x)=3^{100} x+1$.
We need to find $x$ such that $f(g(x))=x$.
Substituting $g(x)$ into $f(x)$,we get:
$f(3^{100} x+1) = x$
$2^{100}(3^{100} x+1) + 1 = x$
$2^{100} \cdot 3^{100} x + 2^{100} + 1 = x$
$6^{100} x + 2^{100} + 1 = x$
$x(6^{100} - 1) = -(2^{100} + 1)$
$x = -\frac{2^{100} + 1}{6^{100} - 1}$
Since there is exactly one value of $x$ that satisfies the equation,the set of real numbers $x$ is a singleton set.