Composition of Functions Questions in English

(C) Given $f(x) = e^x$ and $g(x) = x^2$.
First,find the composite function $f(g(x)) = f(x^2) = e^{x^2}$.
Next,find the composite function $g(f(x)) = g(e^x) = (e^x)^2 = e^{2x}$.
Equating the two,we get $e^{x^2} = e^{2x}$.
Since the exponential function is one-to-one,we can equate the exponents: $x^2 = 2x$.
Rearranging gives $x^2 - 2x = 0$,which factors as $x(x - 2) = 0$.
Thus,the solutions are $x = 0$ and $x = 2$.
Therefore,the number of solutions is $2$.

(B) To find the number of solutions for $f(f(x)) = 2$,let $f(x) = t$. Then the equation becomes $f(t) = 2$.
From the graph,$f(t) = 2$ at $t = -3$,$t = 1/2$,and $t = \alpha$,where $2 < \alpha < 3$.
Now,we solve for $x$ in $f(x) = t$ for each value of $t$:
$1$. $f(x) = -3$: From the graph,the minimum value of $f(x)$ is $-1$. Thus,$f(x) = -3$ has no solutions.
$2$. $f(x) = 1/2$: The horizontal line $y = 1/2$ intersects the graph at $2$ points.
$3$. $f(x) = \alpha$ (where $2 < \alpha < 3$): The horizontal line $y = \alpha$ intersects the graph at $2$ points.
Total number of solutions = $0 + 2 + 2 = 4$.

(C) Given $f\left(x + \frac{1}{x}\right) = x^2 + \frac{1}{x^2}$.
We can rewrite the expression as $f\left(x + \frac{1}{x}\right) = \left(x + \frac{1}{x}\right)^2 - 2$.
Let $t = x + \frac{1}{x}$. Then $f(t) = t^2 - 2$,which implies $f(x) = x^2 - 2$.
Now,we need to find $(f \circ f)(\sqrt{11}) = f(f(\sqrt{11}))$.
First,calculate $f(\sqrt{11}) = (\sqrt{11})^2 - 2 = 11 - 2 = 9$.
Next,calculate $f(9) = 9^2 - 2 = 81 - 2 = 79$.
Therefore,$(f \circ f)(\sqrt{11}) = 79$.

(A) Given $\frac{1}{2} (g \circ f)(x) = 2x^2 - 5x + 2$,we have $(g \circ f)(x) = 4x^2 - 10x + 4$.
Since $g(x) = x^2 + x - 2$,then $g(f(x)) = (f(x))^2 + f(x) - 2$.
Equating the two expressions: $(f(x))^2 + f(x) - 2 = 4x^2 - 10x + 4$.
Rearranging the terms: $(f(x))^2 + f(x) - (4x^2 - 10x + 6) = 0$.
Using the quadratic formula $f(x) = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ where $a=1, b=1, c=-(4x^2 - 10x + 6)$:
$f(x) = \frac{-1 \pm \sqrt{1^2 - 4(1)(-(4x^2 - 10x + 6))}}{2}$
$f(x) = \frac{-1 \pm \sqrt{1 + 16x^2 - 40x + 24}}{2} = \frac{-1 \pm \sqrt{16x^2 - 40x + 25}}{2}$
$f(x) = \frac{-1 \pm \sqrt{(4x - 5)^2}}{2} = \frac{-1 \pm (4x - 5)}{2}$.
Taking the positive root: $f(x) = \frac{-1 + 4x - 5}{2} = \frac{4x - 6}{2} = 2x - 3$.

(D) Given $f(x) = 2^{10}x + 1$ and $g(x) = 3^{10}x - 1$.
We are given $(f \circ g)(x) = x$.
Substituting $g(x)$ into $f(x)$,we get $f(g(x)) = 2^{10}(3^{10}x - 1) + 1 = x$.
Expanding the expression: $2^{10} \cdot 3^{10}x - 2^{10} + 1 = x$.
Since $2^{10} \cdot 3^{10} = (2 \cdot 3)^{10} = 6^{10}$,we have $6^{10}x - x = 2^{10} - 1$.
Factoring out $x$: $x(6^{10} - 1) = 2^{10} - 1$.
Thus,$x = \frac{2^{10} - 1}{6^{10} - 1}$.
To match the options,divide the numerator and denominator by $6^{10}$:
$x = \frac{\frac{2^{10}}{6^{10}} - \frac{1}{6^{10}}}{1 - \frac{1}{6^{10}}} = \frac{3^{-10} - 6^{-10}}{1 - 6^{-10}}$.
Alternatively,divide the numerator and denominator by $2^{10} \cdot 3^{10} = 6^{10}$ or manipulate the expression:
$x = \frac{2^{10} - 1}{6^{10} - 1} = \frac{2^{10}(1 - 2^{-10})}{6^{10}(1 - 6^{-10})} = \frac{3^{-10}(1 - 2^{-10})}{1 - 6^{-10}}$.
Re-evaluating the expression $\frac{2^{10} - 1}{6^{10} - 1}$:
Divide numerator and denominator by $6^{10}$: $\frac{2^{10}/6^{10} - 1/6^{10}}{1 - 1/6^{10}} = \frac{3^{-10} - 6^{-10}}{1 - 6^{-10}}$.
Checking option $D$: $\frac{1 - 2^{-10}}{3^{10} - 2^{-10}} = \frac{(2^{10}-1)/2^{10}}{(3^{10} \cdot 2^{10} - 1)/2^{10}} = \frac{2^{10}-1}{6^{10}-1}$.
Thus,option $D$ is correct.

(C) $f_1(x) = f_{0+1}(x) = f_0(f_0(x)) = \frac{1}{1 - \frac{1}{1 - x}} = \frac{x - 1}{x}$
$f_2(x) = f_{1+1}(x) = f_0(f_1(x)) = \frac{1}{1 - \frac{x - 1}{x}} = x$
$f_3(x) = f_{2+1}(x) = f_0(f_2(x)) = f_0(x) = \frac{1}{1 - x}$
Since $f_3(x) = f_0(x)$,the function repeats with a period of $3$.
$f_{100}(3) = f_{3 \times 33 + 1}(3) = f_1(3) = \frac{3 - 1}{3} = \frac{2}{3}$
$f_1\left( \frac{2}{3} \right) = \frac{\frac{2}{3} - 1}{\frac{2}{3}} = \frac{-\frac{1}{3}}{\frac{2}{3}} = -\frac{1}{2}$
$f_2\left( \frac{3}{2} \right) = \frac{3}{2}$
Therefore,$f_{100}(3) + f_1\left( \frac{2}{3} \right) + f_2\left( \frac{3}{2} \right) = \frac{2}{3} - \frac{1}{2} + \frac{3}{2} = \frac{2}{3} + 1 = \frac{5}{3}$

(B) The function $u(x) = \frac{1}{x - 1}$ is discontinuous at $x = 1$.
The function $f(u) = \frac{1}{u^2 + u - 2} = \frac{1}{(u + 2)(u - 1)}$ is discontinuous at $u = -2$ and $u = 1$.
For the composite function $f(u(x))$,we must consider the points where $u(x)$ is undefined and where $u(x)$ takes the values that make $f(u)$ undefined.
$1$. $u(x)$ is discontinuous at $x = 1$.
$2$. When $u(x) = -2$,we have $\frac{1}{x - 1} = -2$,which implies $x - 1 = -\frac{1}{2}$,so $x = \frac{1}{2}$.
$3$. When $u(x) = 1$,we have $\frac{1}{x - 1} = 1$,which implies $x - 1 = 1$,so $x = 2$.
Thus,the composite function $f(u(x))$ is discontinuous at $x = 1, \frac{1}{2}, 2$.
There are $3$ such points.

(A) Given functions are ${f_1}(x) = \frac{1}{x}$,${f_2}(x) = 1 - x$,and ${f_3}(x) = \frac{1}{1 - x}$.
The given equation is $(f_2 \circ J \circ f_1)(x) = f_3(x)$,which can be written as ${f_2}(J(f_1(x))) = f_3(x)$.
Substituting the expressions for ${f_2}$ and ${f_3}$:
$1 - J(f_1(x)) = \frac{1}{1 - x}$.
Rearranging to solve for $J(f_1(x))$:
$J(f_1(x)) = 1 - \frac{1}{1 - x} = \frac{1 - x - 1}{1 - x} = \frac{-x}{1 - x} = \frac{x}{x - 1}$.
Since ${f_1}(x) = \frac{1}{x}$,let $t = \frac{1}{x}$,which implies $x = \frac{1}{t}$.
Substituting $x = \frac{1}{t}$ into the expression for $J(f_1(x))$:
$J(t) = \frac{\frac{1}{t}}{\frac{1}{t} - 1} = \frac{\frac{1}{t}}{\frac{1 - t}{t}} = \frac{1}{1 - t}$.
Thus,$J(x) = \frac{1}{1 - x}$,which is equal to ${f_3}(x)$.

(C) Given $f(x) = \log_e(\sin x)$ and $g(x) = \sin^{-1}(e^{-x})$.
First,find the composite function $(fog)(x) = f(g(x)) = \log_e(\sin(\sin^{-1}(e^{-x})))$.
Since $\sin(\sin^{-1}(u)) = u$,we have $(fog)(x) = \log_e(e^{-x}) = -x$.
Given $b = (fog)(\alpha)$,we have $b = -\alpha$.
Next,find the derivative $(fog)'(x) = \frac{d}{dx}(-x) = -1$.
Given $a = (fog)'(\alpha)$,we have $a = -1$.
Now,substitute $a = -1$ and $b = -\alpha$ into the options.
Checking option $B$: $a\alpha^2 - b\alpha - a = (-1)\alpha^2 - (-\alpha)\alpha - (-1) = -\alpha^2 + \alpha^2 + 1 = 1 \neq 0$.
Checking option $D$: $a\alpha^2 + b\alpha + a = (-1)\alpha^2 + (-\alpha)\alpha + (-1) = -\alpha^2 - \alpha^2 - 1 = -2\alpha^2 - 1 \neq 0$.
Re-evaluating the expression $a\alpha^2 - b\alpha - a$: Since $a = -1$ and $b = -\alpha$,then $a\alpha^2 - b\alpha - a = (-1)\alpha^2 - (-\alpha)\alpha - (-1) = -\alpha^2 + \alpha^2 + 1 = 1$.
Thus,$a\alpha^2 - b\alpha - a = 1$ is correct,which corresponds to option $C$.

(A) Given functions are $f(x) = \sqrt{x}$,$g(x) = \tan x$,and $h(x) = \frac{1 - x^2}{1 + x^2}$.
First,find $(f \circ g)(x) = f(g(x)) = \sqrt{\tan x}$.
Next,find $\phi(x) = (h \circ (f \circ g))(x) = h(\sqrt{\tan x})$.
Substituting $\sqrt{\tan x}$ into $h(x)$,we get $\phi(x) = \frac{1 - (\sqrt{\tan x})^2}{1 + (\sqrt{\tan x})^2} = \frac{1 - \tan x}{1 + \tan x}$.
Using the trigonometric identity $\tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}$,we can write $\frac{1 - \tan x}{1 + \tan x} = \tan\left( \frac{\pi}{4} - x \right)$.
Now,calculate $\phi\left( \frac{\pi}{3} \right) = \tan\left( \frac{\pi}{4} - \frac{\pi}{3} \right) = \tan\left( \frac{3\pi - 4\pi}{12} \right) = \tan\left( -\frac{\pi}{12} \right) = -\tan\left( \frac{\pi}{12} \right)$.
Since $\tan(\pi - \theta) = -\tan \theta$,we have $-\tan\left( \frac{\pi}{12} \right) = \tan\left( \pi - \frac{\pi}{12} \right) = \tan\left( \frac{11\pi}{12} \right)$.

(B) Given $g(x) = x^{2} + x - 1$ and $(g \circ f)(x) = g(f(x)) = 4x^{2} - 10x + 5$.
Let $f(x) = y$. Then $g(y) = y^{2} + y - 1 = 4x^{2} - 10x + 5$.
$y^{2} + y - 1 = 4x^{2} - 10x + 5$
$y^{2} + y - 6 = 4x^{2} - 10x$.
To solve for $y$,we complete the square for $y^{2} + y$:
$(y + \frac{1}{2})^{2} - \frac{1}{4} - 6 = 4x^{2} - 10x$.
$(y + \frac{1}{2})^{2} = 4x^{2} - 10x + \frac{25}{4} = (2x - \frac{5}{2})^{2}$.
Taking the square root,$y + \frac{1}{2} = \pm(2x - \frac{5}{2})$.
Case $1$: $f(x) = 2x - \frac{5}{2} - \frac{1}{2} = 2x - 3$.
Case $2$: $f(x) = -2x + \frac{5}{2} - \frac{1}{2} = -2x + 2$.
For $f(x) = 2x - 3$,$f(\frac{5}{4}) = 2(\frac{5}{4}) - 3 = \frac{5}{2} - 3 = -\frac{1}{2}$.
For $f(x) = -2x + 2$,$f(\frac{5}{4}) = -2(\frac{5}{4}) + 2 = -\frac{5}{2} + 2 = -\frac{1}{2}$.
Thus,$f(\frac{5}{4}) = -\frac{1}{2}$.

The composition $g \circ f$ is defined as $(g \circ f)(x) = g(f(x))$.
We calculate the values for each element in the domain $\{2, 3, 4, 5\}$:
$(g \circ f)(2) = g(f(2)) = g(3) = 7$
$(g \circ f)(3) = g(f(3)) = g(4) = 7$
$(g \circ f)(4) = g(f(4)) = g(5) = 11$
$(g \circ f)(5) = g(f(5)) = g(5) = 11$
Thus,$g \circ f = \{(2, 7), (3, 7), (4, 11), (5, 11)\}$.

(N/A) Given $f(x) = \cos x$ and $g(x) = 3x^2$.
First,we find $gof(x) = g(f(x)) = g(\cos x) = 3(\cos x)^2 = 3 \cos^2 x$.
Next,we find $fog(x) = f(g(x)) = f(3x^2) = \cos(3x^2)$.
To show $gof \neq fog$,we can test a value,for example,$x = 0$.
$gof(0) = 3 \cos^2(0) = 3(1)^2 = 3$.
$fog(0) = \cos(3(0)^2) = \cos(0) = 1$.
Since $3 \neq 1$,it follows that $gof \neq fog$.

(A) To show $g \circ f = I_{B}$,we calculate:
$g(f(x)) = g\left(\frac{3x+4}{5x-7}\right) = \frac{7\left(\frac{3x+4}{5x-7}\right) + 4}{5\left(\frac{3x+4}{5x-7}\right) - 3}$
$= \frac{\frac{21x + 28 + 20x - 28}{5x - 7}}{\frac{15x + 20 - 15x + 21}{5x - 7}} = \frac{41x}{41} = x$
Thus,$g \circ f(x) = x = I_{B}(x)$ for all $x \in B$.
To show $f \circ g = I_{A}$,we calculate:
$f(g(x)) = f\left(\frac{7x+4}{5x-3}\right) = \frac{3\left(\frac{7x+4}{5x-3}\right) + 4}{5\left(\frac{7x+4}{5x-3}\right) - 7}$
$= \frac{\frac{21x + 12 + 20x - 12}{5x - 3}}{\frac{35x + 20 - 35x + 21}{5x - 3}} = \frac{41x}{41} = x$
Thus,$f \circ g(x) = x = I_{A}(x)$ for all $x \in A$.
Therefore,$f \circ g = I_{A}$ and $g \circ f = I_{B}$.

(N/A) To show that $g \circ f$ is one-one,we assume that $g \circ f(x_1) = g \circ f(x_2)$ for any $x_1, x_2 \in A$.
By the definition of composition of functions,this implies $g(f(x_1)) = g(f(x_2))$.
Since $g: B \rightarrow C$ is given as a one-one function,$g(y_1) = g(y_2) \implies y_1 = y_2$. Therefore,$g(f(x_1)) = g(f(x_2))$ implies $f(x_1) = f(x_2)$.
Since $f: A \rightarrow B$ is also given as a one-one function,$f(x_1) = f(x_2) \implies x_1 = x_2$.
Thus,$g \circ f(x_1) = g \circ f(x_2) \implies x_1 = x_2$,which proves that $g \circ f$ is one-one.

(N/A) Let $z$ be an arbitrary element in $C$.
Since $g: B \rightarrow C$ is an onto function,there exists an element $y \in B$ such that $g(y) = z$.
Since $f: A \rightarrow B$ is also an onto function,for the element $y \in B$,there exists an element $x \in A$ such that $f(x) = y$.
Now,consider the composition $(g \circ f)(x) = g(f(x))$.
Substituting $f(x) = y$,we get $g(f(x)) = g(y)$.
Since $g(y) = z$,we have $(g \circ f)(x) = z$.
Since for every $z \in C$,there exists an $x \in A$ such that $(g \circ f)(x) = z$,the function $g \circ f: A \rightarrow C$ is onto.

(NO) To determine if $f$ and $g$ are necessarily one-one,let us analyze the properties of the composite function $g \circ f$.
If $g \circ f$ is one-one,then for any $x_1, x_2$ in the domain of $f$,$g(f(x_1)) = g(f(x_2))$ implies $f(x_1) = f(x_2)$,which further implies $x_1 = x_2$. This confirms that $f$ must be one-one.
However,$g$ does not necessarily have to be one-one.
Consider the functions $f: \{1, 2, 3, 4\} \rightarrow \{1, 2, 3, 4, 5, 6\}$ defined as $f(x) = x$ for all $x$,and $g: \{1, 2, 3, 4, 5, 6\} \rightarrow \{1, 2, 3, 4, 5, 6\}$ defined as $g(x) = x$ for $x \in \{1, 2, 3, 4\}$ and $g(5) = g(6) = 5$.
Here,$(g \circ f)(x) = x$ for all $x \in \{1, 2, 3, 4\}$,which is clearly one-one.
However,$g$ is not one-one because $g(5) = g(6) = 5$ while $5 \neq 6$.
Therefore,$f$ must be one-one,but $g$ is not necessarily one-one.

(N/A) We have:
$h \circ (g \circ f)(x) = h(g(f(x)))$
$= h(g(2x))$
$= h(3(2x) + 4)$
$= h(6x + 4)$
$= \sin(6x + 4), \forall x \in N$
Also,$((h \circ g) \circ f)(x) = (h \circ g)(f(x))$
$= (h \circ g)(2x)$
$= h(g(2x))$
$= h(3(2x) + 4)$
$= h(6x + 4)$
$= \sin(6x + 4), \forall x \in N$
Since both expressions yield the same result,it is shown that $h \circ (g \circ f) = (h \circ g) \circ f$.
This result holds true in general for the composition of functions.

(A) The functions $f: \{1,3,4\} \rightarrow \{1,2,5\}$ and $g: \{1,2,5\} \rightarrow \{1,3\}$ are defined as $f = \{(1,2), (3,5), (4,1)\}$ and $g = \{(1,3), (2,3), (5,1)\}$.
$(g \circ f)(1) = g(f(1)) = g(2) = 3$ [since $f(1) = 2$ and $g(2) = 3$]
$(g \circ f)(3) = g(f(3)) = g(5) = 1$ [since $f(3) = 5$ and $g(5) = 1$]
$(g \circ f)(4) = g(f(4)) = g(1) = 3$ [since $f(4) = 1$ and $g(1) = 3$]
Therefore,$g \circ f = \{(1,3), (3,1), (4,3)\}$.

To prove: $(f+g)oh = foh + goh$
$LHS = [(f+g)oh](x)$
$= (f+g)[h(x)] = f[h(x)] + g[h(x)]$
$= (foh)(x) + (goh)(x)$
$= \{(foh) + (goh)\}(x) = RHS$
$\therefore \{(f+g)oh\}(x) = \{(foh) + (goh)\}(x)$ for all $x \in R$.
Hence,$(f+g)oh = foh + goh$.
To prove: $(f \cdot g)oh = (foh) \cdot (goh)$
$LHS = [(f \cdot g)oh](x)$
$= (f \cdot g)[h(x)] = f[h(x)] \cdot g[h(x)]$
$= (foh)(x) \cdot (goh)(x)$
$= \{(foh) \cdot (goh)\}(x) = RHS$
$\therefore [(f \cdot g)oh](x) = \{(foh) \cdot (goh)\}(x)$ for all $x \in R$.
Hence,$(f \cdot g)oh = (foh) \cdot (goh)$.

(A) Given functions are $f(x) = |x|$ and $g(x) = |5x-2|.$
To find $g \circ f(x):$
$g \circ f(x) = g(f(x)) = g(|x|).$
Substituting $|x|$ into $g(x),$ we get $g(|x|) = |5|x|-2|.$
To find $f \circ g(x):$
$f \circ g(x) = f(g(x)) = f(|5x-2|).$
Substituting $|5x-2|$ into $f(x),$ we get $f(|5x-2|) = ||5x-2|| = |5x-2|.$
Thus,$g \circ f(x) = |5|x|-2|$ and $f \circ g(x) = |5x-2|.$

(A) Given functions are $f(x) = 8x^3$ and $g(x) = x^{1/3}$.
To find $g \circ f(x)$:
$g \circ f(x) = g(f(x))$
$= g(8x^3)$
$= (8x^3)^{1/3}$
$= (2^3 \cdot x^3)^{1/3}$
$= 2x$.
To find $f \circ g(x)$:
$f \circ g(x) = f(g(x))$
$= f(x^{1/3})$
$= 8(x^{1/3})^3$
$= 8x$.

(D) Given $f(x) = (3 - x^{3})^{\frac{1}{3}}$.
To find $fof(x)$,we compute $f(f(x))$:
$fof(x) = f(f(x)) = f((3 - x^{3})^{\frac{1}{3}})$
Substitute $f(x)$ into the function definition:
$fof(x) = [3 - ((3 - x^{3})^{\frac{1}{3}})^{3}]^{\frac{1}{3}}$
Simplify the expression inside the bracket:
$fof(x) = [3 - (3 - x^{3})]^{\frac{1}{3}}$
$fof(x) = [3 - 3 + x^{3}]^{\frac{1}{3}}$
$fof(x) = (x^{3})^{\frac{1}{3}}$
Thus,$fof(x) = x$.
The correct answer is $D$.

(B) First,we find the composite function $(f \circ g)(x) = f(g(x))$.
Given $g(x) < 0$ when $x < 0$ (since $x^3 < 0$ for $x < 0$),and $g(x) \geq 0$ when $x \geq 0$ (since $x^3 \geq 0$ for $x \in [0, 1)$ and $3x-2 \geq 1$ for $x \geq 1$).
Thus,$(f \circ g)(x) = \begin{cases} g(x)+2, & g(x) < 0 \\ (g(x))^2, & g(x) \geq 0 \end{cases} = \begin{cases} x^3+2, & x < 0 \\ (x^3)^2, & 0 \leq x < 1 \\ (3x-2)^2, & x \geq 1 \end{cases} = \begin{cases} x^3+2, & x < 0 \\ x^6, & 0 \leq x < 1 \\ (3x-2)^2, & x \geq 1 \end{cases}$.
Now,check for continuity and differentiability at the transition points $x=0$ and $x=1$.
At $x=0$: $\lim_{x \to 0^-} (x^3+2) = 2$ and $\lim_{x \to 0^+} (x^6) = 0$. Since $2 \neq 0$,the function is discontinuous at $x=0$,hence not differentiable at $x=0$.
At $x=1$: $\lim_{x \to 1^-} (x^6) = 1$ and $\lim_{x \to 1^+} (3x-2)^2 = (3(1)-2)^2 = 1$. The function is continuous at $x=1$.
Check derivatives at $x=1$: $LHD = \frac{d}{dx}(x^6)|_{x=1} = 6(1)^5 = 6$. $RHD = \frac{d}{dx}(3x-2)^2|_{x=1} = 2(3x-2) \cdot 3|_{x=1} = 6(3-2) = 6$.
Since $LHD = RHD$,the function is differentiable at $x=1$.
Therefore,the only point where $(f \circ g)(x)$ is not differentiable is $x=0$. The number of such points is $1$.

Given $f(x) = x^{2} - 3x + 2$.
To find $f(f(x))$,we substitute $f(x)$ into the function $f$:
$f(f(x)) = f(x^{2} - 3x + 2)$
$= (x^{2} - 3x + 2)^{2} - 3(x^{2} - 3x + 2) + 2$
$= (x^{4} + 9x^{2} + 4 - 6x^{3} + 4x^{2} - 12x) - 3x^{2} + 9x - 6 + 2$
$= x^{4} - 6x^{3} + (9x^{2} + 4x^{2} - 3x^{2}) + (-12x + 9x) + (4 - 6 + 2)$
$= x^{4} - 6x^{3} + 10x^{2} - 3x$

(N/A) Define $f: N \rightarrow Z$ as $f(x) = x$ and $g: Z \rightarrow Z$ as $g(x) = |x|$.
First,we show that $g$ is not injective.
We observe that $g(-1) = |-1| = 1$ and $g(1) = |1| = 1$.
Since $g(-1) = g(1)$ but $-1 \neq 1$,$g$ is not injective.
Now,$g \circ f: N \rightarrow Z$ is defined as $(g \circ f)(x) = g(f(x)) = g(x) = |x|$.
Let $x, y \in N$ such that $(g \circ f)(x) = (g \circ f)(y)$.
This implies $|x| = |y|$.
Since $x, y \in N$,both $x$ and $y$ are positive.
Therefore,$|x| = |y| \Rightarrow x = y$.
Hence,$g \circ f$ is injective.

(N/A) Define $f: N \rightarrow N$ by $f(x) = x + 1$.
Define $g: N \rightarrow N$ by $g(x) = \begin{cases} x - 1, & \text{if } x > 1 \\ 1, & \text{if } x = 1 \end{cases}$.
First,we show that $f$ is not onto. The range of $f$ is $\{2, 3, 4, \dots\}$,which is a proper subset of the codomain $N$. Specifically,the element $1 \in N$ has no preimage in the domain $N$ such that $f(x) = 1$. Thus,$f$ is not onto.
Now,consider the composition $g \circ f: N \rightarrow N$ defined by $(g \circ f)(x) = g(f(x))$.
Since $f(x) = x + 1$,we have $(g \circ f)(x) = g(x + 1)$.
Since $x \in N$,$x \geq 1$,so $x + 1 \geq 2$. Thus,$x + 1 > 1$.
Using the definition of $g$,we get $g(x + 1) = (x + 1) - 1 = x$.
Therefore,$(g \circ f)(x) = x$ for all $x \in N$.
Since for every $y \in N$,there exists $x = y \in N$ such that $(g \circ f)(x) = y$,the function $g \circ f$ is onto.

(N/A) It is given that $f: R \rightarrow R$ is the Signum function and $g: R \rightarrow R$ is the Greatest Integer function $g(x) = [x]$.
For any $x \in (0, 1]$,we analyze the compositions $fog(x)$ and $gof(x)$.
First,consider $fog(x) = f(g(x)) = f([x])$.
If $x = 1$,then $g(1) = [1] = 1$,so $f(g(1)) = f(1) = 1$.
If $x \in (0, 1)$,then $g(x) = [x] = 0$,so $f(g(x)) = f(0) = 0$.
Thus,$fog(x) = \begin{cases} 1, & x = 1 \\ 0, & x \in (0, 1) \end{cases}$.
Next,consider $gof(x) = g(f(x))$.
Since $x \in (0, 1]$,$x > 0$,therefore $f(x) = 1$ for all $x \in (0, 1]$.
Thus,$gof(x) = g(1) = [1] = 1$ for all $x \in (0, 1]$.
Comparing the two,for $x \in (0, 1)$,$fog(x) = 0$ while $gof(x) = 1$.
Therefore,$fog$ and $gof$ do not coincide in $(0, 1]$.

(B) The composite function $(g \circ f)(x)$ is defined as:
$g(f(x)) = \begin{cases} f(x)+1, & f(x) < 0 \\ (f(x)-1)^2+b, & f(x) \geq 0 \end{cases}$
Substituting $f(x)$ into the definition:
For $x < 0$,$f(x) = x+a$. Thus,$g(f(x)) = \begin{cases} x+a+1, & x+a < 0 \\ (x+a-1)^2+b, & x+a \geq 0 \end{cases}$
For $x \geq 0$,$f(x) = |x-1|$. Thus,$g(f(x)) = \begin{cases} |x-1|+1, & |x-1| < 0 \\ (|x-1|-1)^2+b, & |x-1| \geq 0 \end{cases}$
Since $|x-1| \geq 0$ for all $x$,the case $|x-1| < 0$ is impossible. The function simplifies to:
$(g \circ f)(x) = \begin{cases} x+a+1, & x < -a \\ (x+a-1)^2+b, & -a \leq x < 0 \\ (|x-1|-1)^2+b, & x \geq 0 \end{cases}$
For continuity at $x = -a$: $\lim_{x \to -a^-} (x+a+1) = \lim_{x \to -a^+} ((x+a-1)^2+b) \implies 1 = (-1)^2 + b \implies b = 0$.
For continuity at $x = 0$: $\lim_{x \to 0^-} ((x+a-1)^2+b) = \lim_{x \to 0^+} ((|x-1|-1)^2+b) \implies (a-1)^2 + b = (|0-1|-1)^2 + b \implies (a-1)^2 = 0 \implies a = 1$.
Thus,$a+b = 1+0 = 1$.

(C) Given $f(x) = 2x - 1$ and $g(x) = \frac{x - 1/2}{x - 1} = \frac{2x - 1}{2(x - 1)}$.
Calculate the composition $f(g(x))$:
$f(g(x)) = 2(g(x)) - 1 = 2 \left( \frac{2x - 1}{2(x - 1)} \right) - 1$
$= \frac{2x - 1}{x - 1} - 1 = \frac{2x - 1 - (x - 1)}{x - 1} = \frac{x}{x - 1} = 1 + \frac{1}{x - 1}$.
For one-one check:
Let $f(g(x_1)) = f(g(x_2))$.
$1 + \frac{1}{x_1 - 1} = 1 + \frac{1}{x_2 - 1} \implies x_1 - 1 = x_2 - 1 \implies x_1 = x_2$.
Thus,$f(g(x))$ is one-one.
For onto check:
The range of $f(g(x)) = 1 + \frac{1}{x - 1}$ is $R - \{1\}$.
Since the codomain is $R$,the range is not equal to the codomain.
Therefore,$f(g(x))$ is not onto.
Conclusion: $f(g(x))$ is one-one but not onto.

(C) The domain of $(f \circ g)(x) = \sin^{-1}(g(x))$ requires $|g(x)| \leq 1$.
First,we find $g(2) = \lim_{x \to 2} \frac{(x+1)(x-2)}{(2x+3)(x-2)} = \lim_{x \to 2} \frac{x+1}{2x+3} = \frac{3}{7}$.
Thus,we solve $|\frac{x+1}{2x+3}| \leq 1$ for $x \neq 2$.
This implies $-1 \leq \frac{x+1}{2x+3} \leq 1$.
Case $1$: $\frac{x+1}{2x+3} \leq 1 \Rightarrow \frac{x+1 - (2x+3)}{2x+3} \leq 0 \Rightarrow \frac{-x-2}{2x+3} \leq 0 \Rightarrow \frac{x+2}{2x+3} \geq 0$.
The solution is $x \in (-\infty, -2] \cup (-\frac{3}{2}, \infty)$.
Case $2$: $\frac{x+1}{2x+3} \geq -1 \Rightarrow \frac{x+1 + 2x+3}{2x+3} \geq 0 \Rightarrow \frac{3x+4}{2x+3} \geq 0$.
The solution is $x \in (-\infty, -\frac{4}{3}] \cup (-\frac{3}{2}, \infty)$.
Taking the intersection of both cases and excluding $x=2$ (since $g(2)$ is defined as the limit),we get $x \in (-\infty, -2] \cup [-\frac{4}{3}, \infty)$.

(C) Given that $(g \circ f)^{-1}$ exists,the composite function $g \circ f: A \rightarrow C$ must be a bijection (both one-one and onto).
$1$. For $g \circ f$ to be one-one,$f$ must be one-one. If $f$ were not one-one,there would exist $x_1, x_2 \in A$ such that $f(x_1) = f(x_2)$,implying $g(f(x_1)) = g(f(x_2))$,which contradicts the one-one property of $g \circ f$.
$2$. For $g \circ f$ to be onto,$g$ must be onto. If $g$ were not onto,there would exist some $z \in C$ such that no $y \in B$ satisfies $g(y) = z$,meaning no $x \in A$ could satisfy $g(f(x)) = z$,contradicting the onto property of $g \circ f$.
Therefore,$f$ must be one-one and $g$ must be onto.

(C) Given $f(x) = \left[2\left(1 - \frac{x^{25}}{2}\right)\left(2 + x^{25}\right)\right]^{\frac{1}{50}}$.
Simplifying the expression inside the bracket:
$f(x) = \left[\left(2 - x^{25}\right)\left(2 + x^{25}\right)\right]^{\frac{1}{50}} = \left(4 - x^{50}\right)^{\frac{1}{50}}$.
Now,calculate $f(f(x))$:
$f(f(x)) = \left(4 - (f(x))^{50}\right)^{\frac{1}{50}} = \left(4 - (4 - x^{50})\right)^{\frac{1}{50}} = (x^{50})^{\frac{1}{50}} = x$.
Since $f(f(x)) = x$,it follows that $f(f(f(x))) = f(x)$.
Given $g(x) = f(f(f(x))) + f(f(x))$,we substitute the results:
$g(x) = f(x) + x$.
For $x = 1$:
$g(1) = f(1) + 1 = (4 - 1^{50})^{\frac{1}{50}} + 1 = 3^{\frac{1}{50}} + 1$.
Since $1 < 3^{\frac{1}{50}} < 2$ (as $1^{50} < 3 < 2^{50}$),we have $1 < 3^{\frac{1}{50}} + 1 < 2$.
Therefore,the greatest integer less than or equal to $g(1)$ is $\lfloor 3^{\frac{1}{50}} + 1 \rfloor = 2$.

(D) Given $f(x) = x - 1$ and $g(x) = \frac{x^2}{x^2 - 1}$.
$f(g(x)) = g(x) - 1 = \frac{x^2}{x^2 - 1} - 1 = \frac{x^2 - (x^2 - 1)}{x^2 - 1} = \frac{1}{x^2 - 1}$.
Let $h(x) = f(g(x)) = \frac{1}{x^2 - 1}$.
Since $h(x) = h(-x)$,the function is an even function,which implies it is many-one.
To find the range,let $y = \frac{1}{x^2 - 1}$.
Then $x^2 - 1 = \frac{1}{y} \Rightarrow x^2 = \frac{1}{y} + 1 = \frac{1 + y}{y}$.
For $x$ to be real,$x^2 \geq 0$,so $\frac{1 + y}{y} \geq 0$.
This inequality holds for $y \in (-\infty, -1] \cup (0, \infty)$.
Since the range is not equal to the co-domain $(R)$,the function is into.
Thus,$f \circ g$ is neither one-one nor onto.

(B) Given $f(x) = \frac{x-1}{x+1}$.
$f^2(x) = f(f(x)) = \frac{\frac{x-1}{x+1} - 1}{\frac{x-1}{x+1} + 1} = \frac{x-1-x-1}{x-1+x+1} = \frac{-2}{2x} = -\frac{1}{x}$.
$f^3(x) = f(f^2(x)) = f(-\frac{1}{x}) = \frac{-\frac{1}{x} - 1}{-\frac{1}{x} + 1} = \frac{-1-x}{-1+x} = \frac{x+1}{1-x}$.
$f^4(x) = f(f^3(x)) = f(\frac{x+1}{1-x}) = \frac{\frac{x+1}{1-x} - 1}{\frac{x+1}{1-x} + 1} = \frac{x+1-1+x}{x+1+1-x} = \frac{2x}{2} = x$.
Since $f^4(x) = x$,the function is periodic with period $4$.
$f^6(6) = f^2(6) = -\frac{1}{6}$.
$f^7(7) = f^3(7) = \frac{7+1}{1-7} = \frac{8}{-6} = -\frac{4}{3}$.
Therefore,$f^6(6) + f^7(7) = -\frac{1}{6} - \frac{4}{3} = \frac{-1-8}{6} = -\frac{9}{6} = -\frac{3}{2}$.

(D) Given $f(n) = \begin{cases} 2n, & n \in \{1, 2, 3, 4, 5\} \\ 2n - 11, & n \in \{6, 7, 8, 9, 10\} \end{cases}$.
We find $f^{-1}(n)$ by solving $f(x) = n$:
If $n \in \{2, 4, 6, 8, 10\}$,$2x = n \implies x = n/2$.
If $n \in \{1, 3, 5, 7, 9\}$,$2x - 11 = n \implies x = (n + 11)/2$.
Thus,$f^{-1}(n) = \begin{cases} n/2, & n \in \{2, 4, 6, 8, 10\} \\ (n + 11)/2, & n \in \{1, 3, 5, 7, 9\} \end{cases}$.
Given $f(g(n)) = \begin{cases} n + 1, & n \text{ is odd} \\ n - 1, & n \text{ is even} \end{cases}$,we have $g(n) = f^{-1}(f(g(n)))$.
For $n$ odd,$g(n) = f^{-1}(n + 1)$. Since $n+1$ is even,$g(n) = (n + 1)/2$.
For $n$ even,$g(n) = f^{-1}(n - 1)$. Since $n-1$ is odd,$g(n) = (n - 1 + 11)/2 = (n + 10)/2$.
Calculating values:
$g(1) = (1+1)/2 = 1$,$g(2) = (2+10)/2 = 6$,$g(3) = (3+1)/2 = 2$,$g(4) = (4+10)/2 = 7$,$g(5) = (5+1)/2 = 3$,$g(10) = (10+10)/2 = 10$.
Therefore,$g(10) \cdot (g(1) + g(2) + g(3) + g(4) + g(5)) = 10 \cdot (1 + 6 + 2 + 7 + 3) = 10 \cdot 19 = 190$.

(A) Given that $g(x)$ is a polynomial of degree $1$,let $g(x) = ax + b$.
Then $f(g(x)) = f(ax + b) = 8x^2 - 2x$. Since $f$ is of degree $2$,let $f(x) = px^2 + qx + r$.
Substituting $g(x)$ into $f(x)$,we get $p(ax+b)^2 + q(ax+b) + r = 8x^2 - 2x$.
Comparing the coefficients of $x^2$,we have $pa^2 = 8$. Since $g(f(x)) = 4x^2 + 6x + 1$,the leading coefficient of $g(f(x))$ is $a \cdot p = 4$.
Dividing $pa^2 = 8$ by $ap = 4$,we get $a = 2$. Then $p(2)^2 = 8 \implies 4p = 8 \implies p = 2$.
Now,$f(g(x)) = 2(2x+b)^2 + q(2x+b) + r = 2(4x^2 + 4bx + b^2) + 2qx + qb + r = 8x^2 + (8b + 2q)x + (2b^2 + qb + r) = 8x^2 - 2x$.
Comparing coefficients: $8b + 2q = -2$ and $2b^2 + qb + r = 0$.
From $g(f(x)) = a(px^2 + qx + r) + b = 2(2x^2 + qx + r) + b = 4x^2 + 2qx + 2r + b = 4x^2 + 6x + 1$.
Comparing coefficients: $2q = 6 \implies q = 3$. Then $8b + 2(3) = -2 \implies 8b = -8 \implies b = -1$.
Thus,$g(x) = 2x - 1$ and $f(x) = 2x^2 + 3x + r$. Using $2b^2 + qb + r = 0$,we get $2(-1)^2 + 3(-1) + r = 0 \implies 2 - 3 + r = 0 \implies r = 1$.
So,$f(x) = 2x^2 + 3x + 1$.
Now,$f(2) = 2(2)^2 + 3(2) + 1 = 8 + 6 + 1 = 15$ and $g(2) = 2(2) - 1 = 3$.
Therefore,$f(2) + g(2) = 15 + 3 = 18$.

(B) Given $f(x) = \frac{x+1}{x-1}$.
First,calculate $f^2(x) = f(f(x)) = \frac{\frac{x+1}{x-1} + 1}{\frac{x+1}{x-1} - 1} = \frac{x+1+x-1}{x+1-(x-1)} = \frac{2x}{2} = x$.
Since $f^2(x) = x$,we have $f^3(x) = f(f^2(x)) = f(x)$ and $f^4(x) = f^2(f^2(x)) = x$.
Now,calculate the terms:
$f^1(2) = \frac{2+1}{2-1} = 3$.
$f^2(3) = 3$ (since $f^2(x) = x$).
$f^3(4) = f(4) = \frac{4+1}{4-1} = \frac{5}{3}$.
$f^4(5) = 5$ (since $f^4(x) = x$).
Thus,$P = 3 \cdot 3 \cdot \frac{5}{3} \cdot 5 = 75$.
The multiples of $75$ are $75, 150, 225, 300, 375, \dots$.
Comparing with the given options,$375$ is a multiple of $75$.

(D) Given $f^1(x) = \frac{3x + 2}{2x + 3}$.
Calculate $f^2(x) = f^1(f^1(x)) = \frac{3(\frac{3x+2}{2x+3}) + 2}{2(\frac{3x+2}{2x+3}) + 3} = \frac{9x + 6 + 4x + 6}{6x + 4 + 6x + 9} = \frac{13x + 12}{12x + 13}$.
Calculate $f^3(x) = f^1(f^2(x)) = \frac{3(\frac{13x+12}{12x+13}) + 2}{2(\frac{13x+12}{12x+13}) + 3} = \frac{39x + 36 + 24x + 26}{26x + 24 + 36x + 39} = \frac{63x + 62}{62x + 63}$.
Observe the pattern: $f^n(x) = \frac{(2^{2n}-1)x + (2^{2n}-2)}{2(2^{2n}-2)x + (2^{2n}-1)}$ is not quite right,let's look at the coefficients: $a_n = 3, 13, 63, \dots$ where $a_n = \frac{3^n + 2^n}{2}$? No,let $f^n(x) = \frac{A_n x + B_n}{B_n x + A_n}$.
For $n=1, A_1=3, B_1=2$. For $n=2, A_2=13, B_2=12$. For $n=3, A_3=63, B_3=62$.
The recurrence is $A_n = 3A_{n-1} + 2B_{n-1}$ and $B_n = 2A_{n-1} + 3B_{n-1}$.
Adding these: $A_n + B_n = 5(A_{n-1} + B_{n-1})$.
Since $A_1 + B_1 = 3 + 2 = 5$,then $A_n + B_n = 5^n$.
For $n=5$,$A_5 + B_5 = 5^5 = 3125$.

(B) Let $f(x) = \frac{Ax + B}{Cx - A}$,where $A = \tan 1^{\circ}$,$B = \log_{e}(123)$,and $C = \log_{e}(1234)$.
First,we calculate $f(f(x))$:
$f(f(x)) = \frac{A(\frac{Ax + B}{Cx - A}) + B}{C(\frac{Ax + B}{Cx - A}) - A} = \frac{A(Ax + B) + B(Cx - A)}{C(Ax + B) - A(Cx - A)} = \frac{A^2x + AB + BCx - AB}{ACx + BC - ACx + A^2} = \frac{x(A^2 + BC)}{A^2 + BC} = x$.
Since $f(f(x)) = x$ for all $x$ in the domain,we have $f(f(x)) = x$ and $f(f(4/x)) = 4/x$.
Therefore,$f(f(x)) + f(f(4/x)) = x + \frac{4}{x}$.
Using the Arithmetic Mean-Geometric Mean inequality $(AM \geq GM)$,for $x > 0$:
$x + \frac{4}{x} \geq 2 \sqrt{x \cdot \frac{4}{x}} = 2 \sqrt{4} = 4$.
The least value is $4$.

(B) First, we simplify $f(x)$ as $f(x) = \begin{cases} x+1, & x < 0 \\ 1-x, & 0 \leq x < 1 \\ x-1, & x \geq 1 \end{cases}$.
Now, we find $(g \circ f)(x) = g(f(x))$.
If $f(x) < 0$, then $x+1 < 0 \implies x < -1$. In this case, $g(f(x)) = f(x) + 1 = (x+1) + 1 = x+2$.
If $f(x) \geq 0$, then $x \geq -1$. In this case, $g(f(x)) = 1$.
Thus, $(g \circ f)(x) = \begin{cases} x+2, & x < -1 \\ 1, & x \geq -1 \end{cases}$.
Checking continuity at $x = -1$: $\lim_{x \to -1^-} (x+2) = 1$ and $\lim_{x \to -1^+} (1) = 1$. Since the limits are equal to $g(f(-1)) = 1$, the function is continuous everywhere.
Checking differentiability at $x = -1$: The left-hand derivative is $\frac{d}{dx}(x+2) = 1$, and the right-hand derivative is $\frac{d}{dx}(1) = 0$. Since $1 \neq 0$, it is not differentiable at $x = -1$.

(C) Given $g(x) = \sqrt{x} + 1$.
We are given $(f \circ g)(x) = f(g(x)) = x + 3 - \sqrt{x}$.
Let $u = g(x) = \sqrt{x} + 1$.
Then $\sqrt{x} = u - 1$.
Substituting this into the expression for $(f \circ g)(x)$:
$f(u) = (u - 1)^2 + 3 - (u - 1)$.
$f(u) = (u^2 - 2u + 1) + 3 - u + 1$.
$f(u) = u^2 - 3u + 5$.
Thus,$f(x) = x^2 - 3x + 5$.
To find $f(0)$,substitute $x = 0$:
$f(0) = (0)^2 - 3(0) + 5 = 5$.

(A) The domain of $f \circ g$ consists of all $x$ in the domain of $g$ such that $g(x)$ is in the domain of $f$.
$1$. The domain of $g$ is $R - \{-\frac{5}{2}\}$.
$2$. For $f(g(x))$ to be defined,$g(x)$ must not be equal to $-\frac{1}{2}$ (the value excluded from the domain of $f$).
$3$. Set $g(x) = -\frac{1}{2}$:
$\frac{|x|+1}{2x+5} = -\frac{1}{2}$
$2(|x|+1) = -(2x+5)$
$2|x| + 2 = -2x - 5$
$2|x| = -2x - 7$
Case $I$: If $x \ge 0$,then $2x = -2x - 7 \Rightarrow 4x = -7 \Rightarrow x = -\frac{7}{4}$. Since $x \ge 0$,this is not a solution.
Case $II$: If $x < 0$,then $2(-x) = -2x - 7 \Rightarrow -2x = -2x - 7 \Rightarrow 0 = -7$,which is impossible.
Thus,$g(x)$ is never equal to $-\frac{1}{2}$.
Therefore,the only restriction on the domain of $f \circ g$ is $x \neq -\frac{5}{2}$.
The domain is $R - \{-\frac{5}{2}\}$.

(C) Given $f(x) = \begin{cases} 2+2x, & -1 \leq x < 0 \\ 1-\frac{x}{3}, & 0 \leq x \leq 3 \end{cases}$ and $g(x) = \begin{cases} -x, & -3 \leq x \leq 0 \\ x, & 0 < x \leq 1 \end{cases}$.
We need to find the range of $f(g(x))$.
Case $1$: $-3 \leq x \leq 0$,then $g(x) = -x$. Since $-3 \leq x \leq 0$,we have $0 \leq g(x) \leq 3$.
For $0 \leq g(x) \leq 3$,$f(g(x)) = 1 - \frac{g(x)}{3} = 1 - \frac{-x}{3} = 1 + \frac{x}{3}$.
As $x$ varies from $-3$ to $0$,$f(g(x))$ varies from $1 + \frac{-3}{3} = 0$ to $1 + \frac{0}{3} = 1$.
So,the range for this part is $[0, 1]$.
Case $2$: $0 < x \leq 1$,then $g(x) = x$. Since $0 < x \leq 1$,we have $0 < g(x) \leq 1$.
For $0 < g(x) \leq 3$,$f(g(x)) = 1 - \frac{g(x)}{3} = 1 - \frac{x}{3}$.
As $x$ varies from $0$ to $1$,$f(g(x))$ varies from $1 - \frac{0}{3} = 1$ to $1 - \frac{1}{3} = \frac{2}{3}$.
So,the range for this part is $[\frac{2}{3}, 1)$.
Combining both cases,the range of $f(g(x))$ is $[0, 1]$.

(D) Given $f(x) = \frac{4x+3}{6x-4}$.
First,we find $g(x) = (f \circ f)(x) = f(f(x))$.
$g(x) = f\left(\frac{4x+3}{6x-4}\right) = \frac{4\left(\frac{4x+3}{6x-4}\right) + 3}{6\left(\frac{4x+3}{6x-4}\right) - 4}$.
Multiplying the numerator and denominator by $(6x-4)$:
$g(x) = \frac{4(4x+3) + 3(6x-4)}{6(4x+3) - 4(6x-4)} = \frac{16x + 12 + 18x - 12}{24x + 18 - 24x + 16} = \frac{34x}{34} = x$.
Since $g(x) = x$ for all $x$ in the domain,$g$ is the identity function.
Therefore,$(g \circ g \circ g)(4) = g(g(g(4))) = g(g(4)) = g(4) = 4$.

(B) We are given $f(x) = \begin{cases} \ln x & x > 0 \\ e^{-x} & x \leq 0 \end{cases}$ and $g(x) = \begin{cases} x & x \geq 0 \\ e^x & x < 0 \end{cases}$.
To find $(gof)(x) = g(f(x))$,we analyze the cases for $f(x)$:
Case $1$: $x \leq 0$. Then $f(x) = e^{-x}$. Since $e^{-x} > 0$ for all $x$,$g(f(x)) = f(x) = e^{-x}$.
Case $2$: $x > 0$. Then $f(x) = \ln x$.
Subcase 2a: $0 < x < 1$. Then $f(x) = \ln x < 0$. Thus $g(f(x)) = e^{f(x)} = e^{\ln x} = x$.
Subcase 2b: $x \geq 1$. Then $f(x) = \ln x \geq 0$. Thus $g(f(x)) = f(x) = \ln x$.
Combining these,$(gof)(x) = \begin{cases} e^{-x} & x \leq 0 \\ x & 0 < x < 1 \\ \ln x & x \geq 1 \end{cases}$.
Checking one-one: For $x \leq 0$,$g(f(x)) = e^{-x} \in [1, \infty)$. For $0 < x < 1$,$g(f(x)) = x \in (0, 1)$. For $x \geq 1$,$g(f(x)) = \ln x \in [0, \infty)$.
Since the range of $g(f(x))$ is $[0, \infty)$,it is not onto (as the codomain is $R$).
Also,for $x \leq 0$,$g(f(x)) \geq 1$,and for $x \geq 1$,$g(f(x)) \geq 0$. There exist values in the range that are mapped to by multiple $x$,and the function is not injective. Thus,it is neither one-one nor onto.

(D) Let $f_n(x)$ denote the $n$-th composition of $f(x)$.
$f_1(x) = \frac{2x}{\sqrt{1+9x^2}}$
$f_2(x) = f(f(x)) = \frac{2 \cdot \frac{2x}{\sqrt{1+9x^2}}}{\sqrt{1+9 \cdot \frac{4x^2}{1+9x^2}}} = \frac{4x}{\sqrt{1+9x^2+36x^2}} = \frac{2^2x}{\sqrt{1+9x^2(1+2^2)}}$
$f_3(x) = f(f_2(x)) = \frac{2 \cdot \frac{2^2x}{\sqrt{1+9x^2(1+2^2)}}}{\sqrt{1+9 \cdot \frac{2^4x^2}{1+9x^2(1+2^2)}}} = \frac{2^3x}{\sqrt{1+9x^2(1+2^2+2^4)}}$
By induction,$f_n(x) = \frac{2^nx}{\sqrt{1+9x^2(1+2^2+2^4+\ldots+2^{2n-2})}}$.
For $n=10$,the denominator contains $9\alpha x^2$,where $\alpha = 1+2^2+2^4+\ldots+2^{18}$.
This is a geometric progression with $a=1$,$r=4$,and $n=10$ terms.
$\alpha = \frac{1(4^{10}-1)}{4-1} = \frac{2^{20}-1}{3}$.
Thus,$3\alpha + 1 = 3(\frac{2^{20}-1}{3}) + 1 = 2^{20} - 1 + 1 = 2^{20}$.
Therefore,$\sqrt{3\alpha+1} = \sqrt{2^{20}} = 2^{10} = 1024$.

(A) We are given $f(x) = |x-1|$ and $g(x) = \begin{cases} e^x, & x \geq 0 \\ x+1, & x \leq 0 \end{cases}$.
To find $f(g(x))$,we substitute $g(x)$ into $f(x)$:
$f(g(x)) = |g(x) - 1| = \begin{cases} |e^x - 1|, & x \geq 0 \\ |(x+1) - 1|, & x \leq 0 \end{cases} = \begin{cases} e^x - 1, & x \geq 0 \\ |x|, & x \leq 0 \end{cases} = \begin{cases} e^x - 1, & x \geq 0 \\ -x, & x \leq 0 \end{cases}$.
Now,let $h(x) = f(g(x))$.
For $x \geq 0$,$h(x) = e^x - 1$. As $x$ increases from $0$ to $\infty$,$h(x)$ increases from $0$ to $\infty$.
For $x \leq 0$,$h(x) = -x$. As $x$ decreases from $0$ to $-\infty$,$h(x)$ increases from $0$ to $\infty$.
Since $h(x)$ takes the same positive values for both positive and negative $x$ (e.g.,$h(1) = e-1$ and $h(-(e-1)) = e-1$),the function is not one-one.
Since the range of $h(x)$ is $[0, \infty)$,which is a proper subset of the codomain $R$,the function is not onto.
Therefore,the function is neither one-one nor onto.

(A) Given,$f(x) = x^2$ and $g(x) = \sin x$ for all $x \in R$.
First,calculate $(f \circ g \circ g \circ f)(x)$:
$(f \circ g \circ g \circ f)(x) = f(g(g(f(x)))) = f(g(g(x^2))) = f(g(\sin x^2)) = f(\sin(\sin x^2)) = (\sin(\sin x^2))^2$.
Next,calculate $(g \circ g \circ f)(x)$:
$(g \circ g \circ f)(x) = g(g(f(x))) = g(g(x^2)) = g(\sin x^2) = \sin(\sin x^2)$.
Equating the two expressions:
$(\sin(\sin x^2))^2 = \sin(\sin x^2)$.
Let $u = \sin(\sin x^2)$. Then $u^2 = u$,which implies $u^2 - u = 0$,so $u(u - 1) = 0$.
This gives $u = 0$ or $u = 1$.
Case $1$: $\sin(\sin x^2) = 0$.
This implies $\sin x^2 = n \pi$ for some integer $n$. Since the range of $\sin x^2$ is $[-1, 1]$,the only possible value for $n \pi$ is $0$ (when $n = 0$).
So,$\sin x^2 = 0$,which implies $x^2 = n \pi$ for $n \in \{0, 1, 2, \ldots\}$.
Thus,$x = \pm \sqrt{n \pi}$ for $n \in \{0, 1, 2, \ldots\}$.
Case $2$: $\sin(\sin x^2) = 1$.
This implies $\sin x^2 = \frac{\pi}{2}$. Since $\frac{\pi}{2} \approx 1.57 > 1$,this has no real solution for $x$.
Therefore,the set of all $x$ is $\pm \sqrt{n \pi}$ for $n \in \{0, 1, 2, \ldots\}$.