Composition of Functions Questions in English

(A) Given that $R: A \to B$ and $S: B \to C$.
By the definition of the composition of two relations,the composite relation $S \circ R$ is defined as the set of all ordered pairs $(a, c)$ such that there exists an element $b \in B$ where $(a, b) \in R$ and $(b, c) \in S$.
Therefore,the domain of $S \circ R$ is a subset of $A$ and the codomain is $C$.
Thus,$S \circ R$ is a relation from $A$ to $C$.

(C) Given $A = \{1, 2, 3, 4\}$ and $B = \{1, 3, 5\}$.
$R = \{(a, b) : a \in A, b \in B, a < b\} = \{(1, 3), (1, 5), (2, 3), (2, 5), (3, 5), (4, 5)\}$.
The inverse relation $R^{-1} = \{(b, a) : (a, b) \in R\} = \{(3, 1), (5, 1), (3, 2), (5, 2), (5, 3), (5, 4)\}$.
$R \circ R^{-1}$ is defined as $\{(x, z) : \exists y \text{ such that } (x, y) \in R^{-1} \text{ and } (y, z) \in R\}$.
For $(3, 1) \in R^{-1}$,we look for $y=1$ in $R$: $(1, 3) \in R, (1, 5) \in R \implies (3, 3), (3, 5) \in R \circ R^{-1}$.
For $(5, 1) \in R^{-1}$,we look for $y=1$ in $R$: $(1, 3) \in R, (1, 5) \in R \implies (5, 3), (5, 5) \in R \circ R^{-1}$.
For $(3, 2) \in R^{-1}$,we look for $y=2$ in $R$: $(2, 3) \in R, (2, 5) \in R \implies (3, 3), (3, 5) \in R \circ R^{-1}$.
For $(5, 2) \in R^{-1}$,we look for $y=2$ in $R$: $(2, 3) \in R, (2, 5) \in R \implies (5, 3), (5, 5) \in R \circ R^{-1}$.
For $(5, 3) \in R^{-1}$,we look for $y=3$ in $R$: $(3, 5) \in R \implies (5, 5) \in R \circ R^{-1}$.
For $(5, 4) \in R^{-1}$,we look for $y=4$ in $R$: $(4, 5) \in R \implies (5, 5) \in R \circ R^{-1}$.
Combining these,$R \circ R^{-1} = \{(3, 3), (3, 5), (5, 3), (5, 5)\}$.

(A) First,we find ${R^{ - 1}}$ by reversing the ordered pairs in $R$:
${R^{ - 1}} = \{(5, 4), (4, 1), (6, 4), (6, 7), (7, 3)\}$.
Now,we find the composition ${R^{ - 1}}oR$ by checking all pairs $(x, y) \in R$ and $(y, z) \in {R^{ - 1}}$ such that $(x, z) \in {R^{ - 1}}oR$:
$1$. $(4, 5) \in R$ and $(5, 4) \in {R^{ - 1}} \Rightarrow (4, 4) \in {R^{ - 1}}oR$.
$2$. $(1, 4) \in R$ and $(4, 1) \in {R^{ - 1}} \Rightarrow (1, 1) \in {R^{ - 1}}oR$.
$3$. $(4, 6) \in R$ and $(6, 4) \in {R^{ - 1}} \Rightarrow (4, 4) \in {R^{ - 1}}oR$.
$4$. $(4, 6) \in R$ and $(6, 7) \in {R^{ - 1}} \Rightarrow (4, 7) \in {R^{ - 1}}oR$.
$5$. $(7, 6) \in R$ and $(6, 4) \in {R^{ - 1}} \Rightarrow (7, 4) \in {R^{ - 1}}oR$.
$6$. $(7, 6) \in R$ and $(6, 7) \in {R^{ - 1}} \Rightarrow (7, 7) \in {R^{ - 1}}oR$.
$7$. $(3, 7) \in R$ and $(7, 3) \in {R^{ - 1}} \Rightarrow (3, 3) \in {R^{ - 1}}oR$.
Combining these,we get ${R^{ - 1}}oR = \{(1, 1), (4, 4), (4, 7), (7, 4), (7, 7), (3, 3)\}$.

(C) Given $f(x) = \frac{1 - x}{1 + x}$.
First,calculate $f(\cos 2\theta)$:
$f(\cos 2\theta) = \frac{1 - \cos 2\theta}{1 + \cos 2\theta} = \frac{2\sin^2 \theta}{2\cos^2 \theta} = \tan^2 \theta$.
Now,calculate $f[f(\cos 2\theta)] = f(\tan^2 \theta)$:
$f(\tan^2 \theta) = \frac{1 - \tan^2 \theta}{1 + \tan^2 \theta}$.
Using the trigonometric identity $\cos 2\theta = \frac{1 - \tan^2 \theta}{1 + \tan^2 \theta}$,we get:
$f[f(\cos 2\theta)] = \cos 2\theta$.

(A) Given $f(x) = \frac{x - 3}{x + 1}$.
First,calculate $f[f(x)]$:
$f[f(x)] = \frac{f(x) - 3}{f(x) + 1} = \frac{\frac{x - 3}{x + 1} - 3}{\frac{x - 3}{x + 1} + 1}$
$= \frac{(x - 3) - 3(x + 1)}{(x - 3) + (x + 1)} = \frac{x - 3 - 3x - 3}{2x - 2} = \frac{-2x - 6}{2x - 2} = \frac{-(x + 3)}{x - 1} = \frac{x + 3}{1 - x}$.
Now,calculate $f[f\{f(x)\}] = f\left(\frac{x + 3}{1 - x}\right)$:
$f\left(\frac{x + 3}{1 - x}\right) = \frac{\frac{x + 3}{1 - x} - 3}{\frac{x + 3}{1 - x} + 1}$
$= \frac{(x + 3) - 3(1 - x)}{(x + 3) + (1 - x)} = \frac{x + 3 - 3 + 3x}{x + 3 + 1 - x} = \frac{4x}{4} = x$.
Thus,$f[f\{f(x)\}] = x$.

(A) Given that $g \circ f: A \to C$ is a bijection (both injective and surjective).
$1$. For $g \circ f$ to be injective,$f$ must be injective. If $f(x_1) = f(x_2)$,then $g(f(x_1)) = g(f(x_2))$,which implies $x_1 = x_2$ since $g \circ f$ is injective.
$2$. For $g \circ f$ to be surjective,$g$ must be surjective. For every $z \in C$,there exists $x \in A$ such that $g(f(x)) = z$. This implies that for every $z \in C$,there exists $y = f(x) \in B$ such that $g(y) = z$,which is the definition of $g$ being surjective.
Therefore,$f$ must be injective and $g$ must be surjective.

(C) Given $f(x) = ax + b$ and $g(x) = cx + d$.
We are given the condition $f(g(x)) = g(f(x))$.
First,calculate $f(g(x)) = f(cx + d) = a(cx + d) + b = acx + ad + b$.
Next,calculate $g(f(x)) = g(ax + b) = c(ax + b) + d = cax + cb + d$.
Equating the two expressions: $acx + ad + b = cax + cb + d$.
Since $acx = cax$,we can cancel these terms from both sides:
$ad + b = cb + d$.
Rearranging the terms,we get $ad + b = cb + d$.
Note that $f(d) = ad + b$ and $g(b) = cb + d$.
Therefore,the condition $f(g(x)) = g(f(x))$ is equivalent to $f(d) = g(b)$.

(A) Given the functions $f(x) = [x]$ (greatest integer function) and $g(x) = |x|$ (modulus function).
We need to evaluate $(gof)\left( -\frac{5}{3} \right) - (fog)\left( -\frac{5}{3} \right)$.
First,calculate $(gof)\left( -\frac{5}{3} \right) = g\left( f\left( -\frac{5}{3} \right) \right)$.
Since $f(x) = [x]$,$f\left( -\frac{5}{3} \right) = [-1.666...] = -2$.
Then $g(-2) = |-2| = 2$.
Next,calculate $(fog)\left( -\frac{5}{3} \right) = f\left( g\left( -\frac{5}{3} \right) \right)$.
Since $g(x) = |x|$,$g\left( -\frac{5}{3} \right) = \left| -\frac{5}{3} \right| = \frac{5}{3}$.
Then $f\left( \frac{5}{3} \right) = [1.666...] = 1$.
Finally,$(gof)\left( -\frac{5}{3} \right) - (fog)\left( -\frac{5}{3} \right) = 2 - 1 = 1$.

(C) The composition of functions $(gof)(x)$ is defined as $g(f(x))$.
Given $f(x) = x^2 - 1$ and $g(x) = 3x + 1$.
Substitute $f(x)$ into $g(x)$:
$(gof)(x) = g(x^2 - 1)$
$= 3(x^2 - 1) + 1$
$= 3x^2 - 3 + 1$
$= 3x^2 - 2$.

(B) Let the exponential function be $f(x) = e^x$ and the logarithmic function be $g(x) = \log_e x$.
By the definition of composition of functions,$(fog)(x) = f(g(x))$.
We need to evaluate $(fog)(1) = f(g(1))$.
Since $g(x) = \log_e x$,we have $g(1) = \log_e 1 = 0$.
Now,$f(g(1)) = f(0) = e^0 = 1$.
Since $\log_e e = 1$,the value is $\log_e e$.

(D) Given $f(x) = e^{2x}$ and $g(x) = \log \sqrt{x}$.
To find $fog(x)$,we compute $f(g(x))$.
$fog(x) = f(g(x)) = e^{2g(x)}$.
Substitute $g(x) = \log \sqrt{x}$ into the expression:
$fog(x) = e^{2 \log \sqrt{x}}$.
Using the logarithmic property $a \log b = \log(b^a)$:
$fog(x) = e^{\log(\sqrt{x})^2}$.
Since $(\sqrt{x})^2 = x$ for $x > 0$:
$fog(x) = e^{\log x}$.
Using the property $e^{\log_e x} = x$:
$fog(x) = x$.

(C) The composition of functions $gof(x)$ is defined as $g(f(x))$.
Given $f(x) = |\cos x|$ and $g(x) = [x]$.
Substituting $f(x)$ into $g(x)$,we get $g(f(x)) = g(|\cos x|)$.
Since $g(x) = [x]$,we replace $x$ with $|\cos x|$ to obtain $[|\cos x|]$.
Therefore,$gof(x) = [|\cos x|]$.

(C) Given the function $f(x) = x^2 + 1$.
To find $fof(x)$,we calculate $f(f(x))$.
$fof(x) = f(f(x)) = f(x^2 + 1)$.
Substituting $(x^2 + 1)$ into the function $f(x)$,we get:
$f(x^2 + 1) = (x^2 + 1)^2 + 1$.
Expanding the square using the identity $(a + b)^2 = a^2 + 2ab + b^2$:
$(x^2 + 1)^2 + 1 = (x^4 + 2x^2 + 1) + 1$.
Simplifying the expression:
$x^4 + 2x^2 + 2$.
Thus,$fof(x) = x^4 + 2x^2 + 2$.

(B) Given $f(x) = \frac{x}{\sqrt{1 + x^2}}$.
First,calculate $(fof)(x) = f(f(x)) = f\left(\frac{x}{\sqrt{1 + x^2}}\right)$.
$= \frac{\frac{x}{\sqrt{1 + x^2}}}{\sqrt{1 + \left(\frac{x}{\sqrt{1 + x^2}}\right)^2}} = \frac{\frac{x}{\sqrt{1 + x^2}}}{\sqrt{1 + \frac{x^2}{1 + x^2}}} = \frac{\frac{x}{\sqrt{1 + x^2}}}{\sqrt{\frac{1 + x^2 + x^2}{1 + x^2}}} = \frac{x}{\sqrt{1 + 2x^2}}$.
Now,calculate $(fofof)(x) = f((fof)(x)) = f\left(\frac{x}{\sqrt{1 + 2x^2}}\right)$.
$= \frac{\frac{x}{\sqrt{1 + 2x^2}}}{\sqrt{1 + \left(\frac{x}{\sqrt{1 + 2x^2}}\right)^2}} = \frac{\frac{x}{\sqrt{1 + 2x^2}}}{\sqrt{1 + \frac{x^2}{1 + 2x^2}}} = \frac{\frac{x}{\sqrt{1 + 2x^2}}}{\sqrt{\frac{1 + 2x^2 + x^2}{1 + 2x^2}}} = \frac{x}{\sqrt{1 + 3x^2}}$.

(C) Given functions are $\phi(x) = x^2 + 1$ and $\psi(x) = 3^x$.
First,we calculate the composition $\phi \{ \psi(x) \}$:
$\phi \{ \psi(x) \} = \phi(3^x) = (3^x)^2 + 1 = 3^{2x} + 1$.
Next,we calculate the composition $\psi \{ \phi(x) \}$:
$\psi \{ \phi(x) \} = \psi(x^2 + 1) = 3^{x^2 + 1}$.
Thus,the required values are $3^{2x} + 1$ and $3^{x^2 + 1}$.

(A) Given $\frac{1}{2}g(f(x)) = 2x^2 - 5x + 2$,we have $g(f(x)) = 4x^2 - 10x + 4$.
Since $g(x) = x^2 + x - 2$,substituting $f(x)$ into $g(x)$ gives:
$(f(x))^2 + f(x) - 2 = 4x^2 - 10x + 4$.
Rearranging the terms,we get $(f(x))^2 + f(x) - (4x^2 - 10x + 6) = 0$.
This is a quadratic equation in $f(x)$. Using the quadratic formula $f(x) = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:
$f(x) = \frac{-1 \pm \sqrt{1^2 - 4(1)(-(4x^2 - 10x + 6))}}{2(1)}$
$f(x) = \frac{-1 \pm \sqrt{1 + 16x^2 - 40x + 24}}{2}$
$f(x) = \frac{-1 \pm \sqrt{16x^2 - 40x + 25}}{2}$
$f(x) = \frac{-1 \pm \sqrt{(4x - 5)^2}}{2}$
$f(x) = \frac{-1 \pm (4x - 5)}{2}$.
Taking the positive root,$f(x) = \frac{-1 + 4x - 5}{2} = \frac{4x - 6}{2} = 2x - 3$.

(A) Given $f(x) = \log_a x$ and $F(x) = a^x$.
To find $F[f(x)]$,we substitute $f(x)$ into $F(x)$.
$F[f(x)] = F(\log_a x) = a^{\log_a x}$.
Using the logarithmic identity $a^{\log_a x} = x$,we get $F[f(x)] = x$.

(A) Given functions are $f(x) = \frac{x}{x + 1}$ and $g(x) = \frac{x}{1 - x}$.
To find $(fog)(x)$,we calculate $f(g(x))$.
$(fog)(x) = f\left( \frac{x}{1 - x} \right)$
Substitute $g(x)$ into $f(x)$:
$= \frac{\frac{x}{1 - x}}{\frac{x}{1 - x} + 1}$
$= \frac{\frac{x}{1 - x}}{\frac{x + (1 - x)}{1 - x}}$
$= \frac{x}{x + 1 - x}$
$= \frac{x}{1} = x$.
Thus,$(fog)(x) = x$.

(A) Given functions are $f(x) = (x + 1)^2$ and $g(x) = x^2 + 1$.
We need to find the value of $(fog)(-3)$.
By definition of composition of functions,$(fog)(x) = f(g(x))$.
First,calculate $g(-3)$:
$g(-3) = (-3)^2 + 1 = 9 + 1 = 10$.
Now,substitute this value into $f(x)$:
$(fog)(-3) = f(g(-3)) = f(10)$.
Since $f(x) = (x + 1)^2$,we have:
$f(10) = (10 + 1)^2 = 11^2 = 121$.
Therefore,$(fog)(-3) = 121$.

(B) Given that $g(x) = 1 + \sqrt{x}$ and $f(g(x)) = 3 + 2\sqrt{x} + x$.
We can rewrite the expression for $f(g(x))$ as follows:
$f(g(x)) = 1 + 2 + 2\sqrt{x} + x = 2 + (1 + 2\sqrt{x} + x) = 2 + (1 + \sqrt{x})^2$.
Since $g(x) = 1 + \sqrt{x}$,we substitute $g(x)$ into the expression:
$f(g(x)) = 2 + (g(x))^2$.
Therefore,replacing $g(x)$ with $x$,we get $f(x) = 2 + x^2$.

(C) Given the functions $f:R \to R$ where $f(x) = \sin x$ and $g:R \to R$ where $g(x) = x^2$.
By the definition of composite mapping,$(fog)(x) = f(g(x))$.
Substituting the value of $g(x)$,we get $(fog)(x) = f(x^2)$.
Since $f(x) = \sin x$,substituting $x^2$ into $f$ gives $f(x^2) = \sin(x^2)$.
Therefore,the composite mapping $(fog)(x) = \sin(x^2)$.

(B) Given $f(x) = ax + b$ and $g(x) = cx + d$.
We are given $a = 1$ and $b = 2$,so $f(x) = x + 2$.
Now,calculate the composition $(fog)(x) = f(g(x)) = f(cx + d) = (cx + d) + 2 = cx + d + 2$.
Next,calculate the composition $(gof)(x) = g(f(x)) = g(x + 2) = c(x + 2) + d = cx + 2c + d$.
Since $(fog)(x) = (gof)(x)$ for all $x$,we equate the two expressions:
$cx + d + 2 = cx + 2c + d$.
Subtracting $cx$ and $d$ from both sides,we get $2 = 2c$,which implies $c = 1$.
Since $d$ cancels out from both sides,$d$ can be any real number (arbitrary).
Thus,$c = 1$ and $d$ is arbitrary.

(D) Given $f(x) = \frac{\alpha x}{x + 1}$.
We need to find $\alpha$ such that $f(f(x)) = x$.
First,calculate $f(f(x)) = \frac{\alpha f(x)}{f(x) + 1}$.
Substitute $f(x) = \frac{\alpha x}{x + 1}$ into the expression:
$f(f(x)) = \frac{\alpha \left( \frac{\alpha x}{x + 1} \right)}{\frac{\alpha x}{x + 1} + 1} = \frac{\frac{\alpha^2 x}{x + 1}}{\frac{\alpha x + x + 1}{x + 1}} = \frac{\alpha^2 x}{(\alpha + 1)x + 1}$.
We are given $f(f(x)) = x$,so:
$\frac{\alpha^2 x}{(\alpha + 1)x + 1} = x$.
Assuming $x \neq 0$,we have $\frac{\alpha^2}{(\alpha + 1)x + 1} = 1$.
$\alpha^2 = (\alpha + 1)x + 1$.
For this to hold for all $x$ in the domain,the coefficients of $x$ must be zero and the constant terms must be equal:
$\alpha + 1 = 0 \implies \alpha = -1$.
Checking the constant term: $\alpha^2 = 1$. If $\alpha = -1$,then $(-1)^2 = 1$,which is true.
Thus,the value of $\alpha$ is $-1$.

(D) Given $f(x) = \frac{2x + 1}{3x - 2}$.
First,calculate $f(2)$:
$f(2) = \frac{2(2) + 1}{3(2) - 2} = \frac{4 + 1}{6 - 2} = \frac{5}{4}$.
Now,calculate $(fof)(2) = f(f(2)) = f\left(\frac{5}{4}\right)$:
$f\left(\frac{5}{4}\right) = \frac{2\left(\frac{5}{4}\right) + 1}{3\left(\frac{5}{4}\right) - 2} = \frac{\frac{5}{2} + 1}{\frac{15}{4} - 2} = \frac{\frac{7}{2}}{\frac{15 - 8}{4}} = \frac{\frac{7}{2}}{\frac{7}{4}} = \frac{7}{2} \times \frac{4}{7} = 2$.
Thus,$(fof)(2) = 2$.

(B) Given that $f(x) = \sin^2 x$ and the composite function $g(f(x)) = |\sin x|$.
Substitute $f(x)$ into the composite function expression:
$g(\sin^2 x) = |\sin x|$.
Let $t = \sin^2 x$. Since $\sin^2 x \ge 0$,we have $t \ge 0$.
Then $|\sin x| = \sqrt{\sin^2 x} = \sqrt{t}$.
Therefore,replacing $t$ with $x$,we get $g(x) = \sqrt{x}$ for $x \ge 0$.

(C) Given $f(x) = (a - x^n)^{1/n}$.
To find $f[f(x)]$,we substitute $f(x)$ into the function $f$:
$f[f(x)] = (a - [f(x)]^n)^{1/n}$
Substitute the expression for $f(x)$:
$f[f(x)] = (a - [(a - x^n)^{1/n}]^n)^{1/n}$
Simplify the inner term:
$f[f(x)] = (a - (a - x^n))^{1/n}$
$f[f(x)] = (a - a + x^n)^{1/n}$
$f[f(x)] = (x^n)^{1/n}$
$f[f(x)] = x$.

(D) Given $f(x) = \frac{\alpha x}{x + 1}$.
We need to find $f(f(x))$:
$f(f(x)) = f\left( \frac{\alpha x}{x + 1} \right) = \frac{\alpha \left( \frac{\alpha x}{x + 1} \right)}{\frac{\alpha x}{x + 1} + 1}$
Simplify the expression:
$f(f(x)) = \frac{\frac{\alpha^2 x}{x + 1}}{\frac{\alpha x + x + 1}{x + 1}} = \frac{\alpha^2 x}{\alpha x + x + 1}$
Given $f(f(x)) = x$,so:
$\frac{\alpha^2 x}{(\alpha + 1)x + 1} = x$
For this to hold for all $x$ (where $f$ is defined),we equate the coefficients or solve for $x \neq 0$:
$\alpha^2 x = x((\alpha + 1)x + 1)$
$\alpha^2 x = (\alpha + 1)x^2 + x$
For this to be an identity $f(f(x)) = x$,the coefficient of $x^2$ must be $0$ and the coefficient of $x$ must be $1$:
$1) \alpha + 1 = 0 \implies \alpha = -1$
$2) \alpha^2 = 1 \implies \alpha = \pm 1$
Checking $\alpha = -1$:
If $\alpha = -1$,then $f(f(x)) = \frac{(-1)^2 x}{(-1 + 1)x + 1} = \frac{x}{1} = x$.
Thus,the correct value is $\alpha = -1$.

(B) Given $g(x) = 1 + x - [x]$.
We know that the fractional part function is defined as $\{x\} = x - [x]$.
Therefore,$g(x) = 1 + \{x\}$.
Since $0 \le \{x\} < 1$,it follows that $1 \le 1 + \{x\} < 2$.
Thus,$1 \le g(x) < 2$.
Now,we evaluate $f(g(x))$. Since $g(x) > 0$ for all real values of $x$,we look at the definition of $f(x)$ for $x > 0$.
According to the definition,$f(x) = 1$ for all $x > 0$.
Since $g(x)$ is always in the interval $[1, 2)$,which is strictly greater than $0$,$f(g(x)) = 1$ for all $x$.

(A) To find $\lim_{x \to 0} g(f(x))$,we examine the behavior of $f(x)$ as $x \to 0$.
As $x \to 0$ and $x \neq 0$,$f(x) = \sin x$.
As $x \to 0$,$f(x) = \sin x \to 0$.
Let $u = f(x)$. As $x \to 0$,$u \to 0$.
In the limit $\lim_{u \to 0} g(u)$,we consider values of $u$ near $0$ but $u \neq 0$.
For $u \neq 0$ and $u \neq 2$,$g(u) = u^2 + 1$.
Therefore,$\lim_{u \to 0} g(u) = \lim_{u \to 0} (u^2 + 1) = 0^2 + 1 = 1$.
Thus,$\lim_{x \to 0} g(f(x)) = 1$.

(D) To find $\lim_{x \to 0} g(f(x))$,we examine the left-hand and right-hand limits.
For $x \to 0$,$x \ne 0$,so $f(x) = \sin x$.
As $x \to 0$,$f(x) = \sin x \to 0$.
Let $u = f(x)$. As $x \to 0$,$u \to 0$ through values close to $0$ but not equal to $0$ (since $\sin x \ne 0$ for $x$ near $0$ except at $x=0$).
Thus,$\lim_{x \to 0} g(f(x)) = \lim_{u \to 0} g(u)$.
Since $u \to 0$ but $u \ne 0$,we use the definition $g(u) = u^2 + 1$.
Therefore,$\lim_{u \to 0} (u^2 + 1) = 0^2 + 1 = 1$.

(D) Given $f(x) = |x|$ and $g(x) = |x|$ for all $x \in R$.
We need to find the set of $x$ such that $g(f(x)) \le f(g(x))$.
Substitute the functions: $g(|x|) \le f(|x|)$.
Since $g(x) = |x|$,$g(|x|) = ||x|| = |x|$.
Since $f(x) = |x|$,$f(|x|) = ||x|| = |x|$.
The inequality becomes $|x| \le |x|$.
This inequality holds true for all real numbers $x \in R$ because any real number is less than or equal to itself.
Therefore,the set is $R$.

(B) The domain of $(g \circ f)(x)$ is the set of all $x$ in the domain of $f$ such that $f(x)$ is in the domain of $g$.
First,find the domain of $g(x) = \sqrt{3 + 4x - 4x^2}$.
For $g(x)$ to be defined,$3 + 4x - 4x^2 \ge 0$,which implies $4x^2 - 4x - 3 \le 0$.
Factoring the quadratic: $(2x - 3)(2x + 1) \le 0$.
This inequality holds for $x \in \left[ -\frac{1}{2}, \frac{3}{2} \right]$.
Now,we require $f(x) \in \left[ -\frac{1}{2}, \frac{3}{2} \right]$,so $-\frac{1}{2} \le \frac{1}{2} - \tan \left( \frac{\pi x}{2} \right) \le \frac{3}{2}$.
Subtracting $\frac{1}{2}$ from all parts: $-1 \le -\tan \left( \frac{\pi x}{2} \right) \le 1$.
Multiplying by $-1$ (reversing the inequality): $-1 \le \tan \left( \frac{\pi x}{2} \right) \le 1$.
Applying the inverse tangent function: $-\frac{\pi}{4} \le \frac{\pi x}{2} \le \frac{\pi}{4}$.
Dividing by $\frac{\pi}{2}$: $-\frac{1}{2} \le x \le \frac{1}{2}$.
Since this interval is within the given domain of $f$ $(-1 < x < 1)$,the domain of $(g \circ f)$ is $\left[ -\frac{1}{2}, \frac{1}{2} \right]$.

(D) Given $f(x) = \sin^2 x + \sin^2(x + \frac{\pi}{3}) + \cos x \cos(x + \frac{\pi}{3})$.
Using the identity $\sin^2 \theta = \frac{1 - \cos 2\theta}{2}$,we get:
$f(x) = \frac{1 - \cos 2x}{2} + \frac{1 - \cos(2x + \frac{2\pi}{3})}{2} + \cos x \cos(x + \frac{\pi}{3})$
$f(x) = 1 - \frac{1}{2} [\cos 2x + \cos(2x + \frac{2\pi}{3})] + \cos x \cos(x + \frac{\pi}{3})$
Using $\cos A + \cos B = 2 \cos(\frac{A+B}{2}) \cos(\frac{A-B}{2})$:
$f(x) = 1 - \frac{1}{2} [2 \cos(2x + \frac{\pi}{3}) \cos(-\frac{\pi}{3})] + \cos x \cos(x + \frac{\pi}{3})$
Since $\cos(\frac{\pi}{3}) = \frac{1}{2}$,we have:
$f(x) = 1 - \frac{1}{2} \cos(2x + \frac{\pi}{3}) + \cos x \cos(x + \frac{\pi}{3})$
Using $2 \cos A \cos B = \cos(A+B) + \cos(A-B)$:
$f(x) = 1 - \frac{1}{2} \cos(2x + \frac{\pi}{3}) + \frac{1}{2} [\cos(2x + \frac{\pi}{3}) + \cos(-\frac{\pi}{3})]$
$f(x) = 1 + \frac{1}{2} \cos(\frac{\pi}{3}) = 1 + \frac{1}{2} \cdot \frac{1}{2} = 1 + \frac{1}{4} = \frac{5}{4}$.
Since $f(x) = \frac{5}{4}$ for all $x \in R$,then $(g \circ f)(x) = g(f(x)) = g(\frac{5}{4})$.
Given $g(\frac{5}{4}) = 1$,therefore $(g \circ f)(x) = 1$.

(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 $g(x) = \sqrt{x}$.
Step $1$: Calculate $g(f(x))$.
$g(f(x)) = g(\sin^2 x) = \sqrt{\sin^2 x} = |\sin x|$. This matches the given condition.
Step $2$: Calculate $f(g(x))$.
$f(g(x)) = f(\sqrt{x}) = (\sin \sqrt{x})^2$. This also matches the given condition.
Therefore,the correct option is $A$.

(A) Given $f(x) = 3x + 10$ and $g(x) = x^2 - 1$.
First,find the composition $(fog)(x) = f(g(x))$.
$(fog)(x) = 3(g(x)) + 10 = 3(x^2 - 1) + 10 = 3x^2 - 3 + 10 = 3x^2 + 7$.
Let $y = (fog)(x) = 3x^2 + 7$.
To find the inverse $(fog)^{-1}(x)$,we solve for $x$ in terms of $y$:
$y = 3x^2 + 7$
$y - 7 = 3x^2$
$x^2 = \frac{y - 7}{3}$
$x = (\frac{y - 7}{3})^{1/2}$.
Replacing $y$ with $x$,we get the inverse function:
$(fog)^{-1}(x) = (\frac{x - 7}{3})^{1/2}$.

(C) Given $f(x) = \frac{1}{1 - x}$.
First,find $f\{ f(x)\} = f\left( \frac{1}{1 - x} \right) = \frac{1}{1 - \frac{1}{1 - x}} = \frac{1}{\frac{1 - x - 1}{1 - x}} = \frac{1 - x}{-x} = \frac{x - 1}{x}$.
Next,find $f[f\{ f(x)\} ] = f\left( \frac{x - 1}{x} \right) = \frac{1}{1 - \frac{x - 1}{x}} = \frac{1}{\frac{x - (x - 1)}{x}} = \frac{1}{\frac{1}{x}} = x$.
Therefore,the derivative of $f[f\{ f(x)\} ]$ with respect to $x$ is $\frac{d}{dx}(x) = 1$.

(A) Let $x_1, x_2$ be in the domain such that $x_1 < x_2$.
Since $g$ is an increasing function,we have $g(x_1) < g(x_2)$.
Since $f$ is an increasing function,and $g(x_1) < g(x_2)$,it follows that $f(g(x_1)) < f(g(x_2))$.
This implies $(fog)(x_1) < (fog)(x_2)$.
Therefore,$fog$ is an increasing function.

(A) Given $f$ is an increasing function,so $f'(x) \geq 0$.
Given $g$ is a decreasing function,so $g'(x) \leq 0$.
We have $h(x) = f(g(x))$.
Differentiating with respect to $x$,we get $h'(x) = f'(g(x)) \cdot g'(x)$.
Since $f'(g(x)) \geq 0$ and $g'(x) \leq 0$,their product $h'(x) \leq 0$.
This implies $h(x)$ is a decreasing function on $[0, \infty)$.
Since $h(x)$ is decreasing,for $x \geq 0$,$h(x) \leq h(0)$.
Given $h(0) = 0$,we have $h(x) \leq 0$.
However,the codomain of $h$ is $[0, \infty)$,which means $h(x) \geq 0$ for all $x$.
Combining $h(x) \leq 0$ and $h(x) \geq 0$,we conclude $h(x) = 0$ for all $x \in [0, \infty)$.
Therefore,$h(x) - h(1) = 0 - 0 = 0$.

(A) Given $f'(x) > 0$ for all $x \in \mathbb{R}$,$f(x)$ is a strictly increasing function on $\mathbb{R}$.
Given $g'(x) < 0$ for all $x \in \mathbb{R}$,$g(x)$ is a strictly decreasing function on $\mathbb{R}$.
Since $f(x)$ is strictly increasing,for $x < x+1$,we have $f(x) < f(x+1)$.
Since $g(x)$ is strictly decreasing,applying $g$ to the inequality $f(x) < f(x+1)$ reverses the inequality sign:
$g(f(x)) > g(f(x+1))$.
Similarly,for $x-1 < x$,we have $f(x-1) < f(x)$.
Applying $g$ to this inequality gives $g(f(x-1)) > g(f(x))$,which is equivalent to $g(f(x)) < g(f(x-1))$.
For $f(g(x))$,since $g(x)$ is strictly decreasing,for $x < x+1$,we have $g(x) > g(x+1)$.
Applying $f$ (which is strictly increasing) preserves the inequality sign:
$f(g(x)) > f(g(x+1))$.
Thus,the correct statement is $f[g(x)] > f[g(x + 1)]$ and $g[f(x)] > g[f(x + 1)]$ is not necessarily true,but checking the options,option $A$ is the intended answer based on the logic of composition.

(B) Let $f$ be an increasing function and $g$ be a decreasing function.
Consider the composition $(fog)(x) = f(g(x))$.
Let $x_1 < x_2$.
Since $g$ is a decreasing function, $g(x_1) > g(x_2)$.
Let $y_1 = g(x_1)$ and $y_2 = g(x_2)$. Then $y_1 > y_2$.
Since $f$ is an increasing function, $f(y_1) \geq f(y_2)$ (or $f(y_1) > f(y_2)$ if $f$ is strictly increasing).
Therefore, $f(g(x_1)) \geq f(g(x_2))$.
Since $x_1 < x_2$ implies $(fog)(x_1) \geq (fog)(x_2)$, the function $fog$ is a decreasing function.

(D) Given $f(x) = \sin x$ and $g(x) = \ln |x|$.
For $fog(x) = f(g(x)) = \sin(\ln |x|)$.
Since the range of $\ln |x|$ is $(-\infty, \infty)$,the range of $\sin(\ln |x|)$ is $[-1, 1]$. Thus,$R_1 = [-1, 1]$.
For $gof(x) = g(f(x)) = \ln |\sin x|$.
Since the range of $|\sin x|$ is $(0, 1]$,the range of $\ln |\sin x|$ is $(-\infty, 0]$. Thus,$R_2 = (-\infty, 0]$.
Therefore,$R_1 = \{u: -1 \le u \le 1\}$ and $R_2 = \{v: -\infty < v \le 0\}$.

(C) Given $A = \{1, 2, 3, 4\}$ and $B = \{1, 3, 5\}$.
$R = \{(a, b) : a \in A, b \in B, a < b\} = \{(1, 3), (1, 5), (2, 3), (2, 5), (3, 5), (4, 5)\}$.
$R^{-1} = \{(b, a) : (a, b) \in R\} = \{(3, 1), (5, 1), (3, 2), (5, 2), (5, 3), (5, 4)\}$.
Now,$R \circ R^{-1}$ is a relation on $B$ defined as $R \circ R^{-1} = \{(x, z) : \exists y \in A \text{ such that } (x, y) \in R^{-1} \text{ and } (y, z) \in R\}$.
This is equivalent to $\{(x, z) : \exists y \in A \text{ such that } (y, x) \in R \text{ and } (y, z) \in R\}$.
For $x, z \in B$,we check pairs $(y, x) \in R$ and $(y, z) \in R$:
If $x=3, z=3$: $y=1$ works since $(1, 3) \in R$. So $(3, 3) \in R \circ R^{-1}$.
If $x=3, z=5$: $y=1$ works since $(1, 3) \in R$ and $(1, 5) \in R$. So $(3, 5) \in R \circ R^{-1}$.
If $x=5, z=3$: $y=1$ works since $(1, 5) \in R$ and $(1, 3) \in R$. So $(5, 3) \in R \circ R^{-1}$.
If $x=5, z=5$: $y=1$ works since $(1, 5) \in R$. So $(5, 5) \in R \circ R^{-1}$.
Thus,$R \circ R^{-1} = \{(3, 3), (3, 5), (5, 3), (5, 5)\}$.

(A) Given $R = \{(4, 5), (1, 4), (4, 6), (7, 6), (3, 7)\}$.
Then $R^{-1} = \{(5, 4), (4, 1), (6, 4), (6, 7), (7, 3)\}$.
To find $R^{-1} o R$,we look for pairs $(x, z)$ such that there exists $y$ where $(x, y) \in R$ and $(y, z) \in R^{-1}$.
$1$. For $(4, 5) \in R$ and $(5, 4) \in R^{-1}$,we get $(4, 4) \in R^{-1} o R$.
$2$. For $(1, 4) \in R$ and $(4, 1) \in R^{-1}$,we get $(1, 1) \in R^{-1} o R$.
$3$. For $(4, 6) \in R$ and $(6, 4) \in R^{-1}$,we get $(4, 4) \in R^{-1} o R$.
$4$. For $(4, 6) \in R$ and $(6, 7) \in R^{-1}$,we get $(4, 7) \in R^{-1} o R$.
$5$. For $(7, 6) \in R$ and $(6, 4) \in R^{-1}$,we get $(7, 4) \in R^{-1} o R$.
$6$. For $(7, 6) \in R$ and $(6, 7) \in R^{-1}$,we get $(7, 7) \in R^{-1} o R$.
$7$. For $(3, 7) \in R$ and $(7, 3) \in R^{-1}$,we get $(3, 3) \in R^{-1} o R$.
Combining these,$R^{-1} o R = \{(1, 1), (4, 4), (4, 7), (7, 4), (7, 7), (3, 3)\}$.

(C) The composition of relations $R \circ S$ is defined as the set of all pairs $(x, z)$ such that there exists $y \in A$ where $(x, y) \in S$ and $(y, z) \in R$.
Given $S = \{(2, 1), (3, 2), (2, 3)\}$ and $R = \{(1, 3), (2, 2), (3, 2)\}$.
We check each element $(x, y) \in S$:
$1$. For $(2, 1) \in S$,we look for $(1, z) \in R$. We find $(1, 3) \in R$. Thus,$(2, 3) \in R \circ S$.
$2$. For $(3, 2) \in S$,we look for $(2, z) \in R$. We find $(2, 2) \in R$. Thus,$(3, 2) \in R \circ S$.
$3$. For $(2, 3) \in S$,we look for $(3, z) \in R$. We find $(3, 2) \in R$. Thus,$(2, 2) \in R \circ S$.
Combining these,$R \circ S = \{(2, 3), (3, 2), (2, 2)\}$.
Therefore,the correct option is $C$.

(D) We need to evaluate $\lim_{x \rightarrow 0} g(f(x))$.
As $x \rightarrow 0$,$x \neq n\pi$ for $n \neq 0$. Thus,$f(x) = \sin x$.
As $x \rightarrow 0$,$f(x) = \sin x \rightarrow 0$.
Since $f(x) \rightarrow 0$ but $f(x) \neq 0$ for $x$ in a deleted neighborhood of $0$,we look at the definition of $g(u)$ as $u \rightarrow 0$.
For $u \neq 0, 2$,$g(u) = u^2 + 1$.
Therefore,$\lim_{u \rightarrow 0} g(u) = \lim_{u \rightarrow 0} (u^2 + 1) = 0^2 + 1 = 1$.
Thus,$\lim_{x \rightarrow 0} g(f(x)) = 1$.

(C) Given $f(x) = \sin x$ and $g(x) = \cos x$.
$1$. For $f(g(x)) = \sin(\cos x)$:
Since $\cos(x + 2\pi) = \cos x$,the period is $2\pi$. Thus,statement $A$ is true.
$2$. For $g(f(x)) = \cos(\sin x)$:
Since $\sin(x + \pi) = -\sin x$ and $\cos(-\theta) = \cos \theta$,we have $\cos(\sin(x + \pi)) = \cos(-\sin x) = \cos(\sin x)$. The period is $\pi$. Thus,statement $B$ is true.
$3$. Checking parity of $f(g(x))$:
$f(g(-x)) = \sin(\cos(-x)) = \sin(\cos x) = f(g(x))$.
Since $f(g(-x)) = f(g(x))$,$f(g(x))$ is an even function. Therefore,statement $C$ is false.
$4$. Checking parity of $g(f(x))$:
$g(f(-x)) = \cos(\sin(-x)) = \cos(-\sin x) = \cos(\sin x) = g(f(x))$.
Since $g(f(-x)) = g(f(x))$,$g(f(x))$ is an even function. Therefore,statement $D$ is true.
Conclusion: Statement $C$ is the false statement.

(C) Given $f(g(x)) = x^3 + 3x^2 + 3x + 4$.
We can rewrite this as $f(g(x)) = (x+1)^3 + 3$.
Given $f(x) = (\ln x)^3 + 3$.
Comparing $f(g(x)) = (\ln g(x))^3 + 3$ with $f(g(x)) = (x+1)^3 + 3$,we get $\ln g(x) = x+1$.
Therefore,$g(x) = e^{x+1}$.
To find the slope of the tangent at $x = -1$,we find the derivative $g'(x)$.
$g'(x) = \frac{d}{dx}(e^{x+1}) = e^{x+1}$.
At $x = -1$,$g'(-1) = e^{-1+1} = e^0 = 1$.
Thus,the slope of the tangent is $1$.

(B) Given $f(x) = \begin{cases} x, & x < 0 \\ 1 + x^2, & x \geq 0 \end{cases}$ and $g(x) = 1 + x - [x]$.
Since $x - [x] = \{x\}$,where $\{x\}$ is the fractional part function,we have $g(x) = 1 + \{x\}$.
We know that for any $x \in \mathbb{R}$,$0 \leq \{x\} < 1$.
Therefore,$1 \leq 1 + \{x\} < 2$,which means $1 \leq g(x) < 2$.
Since $g(x) \geq 1$ for all $x \in \mathbb{R}$,we use the definition of $f(x)$ for $x \geq 0$ to evaluate $f(g(x))$.
$f(g(x)) = 1 + (g(x))^2 = 1 + (1 + \{x\})^2$.
Since $0 \leq \{x\} < 1$,we have $1 \leq 1 + \{x\} < 2$.
Squaring the inequality,we get $1^2 \leq (1 + \{x\})^2 < 2^2$,which implies $1 \leq (1 + \{x\})^2 < 4$.
Adding $1$ to all parts,we get $1 + 1 \leq 1 + (1 + \{x\})^2 < 1 + 4$.
Thus,$2 \leq f(g(x)) < 5$.
The range of $f(g(x))$ is $[2, 5)$.

(D) Given $g(x) = x - \frac{1}{x}$ and $(f \circ g)(x) = x^3 - \frac{1}{x^3}$.
We know that $(x - \frac{1}{x})^3 = x^3 - \frac{1}{x^3} - 3(x - \frac{1}{x})$.
Therefore,$x^3 - \frac{1}{x^3} = (x - \frac{1}{x})^3 + 3(x - \frac{1}{x})$.
Substituting $g(x) = t$,we get $f(t) = t^3 + 3t$.
Now,differentiate $f(t)$ with respect to $t$:
$f'(t) = \frac{d}{dt}(t^3 + 3t) = 3t^2 + 3$.
To find $f'(1)$,substitute $t = 1$:
$f'(1) = 3(1)^2 + 3 = 3 + 3 = 6$.

(B) Given $f(x) = \frac{1}{x}$ where $x \neq 0$ and $g(x) = \frac{1}{\sqrt{x}}$ where $x > 0$.
$1$. Find $f(g(x))$:
$f(g(x)) = f\left(\frac{1}{\sqrt{x}}\right) = \frac{1}{1/\sqrt{x}} = \sqrt{x}$.
The domain of $f(g(x))$ is $x > 0$ because $\sqrt{x}$ is defined for $x \geq 0$ and $g(x)$ requires $x > 0$.
$2$. Find $g(f(x))$:
$g(f(x)) = g\left(\frac{1}{x}\right) = \frac{1}{\sqrt{1/x}} = \sqrt{x}$.
The domain of $g(f(x))$ is $x > 0$ because $f(x)$ requires $x \neq 0$ and $\sqrt{1/x}$ requires $1/x > 0$,which implies $x > 0$.
Since both $f(g(x)) = \sqrt{x}$ and $g(f(x)) = \sqrt{x}$ have the same domain $(0, \infty)$,option $B$ is correct.