Inverse Function Questions in English

(C) The inverse relation ${R^{-1}}$ is defined as the set of all ordered pairs $(y, x)$ such that $(x, y) \in R$.
Given $R = \{(1, 3), (1, 5), (2, 1)\}$.
By swapping the elements of each ordered pair in $R$,we get:
$(1, 3) \to (3, 1)$
$(1, 5) \to (5, 1)$
$(2, 1) \to (1, 2)$
Therefore,${R^{-1}} = \{(3, 1), (5, 1), (1, 2)\}$.
Thus,the correct option is $C$.

(D) Given the function $y = f(x) = \frac{ax + b}{cx - a}$.
To find $x$ in terms of $y$,we perform the following steps:
$y(cx - a) = ax + b$
$cxy - ay = ax + b$
$cxy - ax = ay + b$
$x(cy - a) = ay + b$
$x = \frac{ay + b}{cy - a}$
Since $f(x) = \frac{ax + b}{cx - a}$,replacing $x$ with $y$ gives $f(y) = \frac{ay + b}{cy - a}$.
Therefore,$x = f(y)$.

(A) function $f: A \to B$ is invertible if and only if it is a bijection,meaning it is both one-one (injective) and onto (surjective).
$1$. For $f(x) = 2^x$,if we consider the domain as $\mathbb{R}$ and codomain as $(0, \infty)$,the function is strictly increasing (one-one) and covers all positive real values (onto). Thus,it is invertible.
$2$. For $f(x) = x^3 - x$,the derivative $f'(x) = 3x^2 - 1$ changes sign at $x = \pm \frac{1}{\sqrt{3}}$,so it is not monotonic and not one-one on $\mathbb{R}$.
$3$. For $f(x) = x^2$,$f(1) = f(-1) = 1$,so it is not one-one.
Therefore,$f(x) = 2^x$ is the correct answer.

(A) Given the function $y = \frac{x + 2}{x - 1}$.
To find $x$ in terms of $y$,we perform the following steps:
$y(x - 1) = x + 2$
$yx - y = x + 2$
$yx - x = y + 2$
$x(y - 1) = y + 2$
$x = \frac{y + 2}{y - 1}$
Since $f(y) = \frac{y + 2}{y - 1}$,we conclude that $x = f(y)$.

(A) function $f$ is the inverse of itself if $f(f(x)) = x$ for all $x$ in its domain.
For option $A$,let $f(x) = \frac{1 - x}{1 + x}$.
Then $f(f(x)) = f\left(\frac{1 - x}{1 + x}\right) = \frac{1 - \left(\frac{1 - x}{1 + x}\right)}{1 + \left(\frac{1 - x}{1 + x}\right)}$.
Multiplying the numerator and denominator by $(1 + x)$,we get:
$f(f(x)) = \frac{(1 + x) - (1 - x)}{(1 + x) + (1 - x)} = \frac{1 + x - 1 + x}{1 + x + 1 - x} = \frac{2x}{2} = x$.
Since $f(f(x)) = x$,the function $f(x) = \frac{1 - x}{1 + x}$ is the inverse of itself.

(B) Let $y = f(x) = \frac{e^x - e^{-x}}{e^x + e^{-x}} + 2$.
Multiply the numerator and denominator of the fraction by $e^x$:
$y = \frac{e^{2x} - 1}{e^{2x} + 1} + 2$.
Subtract $2$ from both sides:
$y - 2 = \frac{e^{2x} - 1}{e^{2x} + 1}$.
Let $u = e^{2x}$. Then $y - 2 = \frac{u - 1}{u + 1}$.
$(y - 2)(u + 1) = u - 1$
$yu + y - 2u - 2 = u - 1$
$u(y - 2 - 1) = -1 - y + 2$
$u(y - 3) = 1 - y$
$u = \frac{1 - y}{y - 3} = \frac{y - 1}{3 - y}$.
Since $u = e^{2x}$,we have $e^{2x} = \frac{y - 1}{3 - y}$.
Taking the natural logarithm on both sides:
$2x = \log_e \left( \frac{y - 1}{3 - y} \right)$
$x = \frac{1}{2} \log_e \left( \frac{y - 1}{3 - y} \right) = \log_e \left( \frac{y - 1}{3 - y} \right)^{1/2}$.
Thus,the inverse function is $f^{-1}(x) = \log_e \left( \frac{x - 1}{3 - x} \right)^{1/2}$.

(B) Given $f(x) = 2^{x(x - 1)}$.
To find the inverse,let $y = f(x) = 2^{x(x - 1)}$.
Taking $\log_2$ on both sides,we get $\log_2 y = x(x - 1)$.
This simplifies to the quadratic equation $x^2 - x - \log_2 y = 0$.
Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$,we get $x = \frac{1 \pm \sqrt{1 + 4\log_2 y}}{2}$.
Since the domain of $f$ is $[1, \infty)$,we must have $x \ge 1$.
If we choose the negative sign,$x = \frac{1 - \sqrt{1 + 4\log_2 y}}{2} < 1$ for $y \ge 1$.
Thus,we must choose the positive sign: $x = \frac{1 + \sqrt{1 + 4\log_2 y}}{2}$.
Replacing $y$ with $x$,we get $f^{-1}(x) = \frac{1}{2}(1 + \sqrt{1 + 4\log_2 x})$.

(B) Let $f(x) = y$.
Since $f(x) = 3x - 5$,we have $y = 3x - 5$.
To find the inverse,we solve for $x$ in terms of $y$:
$y + 5 = 3x$
$x = \frac{y + 5}{3}$.
By definition,$x = f^{-1}(y)$,so $f^{-1}(y) = \frac{y + 5}{3}$.
Replacing $y$ with $x$,we get $f^{-1}(x) = \frac{x + 5}{3}$.
Since $f(x) = 3x - 5$ is a linear function,it is both one-one (injective) and onto (surjective),therefore $f$ is invertible.

(B) Given $f(x) = 3x - 4$.
To find the inverse function ${f^{ - 1}}(x)$,let $y = f(x)$.
Then $y = 3x - 4$.
Solving for $x$ in terms of $y$:
$y + 4 = 3x$
$x = \frac{y + 4}{3}$.
By definition,if $y = f(x)$,then $x = {f^{ - 1}}(y)$.
Therefore,${f^{ - 1}}(y) = \frac{y + 4}{3}$.
Replacing $y$ with $x$,we get ${f^{ - 1}}(x) = \frac{x + 4}{3}$.

(D) Given $f(x) = \frac{x}{1 + x}$.
Let $y = f(x)$,then $x = {f^{-1}}(y)$.
Substituting $y$ for $f(x)$,we have $y = \frac{x}{1 + x}$.
Cross-multiplying gives $y(1 + x) = x$,which simplifies to $y + yx = x$.
Rearranging the terms to solve for $x$,we get $y = x - yx = x(1 - y)$.
Thus,$x = \frac{y}{1 - y}$.
Since $x = {f^{-1}}(y)$,we have ${f^{-1}}(y) = \frac{y}{1 - y}$.
Replacing $y$ with $x$,we obtain ${f^{-1}}(x) = \frac{x}{1 - x}$.

(A) function is invertible if and only if it is a bijection (both one-to-one and onto).
$1$. For $f(x) = \frac{1}{x - 1}$,the domain is $x \in \mathbb{R} \setminus \{1\}$ and the range is $y \in \mathbb{R} \setminus \{0\}$. This function is one-to-one and onto,hence it is invertible.
$2$. For $f(x) = x^2$ defined for all $x \in \mathbb{R}$,it is not one-to-one because $f(1) = f(-1) = 1$. Thus,it is not invertible.
$3$. For $f(x) = x^2$ with $x \ge 0$ or $x \le 0$,the function becomes a bijection on its restricted domain and codomain,but the standard definition of an invertible function usually refers to the function being bijective over its natural domain. However,among the given choices,$f(x) = \frac{1}{x - 1}$ is the most standard example of a function that is bijective on its entire domain.

(A) Let $y = \frac{10^x - 10^{-x}}{10^x + 10^{-x}}$.
To find the inverse,we solve for $x$ in terms of $y$.
Multiply the numerator and denominator by $10^x$:
$y = \frac{10^{2x} - 1}{10^{2x} + 1}$.
$y(10^{2x} + 1) = 10^{2x} - 1$.
$y \cdot 10^{2x} + y = 10^{2x} - 1$.
$1 + y = 10^{2x}(1 - y)$.
$10^{2x} = \frac{1+y}{1-y}$.
Taking $\log_{10}$ on both sides:
$2x = \log_{10} \left( \frac{1+y}{1-y} \right)$.
$x = \frac{1}{2} \log_{10} \left( \frac{1+y}{1-y} \right)$.
Replacing $y$ with $x$ to get the inverse function:
$f^{-1}(x) = \frac{1}{2} \log_{10} \left( \frac{1+x}{1-x} \right)$.

(A) Given the function $y = 2x - 3$.
To find the inverse,we solve for $x$ in terms of $y$:
$y + 3 = 2x$
$x = \frac{y + 3}{2}$
Now,replace $y$ with $x$ to express the inverse function $f^{-1}(x)$:
$f^{-1}(x) = \frac{x + 3}{2}$
Thus,the correct option is $A$.

(A) To find the inverse of the function $f(x) = \frac{2x + 1}{1 - 3x}$,we set $y = f(x)$.
$y = \frac{2x + 1}{1 - 3x}$
Multiply both sides by $(1 - 3x)$:
$y(1 - 3x) = 2x + 1$
$y - 3xy = 2x + 1$
Rearrange the terms to isolate $x$:
$y - 1 = 2x + 3xy$
$y - 1 = x(2 + 3y)$
$x = \frac{y - 1}{3y + 2}$
Since $f^{-1}(y) = x$,we have $f^{-1}(y) = \frac{y - 1}{3y + 2}$.
Replacing $y$ with $x$,we get $f^{-1}(x) = \frac{x - 1}{3x + 2}$.

(D) Given $f(x) = x^2 + 1$.
To find the inverse,let $y = f(x) = x^2 + 1$.
Solving for $x$,we get $x^2 = y - 1$,which implies $x = \pm \sqrt{y - 1}$.
Thus,$f^{-1}(y) = \pm \sqrt{y - 1}$.
For $f^{-1}(17)$,we substitute $y = 17$: $f^{-1}(17) = \pm \sqrt{17 - 1} = \pm \sqrt{16} = \pm 4$.
For $f^{-1}(-3)$,we substitute $y = -3$: $f^{-1}(-3) = \pm \sqrt{-3 - 1} = \pm \sqrt{-4}$.
Since $\sqrt{-4}$ is not a real number,$f^{-1}(-3)$ is not defined in the set of real numbers.
Therefore,the correct option is $D$ (None of these).

(C) Given $f(x) = \sin x + \cos x$ and $g(x) = x^2 - 1$.
First,we find the composition $g(f(x))$:
$g(f(x)) = (\sin x + \cos x)^2 - 1$
$= (\sin^2 x + \cos^2 x + 2 \sin x \cos x) - 1$
$= (1 + \sin 2x) - 1 = \sin 2x$.
For a function to be invertible,it must be bijective (one-to-one and onto).
The function $h(x) = \sin 2x$ is invertible in the interval where the argument $2x$ lies within the principal branch of the inverse sine function,i.e.,$[ -\frac{\pi}{2}, \frac{\pi}{2} ]$.
Thus,we require:
$-\frac{\pi}{2} \le 2x \le \frac{\pi}{2}$
Dividing by $2$,we get:
$-\frac{\pi}{4} \le x \le \frac{\pi}{4}$.
Therefore,$g(f(x))$ is invertible for $x \in [ -\frac{\pi}{4}, \frac{\pi}{4} ]$.

(B) Let $f(x) = y$,which implies $x = f^{-1}(y)$.
Given $y = \frac{2x - 1}{x + 5}$,where $x \ne -5$.
Multiply both sides by $(x + 5)$:
$y(x + 5) = 2x - 1$
$xy + 5y = 2x - 1$
Rearrange the terms to isolate $x$:
$5y + 1 = 2x - xy$
$5y + 1 = x(2 - y)$
Solving for $x$:
$x = \frac{5y + 1}{2 - y}$
Since $x = f^{-1}(y)$,we have:
$f^{-1}(y) = \frac{5y + 1}{2 - y}$
Replacing $y$ with $x$,we get:
$f^{-1}(x) = \frac{5x + 1}{2 - x}$,where $x \ne 2$.

(C) For a function $f: A \to B$ to have an inverse,it must be a bijection,which means it must be both one-to-one (injective) and onto (surjective).
$A$ function that is strictly monotonic (either strictly increasing or strictly decreasing) is always one-to-one.
If a function is continuous and strictly monotonic on its domain,it maps its domain to its range in a one-to-one manner,ensuring the existence of an inverse function $f^{-1}$.

(B) Given that ${e^x} = y + \sqrt{1 + y^2}$.
Rearranging the terms,we get ${e^x} - y = \sqrt{1 + y^2}$.
Squaring both sides,we obtain $({e^x} - y)^2 = 1 + y^2$.
Expanding the left side,${e^{2x}} + y^2 - 2y{e^x} = 1 + y^2$.
Subtracting $y^2$ from both sides,${e^{2x}} - 2y{e^x} = 1$.
Rearranging to solve for $y$,$2y{e^x} = {e^{2x}} - 1$.
Dividing by $2{e^x}$,$y = \frac{{e^{2x}} - 1}{2{e^x}}$.
Simplifying the expression,$y = \frac{1}{2} \left( \frac{e^{2x}}{e^x} - \frac{1}{e^x} \right) = \frac{e^x - e^{-x}}{2}$.

(D) Given the function $f:(2, 3) \to (0, 1)$ defined by $f(x) = x - [x]$.
For $x \in (2, 3)$,the greatest integer function $[x] = 2$.
Therefore,$f(x) = x - 2$.
Let $y = f(x)$,so $y = x - 2$.
To find the inverse,solve for $x$ in terms of $y$:
$x = y + 2$.
Replacing $y$ with $x$,we get ${f^{ - 1}}(x) = x + 2$.

(D) Given $f(x) = (x + 1)^2 - 1$ for $x \ge -1$.
To find the set $S = \{ x : f(x) = f^{-1}(x) \}$,we note that the solutions to $f(x) = f^{-1}(x)$ are the same as the solutions to $f(x) = x$ and the points where the graph of $f(x)$ intersects its inverse.
Setting $f(x) = x$,we get $(x + 1)^2 - 1 = x$.
$x^2 + 2x + 1 - 1 = x \Rightarrow x^2 + x = 0 \Rightarrow x(x + 1) = 0$.
Thus,$x = 0$ and $x = -1$ are solutions.
However,the equation $f(x) = f^{-1}(x)$ is equivalent to $f(f(x)) = x$ for $x$ in the domain.
$( (x + 1)^2 - 1 + 1 )^2 - 1 = x \Rightarrow ((x + 1)^2)^2 - 1 = x$.
$(x + 1)^4 - 1 = x \Rightarrow (x + 1)^4 - (x + 1) = 0$.
$(x + 1) [ (x + 1)^3 - 1 ] = 0$.
This gives $x + 1 = 0 \Rightarrow x = -1$ or $(x + 1)^3 = 1$.
The roots of $(x + 1)^3 = 1$ are $x + 1 = 1, \omega, \omega^2$,where $\omega = \frac{-1 + i\sqrt{3}}{2}$.
$x + 1 = 1 \Rightarrow x = 0$.
$x + 1 = \frac{-1 + i\sqrt{3}}{2} \Rightarrow x = \frac{-3 + i\sqrt{3}}{2}$.
$x + 1 = \frac{-1 - i\sqrt{3}}{2} \Rightarrow x = \frac{-3 - i\sqrt{3}}{2}$.
Thus,the set $S = \{ 0, -1, \frac{-3 + i\sqrt{3}}{2}, \frac{-3 - i\sqrt{3}}{2} \}$.

(B) Since $g(x)$ is the inverse of the function $f(x)$,we have $g(f(x)) = x$ for all $x$ in the domain.
Differentiating both sides with respect to $x$ using the chain rule:
$\frac{d}{dx}[g(f(x))] = \frac{d}{dx}(x)$
$g'(f(x)) \cdot f'(x) = 1$
Assuming $f'(x) \neq 0$,we can write:
$g'(f(x)) = \frac{1}{f'(x)}$
Substituting $x = c$ into the equation:
$g'(f(c)) = \frac{1}{f'(c)}$
Thus,the correct option is $B$.

(C) Given that $g(x)$ is the inverse of $f(x)$,we have $f(g(x)) = x$.
Differentiating both sides with respect to $x$ using the chain rule:
$f'(g(x)) \cdot g'(x) = 1$
Therefore,$g'(x) = \frac{1}{f'(g(x))}$.
Given $f'(x) = \frac{1}{1 + x^3}$,we substitute $g(x)$ for $x$ in the derivative expression:
$f'(g(x)) = \frac{1}{1 + (g(x))^3}$.
Substituting this back into the expression for $g'(x)$:
$g'(x) = \frac{1}{\frac{1}{1 + (g(x))^3}} = 1 + (g(x))^3$.
Thus,the correct option is $C$.

(C) function $f: A \to B$ is invertible if and only if it is a bijection,meaning it is both one-to-one (injective) and onto (surjective).
For a real-valued function $f(x)$ to be a bijection on its domain,it must be strictly monotonic (either strictly increasing or strictly decreasing) and continuous.
If a function is strictly monotonic,it is guaranteed to be one-to-one.
If it is also continuous on its domain,it maps the domain to its range,satisfying the onto condition.
Therefore,the correct condition is that the function must be strictly monotonic and continuous in its domain.

(A) Given that $g$ is the inverse of $f$,we have $f(g(x)) = x$.
Differentiating both sides with respect to $x$ using the chain rule,we get $f'(g(x)) \cdot g'(x) = 1$.
Therefore,$g'(x) = \frac{1}{f'(g(x))}$.
Given $f'(x) = \frac{1}{1 + x^5}$,substituting $g(x)$ for $x$ gives $f'(g(x)) = \frac{1}{1 + (g(x))^5}$.
Thus,$g'(x) = \frac{1}{\frac{1}{1 + (g(x))^5}} = 1 + (g(x))^5$.

(D) To show that $f$ is invertible,we need to find a function $g: Y \to N$ such that $g \circ f = I_N$ and $f \circ g = I_Y$.
Given $f(x) = 4x + 3$,let $y = f(x) = 4x + 3$.
Solving for $x$ in terms of $y$,we get $y - 3 = 4x$,which implies $x = \frac{y - 3}{4}$.
Define a function $g: Y \to N$ by $g(y) = \frac{y - 3}{4}$.
Now,check the composition $g \circ f(x) = g(f(x)) = g(4x + 3) = \frac{(4x + 3) - 3}{4} = \frac{4x}{4} = x = I_N(x)$.
Next,check the composition $f \circ g(y) = f(g(y)) = f\left(\frac{y - 3}{4}\right) = 4\left(\frac{y - 3}{4}\right) + 3 = (y - 3) + 3 = y = I_Y(y)$.
Since $g \circ f = I_N$ and $f \circ g = I_Y$,the function $f$ is invertible and its inverse is $g(y) = \frac{y - 3}{4}$.

(A) Given $f(x) = (x+1)^2 - 1$ for $x \geq -1$.
To find $f^{-1}(x)$,set $y = (x+1)^2 - 1$. Since $x \geq -1$,$y \geq -1$.
Solving for $x$: $(x+1)^2 = y+1 \Rightarrow x+1 = \sqrt{y+1} \Rightarrow x = \sqrt{y+1} - 1$.
Thus,$f^{-1}(x) = \sqrt{x+1} - 1$. Since $f$ is strictly increasing on $[-1, \infty)$,it is a bijection (one-one and onto). So,Statement-$2$ is true.
To find $S = \{x : f(x) = f^{-1}(x)\}$,we solve $f(x) = x$ because $f$ is increasing.
$(x+1)^2 - 1 = x \Rightarrow x^2 + 2x + 1 - 1 = x \Rightarrow x^2 + x = 0 \Rightarrow x(x+1) = 0$.
Thus,$x = 0$ or $x = -1$. So,$S = \{0, -1\}$. Statement-$1$ is true.
Since $f(x) = f^{-1}(x)$ is equivalent to $f(x) = x$ for increasing functions,Statement-$2$ explains Statement-$1$.

(D) Given $f(x) = (x - 1)^2 + 1$ for $x \ge 1$.
To find $f^{-1}(x)$,set $y = (x - 1)^2 + 1$.
Then $y - 1 = (x - 1)^2$. Since $x \ge 1$,we have $x - 1 = \sqrt{y - 1}$,so $x = 1 + \sqrt{y - 1}$.
Thus,$f^{-1}(x) = 1 + \sqrt{x - 1}$ for $x \ge 1$. So,Statement-$2$ is true.
To find $S = \{x : f(x) = f^{-1}(x)\}$,we solve $f(x) = x$.
$(x - 1)^2 + 1 = x \implies x^2 - 2x + 1 + 1 = x \implies x^2 - 3x + 2 = 0$.
$(x - 1)(x - 2) = 0$,so $x = 1$ or $x = 2$.
Since both values satisfy $x \ge 1$,$S = \{1, 2\}$. So,Statement-$1$ is true.
Since $f(x) = f^{-1}(x)$ is equivalent to $f(x) = x$ for increasing functions,Statement-$2$ explains Statement-$1$.

(D) The graph of $f(x)$ is given by the equation $y = (x + 1)^2$ for $x \ge -1$.
To find the reflection of the graph of $f(x)$ with respect to the line $y = x$,we interchange $x$ and $y$.
Thus,the equation for the graph of $g(x)$ becomes $x = (y + 1)^2$ for $y \ge -1$.
Solving for $y$:
$y + 1 = \pm\sqrt{x}$
Since $y \ge -1$,we have $y + 1 \ge 0$,so we take the positive root:
$y + 1 = \sqrt{x}$
$y = \sqrt{x} - 1$
Since the domain of $f(x)$ is $x \ge -1$ and its range is $y \ge 0$,the domain of $g(x)$ is $x \ge 0$ and its range is $y \ge -1$.
Therefore,$g(x) = \sqrt{x} - 1$ for $x \ge 0$.

(A) Given $y = f(x) = x + \frac{1}{x}$ where $x \ge 1$ and $y \ge 2$.
Multiplying by $x$,we get $yx = x^2 + 1$,which rearranges to the quadratic equation $x^2 - yx + 1 = 0$.
Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$,we get $x = \frac{y \pm \sqrt{y^2 - 4}}{2}$.
Since the domain of $f$ is $[1, +\infty)$,we must have $x \ge 1$.
If we take the negative sign,$x = \frac{y - \sqrt{y^2 - 4}}{2}$. For $y \ge 2$,this value is $\le 1$. Specifically,as $y \to \infty$,this expression approaches $0$.
Thus,to satisfy $x \ge 1$,we must take the positive sign: $x = \frac{y + \sqrt{y^2 - 4}}{2}$.
Replacing $y$ with $x$,we obtain ${f^{-1}}(x) = \frac{x + \sqrt{x^2 - 4}}{2}$.

(C) The inverse relation ${R^{-1}}$ is defined as the set of all ordered pairs $(y, x)$ such that $(x, y) \in R$.
Given $R = \{(1, 3), (1, 5), (2, 1)\}$.
By swapping the elements of each ordered pair in $R$,we get:
$(1, 3) \to (3, 1)$
$(1, 5) \to (5, 1)$
$(2, 1) \to (1, 2)$
Therefore,${R^{-1}} = \{(3, 1), (5, 1), (1, 2)\}$.

(B) Given the relation $R = \{(a, b) : a = 2b\}$ where $a, b \in \mathbb{N}$.
By substituting values for $b$,we get $R = \{(2, 1), (4, 2), (6, 3), \dots\}$.
The inverse relation ${R^{-1}}$ is obtained by interchanging the elements of the ordered pairs in $R$.
Thus,${R^{-1}} = \{(b, a) : (a, b) \in R\} = \{(1, 2), (2, 4), (3, 6), \dots\}$.

(A) The relation $R$ is given by $R = \{(1, 3), (2, 5), (3, 3)\}$.
By definition,the inverse relation $R^{-1}$ is obtained by interchanging the elements of the ordered pairs in $R$.
That is,if $(a, b) \in R$,then $(b, a) \in R^{-1}$.
Applying this to each element of $R$:
$(1, 3) \in R \implies (3, 1) \in R^{-1}$
$(2, 5) \in R \implies (5, 2) \in R^{-1}$
$(3, 3) \in R \implies (3, 3) \in R^{-1}$
Therefore,$R^{-1} = \{(3, 1), (5, 2), (3, 3)\}$.
Rearranging the set,we get $R^{-1} = \{(3, 3), (3, 1), (5, 2)\}$.
Thus,the correct option is $A$.

(C) Given $f(x) = \int\limits_2^x \frac{dt}{\sqrt{1 + t^4}}$.
By the Fundamental Theorem of Calculus,$f'(x) = \frac{1}{\sqrt{1 + x^4}}$.
Since $g$ is the inverse of $f$,we have $g(f(x)) = x$. Differentiating both sides with respect to $x$,we get $g'(f(x)) \cdot f'(x) = 1$,which implies $g'(f(x)) = \frac{1}{f'(x)} = \sqrt{1 + x^4}$.
To find $g'(0)$,we need to find $x$ such that $f(x) = 0$.
$f(x) = \int\limits_2^x \frac{dt}{\sqrt{1 + t^4}} = 0$ implies $x = 2$.
Therefore,$g'(0) = g'(f(2)) = \sqrt{1 + 2^4} = \sqrt{1 + 16} = \sqrt{17}$.

(A) Given that $g$ is the inverse of $f$,we have $f(g(x)) = x$.
Differentiating both sides with respect to $x$ using the chain rule:
$f'(g(x)) \cdot g'(x) = 1$
We are given $f'(x) = \frac{1}{1 + x^5}$. Therefore,$f'(g(x)) = \frac{1}{1 + [g(x)]^5}$.
Substituting this into the differentiated equation:
$\frac{1}{1 + [g(x)]^5} \cdot g'(x) = 1$
Solving for $g'(x)$:
$g'(x) = 1 + [g(x)]^5$.

(B) Since $g$ is the inverse of $f$,we have $f(g(x)) = x$.
Differentiating both sides with respect to $x$ using the chain rule,we get $f'(g(x)) \cdot g'(x) = 1$.
Substituting $x = 2$,we get $f'(g(2)) \cdot g'(2) = 1$.
Given $g(2) = a$,this becomes $f'(a) \cdot g'(2) = 1$,which implies $g'(2) = \frac{1}{f'(a)}$.
We are given $f'(x) = \frac{x^{10}}{1 + x^2}$,so $f'(a) = \frac{a^{10}}{1 + a^2}$.
Therefore,$g'(2) = \frac{1}{\frac{a^{10}}{1 + a^2}} = \frac{1 + a^2}{a^{10}}$.

(C) Given $e^{f(x)} = \ln x$. Taking the natural logarithm on both sides,we get $f(x) = \ln(\ln x)$.
Since $g(x)$ is the inverse function of $f(x)$,we have $f(g(x)) = x$.
Substituting $g(x)$ into the expression for $f$,we get $\ln(\ln(g(x))) = x$.
Exponentiating both sides,we get $\ln(g(x)) = e^x$.
Exponentiating again,we get $g(x) = e^{e^x}$.
Now,we differentiate $g(x)$ with respect to $x$ using the chain rule:
$g'(x) = \frac{d}{dx}(e^{e^x}) = e^{e^x} \cdot \frac{d}{dx}(e^x) = e^{e^x} \cdot e^x$.
Using the property of exponents $a^m \cdot a^n = a^{m+n}$,we get $g'(x) = e^{e^x + x}$.
Thus,the correct option is $C$.

(B) Given $f(x) = e^x + x$.
We need to find $(f^{-1})'(f(\ln 2))$.
By the formula for the derivative of an inverse function,$(f^{-1})'(y) = \frac{1}{f'(x)}$,where $y = f(x)$.
Here,$y = f(\ln 2)$,so $x = \ln 2$.
First,find $f'(x)$:
$f'(x) = \frac{d}{dx}(e^x + x) = e^x + 1$.
Now,evaluate $f'(x)$ at $x = \ln 2$:
$f'(\ln 2) = e^{\ln 2} + 1 = 2 + 1 = 3$.
Therefore,$(f^{-1})'(f(\ln 2)) = \frac{1}{f'(\ln 2)} = \frac{1}{3}$.

(D) To determine if a function $f(x)$ has an inverse,we check if it is strictly monotonic (either strictly increasing or strictly decreasing).
Given $f(x) = x^3 + 8x + 3$.
The derivative is $f'(x) = \frac{d}{dx}(x^3 + 8x + 3) = 3x^2 + 8$.
Since $x^2 \ge 0$ for all real $x$,it follows that $3x^2 \ge 0$.
Therefore,$f'(x) = 3x^2 + 8 \ge 8$.
Since $f'(x) > 0$ for all $x \in \mathbb{R}$,the function $f(x)$ is strictly increasing on its entire domain.
$A$ strictly monotonic function is always one-to-one (injective) and thus possesses an inverse.
Therefore,the property that $f'(x)$ is always positive allows us to conclude that $f(x)$ has an inverse.

(A) Given the function $f(x) = x^3 - 3x^2 + 3x - 2$.
We can rewrite the function as $f(x) = (x^3 - 3x^2 + 3x - 1) - 1$.
This simplifies to $f(x) = (x - 1)^3 - 1$.
To find the inverse $f^{-1}(x)$,let $y = f(x)$.
So,$y = (x - 1)^3 - 1$.
Add $1$ to both sides: $y + 1 = (x - 1)^3$.
Take the cube root of both sides: $\sqrt[3]{y + 1} = x - 1$.
Add $1$ to both sides: $x = 1 + \sqrt[3]{y + 1}$.
Replacing $y$ with $x$,we get $f^{-1}(x) = 1 + \sqrt[3]{x + 1}$.

(D) Given $f(x) = x^{11} + \sin^3(35x) + 111x$.
Since $x^{11}$,$\sin^3(35x)$,and $111x$ are all odd functions,their sum $f(x)$ is an odd function,i.e.,$f(-x) = -f(x)$.
If $f(x)$ is an odd function,then its inverse $f^{-1}(x)$ is also an odd function,i.e.,$f^{-1}(-y) = -f^{-1}(y)$.
We need to calculate $S = f^{-1}(\sin \frac{\pi}{5}) + f^{-1}(\sin \frac{6\pi}{5}) + f^{-1}(\sin \frac{\pi}{7}) + f^{-1}(\sin \frac{8\pi}{7})$.
Note that $\sin \frac{6\pi}{5} = \sin(\pi + \frac{\pi}{5}) = -\sin \frac{\pi}{5}$.
Similarly,$\sin \frac{8\pi}{7} = \sin(\pi + \frac{\pi}{7}) = -\sin \frac{\pi}{7}$.
Substituting these into the expression:
$S = f^{-1}(\sin \frac{\pi}{5}) + f^{-1}(-\sin \frac{\pi}{5}) + f^{-1}(\sin \frac{\pi}{7}) + f^{-1}(-\sin \frac{\pi}{7})$.
Using the property $f^{-1}(-y) = -f^{-1}(y)$:
$S = f^{-1}(\sin \frac{\pi}{5}) - f^{-1}(\sin \frac{\pi}{5}) + f^{-1}(\sin \frac{\pi}{7}) - f^{-1}(\sin \frac{\pi}{7}) = 0$.
Since $f(0) = 0^{11} + \sin^3(0) + 111(0) = 0$,the result $0$ is equal to $f(0)$.

(A) Given $f(x) = -4e^{\left(\frac{1-x}{2}\right)} + 1 + x + \frac{x^2}{2} + \frac{x^3}{3}$.
Since $g(x) = f^{-1}(x)$,we have $g(f(x)) = x$.
Differentiating both sides with respect to $x$,we get $g'(f(x)) \cdot f'(x) = 1$,which implies $g'(f(x)) = \frac{1}{f'(x)}$.
First,we find $x$ such that $f(x) = -\frac{7}{6}$. By inspection,if $x = 1$,then $f(1) = -4e^0 + 1 + 1 + \frac{1}{2} + \frac{1}{3} = -4 + 2 + \frac{3+2}{6} = -2 + \frac{5}{6} = -\frac{7}{6}$.
Now,calculate $f'(x) = -4e^{\left(\frac{1-x}{2}\right)} \cdot \left(-\frac{1}{2}\right) + 1 + x + x^2 = 2e^{\left(\frac{1-x}{2}\right)} + 1 + x + x^2$.
At $x = 1$,$f'(1) = 2e^0 + 1 + 1 + 1 = 2 + 3 = 5$.
Therefore,$g'(-\frac{7}{6}) = g'(f(1)) = \frac{1}{f'(1)} = \frac{1}{5}$.

(B) function $f(x)$ is invertible if and only if it is strictly monotonic (either strictly increasing or strictly decreasing) on its domain.
For a cubic polynomial $f(x)$,this occurs when $f'(x) \ge 0$ or $f'(x) \le 0$ for all $x \in R$.
First,find the derivative: $f'(x) = 6x^2 - 6(a + 2)x + 12a$.
Factor the derivative: $f'(x) = 6(x^2 - (a + 2)x + 2a) = 6(x - a)(x - 2)$.
For $f(x)$ to be monotonic,$f'(x)$ must not change sign. However,since $f'(x)$ is a quadratic with two roots $a$ and $2$,it changes sign unless the roots are equal.
Thus,we must have $a = 2$.
If $a = 2$,$f'(x) = 6(x - 2)^2 \ge 0$ for all $x$,so the function is strictly increasing and thus invertible.
Since $a = 2$ is the only integral value in the interval $[-4, 6]$ that satisfies the condition,the number of such values is $1$.

(B) Given $f(x) = (2x - 3\pi)^5 + \frac{4}{3}x + \cos x$.
Since $g$ is the inverse of $f$,we have $g(f(x)) = x$.
Differentiating both sides with respect to $x$,we get $g'(f(x)) \cdot f'(x) = 1$,which implies $g'(f(x)) = \frac{1}{f'(x)}$.
We need to find $g'(2\pi)$. Let $f(x) = 2\pi$.
$(2x - 3\pi)^5 + \frac{4}{3}x + \cos x = 2\pi$.
By observation,if $x = \frac{3\pi}{2}$,then $f(\frac{3\pi}{2}) = (2(\frac{3\pi}{2}) - 3\pi)^5 + \frac{4}{3}(\frac{3\pi}{2}) + \cos(\frac{3\pi}{2}) = (0)^5 + 2\pi + 0 = 2\pi$.
Now,find $f'(x)$: $f'(x) = 5(2x - 3\pi)^4 \cdot 2 + \frac{4}{3} - \sin x = 10(2x - 3\pi)^4 + \frac{4}{3} - \sin x$.
At $x = \frac{3\pi}{2}$,$f'(\frac{3\pi}{2}) = 10(0)^4 + \frac{4}{3} - \sin(\frac{3\pi}{2}) = 0 + \frac{4}{3} - (-1) = \frac{4}{3} + 1 = \frac{7}{3}$.
Therefore,$g'(2\pi) = \frac{1}{f'(\frac{3\pi}{2})} = \frac{1}{7/3} = \frac{3}{7}$.

(D) Given $f(x) = e^{2x^3 + 3x^2 + 6x}$.
Since $g(x)$ is the inverse of $f(x)$,we have $g(f(x)) = x$.
Differentiating both sides with respect to $x$,we get $g'(f(x)) \cdot f'(x) = 1$,which implies $g'(f(x)) = \frac{1}{f'(x)}$.
First,find $f'(x)$:
$f'(x) = e^{2x^3 + 3x^2 + 6x} \cdot \frac{d}{dx}(2x^3 + 3x^2 + 6x) = e^{2x^3 + 3x^2 + 6x} \cdot (6x^2 + 6x + 6)$.
We need to find $g'(e^{11})$. Set $f(x) = e^{11}$,so $e^{2x^3 + 3x^2 + 6x} = e^{11}$.
This implies $2x^3 + 3x^2 + 6x = 11$,or $2x^3 + 3x^2 + 6x - 11 = 0$.
By inspection,$x = 1$ is a root because $2(1)^3 + 3(1)^2 + 6(1) - 11 = 2 + 3 + 6 - 11 = 0$.
Thus,$f(1) = e^{11}$.
Using the formula $g'(f(1)) = \frac{1}{f'(1)}$:
$f'(1) = e^{11} \cdot (6(1)^2 + 6(1) + 6) = e^{11} \cdot (6 + 6 + 6) = 18e^{11}$.
Therefore,$g'(e^{11}) = \frac{1}{18e^{11}}$.

(B) Given $f(x) = \ln(x + \sqrt{x^2 + 1})$.
Let $y = \ln(x + \sqrt{x^2 + 1})$.
Then $e^y = x + \sqrt{x^2 + 1}$.
Also,$e^{-y} = \frac{1}{x + \sqrt{x^2 + 1}} = \sqrt{x^2 + 1} - x$.
Subtracting the two equations: $e^y - e^{-y} = 2x$,which implies $x = \frac{e^y - e^{-y}}{2} = \sinh(y)$.
Thus,$f^{-1}(x) = \sinh(x) = \frac{e^x - e^{-x}}{2}$.
We need to find the number of solutions to $|f^{-1}(x)| = e^{-|x|}$,which is $|sinh(x)| = e^{-|x|}$.
Since both sides are even functions,we can analyze for $x \ge 0$: $\sinh(x) = e^{-x}$.
At $x = 0$,$\sinh(0) = 0$ and $e^0 = 1$,so $0 \neq 1$.
As $x$ increases,$\sinh(x)$ is strictly increasing from $0$ to $\infty$,and $e^{-x}$ is strictly decreasing from $1$ to $0$.
Therefore,there is exactly one intersection point for $x > 0$.
By symmetry,there is also exactly one intersection point for $x < 0$.
Thus,there are $2$ solutions in total.

(B) Given $f(x) = 5^{x(x - 4)} = y$.
Taking $\log_5$ on both sides,we get $\log_5 y = x(x - 4)$.
$\log_5 y = x^2 - 4x$.
To solve for $x$,we complete the square: $x^2 - 4x + 4 = \log_5 y + 4$.
$(x - 2)^2 = \log_5 y + 4$.
$x - 2 = \pm \sqrt{\log_5 y + 4}$.
Since the domain is $x \in [4, \infty)$,we must have $x - 2 \ge 2$,so we take the positive root: $x = 2 + \sqrt{\log_5 y + 4}$.
Replacing $y$ with $x$,we get $f^{-1}(x) = 2 + \sqrt{4 + \log_5 x}$.

(A) Given $f(x) = x^5 + e^{x/5}$ and $g(x) = f^{-1}(x)$.
We need to find the value of $\frac{1}{g'(1 + e^{1/5})}$.
By the property of inverse functions,$g'(y) = \frac{1}{f'(x)}$ where $y = f(x)$.
Here,$y = 1 + e^{1/5}$.
Setting $f(x) = x^5 + e^{x/5} = 1 + e^{1/5}$,we observe that $x = 1$ satisfies the equation because $1^5 + e^{1/5} = 1 + e^{1/5}$.
Now,find the derivative $f'(x) = \frac{d}{dx}(x^5 + e^{x/5}) = 5x^4 + \frac{1}{5}e^{x/5}$.
At $x = 1$,$f'(1) = 5(1)^4 + \frac{1}{5}e^{1/5} = 5 + \frac{e^{1/5}}{5}$.
Since $g'(y) = \frac{1}{f'(x)}$,we have $g'(1 + e^{1/5}) = \frac{1}{f'(1)} = \frac{1}{5 + \frac{e^{1/5}}{5}}$.
Therefore,$\frac{1}{g'(1 + e^{1/5})} = f'(1) = 5 + \frac{e^{1/5}}{5}$.

(B) Given $f(x) = \int\limits_0^x \frac{1}{\sqrt{1+t^3}} dt$. By the Fundamental Theorem of Calculus,$f'(x) = \frac{1}{\sqrt{1+x^3}}$.
Since $h(x)$ is the inverse of $f(x)$,we have $f(h(x)) = x$.
Differentiating both sides with respect to $x$,we get $f'(h(x)) \cdot h'(x) = 1$,which implies $h'(x) = \frac{1}{f'(h(x))} = \sqrt{1 + h^3(x)}$.
Now,differentiate $h'(x) = (1 + h^3(x))^{1/2}$ with respect to $x$:
$h''(x) = \frac{1}{2}(1 + h^3(x))^{-1/2} \cdot (3h^2(x) \cdot h'(x))$.
Substitute $h'(x) = \sqrt{1 + h^3(x)}$ into the expression:
$h''(x) = \frac{3h^2(x) \cdot \sqrt{1 + h^3(x)}}{2 \sqrt{1 + h^3(x)}} = \frac{3}{2} h^2(x)$.
Therefore,$\frac{h''(x)}{h^2(x)} = \frac{3}{2}$.