Composition of Functions Questions in English

(A) Given $f(x) = x^{2} + 2x - 3$ and $g(x) = x + 1$.
We need to find $x$ such that $f(g(x)) = g(f(x))$.
First, calculate $f(g(x))$:
$f(g(x)) = f(x + 1) = (x + 1)^{2} + 2(x + 1) - 3 = (x^{2} + 2x + 1) + 2x + 2 - 3 = x^{2} + 4x$.
Next, calculate $g(f(x))$:
$g(f(x)) = g(x^{2} + 2x - 3) = (x^{2} + 2x - 3) + 1 = x^{2} + 2x - 2$.
Equating both expressions:
$x^{2} + 4x = x^{2} + 2x - 2$.
Subtracting $x^{2}$ from both sides:
$4x = 2x - 2$.
$4x - 2x = -2$.
$2x = -2$.
$x = -1$.

(A) Given functions are $f(x) = 5 - x^2$ and $g(x) = 3x - 4$.
To find $(f \circ g)(-1)$, we use the definition of composition of functions: $(f \circ g)(x) = f(g(x))$.
First, calculate $g(-1)$:
$g(-1) = 3(-1) - 4 = -3 - 4 = -7$.
Now, substitute this value into $f(x)$:
$(f \circ g)(-1) = f(g(-1)) = f(-7)$.
Using the definition of $f(x)$:
$f(-7) = 5 - (-7)^2 = 5 - 49 = -44$.
Thus, the value of $(f \circ g)(-1)$ is $-44$.

(C) Let $x_1, x_2 \in S$ such that $f(x_1) = f(x_2)$.
Applying $g$ to both sides, we get $g(f(x_1)) = g(f(x_2))$.
This is equivalent to $(g \circ f)(x_1) = (g \circ f)(x_2)$.
Since $g \circ f$ is given as an injective mapping, $(g \circ f)(x_1) = (g \circ f)(x_2)$ implies $x_1 = x_2$.
Since $f(x_1) = f(x_2)$ leads to $x_1 = x_2$, the function $f$ must be injective.
Therefore, $f$ is necessarily injective.

(B) We are given that $g \circ f: S \rightarrow U$ is a surjective (onto) function.
By definition of a surjective function, for every element $z \in U$, there exists at least one element $x \in S$ such that $(g \circ f)(x) = z$.
This can be written as $g(f(x)) = z$.
Let $y = f(x)$. Since $x \in S$ and $f: S \rightarrow T$, it follows that $y \in T$.
Substituting this into the equation, we get $g(y) = z$.
Since for every $z \in U$, we have found an element $y \in T$ such that $g(y) = z$, it follows that $g: T \rightarrow U$ is a surjective function.
However, $f$ does not necessarily have to be surjective because the elements in $T$ that are not in the image of $f$ do not affect the surjectivity of $g \circ f$ as long as the image of $f$ covers enough elements in $T$ to map onto all of $U$ through $g$.

(C) Given $f: R \rightarrow R$ where $f(x) = e^{x}$ and $g: R \rightarrow R$ where $g(x) = x^{2}$.
We calculate the composite function $(g \circ f)(x) = g(f(x)) = g(e^{x}) = (e^{x})^{2} = e^{2x}$.
For $(g \circ f)(x) = e^{2x}$, if $(g \circ f)(x_{1}) = (g \circ f)(x_{2})$, then $e^{2x_{1}} = e^{2x_{2}}$, which implies $2x_{1} = 2x_{2}$, so $x_{1} = x_{2}$. Thus, $g \circ f$ is injective.
However, the range of $g \circ f$ is $(0, \infty)$, which is not equal to the codomain $R$, so $g \circ f$ is not surjective.
For $g(x) = x^{2}$, $g(-1) = 1$ and $g(1) = 1$, so $g$ is not injective. Also, the range of $g$ is $[0, \infty)$, so $g$ is not surjective.
Therefore, $g \circ f$ is injective but $g$ is not bijective.

(A) Given $f(x) = \frac{x}{x+1}$.
Calculating the first few terms:
$f_1(x) = \frac{x}{x+1}$
$f_2(x) = f(f(x)) = \frac{\frac{x}{x+1}}{\frac{x}{x+1} + 1} = \frac{x}{x + x + 1} = \frac{x}{2x+1}$
$f_3(x) = f(f_2(x)) = \frac{\frac{x}{2x+1}}{\frac{x}{2x+1} + 1} = \frac{x}{x + 2x + 1} = \frac{x}{3x+1}$
By induction, $f_n(x) = \frac{x}{nx+1}$.
Evaluating at $x = -2$:
$f_n(-2) = \frac{-2}{n(-2)+1} = \frac{-2}{-2n+1} = \frac{2}{2n-1}$.
The product is $P = f_1(-2) \cdot f_2(-2) \cdot \ldots \cdot f_n(-2) = \prod_{k=1}^{n} \frac{2}{2k-1} = \frac{2^n}{1 \cdot 3 \cdot 5 \cdot \ldots \cdot (2n-1)}$.

(C) Given $g(x) = 3x^{2} + 2x - 3$. First, calculate $g(2)$:
$g(2) = 3(2)^{2} + 2(2) - 3 = 12 + 4 - 3 = 13$.
We need to find $f(g(2)) = f(13)$.
Given $4g(f(x)) = 3x^{2} - 32x + 72$, substitute $g(f(x)) = 3(f(x))^{2} + 2f(x) - 3$:
$4[3(f(x))^{2} + 2f(x) - 3] = 3x^{2} - 32x + 72$
$12(f(x))^{2} + 8f(x) - 12 = 3x^{2} - 32x + 72$
$12(f(x))^{2} + 8f(x) - (3x^{2} - 32x + 84) = 0$.
Using the quadratic formula for $f(x)$:
$f(x) = \frac{-8 \pm \sqrt{64 - 4(12)(-(3x^{2} - 32x + 84))}}{24} = \frac{-8 \pm \sqrt{64 + 48(3x^{2} - 32x + 84)}}{24}$
$f(x) = \frac{-8 \pm \sqrt{144x^{2} - 1536x + 4096}}{24} = \frac{-8 \pm \sqrt{(12x - 64)^{2}}}{24} = \frac{-8 \pm (12x - 64)}{24}$.
Since $f(0) = -3$, we test the signs at $x=0$:
If we take the positive sign: $f(0) = \frac{-8 + (-64)}{24} = -3$ (Correct).
So, $f(x) = \frac{-8 + 12x - 64}{24} = \frac{12x - 72}{24} = \frac{x - 6}{2}$.
Finally, $f(13) = \frac{13 - 6}{2} = \frac{7}{2}$.

(A) Given $g(x) = x^2 + x - 2$ and $(g \circ f)(x) = g(f(x)) = 2x^2 - 5x + 2$.
Let $f(x) = y$. Then $g(y) = y^2 + y - 2 = 2x^2 - 5x + 2$.
$y^2 + y - 2 - (2x^2 - 5x + 2) = 0$.
$y^2 + y - (2x^2 - 5x + 4) = 0$.
Using the quadratic formula $y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ where $a=1, b=1, c=-(2x^2 - 5x + 4)$:
$y = \frac{-1 \pm \sqrt{1^2 - 4(1)(-(2x^2 - 5x + 4))}}{2}$.
$y = \frac{-1 \pm \sqrt{1 + 8x^2 - 20x + 16}}{2} = \frac{-1 \pm \sqrt{8x^2 - 20x + 17}}{2}$.
Since the options provided do not match this result, we test $f(x) = 2x - 3$ in $g(f(x))$:
$g(2x - 3) = (2x - 3)^2 + (2x - 3) - 2 = (4x^2 - 12x + 9) + 2x - 3 - 2 = 4x^2 - 10x + 4 \neq 2x^2 - 5x + 2$.
Testing $f(x) = 2x - 3$ was incorrect. Let's re-evaluate $g(f(x)) = (f(x)+2)(f(x)-1) = 2x^2 - 5x + 2 = (2x-1)(x-2)$.
This implies $f(x) = 2x-1$ or $f(x) = x-2$. None of the options match. The question options are incorrect.

(A) Step $1$: Calculate $(f \circ f)(x)$.
$(f \circ f)(x) = f(f(x)) = \frac{4(\frac{4x+3}{6x-4}) + 3}{6(\frac{4x+3}{6x-4}) - 4}$.
Step $2$: Simplify the expression.
$= \frac{16x + 12 + 18x - 12}{24x + 18 - 24x + 16} = \frac{34x}{34} = x$.
Step $3$: Since $(f \circ f)(x) = x$, then $g(x) = x$.
Step $4$: The composition $(g \circ g \circ g \circ g \circ g)(x)$ is equivalent to $g(g(g(g(g(x)))))$.
Since $g(x) = x$, then $g(g(g(g(g(x))))) = x$.
Step $5$: Evaluate at $x = 3$.
$(g \circ g \circ g \circ g \circ g)(3) = 3$.

(D) Step $1$: Calculate $f(x)$ and $g(x)$ for specific values.
For $x \ge 0$, $f(x) = 2x - x = x$ and $g(x) = 2x + x = 3x$.
For $x < 0$, $f(x) = 2x - (-x) = 3x$ and $g(x) = 2x + (-x) = x$.
Step $2$: Evaluate $(fog)(2)$ and $(gof)(2)$.
$(fog)(2) = f(g(2)) = f(3(2)) = f(6) = 6$.
$(gof)(2) = g(f(2)) = g(2) = 3(2) = 6$.
Step $3$: Evaluate $(fog)(-2)$ and $(gof)(-2)$.
$(fog)(-2) = f(g(-2)) = f(-2) = 3(-2) = -6$.
$(gof)(-2) = g(f(-2)) = g(3(-2)) = g(-6) = -6$.
Step $4$: Check the options.
Option $(B)$: $(fog)(2) - (gof)(-2) = 6 - (-6) = 12 \neq 0$.
Option $(C)$: $(fog)(2) - (fog)(-2) = 6 - (-6) = 12 \neq 0$.
Option $(D)$: $(gof)(2) + (gof)(-2) = 6 + (-6) = 0$. Thus, option $(D)$ is correct.

(A) Given $f(x) = \frac{1 - x}{1 + x}$.
First, calculate $f(\cos x) = \frac{1 - \cos x}{1 + \cos x}$.
Using trigonometric identities $1 - \cos x = 2 \sin^2 \frac{x}{2}$ and $1 + \cos x = 2 \cos^2 \frac{x}{2}$, we get $f(\cos x) = \frac{2 \sin^2 (x/2)}{2 \cos^2 (x/2)} = \tan^2 \frac{x}{2}$.
Now, calculate $f(f(\cos x)) = f(\tan^2 \frac{x}{2}) = \frac{1 - \tan^2 (x/2)}{1 + \tan^2 (x/2)}$.
Using the identity $\cos 2\theta = \frac{1 - \tan^2 \theta}{1 + \tan^2 \theta}$, where $\theta = \frac{x}{2}$, we get $f(f(\cos x)) = \cos(2 \cdot \frac{x}{2}) = \cos x$.

(C) Step $1$: Compute $f(g(x)) = f(cx + d) = a(cx + d) + b = acx + ad + b$.
Step $2$: Compute $g(f(x)) = g(ax + b) = c(ax + b) + d = cax + cb + d$.
Step $3$: Equate the two expressions: $acx + ad + b = cax + cb + d$.
Step $4$: Since this holds for all $x$, the constant terms must be equal: $ad + b = cb + d$.
Step $5$: Rearrange the equation: $ad - d = cb - b$, which is $d(a - 1) = b(c - 1)$.
Step $6$: Alternatively, observe that $f(d) = ad + b$ and $g(b) = cb + d$. From the equality $ad + b = cb + d$, we see that $f(d) = g(b)$.

(C) Step $1$: Calculate $(f \circ g)(\frac{3}{2}) = f(g(\frac{3}{2}))$.
$g(\frac{3}{2}) = [(\frac{3}{2})^2] = [\frac{9}{4}] = [2.25] = 2$.
$f(g(\frac{3}{2})) = f(2) = 2^2 = 4$.
Step $2$: Calculate $(g \circ f)(\frac{3}{2}) = g(f(\frac{3}{2}))$.
$f(\frac{3}{2}) = (\frac{3}{2})^2 = \frac{9}{4} = 2.25$.
$g(f(\frac{3}{2})) = [2.25^2] = [5.0625] = 5$.
Step $3$: Add the results.
$(f \circ g)(\frac{3}{2}) + (g \circ f)(\frac{3}{2}) = 4 + 5 = 9$.

(B) Given $g(x) = 1 - \sqrt{x}$.
We want to find $f(6)$, so set $g(x) = 6$.
$1 - \sqrt{x} = 6 \implies -\sqrt{x} = 5 \implies \sqrt{x} = -5$.
Since $\sqrt{x}$ must be non-negative, this implies $x = 25$.
Substitute $\sqrt{x} = -5$ into the expression for $f(g(x))$:
$f(6) = 5 + 4(-5) + (-5)^2$.
$f(6) = 5 - 20 + 25$.
$f(6) = 10$.