Maxima and Minima Questions in English

(A) Consider the function $f(x) = x^{1/x}$ for $x > 0$.
Taking the natural logarithm,$\ln(f(x)) = \frac{1}{x} \ln(x)$.
Differentiating with respect to $x$,$\frac{f'(x)}{f(x)} = \frac{x(\frac{1}{x}) - \ln(x)(1)}{x^2} = \frac{1 - \ln(x)}{x^2}$.
Thus,$f'(x) = x^{1/x} \left( \frac{1 - \ln(x)}{x^2} \right)$.
$f'(x) > 0$ when $1 - \ln(x) > 0$,i.e.,$\ln(x) < 1$,which means $0 < x < e$.
$f'(x) < 0$ when $1 - \ln(x) < 0$,i.e.,$\ln(x) > 1$,which means $x > e$.
So,Statement-$II$ is true.
Now,$e \approx 2.718$. Since $2 < e < 3$,the function $f(x)$ increases on $(0, e)$ and decreases on $(e, \infty)$.
Comparing $2^{1/2}$ and $3^{1/3}$: Since $2 < 3 < e$,$f(2) < f(3)$,so $2^{1/2} < 3^{1/3}$.
Comparing $3^{1/3}$ and $4^{1/4}$: Since $3 < e < 4$,$f(3)$ is the maximum value of the function $f(x)$ for integer values of $x$.
Since $f(x)$ decreases for $x > e$,$f(3) > f(4) > f(5) > f(6) > f(7)$.
Thus,$3^{1/3}$ is indeed the maximum value among the given numbers.
Therefore,Statement-$I$ is true and Statement-$II$ is the correct explanation for Statement-$I$.

(B) Given the function $f(x) = \sqrt{x^2 + x} + \frac{\tan^2 \alpha}{\sqrt{x^2 + x}}$.
Since $x > 0$,the term $\sqrt{x^2 + x}$ is always positive.
We use the Arithmetic Mean-Geometric Mean $(AM \ge GM)$ inequality,which states that for positive real numbers $a$ and $b$,$\frac{a + b}{2} \ge \sqrt{ab}$,or $a + b \ge 2\sqrt{ab}$.
Let $a = \sqrt{x^2 + x}$ and $b = \frac{\tan^2 \alpha}{\sqrt{x^2 + x}}$.
Then $f(x) = a + b \ge 2\sqrt{a \cdot b}$.
$f(x) \ge 2\sqrt{\sqrt{x^2 + x} \cdot \frac{\tan^2 \alpha}{\sqrt{x^2 + x}}}$.
$f(x) \ge 2\sqrt{\tan^2 \alpha}$.
Since $\alpha \in (0, \pi/2)$,$\tan \alpha > 0$,so $f(x) \ge 2 \tan \alpha$.
The minimum value of $f(x)$ is $2 \tan \alpha$.

(C) Let $f(x) = x^5 - 5x^4 + 5x^3 - 10$.
First,find the derivative: $f'(x) = 5x^4 - 20x^3 + 15x^2 = 5x^2(x^2 - 4x + 3) = 5x^2(x - 1)(x - 3)$.
For critical points,set $f'(x) = 0$,which gives $x = 0, 1, 3$.
Now,find the second derivative: $f''(x) = 20x^3 - 60x^2 + 30x$.
Check the nature of critical points:
At $x = 1$: $f''(1) = 20(1)^3 - 60(1)^2 + 30(1) = 20 - 60 + 30 = -10 < 0$. Since $f''(1) < 0$,$x = 1$ is a point of local maxima. Thus,$p = 1$.
At $x = 3$: $f''(3) = 20(3)^3 - 60(3)^2 + 30(3) = 540 - 540 + 90 = 90 > 0$. Since $f''(3) > 0$,$x = 3$ is a point of local minima. Thus,$q = 3$.
At $x = 0$: $f''(0) = 0$. Using the first derivative test,$f'(x)$ does not change sign around $x = 0$ (it is positive on both sides),so $x = 0$ is a point of inflection,not a local extremum.
Therefore,$p = 1$ and $q = 3$.

(B) Given $f(x) = (x + 1)^{\frac{1}{3}} - (x - 1)^{\frac{1}{3}}$.
First,find the derivative $f'(x)$:
$f'(x) = \frac{1}{3}(x + 1)^{-\frac{2}{3}} - \frac{1}{3}(x - 1)^{-\frac{2}{3}} = \frac{1}{3} \left[ \frac{1}{(x + 1)^{\frac{2}{3}}} - \frac{1}{(x - 1)^{\frac{2}{3}}} \right]$.
For critical points,set $f'(x) = 0$:
$\frac{1}{(x + 1)^{\frac{2}{3}}} = \frac{1}{(x - 1)^{\frac{2}{3}}} \implies (x + 1)^2 = (x - 1)^2$.
$x^2 + 2x + 1 = x^2 - 2x + 1 \implies 4x = 0 \implies x = 0$.
Now,evaluate $f(x)$ at the critical point and the endpoints of the interval $[0, 1]$:
At $x = 0$: $f(0) = (0 + 1)^{\frac{1}{3}} - (0 - 1)^{\frac{1}{3}} = 1 - (-1) = 2$.
At $x = 1$: $f(1) = (1 + 1)^{\frac{1}{3}} - (1 - 1)^{\frac{1}{3}} = 2^{\frac{1}{3}} - 0 = \sqrt[3]{2} \approx 1.26$.
Comparing the values $f(0) = 2$ and $f(1) = \sqrt[3]{2}$,the maximum value is $2$.

(A) Given the function $f(x) = x^3 - px + q$.
To find the critical points,we find the first derivative: $f'(x) = 3x^2 - p$.
Setting $f'(x) = 0$,we get $3x^2 = p$,which implies $x^2 = \frac{p}{3}$,so $x = \pm \sqrt{\frac{p}{3}}$.
Now,we find the second derivative to determine the nature of these points: $f''(x) = 6x$.
At $x = \sqrt{\frac{p}{3}}$,$f''(\sqrt{\frac{p}{3}}) = 6\sqrt{\frac{p}{3}} > 0$ (since $p > 0$). Thus,$f(x)$ has a local minimum at $x = \sqrt{\frac{p}{3}}$.
At $x = -\sqrt{\frac{p}{3}}$,$f''(-\sqrt{\frac{p}{3}}) = -6\sqrt{\frac{p}{3}} < 0$. Thus,$f(x)$ has a local maximum at $x = -\sqrt{\frac{p}{3}}$.
Therefore,option $A$ is correct.

(C) Let $y = x^x$. Taking the natural logarithm on both sides,we get $\ln y = x \ln x$.
Differentiating with respect to $x$,we have $\frac{1}{y} \frac{dy}{dx} = 1 \cdot \ln x + x \cdot \frac{1}{x} = \ln x + 1$.
Thus,$\frac{dy}{dx} = y(\ln x + 1) = x^x(\ln x + 1)$.
For critical points,set $\frac{dy}{dx} = 0$. Since $x^x > 0$ for $x > 0$,we have $\ln x + 1 = 0$,which implies $\ln x = -1$,so $x = e^{-1}$.
To check for the minimum,we find the second derivative or use the first derivative test. For $x < e^{-1}$,$\ln x < -1$,so $\frac{dy}{dx} < 0$. For $x > e^{-1}$,$\ln x > -1$,so $\frac{dy}{dx} > 0$.
Since the derivative changes sign from negative to positive at $x = e^{-1}$,the function attains its minimum value at $x = e^{-1}$.

(B) Let $f(x) = (1/x)^x = x^{-x}$.
Taking the natural logarithm on both sides,$\ln(f(x)) = -x \ln(x)$.
Differentiating with respect to $x$,$\frac{f'(x)}{f(x)} = -(\ln(x) + x \cdot \frac{1}{x}) = -(\ln(x) + 1)$.
So,$f'(x) = -(1/x)^x (\ln(x) + 1)$.
For critical points,set $f'(x) = 0$,which implies $\ln(x) + 1 = 0$,so $\ln(x) = -1$,which gives $x = 1/e$.
To check for maxima,we evaluate the second derivative or observe the sign change of $f'(x)$.
For $x < 1/e$,$\ln(x) < -1$,so $f'(x) > 0$.
For $x > 1/e$,$\ln(x) > -1$,so $f'(x) < 0$.
Since $f'(x)$ changes from positive to negative at $x = 1/e$,$f(x)$ has a local maximum at $x = 1/e$.
The maximum value is $f(1/e) = (1/(1/e))^{1/e} = e^{1/e}$.

(C) Given $f(x) = \int_{0}^{x} \frac{\sin t}{t} dt$.
By the Fundamental Theorem of Calculus,$f'(x) = \frac{\sin x}{x}$ for $x > 0$.
To find critical points,set $f'(x) = 0$,which implies $\sin x = 0$,so $x = n\pi$ for $n = 1, 2, 3, \dots$.
The second derivative is $f''(x) = \frac{x \cos x - \sin x}{x^2}$.
At $x = n\pi$,$f''(n\pi) = \frac{n\pi \cos(n\pi) - \sin(n\pi)}{(n\pi)^2} = \frac{n\pi (-1)^n - 0}{(n\pi)^2} = \frac{(-1)^n}{n\pi}$.
If $n$ is odd,$f''(n\pi) = \frac{-1}{n\pi} < 0$,which indicates a local maximum.
If $n$ is even,$f''(n\pi) = \frac{1}{n\pi} > 0$,which indicates a local minimum.
Thus,the function has a maximum at $x = n\pi$ where $n$ is odd.

(C) Let $f(x) = \sin x(1 + \cos x) = \sin x + \sin x \cos x = \sin x + \frac{1}{2} \sin 2x$.
To find the critical points,we differentiate $f(x)$ with respect to $x$:
$f'(x) = \cos x + \cos 2x$.
Set $f'(x) = 0$ for extrema:
$\cos x + \cos 2x = 0 \implies \cos 2x = -\cos x$.
Using the identity $-\cos x = \cos(\pi - x)$:
$\cos 2x = \cos(\pi - x) \implies 2x = \pi - x \implies 3x = \pi \implies x = \pi / 3$.
Now,find the second derivative $f''(x)$ to check for maxima:
$f''(x) = -\sin x - 2 \sin 2x$.
At $x = \pi / 3$:
$f''(\pi / 3) = -\sin(\pi / 3) - 2 \sin(2\pi / 3) = -\frac{\sqrt{3}}{2} - 2(\frac{\sqrt{3}}{2}) = -\frac{\sqrt{3}}{2} - \sqrt{3} = -\frac{3\sqrt{3}}{2}$.
Since $f''(\pi / 3) < 0$,the function has a maximum value at $x = \pi / 3$.

(D) Given $f(x) = x^{25}(1 - x)^{75}$.
To find the maximum,we calculate the derivative $f'(x)$ using the product rule:
$f'(x) = 25x^{24}(1 - x)^{75} + x^{25} \cdot 75(1 - x)^{74} \cdot (-1)$
$f'(x) = 25x^{24}(1 - x)^{74} [(1 - x) - 3x]$
$f'(x) = 25x^{24}(1 - x)^{74} (1 - 4x)$
Setting $f'(x) = 0$ for critical points in $(0, 1)$:
$1 - 4x = 0 \implies x = 1/4$.
Since $f(0) = 0$ and $f(1) = 0$,and $f(x) > 0$ for $x \in (0, 1)$,the function must attain its maximum at $x = 1/4$.

(D) Given $y = a \log x + bx^2 + x$.
Find the derivative: $\frac{dy}{dx} = \frac{a}{x} + 2bx + 1$.
Since the function has extreme values at $x = 1$ and $x = 2$,the derivative must be zero at these points.
At $x = 1$: $\frac{a}{1} + 2b(1) + 1 = 0 \implies a + 2b = -1$ (Equation $1$).
At $x = 2$: $\frac{a}{2} + 2b(2) + 1 = 0 \implies \frac{a}{2} + 4b = -1 \implies a + 8b = -2$ (Equation $2$).
Subtracting Equation $1$ from Equation $2$: $(a + 8b) - (a + 2b) = -2 - (-1) \implies 6b = -1 \implies b = -\frac{1}{6}$.
Substitute $b = -\frac{1}{6}$ into Equation $1$: $a + 2(-\frac{1}{6}) = -1 \implies a - \frac{1}{3} = -1 \implies a = -1 + \frac{1}{3} = -\frac{2}{3}$.
Thus,$(a, b) = \left( -\frac{2}{3}, -\frac{1}{6} \right)$.

(C) Given function is $f(x) = x^3 - 18x^2 + 96x$.
Find the first derivative: $f'(x) = 3x^2 - 36x + 96$.
Set $f'(x) = 0$ to find critical points: $3(x^2 - 12x + 32) = 0$.
Factor the quadratic: $3(x - 4)(x - 8) = 0$.
Thus,the critical points are $x = 4$ and $x = 8$.
Find the second derivative: $f''(x) = 6x - 36$.
Check for local maxima at $x = 4$: $f''(4) = 6(4) - 36 = 24 - 36 = -12 < 0$.
Since $f''(4) < 0$,$x = 4$ is a point of local maxima.
Calculate the maximum value: $f(4) = (4)^3 - 18(4)^2 + 96(4) = 64 - 18(16) + 384 = 64 - 288 + 384 = 160$.
Therefore,the maximum value in the interval $(0, 9)$ is $160$.

(B) Given the curve $y = f(x) = -x^3 + 3x^2 + 9x - 27$.
The slope of the curve is given by the derivative $f'(x) = -3x^2 + 6x + 9$.
Let $g(x) = f'(x) = -3x^2 + 6x + 9$ represent the slope function.
To find the maximum slope,we differentiate $g(x)$ with respect to $x$:
$g'(x) = -6x + 6$.
Setting $g'(x) = 0$ gives $-6x + 6 = 0$,which implies $x = 1$.
Now,we check the second derivative $g''(x) = -6$. Since $g''(x) < 0$,the function $g(x)$ attains a maximum at $x = 1$.
The maximum slope is $g(1) = -3(1)^2 + 6(1) + 9 = -3 + 6 + 9 = 12$.

(D) Let $f(x) = \sin x - x \cos x$.
First,find the derivative: $f'(x) = \cos x - (\cos x - x \sin x) = x \sin x$.
For critical points,set $f'(x) = 0$,which gives $x \sin x = 0$. Since $x = n\pi$,this is satisfied for any integer $n$.
Next,find the second derivative: $f''(x) = \sin x + x \cos x$.
Evaluate $f''(x)$ at $x = n\pi$: $f''(n\pi) = \sin(n\pi) + n\pi \cos(n\pi) = 0 + n\pi (-1)^n = n\pi (-1)^n$.
For $f(x)$ to have a maximum at $x = n\pi$,we must have $f''(n\pi) < 0$.
Thus,$n\pi (-1)^n < 0$. Since $\pi > 0$,we require $n(-1)^n < 0$.
Case $1$: If $n$ is an odd positive integer (e.g.,$n=1, 3, 5$),then $n(-1)^n = n(-1) = -n < 0$. This satisfies the condition.
Case $2$: If $n$ is an even negative integer (e.g.,$n=-2, -4$),then $n(-1)^n = n(1) = n < 0$. This also satisfies the condition.
Therefore,$n$ is an odd positive integer or an even negative integer.

(A) Consider the function $f(x) = x^{1/x}$ for $x > 0$.
Taking the natural logarithm,we get $\ln(f(x)) = \frac{1}{x} \ln(x)$.
Differentiating with respect to $x$,we get $\frac{f'(x)}{f(x)} = \frac{x(\frac{1}{x}) - \ln(x)(1)}{x^2} = \frac{1 - \ln(x)}{x^2}$.
Setting $f'(x) = 0$,we find $1 - \ln(x) = 0$,which implies $\ln(x) = 1$,so $x = e$.
For $x < e$,$f'(x) > 0$ and for $x > e$,$f'(x) < 0$. Thus,$f(x)$ has a global maximum at $x = e$. This proves Statement-$II$ is true.
Since $f(x)$ has a global maximum at $x = e$,for any $x \neq e$,$f(e) > f(x)$.
Substituting $x = \pi$,we have $e^{1/e} > \pi^{1/\pi}$.
Raising both sides to the power of $e\pi$,we get $(e^{1/e})^{e\pi} > (\pi^{1/\pi})^{e\pi}$,which simplifies to $e^{\pi} > \pi^e$. This proves Statement-$I$ is true.
Since Statement-$II$ directly leads to the inequality in Statement-$I$,Statement-$II$ is the correct explanation for Statement-$I$.

(D) Given the equation $x + 2y = 8$,we can express $x$ as $x = 8 - 2y$.
Let $f(y) = xy = (8 - 2y)y = 8y - 2y^2$.
To find the maximum value,we differentiate $f(y)$ with respect to $y$:
$f'(y) = \frac{d}{dy}(8y - 2y^2) = 8 - 4y$.
Setting the first derivative to zero for critical points:
$8 - 4y = 0 \implies y = 2$.
Now,we find the second derivative to check for maxima:
$f''(y) = -4$.
Since $f''(2) = -4 < 0$,the function $f(y)$ has a local maximum at $y = 2$.
Substituting $y = 2$ back into the expression for $f(y)$:
$f(2) = 8(2) - 2(2)^2 = 16 - 8 = 8$.
Thus,the maximum value of $xy$ is $8$.

(B) Let $f(x) = e^x \sin x$. The slope of the tangent is given by $f'(x) = e^x \sin x + e^x \cos x = e^x(\sin x + \cos x)$.
Let $g(x) = f'(x) = e^x(\sin x + \cos x)$.
To find the maximum slope,we find $g'(x) = e^x(\sin x + \cos x) + e^x(\cos x - \sin x) = 2e^x \cos x$.
Setting $g'(x) = 0$ for critical points,we get $2e^x \cos x = 0$. Since $e^x \neq 0$,we have $\cos x = 0$.
In the interval $[0, 2\pi]$,$\cos x = 0$ at $x = \pi / 2$ and $x = 3\pi / 2$.
Checking the second derivative $g''(x) = 2e^x \cos x - 2e^x \sin x = 2e^x(\cos x - \sin x)$.
At $x = \pi / 2$,$g''(\pi / 2) = 2e^{\pi / 2}(0 - 1) = -2e^{\pi / 2} < 0$,which indicates a local maximum.
At $x = 3\pi / 2$,$g''(3\pi / 2) = 2e^{3\pi / 2}(0 - (-1)) = 2e^{3\pi / 2} > 0$,which indicates a local minimum.
Thus,the slope is maximum at $x = \pi / 2$.

(B) Let $f(x) = ax + \frac{b}{x} - c$,where $x > 0$ and $a, b > 0$.
To find the minimum value,we differentiate $f(x)$ with respect to $x$:
$f'(x) = a - \frac{b}{x^2} = \frac{ax^2 - b}{x^2}$.
Setting $f'(x) = 0$,we get $ax^2 = b$,which implies $x = \sqrt{\frac{b}{a}}$ (since $x > 0$).
The function $f(x)$ attains its minimum value at this point.
Substituting $x = \sqrt{\frac{b}{a}}$ into $f(x)$:
$f\left(\sqrt{\frac{b}{a}}\right) = a\sqrt{\frac{b}{a}} + \frac{b}{\sqrt{b/a}} - c = \sqrt{ab} + \sqrt{ab} - c = 2\sqrt{ab} - c$.
According to the problem,$f(x) \ge 0$ for all $x > 0$,so the minimum value must be $\ge 0$.
Therefore,$2\sqrt{ab} - c \ge 0$,which means $2\sqrt{ab} \ge c$.
Squaring both sides,we get $4ab \ge c^2$,or $ab \ge \frac{c^2}{4}$.

(B) Given the function $y = a \log|x| + bx^2 + x$.
The derivative is $\frac{dy}{dx} = \frac{a}{x} + 2bx + 1$.
Since $x = -1$ and $x = 2$ are extreme points,the derivative must be zero at these points.
For $x = -1$: $\frac{a}{-1} + 2b(-1) + 1 = 0 \implies -a - 2b + 1 = 0 \implies a + 2b = 1$.
For $x = 2$: $\frac{a}{2} + 2b(2) + 1 = 0 \implies \frac{a}{2} + 4b + 1 = 0 \implies a + 8b = -2$.
Subtracting the first equation from the second: $(a + 8b) - (a + 2b) = -2 - 1 \implies 6b = -3 \implies b = -1/2$.
Substituting $b = -1/2$ into $a + 2b = 1$: $a + 2(-1/2) = 1 \implies a - 1 = 1 \implies a = 2$.
Thus,$a = 2$ and $b = -1/2$.

(C) Given $f(x) = x^3 + ax^2 + bx + c$.
Find the derivative: $f'(x) = 3x^2 + 2ax + b$.
Since the function has extrema at $x = 3$ and $x = -1$,$f'(x) = 0$ at these points.
Thus,the roots of $3x^2 + 2ax + b = 0$ are $x = 3$ and $x = -1$.
This implies $f'(x) = k(x - 3)(x + 1) = k(x^2 - 2x - 3) = kx^2 - 2kx - 3k$.
Comparing this with $3x^2 + 2ax + b$,we get $k = 3$.
So,$f'(x) = 3(x^2 - 2x - 3) = 3x^2 - 6x - 9$.
Comparing coefficients: $2a = -6 \implies a = -3$ and $b = -9$.
Since $c$ is a constant of integration,it can be any real number,$c \in \mathbb{R}$.

(A) Let the first number be $(3 - x)$ and the second number be $x$.
Define the function $f(x) = (3 - x)x^2 = 3x^2 - x^3$.
To find the critical points,calculate the first derivative:
$f'(x) = 6x - 3x^2$.
Set $f'(x) = 0$:
$3x(2 - x) = 0 \Rightarrow x = 0$ or $x = 2$.
Find the second derivative to check for maxima:
$f''(x) = 6 - 6x$.
Evaluate at $x = 2$:
$f''(2) = 6 - 6(2) = -6 < 0$.
Since the second derivative is negative,$x = 2$ gives the maximum value.
Maximum value = $(3 - 2) \times 2^2 = 1 \times 4 = 4$.

(A) Given function: $f(x) = 2x^3 - 21x^2 + 36x - 20$.
Find the first derivative: $f'(x) = 6x^2 - 42x + 36$.
Set $f'(x) = 0$ to find critical points: $6(x^2 - 7x + 6) = 0 \Rightarrow 6(x - 1)(x - 6) = 0$.
Thus,the critical points are $x = 1$ and $x = 6$.
Find the second derivative: $f''(x) = 12x - 42$.
Evaluate at critical points:
For $x = 1$,$f''(1) = 12(1) - 42 = -30 < 0$ (Local maximum).
For $x = 6$,$f''(6) = 12(6) - 42 = 72 - 42 = 30 > 0$ (Local minimum).
Calculate the minimum value at $x = 6$:
$f(6) = 2(6)^3 - 21(6)^2 + 36(6) - 20$
$f(6) = 2(216) - 21(36) + 216 - 20$
$f(6) = 432 - 756 + 216 - 20 = -128$.
Therefore,the minimum value is $-128$.

(C) To check the behavior of $f(x) = \frac{x^2}{x^3 + 200}$,we find its derivative $f'(x)$.
Using the quotient rule: $f'(x) = \frac{(x^3 + 200)(2x) - (x^2)(3x^2)}{(x^3 + 200)^2} = \frac{2x^4 + 400x - 3x^4}{(x^3 + 200)^2} = \frac{400x - x^4}{(x^3 + 200)^2} = \frac{x(400 - x^3)}{(x^3 + 200)^2}$.
For local extrema,set $f'(x) = 0$,which gives $x = 0$ or $x^3 = 400$,so $x = \sqrt[3]{400} \approx 7.368$.
Since $f'(x) > 0$ for $x < \sqrt[3]{400}$ and $f'(x) < 0$ for $x > \sqrt[3]{400}$,the function has a local maximum at $x = \sqrt[3]{400}$.
For the sequence $a_n$,we compare $a_7$ and $a_8$.
$a_7 = \frac{49}{343 + 200} = \frac{49}{543} \approx 0.0902$.
$a_8 = \frac{64}{512 + 200} = \frac{64}{712} = \frac{8}{89} \approx 0.0898$.
Since $a_7 > a_8$ and $a_7 > a_6$ (as $f(x)$ is increasing until $x \approx 7.368$),$a_7$ is indeed the largest term.
However,Statement-$II$ claims the maximum is at $x = 7$,which is incorrect as the maximum is at $x = \sqrt[3]{400} \approx 7.368$.

(D) Let $r$ be the radius and $\theta$ be the central angle in radians.
The perimeter of the sector is given by $P = 2r + r\theta = p$.
From this,we have $\theta = \frac{p - 2r}{r} = \frac{p}{r} - 2$.
The area $A$ of the sector is given by $A = \frac{1}{2} r^2 \theta$.
Substituting the value of $\theta$,we get $A = \frac{1}{2} r^2 \left( \frac{p - 2r}{r} \right) = \frac{1}{2} r(p - 2r) = \frac{1}{2} pr - r^2$.
To find the maximum area,we differentiate $A$ with respect to $r$:
$\frac{dA}{dr} = \frac{1}{2} p - 2r$.
Setting $\frac{dA}{dr} = 0$,we get $\frac{1}{2} p = 2r$,which implies $r = \frac{p}{4}$.
To verify the maximum,we check the second derivative: $\frac{d^2A}{dr^2} = -2 < 0$,which confirms that the area is maximum at $r = \frac{p}{4}$.

(D) For a function $f(x)$ to have a local minimum at $x = a$,the value $f(a)$ must be less than or equal to the values of $f(x)$ in the immediate neighborhood of $a$.
Specifically,for $x = -1$,we require $f(-1) \leqslant f(-1 + h)$ and $f(-1) \leqslant f(-1 - h)$ for small $h > 0$.
First,calculate $f(-1)$ using the first part of the definition: $f(-1) = k - 2(-1) = k + 2$.
For $x > -1$,$f(x) = 2x + 3$. As $x \to -1^+$,$f(x) \to 2(-1) + 3 = 1$.
For $x \leqslant -1$,$f(x) = k - 2x$. As $x \to -1^-$,$f(x) \to k + 2$.
For $x = -1$ to be a local minimum,we must have $f(-1) \leqslant \lim_{x \to -1^+} f(x)$ and $f(-1) \leqslant \lim_{x \to -1^-} f(x)$.
Thus,$k + 2 \leqslant 1$ and $k + 2 \leqslant k + 2$.
Solving $k + 2 \leqslant 1$,we get $k \leqslant -1$.
Among the given options,the value that satisfies this condition is $k = -1$.

(D) Given the function $f(x) = \frac{x}{2} + \frac{2}{x}$.
To find the local extrema,we first find the derivative $f'(x)$:
$f'(x) = \frac{1}{2} - \frac{2}{x^2}$.
Setting $f'(x) = 0$ to find critical points:
$\frac{1}{2} - \frac{2}{x^2} = 0 \Rightarrow \frac{2}{x^2} = \frac{1}{2} \Rightarrow x^2 = 4 \Rightarrow x = 2, -2$.
Now,we find the second derivative $f''(x)$ to test for local minima:
$f''(x) = \frac{d}{dx}(\frac{1}{2} - 2x^{-2}) = 0 - 2(-2)x^{-3} = \frac{4}{x^3}$.
Evaluating at $x = 2$:
$f''(2) = \frac{4}{2^3} = \frac{4}{8} = \frac{1}{2} > 0$.
Since $f''(2) > 0$,the function has a local minimum at $x = 2$.

(C) Let the triangular park be $\Delta ABC$,where $AB = AC = x$. Let $AT$ be the altitude from $A$ to the river bank $BC$. Let $\angle ABT = \theta$.
Then $AT = x \sin \theta$ and $BT = x \cos \theta$.
Since the triangle is isosceles with $AB=AC$,the altitude $AT$ bisects $BC$,so $BC = 2BT = 2x \cos \theta$.
The area $A$ of the triangle is given by $A = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times (2x \cos \theta) \times (x \sin \theta) = x^2 \sin \theta \cos \theta = \frac{1}{2} x^2 \sin(2\theta)$.
To maximize the area,we need to maximize $\sin(2\theta)$. The maximum value of $\sin(2\theta)$ is $1$,which occurs when $2\theta = 90^\circ$ or $\theta = 45^\circ$.
Thus,the maximum area is $\frac{1}{2} x^2 (1) = \frac{1}{2} x^2$.

(D) Let $f(x) = x^3 - px + q$.
To find the critical points,we calculate the first derivative: $f'(x) = 3x^2 - p$.
Setting $f'(x) = 0$,we get $3x^2 = p$,which implies $x = \pm \sqrt{\frac{p}{3}}$.
Now,we find the second derivative: $f''(x) = 6x$.
Evaluating at the critical points:
For $x = -\sqrt{\frac{p}{3}}$,$f''(-\sqrt{\frac{p}{3}}) = -6\sqrt{\frac{p}{3}} < 0$ (since $p > 0$). Thus,there is a local maximum at $x = -\sqrt{\frac{p}{3}}$.
For $x = \sqrt{\frac{p}{3}}$,$f''(\sqrt{\frac{p}{3}}) = 6\sqrt{\frac{p}{3}} > 0$ (since $p > 0$). Thus,there is a local minimum at $x = \sqrt{\frac{p}{3}}$.
Therefore,the cubic has a maximum at $-\sqrt{\frac{p}{3}}$ and a minimum at $\sqrt{\frac{p}{3}}$.

(B) Given $P(x) = x^4 + ax^3 + bx^2 + cx + d$.
$P'(x) = 4x^3 + 3ax^2 + 2bx + c$.
Since $x=0$ is the only real root of $P'(x)=0$,we have $P'(0) = 0$,which implies $c=0$.
Thus,$P'(x) = x(4x^2 + 3ax + 2b)$.
Since $x=0$ is the only real root,the quadratic factor $4x^2 + 3ax + 2b$ must have no real roots,meaning its discriminant $D < 0$.
$D = (3a)^2 - 4(4)(2b) = 9a^2 - 32b < 0$.
Since the leading coefficient $4 > 0$,the quadratic $4x^2 + 3ax + 2b > 0$ for all $x \in \mathbb{R}$.
Therefore,the sign of $P'(x)$ is determined by the sign of $x$.
For $x \in [-1, 0)$,$P'(x) < 0$,so $P(x)$ is strictly decreasing.
For $x \in (0, 1]$,$P'(x) > 0$,so $P(x)$ is strictly increasing.
Since $P(x)$ decreases on $[-1, 0]$ and increases on $[0, 1]$,the global minimum on $[-1, 1]$ occurs at $x=0$.
Thus,$P(-1)$ is not the minimum.
Since $P(x)$ is increasing on $(0, 1]$ and $P(-1) < P(1)$,the maximum value on $[-1, 1]$ must be $P(1)$.
Therefore,$P(-1)$ is not the minimum,but $P(1)$ is the maximum.

(A) Given $f(x) = \alpha \log |x| + \beta x^2 + x$.
First,find the derivative $f'(x) = \frac{\alpha}{x} + 2\beta x + 1$.
Since $x = -1$ and $x = 2$ are extreme points,$f'(x) = 0$ at these points.
For $x = -1$: $\frac{\alpha}{-1} + 2\beta(-1) + 1 = 0 \Rightarrow -\alpha - 2\beta + 1 = 0 \Rightarrow \alpha + 2\beta = 1$ (Equation $1$).
For $x = 2$: $\frac{\alpha}{2} + 2\beta(2) + 1 = 0 \Rightarrow \frac{\alpha}{2} + 4\beta + 1 = 0 \Rightarrow \alpha + 8\beta = -2$ (Equation $2$).
Subtracting Equation $1$ from Equation $2$: $(\alpha + 8\beta) - (\alpha + 2\beta) = -2 - 1 \Rightarrow 6\beta = -3 \Rightarrow \beta = -\frac{1}{2}$.
Substitute $\beta = -\frac{1}{2}$ into Equation $1$: $\alpha + 2(-\frac{1}{2}) = 1 \Rightarrow \alpha - 1 = 1 \Rightarrow \alpha = 2$.
Thus,$(\alpha, \beta) = (2, -\frac{1}{2})$.

(D) Given $\mathop {\lim }\limits_{x \to 0} \left[ {1 + \frac{{f(x)}}{{{x^2}}}} \right] = 3$,this implies $\mathop {\lim }\limits_{x \to 0} \frac{{f(x)}}{{{x^2}}} = 2$.
Since $f(x)$ is a polynomial of degree $4$,let $f(x) = ax^4 + bx^3 + cx^2 + dx + e$.
For the limit to exist and equal $2$,we must have $e=0$ and $d=0$,and $c=2$.
Thus,$f(x) = ax^4 + bx^3 + 2x^2$.
Then $f'(x) = 4ax^3 + 3bx^2 + 4x$.
Since $f(x)$ has extreme values at $x=1$ and $x=2$,$f'(1) = 0$ and $f'(2) = 0$.
$f'(1) = 4a + 3b + 4 = 0 \Rightarrow 4a + 3b = -4$.
$f'(2) = 4a(8) + 3b(4) + 4(2) = 32a + 12b + 8 = 0 \Rightarrow 8a + 3b = -2$.
Subtracting the first equation from the second: $(8a + 3b) - (4a + 3b) = -2 - (-4) \Rightarrow 4a = 2 \Rightarrow a = \frac{1}{2}$.
Substituting $a = \frac{1}{2}$ into $4a + 3b = -4$: $4(\frac{1}{2}) + 3b = -4 \Rightarrow 2 + 3b = -4 \Rightarrow 3b = -6 \Rightarrow b = -2$.
So,$f(x) = \frac{1}{2}x^4 - 2x^3 + 2x^2$.
Calculating $f(2)$: $f(2) = \frac{1}{2}(16) - 2(8) + 2(4) = 8 - 16 + 8 = 0$.

(D) Let $r$ be the radius and $\theta$ be the central angle of the circular sector in radians.
The perimeter of the circular sector is given by $P = r + r + r\theta = 2r + r\theta$.
Given that the total length of the wire is $20 \ m$,we have:
$2r + r\theta = 20$
$\Rightarrow r\theta = 20 - 2r$
$\Rightarrow \theta = \frac{20 - 2r}{r}$
The area $A$ of the circular sector is given by:
$A = \frac{1}{2} r^2 \theta$
Substituting the value of $\theta$:
$A = \frac{1}{2} r^2 \left( \frac{20 - 2r}{r} \right) = \frac{1}{2} r (20 - 2r) = 10r - r^2$
To find the maximum area,we differentiate $A$ with respect to $r$ and set it to zero:
$\frac{dA}{dr} = 10 - 2r = 0$
$\Rightarrow r = 5 \ m$
To verify the maximum,we check the second derivative:
$\frac{d^2A}{dr^2} = -2 < 0$
Since the second derivative is negative,the area is maximum at $r = 5$.
Substituting $r = 5$ into the area formula:
$A_{max} = 10(5) - (5)^2 = 50 - 25 = 25 \ m^2$.

(C) Given $h(x) = \frac{x^2 + \frac{1}{x^2}}{x - \frac{1}{x}}$.
We can rewrite the numerator as $(x - \frac{1}{x})^2 + 2$.
So,$h(x) = \frac{(x - \frac{1}{x})^2 + 2}{x - \frac{1}{x}} = (x - \frac{1}{x}) + \frac{2}{x - \frac{1}{x}}$.
Let $t = x - \frac{1}{x}$. Since $x \in R - \{-1, 1, 0\}$,$t$ can take any real value except $0$.
Then $h(t) = t + \frac{2}{t}$.
For $t > 0$,by the $AM$-$GM$ inequality,$t + \frac{2}{t} \ge 2\sqrt{t \cdot \frac{2}{t}} = 2\sqrt{2}$.
The equality holds when $t = \frac{2}{t}$,i.e.,$t^2 = 2$,so $t = \sqrt{2}$ (since $t > 0$).
For $t < 0$,let $u = -t$,where $u > 0$. Then $h(t) = -u - \frac{2}{u} = -(u + \frac{2}{u}) \le -2\sqrt{2}$.
Thus,the local minimum value of $h(x)$ is $2\sqrt{2}$.

(A) Given the curve equation: $y = x(x - 2)(x - 4)$.
Expanding the expression: $y = x(x^2 - 6x + 8) = x^3 - 6x^2 + 8x$.
To find the slope of the tangent,we differentiate $y$ with respect to $x$: $\frac{dy}{dx} = 3x^2 - 12x + 8$.
For the tangent to be parallel to the $x$-axis,the slope must be zero: $\frac{dy}{dx} = 0$.
Setting the derivative to zero: $3x^2 - 12x + 8 = 0$.
Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:
$x = \frac{12 \pm \sqrt{(-12)^2 - 4(3)(8)}}{2(3)} = \frac{12 \pm \sqrt{144 - 96}}{6} = \frac{12 \pm \sqrt{48}}{6}$.
Simplifying the expression: $x = \frac{12 \pm 4\sqrt{3}}{6} = 2 \pm \frac{2\sqrt{3}}{3} = 2 \pm \frac{2}{\sqrt{3}}$.

(B) Given $y = a \log |x| + b x^2 + x$.
Since $\frac{d}{dx} \log |x| = \frac{1}{x}$ for all $x \neq 0$,we differentiate $y$ with respect to $x$:
$\frac{dy}{dx} = \frac{a}{x} + 2bx + 1$.
Since the function has extremum values at $x = -1$ and $x = 2$,the derivative must be zero at these points.
For $x = -1$: $\frac{a}{-1} + 2b(-1) + 1 = 0 \implies -a - 2b + 1 = 0 \implies a + 2b = 1$ (Equation $1$).
For $x = 2$: $\frac{a}{2} + 2b(2) + 1 = 0 \implies \frac{a}{2} + 4b + 1 = 0 \implies a + 8b = -2$ (Equation $2$).
Subtracting Equation $1$ from Equation $2$: $(a + 8b) - (a + 2b) = -2 - 1 \implies 6b = -3 \implies b = -\frac{1}{2}$.
Substituting $b = -\frac{1}{2}$ into Equation $1$: $a + 2(-\frac{1}{2}) = 1 \implies a - 1 = 1 \implies a = 2$.
Thus,$a = 2$ and $b = -\frac{1}{2}$.

(A) The function is defined as $f(x) = |x|$ for $x \in [-2, 0) \cup (0, 2]$ and $f(0) = 1$.
For any small neighborhood $(0 - h, 0 + h)$ around $x = 0$ (where $h > 0$ is very small),we have:
$f(0) = 1$
For $x \neq 0$ in this neighborhood,$f(x) = |x|$. Since $x$ is very close to $0$,$|x| < 1$.
Thus,$f(x) < f(0)$ for all $x$ in the neighborhood of $0$.
By the definition of a local maximum,if $f(x) \le f(a)$ for all $x$ in some neighborhood of $a$,then $f$ has a local maximum at $x = a$.
Since $f(x) < f(0)$ for all $x \in (-h, h) \setminus \{0\}$,the function $f$ has a local maximum at $x = 0$.

(C) Given $f(x) = \int\limits_0^x t(t - 1)(t - 2) dt$.
By the Fundamental Theorem of Calculus,$f'(x) = x(x - 1)(x - 2)$.
To find the critical points,set $f'(x) = 0$,which gives $x = 0, 1, 2$.
We analyze the sign of $f'(x)$ around these points:
For $x < 0$,$f'(x) < 0$.
For $0 < x < 1$,$f'(x) > 0$.
For $1 < x < 2$,$f'(x) < 0$.
For $x > 2$,$f'(x) > 0$.
At $x = 0$,$f'(x)$ changes from negative to positive,so $x = 0$ is a local minimum.
At $x = 1$,$f'(x)$ changes from positive to negative,so $x = 1$ is a local maximum.
At $x = 2$,$f'(x)$ changes from negative to positive,so $x = 2$ is a local minimum.
Thus,the function takes its minimum value at $x = 0$ and $x = 2$.

(B) Using Leibniz's rule for differentiation under the integral sign:
$f'(x) = (1 - \sin x + 2\sin^3 x)(\cos x) - (1 - \cos x + 2\cos^3 x)(-\sin x)$
$f'(x) = \cos x - \sin x \cos x + 2\sin^3 x \cos x + \sin x - \sin x \cos x + 2\cos^3 x \sin x$
$f'(x) = \cos x + \sin x - 2\sin x \cos x + 2\sin x \cos x (\sin^2 x + \cos^2 x)$
$f'(x) = \cos x + \sin x - 2\sin x \cos x + 2\sin x \cos x = \cos x + \sin x$
For critical points,set $f'(x) = 0$:
$\cos x + \sin x = 0 \implies \tan x = -1$
In $[0, 2\pi]$,$x = \frac{3\pi}{4}$ and $x = \frac{7\pi}{4}$
$f''(x) = -\sin x + \cos x$
At $x = \frac{3\pi}{4}$,$f''(\frac{3\pi}{4}) = -\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} = -\sqrt{2} < 0$ (Maximum)
At $x = \frac{7\pi}{4}$,$f''(\frac{7\pi}{4}) = -(-\frac{1}{\sqrt{2}}) + \frac{1}{\sqrt{2}} = \sqrt{2} > 0$ (Minimum)
Thus,the function has a maximum at $\frac{3\pi}{4}$ and a minimum at $\frac{7\pi}{4}$.

(D) Given $f(x) = \int_0^x \frac{\sin t}{t} dt$.
By the Fundamental Theorem of Calculus,$f'(x) = \frac{\sin x}{x}$.
For critical points,set $f'(x) = 0$,which implies $\sin x = 0$ (since $x > 0$),so $x = n\pi$ for $n = 1, 2, 3, \dots$.
Now,$f''(x) = \frac{x \cos x - \sin x}{x^2}$.
At $x = n\pi$,$f''(n\pi) = \frac{n\pi \cos(n\pi) - \sin(n\pi)}{(n\pi)^2} = \frac{n\pi (-1)^n - 0}{n^2 \pi^2} = \frac{(-1)^n}{n\pi}$.
If $n$ is odd $(n = 1, 3, 5, \dots)$,$f''(n\pi) = \frac{-1}{n\pi} < 0$,so $f(x)$ has a local maximum.
If $n$ is even $(n = 2, 4, 6, \dots)$,$f''(n\pi) = \frac{1}{n\pi} > 0$,so $f(x)$ has a local minimum.
Thus,both $(A)$ and $(B)$ are correct.

(C) The curve is $y = x^2 + 1$. The point on the curve is $(x_1, x_1^2 + 1)$.
The slope of the tangent at $(x_1, x_1^2 + 1)$ is $\frac{dy}{dx} = 2x_1$.
The equation of the tangent is $y - (x_1^2 + 1) = 2x_1(x - x_1)$,which simplifies to $y = 2x_1x - x_1^2 + 1$.
Let the tangent intersect the lines $x=1$ and $x=2$ at points $P$ and $Q$ respectively.
At $x=1$,$y_P = 2x_1(1) - x_1^2 + 1 = -x_1^2 + 2x_1 + 1$.
At $x=2$,$y_Q = 2x_1(2) - x_1^2 + 1 = -x_1^2 + 4x_1 + 1$.
The area of the trapezium formed by the lines $x=1, x=2$,the $x$-axis,and the tangent is $A = \frac{y_P + y_Q}{2} \times (2 - 1) = \frac{(-x_1^2 + 2x_1 + 1) + (-x_1^2 + 4x_1 + 1)}{2} = \frac{-2x_1^2 + 6x_1 + 2}{2} = -x_1^2 + 3x_1 + 1$.
To maximize the area,we find $\frac{dA}{dx_1} = -2x_1 + 3$. Setting $\frac{dA}{dx_1} = 0$ gives $x_1 = \frac{3}{2}$.
For $x_1 = \frac{3}{2}$,$y_1 = \left(\frac{3}{2}\right)^2 + 1 = \frac{9}{4} + 1 = \frac{13}{4}$.
Thus,the point is $\left( \frac{3}{2}, \frac{13}{4} \right)$.

(D) The domain of $\sin^{-1}(u)$ is $[-1, 1]$. Here,$u = a^2 - 8a + 17 = (a-4)^2 + 1$. For this to be defined,$(a-4)^2 + 1 \le 1$,which implies $(a-4)^2 \le 0$. Since a square is always non-negative,this is only possible if $a = 4$.
Substituting $a = 4$ into $f(x)$:
$f(x) = -\frac{x^3}{3} + x^2 \sin(6) - x \sin(4) \sin(8) - 5 \sin^{-1}(1)$.
Now,differentiate $f(x)$ with respect to $x$:
$f'(x) = -x^2 + 2x \sin(6) - \sin(4) \sin(8)$.
Evaluate $f'( \sin(8) )$:
$f'( \sin(8) ) = -\sin^2(8) + 2 \sin(8) \sin(6) - \sin(4) \sin(8)$.
$f'( \sin(8) ) = \sin(8) [ -\sin(8) + 2 \sin(6) - \sin(4) ]$.
Using the identity $\sin(A) + \sin(B) = 2 \sin(\frac{A+B}{2}) \cos(\frac{A-B}{2})$:
$f'( \sin(8) ) = -\sin(8) [ \sin(8) + \sin(4) - 2 \sin(6) ]$.
$f'( \sin(8) ) = -\sin(8) [ 2 \sin(6) \cos(2) - 2 \sin(6) ]$.
$f'( \sin(8) ) = -2 \sin(8) \sin(6) [ \cos(2) - 1 ] = 2 \sin(8) \sin(6) [ 1 - \cos(2) ]$.
Since $\sin(8) \approx 0.989$ (positive),$\sin(6) \approx -0.279$ (negative),and $1 - \cos(2) > 0$,the product is negative.
Thus,$f'( \sin(8) ) < 0$.

(B) Given $f(x) = 2x^3 - 9ax^2 + 12a^2x + 1$.
Find the derivative: $f'(x) = 6x^2 - 18ax + 12a^2 = 6(x^2 - 3ax + 2a^2) = 6(x - a)(x - 2a)$.
Setting $f'(x) = 0$ gives critical points $x = a$ and $x = 2a$.
Find the second derivative: $f''(x) = 12x - 18a$.
Check the nature of critical points:
$f''(a) = 12a - 18a = -6a$
$f''(2a) = 12(2a) - 18a = 6a$
Case $1$: If $a > 0$,then $f''(a) < 0$ (maximum at $x_1 = a$) and $f''(2a) > 0$ (minimum at $x_2 = 2a$).
Given $x_1^2 = x_2$,we have $a^2 = 2a$. Since $a > 0$,$a = 2$.
Case $2$: If $a < 0$,then $f''(a) > 0$ (minimum at $x_2 = a$) and $f''(2a) < 0$ (maximum at $x_1 = 2a$).
Given $x_1^2 = x_2$,we have $(2a)^2 = a$,which means $4a^2 = a$. Since $a < 0$,this has no solution.
Thus,the value of $a$ is $2$.

(D) The vertices of the triangle $AOB$ are $O(0, 0)$,$B(x, 0)$,and $A(x, e^{-x^2})$.
Since the triangle is right-angled at $B$,the area $A(x)$ is given by:
$A(x) = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times x \times e^{-x^2} = \frac{x e^{-x^2}}{2}$.
To find the maximum area,we differentiate $A(x)$ with respect to $x$:
$A'(x) = \frac{1}{2} [1 \cdot e^{-x^2} + x \cdot e^{-x^2} \cdot (-2x)] = \frac{e^{-x^2}}{2} [1 - 2x^2]$.
Setting $A'(x) = 0$ for critical points:
$1 - 2x^2 = 0 \implies x^2 = \frac{1}{2} \implies x = \frac{1}{\sqrt{2}}$ (since $x > 0$).
Now,calculate the maximum area at $x = \frac{1}{\sqrt{2}}$:
$A_{max} = \frac{1}{2} \cdot \frac{1}{\sqrt{2}} \cdot e^{-(1/\sqrt{2})^2} = \frac{1}{2\sqrt{2}} \cdot e^{-1/2} = \frac{1}{2\sqrt{2} \sqrt{e}} = \frac{1}{\sqrt{8e}}$.
Thus,the maximum area is $\frac{1}{\sqrt{8e}}$.

(B) Let $f(x) = \frac{x^2}{x^3 + 200}$. To find the maximum,we differentiate with respect to $x$:
$f'(x) = \frac{(x^3 + 200)(2x) - x^2(3x^2)}{(x^3 + 200)^2} = \frac{2x^4 + 400x - 3x^4}{(x^3 + 200)^2} = \frac{x(400 - x^3)}{(x^3 + 200)^2}$.
For $f(x)$ to be increasing,$f'(x) > 0$,which implies $400 - x^3 > 0$,or $x < (400)^{1/3}$.
Since $7^3 = 343$ and $8^3 = 512$,we have $7 < (400)^{1/3} < 8$.
Thus,the function increases until $x = (400)^{1/3}$ and decreases thereafter.
We compare $a_7$ and $a_8$:
$a_7 = \frac{7^2}{7^3 + 200} = \frac{49}{343 + 200} = \frac{49}{543} \approx 0.0902$.
$a_8 = \frac{8^2}{8^3 + 200} = \frac{64}{512 + 200} = \frac{64}{712} = \frac{8}{89} \approx 0.0898$.
Since $a_7 > a_8$,the largest term is $a_7$.

(C) Let $f(x) = x(x + 1)(x + 2)(x + 3)$.
We can group the terms as follows:
$f(x) = [x(x + 3)] \cdot [(x + 1)(x + 2)]$
$f(x) = (x^2 + 3x)(x^2 + 3x + 2)$
Let $z = x^2 + 3x$. Then the expression becomes:
$f(x) = z(z + 2) = z^2 + 2z$
To find the minimum value,we complete the square:
$f(x) = (z^2 + 2z + 1) - 1 = (z + 1)^2 - 1$
Since $(z + 1)^2 \ge 0$,the minimum value of $f(x)$ is $-1$,which occurs when $z + 1 = 0$,i.e.,$z = -1$.
Substituting back $z = x^2 + 3x$,we have $x^2 + 3x + 1 = 0$,which has real roots for $x$,so the minimum value $-1$ is attainable.

(A) Given $f(x) = \sin(2x) - x$. To find the critical points,we find $f'(x) = 2\cos(2x) - 1$.
Setting $f'(x) = 0$,we get $2\cos(2x) = 1$,so $\cos(2x) = \frac{1}{2}$.
In the interval $\left[ -\frac{\pi}{2}, \frac{\pi}{2} \right]$,$2x \in [-\pi, \pi]$,so $2x = \pm \frac{\pi}{3}$,which gives $x = \pm \frac{\pi}{6}$.
Now,we evaluate $f(x)$ at the critical points and endpoints:
$f\left(-\frac{\pi}{2}\right) = \sin(-\pi) - (-\frac{\pi}{2}) = 0 + \frac{\pi}{2} = \frac{\pi}{2} \approx 1.57$.
$f\left(-\frac{\pi}{6}\right) = \sin(-\frac{\pi}{3}) - (-\frac{\pi}{6}) = -\frac{\sqrt{3}}{2} + \frac{\pi}{6} \approx -0.866 + 0.523 = -0.343$.
$f\left(\frac{\pi}{6}\right) = \sin(\frac{\pi}{3}) - \frac{\pi}{6} = \frac{\sqrt{3}}{2} - \frac{\pi}{6} \approx 0.866 - 0.523 = 0.343$.
$f\left(\frac{\pi}{2}\right) = \sin(\pi) - \frac{\pi}{2} = 0 - \frac{\pi}{2} = -\frac{\pi}{2} \approx -1.57$.
The greatest value is $\frac{\pi}{2}$ and the least value is $-\frac{\pi}{2}$.
The difference is $\frac{\pi}{2} - (-\frac{\pi}{2}) = \pi$.

(B) Given the function $f(x) = x(\ln x - 2)$ on the interval $[1, e^2]$.
To find the critical points,we find the derivative $f'(x)$:
$f'(x) = x \cdot \frac{d}{dx}(\ln x - 2) + (\ln x - 2) \cdot \frac{d}{dx}(x)$
$f'(x) = x \cdot \frac{1}{x} + \ln x - 2 = 1 + \ln x - 2 = \ln x - 1$.
Setting $f'(x) = 0$ gives $\ln x - 1 = 0$,so $\ln x = 1$,which implies $x = e$.
Now,we evaluate the function at the critical point and the endpoints of the interval:
$f(1) = 1(\ln 1 - 2) = 1(0 - 2) = -2$.
$f(e) = e(\ln e - 2) = e(1 - 2) = -e$.
$f(e^2) = e^2(\ln e^2 - 2) = e^2(2 - 2) = 0$.
Comparing the values $\{-2, -e, 0\}$,since $e \approx 2.718$,the least value is $-e$ and the greatest value is $0$.
The difference between the greatest and the least values is $0 - (-e) = e$.

(C) Let the hypotenuse be $h$ and one side be $x$. Given $h + x = c$ (constant),so $h = c - x$.
In a right-angled triangle,the third side is $\sqrt{h^2 - x^2} = \sqrt{(c-x)^2 - x^2} = \sqrt{c^2 - 2cx}$.
The area $A$ of the triangle is $A = \frac{1}{2} \times x \times \sqrt{c^2 - 2cx}$.
To maximize $A$,we maximize $A^2 = \frac{1}{4} x^2 (c^2 - 2cx) = \frac{1}{4} (c^2 x^2 - 2cx^3)$.
Let $f(x) = c^2 x^2 - 2cx^3$. Differentiating with respect to $x$:
$f'(x) = 2c^2 x - 6cx^2 = 2cx(c - 3x)$.
Setting $f'(x) = 0$,we get $x = \frac{c}{3}$ (since $x \neq 0$).
Now,$h = c - \frac{c}{3} = \frac{2c}{3}$.
Let $\theta$ be the angle between the hypotenuse and the side $x$. Then $\cos \theta = \frac{x}{h} = \frac{c/3}{2c/3} = \frac{1}{2}$.
Therefore,$\theta = \frac{\pi}{3}$.

(A) The area $A$ of a triangle with vertices $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ is given by $A = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$.
Substituting the given vertices $(0, 0), (x, \cos x), (\sin^3 x, 0)$,we get:
$A(x) = \frac{1}{2} |0(\cos x - 0) + x(0 - 0) + \sin^3 x(0 - \cos x)| = \frac{1}{2} |-\sin^3 x \cos x| = \frac{1}{2} \sin^3 x \cos x$.
To find the maximum area,we differentiate $A(x)$ with respect to $x$:
$A'(x) = \frac{1}{2} [3 \sin^2 x \cos x \cdot \cos x + \sin^3 x(-\sin x)] = \frac{1}{2} [3 \sin^2 x \cos^2 x - \sin^4 x]$.
Setting $A'(x) = 0$,we get $3 \sin^2 x \cos^2 x = \sin^4 x$.
Since $0 < x < \frac{\pi}{2}$,$\sin x \neq 0$,so $3 \cos^2 x = \sin^2 x$,which implies $\tan^2 x = 3$,so $\tan x = \sqrt{3}$,hence $x = \frac{\pi}{3}$.
Substituting $x = \frac{\pi}{3}$ into $A(x)$:
$A(\frac{\pi}{3}) = \frac{1}{2} \sin^3(\frac{\pi}{3}) \cos(\frac{\pi}{3}) = \frac{1}{2} (\frac{\sqrt{3}}{2})^3 (\frac{1}{2}) = \frac{1}{2} \cdot \frac{3\sqrt{3}}{8} \cdot \frac{1}{2} = \frac{3\sqrt{3}}{32}$.

(D) Let $f(x) = \frac{4}{\sin x} + \frac{1}{1 - \sin x}$. Let $t = \sin x$. Since $x \in (0, \pi/2)$,$t \in (0, 1)$.
We define $g(t) = \frac{4}{t} + \frac{1}{1 - t}$ for $t \in (0, 1)$.
To find the minimum value,we differentiate $g(t)$ with respect to $t$:
$g'(t) = -\frac{4}{t^2} + \frac{1}{(1 - t)^2}$.
Setting $g'(t) = 0$ gives $\frac{1}{(1 - t)^2} = \frac{4}{t^2}$,which implies $(1 - t)^2 = \frac{t^2}{4}$.
Taking the square root,$1 - t = \frac{t}{2}$ or $1 - t = -\frac{t}{2}$.
For $t \in (0, 1)$,$1 - t = \frac{t}{2} \implies 1 = \frac{3t}{2} \implies t = \frac{2}{3}$.
The value of the function at $t = \frac{2}{3}$ is $g(2/3) = \frac{4}{2/3} + \frac{1}{1 - 2/3} = 6 + 3 = 9$.
Since $g(t) \to \infty$ as $t \to 0^+$ and $t \to 1^-$,the range of $g(t)$ is $[9, \infty)$.
Thus,the least value of $a$ for which the equation has at least one solution is $9$.