Maxima and Minima Questions in English

(D) Given $f(x) = 3^{(x^2-2)^3+4} = 81 \cdot 3^{(x^2-2)^3}$.
First derivative: $f'(x) = 81 \cdot 3^{(x^2-2)^3} \cdot \ln 3 \cdot 3(x^2-2)^2 \cdot 2x = (486 \ln 3) \cdot 3^{(x^2-2)^3} \cdot x(x^2-2)^2$.
For $P$: $f'(x) = 0$ at $x=0$ and $x=\pm \sqrt{2}$. For $x$ near $0$,$x(x^2-2)^2$ changes sign from negative to positive,so $x=0$ is a point of local minima. Thus,$P$ is true.
For $Q$: $f''(x) = 0$ at $x=\sqrt{2}$. Since $f''(x)$ changes sign at $x=\sqrt{2}$ (as $(x^2-2)$ is a factor),$x=\sqrt{2}$ is a point of inflection. Thus,$Q$ is true.
For $R$: For $x > \sqrt{2}$,$f'(x) > 0$. Analyzing $f''(x)$,for $x > \sqrt{2}$,$f''(x) > 0$,which implies $f'(x)$ is increasing. Thus,$R$ is true.
Therefore,all statements $P, Q,$ and $R$ are true.

(B) Given the function $f(x) = e^x + x \ln x$ on the interval $[1, 2]$.
First,we find the derivative of the function:
$f'(x) = \frac{d}{dx}(e^x + x \ln x) = e^x + (1 \cdot \ln x + x \cdot \frac{1}{x}) = e^x + \ln x + 1$.
For $x \in [1, 2]$,we observe that $e^x > 0$,$\ln x \geq 0$,and $1 > 0$. Thus,$f'(x) > 0$ for all $x \in [1, 2]$.
Since the derivative $f'(x)$ is always positive on the interval,the function $f(x)$ is strictly increasing on $[1, 2]$.
Therefore,the maximum value occurs at the right endpoint of the interval,which is $x = 2$.
Calculating $f(2)$:
$f(2) = e^2 + 2 \ln 2$.
Thus,the correct option is $B$.

(B) The curves are $y_1 = x^2-4$ and $y_2 = \frac{4-x^2}{2}$.
Let the vertices of the rectangle be $C(h, y_2)$ and $D(-h, y_2)$ on the upper curve,and $B(h, y_1)$ and $A(-h, y_1)$ on the lower curve,where $h > 0$.
The width of the rectangle is $2h$ and the height is $y_2 - y_1 = \frac{4-x^2}{2} - (x^2-4) = 2 - \frac{x^2}{2} - x^2 + 4 = 6 - \frac{3x^2}{2}$.
Substituting $h$ for $x$,the area $A(h) = 2h \times (6 - \frac{3h^2}{2}) = 12h - 3h^3$.
To find the maximum area,differentiate with respect to $h$: $\frac{dA}{dh} = 12 - 9h^2$.
Setting $\frac{dA}{dh} = 0$,we get $9h^2 = 12$,so $h^2 = \frac{4}{3}$,which means $h = \frac{2}{\sqrt{3}}$.
The maximum area is $A = 12(\frac{2}{\sqrt{3}}) - 3(\frac{2}{\sqrt{3}})^3 = \frac{24}{\sqrt{3}} - 3(\frac{8}{3\sqrt{3}}) = \frac{24}{\sqrt{3}} - \frac{8}{\sqrt{3}} = \frac{16}{\sqrt{3}} = \frac{16\sqrt{3}}{3}$.
Using $\sqrt{3} \approx 1.732$,$A \approx \frac{16 \times 1.732}{3} \approx 9.237$.
The integer closest to $9.237$ is $9$.

(B) Let the radius of the metallic disc be $R$. When a sector is removed and the remaining part is bent to form a cone,the radius $R$ of the disc becomes the slant height $l$ of the cone.
Let the radius of the base of the cone be $x$ and its height be $h$.
We have the relation $R^2 = x^2 + h^2$,where $R = l$.
The volume of the cone is given by $V = \frac{1}{3} \pi x^2 h = 2 \sqrt{3} \pi$.
Thus,$x^2 h = 6 \sqrt{3}$,which implies $x^2 = \frac{6 \sqrt{3}}{h}$.
Substituting this into the expression for $R^2$:
$R^2 = \frac{6 \sqrt{3}}{h} + h^2$.
To find the minimum diameter,we minimize $R^2$ with respect to $h$:
$\frac{d(R^2)}{dh} = -\frac{6 \sqrt{3}}{h^2} + 2h$.
Setting $\frac{d(R^2)}{dh} = 0$,we get $2h = \frac{6 \sqrt{3}}{h^2}$,so $h^3 = 3 \sqrt{3}$.
This gives $h = \sqrt{3}$.
Then $x^2 = \frac{6 \sqrt{3}}{\sqrt{3}} = 6$.
Therefore,$R^2 = 6 + (\sqrt{3})^2 = 6 + 3 = 9$,so $R = 3$.
The diameter of the disc is $2R = 2 \times 3 = 6$.

(A) Let $r$ be the common radius and $h$ be the height of the cylinder.
The total surface area $S$ of the solid is the sum of the curved surface area of the hemisphere $(2\pi r^2)$,the curved surface area of the cylinder $(2\pi rh)$,and the base area of the cylinder $(\pi r^2)$.
$S = 2\pi r^2 + 2\pi rh + \pi r^2 = 3\pi r^2 + 2\pi rh$
From this,we can express $h$ in terms of $S$ and $r$:
$h = \frac{S - 3\pi r^2}{2\pi r}$
The volume $V$ of the solid is the sum of the volume of the hemisphere $(\frac{2}{3}\pi r^3)$ and the volume of the cylinder $(\pi r^2 h)$:
$V = \pi r^2 h + \frac{2}{3}\pi r^3$
Substituting the expression for $h$:
$V = \pi r^2 \left( \frac{S - 3\pi r^2}{2\pi r} \right) + \frac{2}{3}\pi r^3$
$V = \frac{1}{2} (Sr - 3\pi r^3) + \frac{2}{3}\pi r^3 = \frac{1}{2} Sr - \frac{3}{2}\pi r^3 + \frac{2}{3}\pi r^3 = \frac{1}{2} Sr - \frac{5}{6}\pi r^3$
To find the maximum volume,differentiate $V$ with respect to $r$ and set it to $0$:
$\frac{dV}{dr} = \frac{S}{2} - \frac{5}{2}\pi r^2 = 0$
$S = 5\pi r^2$
Now,substitute $S = 5\pi r^2$ back into the expression for $h$:
$h = \frac{5\pi r^2 - 3\pi r^2}{2\pi r} = \frac{2\pi r^2}{2\pi r} = r$
Thus,the ratio of the height of the cylinder to the common radius is $h:r = 1:1$.

(D) Let the radius of the sector be $r$ and the angle subtended by the radii at the centre be $\theta$ (in radians).
The length of the arc is $l = r\theta$.
The perimeter of the sector is $P = 2r + l = 2r + r\theta = r(2 + \theta)$.
From this,we can express the radius as $r = \frac{P}{2 + \theta}$.
The area of the sector is $A = \frac{1}{2}r^2\theta$.
Substituting the expression for $r$ in terms of $P$ and $\theta$:
$A = \frac{1}{2} \left(\frac{P}{2 + \theta}\right)^2 \theta = \frac{P^2}{2} \cdot \frac{\theta}{(2 + \theta)^2}$.
To find the maximum area,we differentiate $A$ with respect to $\theta$ and set it to zero:
$\frac{dA}{d\theta} = \frac{P^2}{2} \cdot \frac{(2 + \theta)^2(1) - \theta(2)(2 + \theta)}{(2 + \theta)^4} = 0$.
This simplifies to:
$(2 + \theta)^2 - 2\theta(2 + \theta) = 0$.
Factoring out $(2 + \theta)$:
$(2 + \theta)(2 + \theta - 2\theta) = 0$.
$(2 + \theta)(2 - \theta) = 0$.
Since $\theta > 0$,we have $\theta = 2$ radians.
Thus,the angle at the centre for maximum area is $2$ radians.

(D) Given $f(x) = \alpha x^2 - 2 + \frac{1}{x} = \frac{\alpha x^3 - 2x + 1}{x}$.
For $f(x) \geq 0$ and $x > 0$,we must have $g(x) = \alpha x^3 - 2x + 1 \geq 0$ for all $x > 0$.
To find the minimum of $g(x)$,we calculate $g'(x) = 3\alpha x^2 - 2$.
Setting $g'(x) = 0$,we get $x^2 = \frac{2}{3\alpha}$,so $x = \sqrt{\frac{2}{3\alpha}}$ (since $x > 0$).
The second derivative $g''(x) = 6\alpha x > 0$ at this point,confirming a local minimum.
Substituting $x = \sqrt{\frac{2}{3\alpha}}$ into $g(x)$:
$g\left(\sqrt{\frac{2}{3\alpha}}\right) = \alpha \left(\frac{2}{3\alpha}\right) \sqrt{\frac{2}{3\alpha}} - 2\sqrt{\frac{2}{3\alpha}} + 1 \geq 0$.
$\sqrt{\frac{2}{3\alpha}} \left( \frac{2}{3} - 2 \right) + 1 \geq 0$.
$1 - \frac{4}{3} \sqrt{\frac{2}{3\alpha}} \geq 0 \Rightarrow 1 \geq \frac{4}{3} \sqrt{\frac{2}{3\alpha}}$.
$\frac{3}{4} \geq \sqrt{\frac{2}{3\alpha}} \Rightarrow \frac{9}{16} \geq \frac{2}{3\alpha}$.
$27\alpha \geq 32 \Rightarrow \alpha \geq \frac{32}{27} = \frac{2^5}{3^3}$.
Thus,the smallest value is $\frac{2^5}{3^3}$.

(C) Let $f(x) = 3x^3 - 25x + n$.
For the cubic equation to have three real roots,the local maximum and local minimum must have opposite signs.
$f'(x) = 9x^2 - 25$.
Setting $f'(x) = 0$,we get $x^2 = \frac{25}{9}$,so $x = \pm \frac{5}{3}$.
Let $x_1 = -\frac{5}{3}$ and $x_2 = \frac{5}{3}$.
$f(x_1) = 3(-\frac{125}{27}) - 25(-\frac{5}{3}) + n = -\frac{125}{9} + \frac{125}{3} + n = n + \frac{250}{9}$.
$f(x_2) = 3(\frac{125}{27}) - 25(\frac{5}{3}) + n = \frac{125}{9} - \frac{125}{3} + n = n - \frac{250}{9}$.
For three real roots,$f(x_1) \cdot f(x_2) < 0$,so $(n + \frac{250}{9})(n - \frac{250}{9}) < 0$.
This implies $-\frac{250}{9} < n < \frac{250}{9}$.
Since $\frac{250}{9} \approx 27.77$,the integers $n$ range from $-27$ to $27$.
The number of such integers is $27 - (-27) + 1 = 55$.

(A) We have $f(x) = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!}$.
Note that $f'(x) = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} = f(x) - \frac{x^4}{4!}$.
Also,$f''(x) = 1 + x + \frac{x^2}{2} = \frac{1}{2}(x^2 + 2x + 1) + \frac{1}{2} = \frac{1}{2}(x+1)^2 + \frac{1}{2} > 0$ for all $x \in \mathbb{R}$.
Since $f''(x) > 0$,$f'(x)$ is a strictly increasing function.
As $x \to -\infty$,$f'(x) \to -\infty$ and as $x \to \infty$,$f'(x) \to \infty$. Thus,$f'(x) = 0$ has exactly one real root,say $x_0$.
Since $f'(x)$ is increasing,$f(x)$ has a global minimum at $x = x_0$.
We observe $f'(x) = 1 + x + \frac{x^2}{2} + \frac{x^3}{6}$.
$f'(-2) = 1 - 2 + 2 - \frac{8}{6} = 1 - 1.33 = -0.33 < 0$.
$f'(-1) = 1 - 1 + \frac{1}{2} - \frac{1}{6} = \frac{1}{3} > 0$.
So,$x_0 \in (-2, -1)$.
At the minimum $x_0$,$f(x_0) = 1 + x_0 + \frac{x_0^2}{2} + \frac{x_0^3}{6} + \frac{x_0^4}{24}$.
Since $f'(x_0) = 0$,we have $1 + x_0 + \frac{x_0^2}{2} + \frac{x_0^3}{6} = 0$,which implies $1 + x_0 + \frac{x_0^2}{2} + \frac{x_0^3}{6} = -\frac{x_0^4}{24}$.
Substituting this into $f(x_0)$,we get $f(x_0) = -\frac{x_0^4}{24} + \frac{x_0^4}{24} + \dots$ wait,$f(x_0) = (1 + x_0 + \frac{x_0^2}{2} + \frac{x_0^3}{6}) + \frac{x_0^4}{24} = 0 + \frac{x_0^4}{24} = \frac{x_0^4}{24}$.
Since $x_0 \neq 0$,$f(x_0) = \frac{x_0^4}{24} > 0$.
Since the global minimum of $f(x)$ is positive,$f(x) = 0$ has no real roots.

(B) Let $f(x) = 3^x + 5^x - (3^x)^2 + (3^x)(5^x) - (5^x)^2$.
Let $a = 3^x$ and $b = 5^x$,where $a, b > 0$.
Then $f(x) = a + b - a^2 + ab - b^2$.
We can rewrite this as $f(x) = a + b - (a^2 - ab + b^2)$.
To find the maximum,consider the expression $-(a^2 - ab + b^2) = -(a - b/2)^2 - 3b^2/4$.
Alternatively,complete the square: $f(x) = -(a^2 - a(1+b) + b^2 + b) = -[(a - \frac{1+b}{2})^2 - \frac{(1+b)^2}{4} + b^2 + b]$.
$f(x) = -(a - \frac{1+b}{2})^2 + \frac{1 + 2b + b^2 - 4b^2 - 4b}{4} = -(a - \frac{1+b}{2})^2 + \frac{1 - 2b - 3b^2}{4}$.
The maximum value for a fixed $b$ occurs at $a = \frac{1+b}{2}$,giving $f_{max}(b) = \frac{1 - 2b - 3b^2}{4}$.
Since $b = 5^x > 0$,the function $g(b) = \frac{1 - 2b - 3b^2}{4}$ is decreasing for $b > 0$.
The maximum value of $g(b)$ as $b \to 0^+$ is $1/4$.
However,checking $x=0$: $f(0) = 3^0 + 5^0 - 9^0 + 15^0 - 25^0 = 1 + 1 - 1 + 1 - 1 = 1$.
Since $1$ is the value at $x=0$,and $0 < 1 < 2$,the condition $0 < M < 2$ is satisfied.

(B) The curves $y=e^x$ and $y=\log_e x$ are inverse functions of each other,meaning they are symmetric about the line $y=x$.
Let $A$ be a point on $y=e^x$ with coordinates $(h, e^h)$. The distance from point $A$ to the line $y=x$ (or $x-y=0$) is given by $AB = \frac{|h-e^h|}{\sqrt{1^2+(-1)^2}} = \frac{|h-e^h|}{\sqrt{2}}$.
Since the curves are symmetric about $y=x$,the minimum distance between the two curves is $AC = 2AB$,where $B$ is the projection of $A$ onto the line $y=x$.
To minimize $AB$,we minimize $f(h) = e^h - h$ (since $e^h > h$ for all $h$).
$f'(h) = e^h - 1$. Setting $f'(h) = 0$ gives $e^h = 1$,so $h=0$.
At $h=0$,the minimum distance $AB = \frac{|0-e^0|}{\sqrt{2}} = \frac{1}{\sqrt{2}}$.
Therefore,the minimum distance between the curves is $AC = 2 \times \frac{1}{\sqrt{2}} = \sqrt{2}$.

(B) Given $f(x) = x^3 + ax^2 + bx + c$.
Since $f(2) = 0$,we have $8 + 4a + 2b + c = 0$ ... $(i)$.
The derivative is $f'(x) = 3x^2 + 2ax + b$.
Since $f(x)$ has critical points at $x = 1$ and $x = -\frac{1}{3}$,$f'(1) = 0$ and $f'(-\frac{1}{3}) = 0$.
$f'(1) = 3 + 2a + b = 0$ ... (ii).
$f'(-\frac{1}{3}) = 3(-\frac{1}{3})^2 + 2a(-\frac{1}{3}) + b = \frac{1}{3} - \frac{2a}{3} + b = 0$,which implies $1 - 2a + 3b = 0$ ... (iii).
Solving (ii) and (iii): $b = -2a - 3$. Substituting into (iii): $1 - 2a + 3(-2a - 3) = 0 \Rightarrow 1 - 2a - 6a - 9 = 0 \Rightarrow -8a = 8 \Rightarrow a = -1$.
Then $b = -2(-1) - 3 = -1$.
From $(i)$,$8 + 4(-1) + 2(-1) + c = 0 \Rightarrow 8 - 4 - 2 + c = 0 \Rightarrow c = -2$.
Thus,$f(x) = x^3 - x^2 - x - 2$.
We need to evaluate $\int_{-1}^1 (x^3 - x^2 - x - 2) dx$.
Since $x^3$ and $-x$ are odd functions,their integral over $[-1, 1]$ is $0$.
So,$\int_{-1}^1 f(x) dx = \int_{-1}^1 (-x^2 - 2) dx = -2 \int_0^1 (x^2 + 2) dx$.
$= -2 [\frac{x^3}{3} + 2x]_0^1 = -2 (\frac{1}{3} + 2) = -2 (\frac{7}{3}) = -\frac{14}{3}$.

(C) From the given graph of $f'(x)$,we observe the points where $f'(x) = 0$ are $x=a, b, c$.
$1$. At $x=a$: The derivative $f'(x)$ changes sign from negative to positive (as it crosses the x-axis from below to above). Thus,$f(x)$ has a local minimum at $x=a$.
$2$. At $x=b$: The derivative $f'(x)$ changes sign from positive to negative (as it crosses the x-axis from above to below). Thus,$f(x)$ has a local maximum at $x=b$.
$3$. At $x=c$: The derivative $f'(x)$ changes sign from negative to positive (as it crosses the x-axis from below to above). Thus,$f(x)$ has a local minimum at $x=c$.
Therefore,$f(x)$ has local minima at $x=a, c$ and a local maximum at $x=b$. The correct option is $C$.

(B) Given $g(x) = \frac{f(x)}{x}$.
Taking the derivative with respect to $x$,we get:
$g'(x) = \frac{x f'(x) - f(x)}{x^2}$.
From the given graph of $f(x)$,we observe that $f(x)$ is a concave downward function with a maximum at $x=c$ where $a < c < b$. Thus,$f'(c) = 0$.
At $x=c$,$g'(c) = \frac{c f'(c) - f(c)}{c^2} = \frac{c(0) - f(c)}{c^2} = -\frac{f(c)}{c^2}$.
Since $f(c) > 0$ and $c > 0$ (as the interval does not contain $0$),$g'(c) < 0$.
This implies that the function $g(x)$ is decreasing at $x=c$. Looking at the provided options,Fig $2$ represents a curve that is concave downward and consistent with the behavior of $g(x) = \frac{f(x)}{x}$ for a positive $f(x)$ and $x > 0$. Therefore,option $(b)$ is correct.

(C) Let $H$ be the height and $R$ be the radius of the given cone $V_1$. The volume is $V_1 = \frac{1}{3} \pi R^2 H$.
Let the inscribed cone $V_2$ have height $h$ and radius $r$. The apex of this cone is at $O$ and its base is at a distance $h$ from $O$. The height of the base of the inscribed cone from the apex $A$ is $H-h$.
By similar triangles,$\frac{r}{R} = \frac{H-h}{H} = 1 - \frac{h}{H}$.
Thus,$\frac{h}{H} = 1 - \frac{r}{R} \Rightarrow h = H(1 - \frac{r}{R})$.
The volume of the inscribed cone is $V_2 = \frac{1}{3} \pi r^2 h = \frac{1}{3} \pi r^2 H(1 - \frac{r}{R}) = \frac{\pi H}{3} (r^2 - \frac{r^3}{R})$.
To maximize $V_2$,we differentiate with respect to $r$: $\frac{dV_2}{dr} = \frac{\pi H}{3} (2r - \frac{3r^2}{R}) = 0$.
This gives $r(2 - \frac{3r}{R}) = 0$. Since $r \neq 0$,we have $r = \frac{2R}{3}$.
Substituting $r = \frac{2R}{3}$ into the expression for $h$: $h = H(1 - \frac{2R/3}{R}) = H(1 - \frac{2}{3}) = \frac{H}{3}$.
Now,$V_2 = \frac{1}{3} \pi (\frac{2R}{3})^2 (\frac{H}{3}) = \frac{1}{3} \pi (\frac{4R^2}{9}) (\frac{H}{3}) = \frac{4}{27} (\frac{1}{3} \pi R^2 H) = \frac{4}{27} V_1$.
Therefore,the ratio $\frac{V_2}{V_1} = \frac{4}{27}$.

(C) Let the parallel sides be $a = 30$ and $b = 30 + 2(30 \cos \theta) = 30 + 60 \cos \theta$. The height of the trapezium is $h = 30 \sin \theta$.
The area $A$ of the trapezium is given by:
$A = \frac{1}{2}(a + b)h = \frac{1}{2}(30 + 30 + 60 \cos \theta)(30 \sin \theta)$
$A = \frac{1}{2}(60 + 60 \cos \theta)(30 \sin \theta) = 900(1 + \cos \theta) \sin \theta = 900(\sin \theta + \sin \theta \cos \theta) = 900(\sin \theta + \frac{1}{2} \sin 2\theta)$.
To maximize the area,differentiate with respect to $\theta$ and set to $0$:
$\frac{dA}{d\theta} = 900(\cos \theta + \cos 2\theta) = 0$.
Since $\cos 2\theta = 2 \cos^2 \theta - 1$,we have:
$2 \cos^2 \theta + \cos \theta - 1 = 0$.
$(2 \cos \theta - 1)(\cos \theta + 1) = 0$.
This gives $\cos \theta = \frac{1}{2}$ or $\cos \theta = -1$.
Since $\theta$ is an angle in a triangle,$\theta = \frac{\pi}{3}$ (as $\theta = \pi$ is not possible).
Thus,the smallest angle is $\frac{\pi}{3}$.

(A) For $x \leq 0$,$f(x) = \int \limits_0^2 e^{t-x} dt = e^{-x}(e^2-1)$.
For $0 < x < 2$,$f(x) = \int \limits_0^x e^{x-t} dt + \int \limits_x^2 e^{t-x} dt = (e^x - 1) + (e^{2-x} - 1) = e^x + e^{2-x} - 2$.
For $x \geq 2$,$f(x) = \int \limits_0^2 e^{x-t} dt = e^{x-2}(e^2-1)$.
For $x \leq 0$,$f(x)$ is decreasing and for $x \geq 2$,$f(x)$ is increasing.
Therefore,the minimum value of $f(x)$ lies in the interval $x \in (0, 2)$.
In the interval $(0, 2)$,$f(x) = e^x + e^{2-x} - 2$.
Using the Arithmetic Mean-Geometric Mean inequality $(AM \geq GM)$:
$e^x + e^{2-x} \geq 2 \sqrt{e^x \cdot e^{2-x}} = 2 \sqrt{e^2} = 2e$.
Thus,the minimum value of $f(x)$ is $2e - 2 = 2(e-1)$.

(A) Given $f(x) = 2x^4 - 18x^2 + 8x + 12$.
First,find the derivative $f'(x) = 8x^3 - 36x + 8$.
Setting $f'(x) = 0$,we get $8x^3 - 36x + 8 = 0$,which simplifies to $2x^3 - 9x + 2 = 0$.
We are given that $x=2$ is a local minimum,so $(x-2)$ is a factor of $2x^3 - 9x + 2$.
Dividing $2x^3 - 9x + 2$ by $(x-2)$,we get $(x-2)(2x^2 + 4x - 1) = 0$.
The roots are $x=2$ and $x = \frac{-4 \pm \sqrt{16 - 4(2)(-1)}}{2(2)} = \frac{-4 \pm \sqrt{24}}{4} = -1 \pm \frac{\sqrt{6}}{2}$.
The critical points are $x=2$,$x = -1 + \frac{\sqrt{6}}{2}$,and $x = -1 - \frac{\sqrt{6}}{2}$.
Using the second derivative test $f''(x) = 24x^2 - 36$:
For $x=2$,$f''(2) = 24(4) - 36 = 60 > 0$ (Local Minima).
For $x = -1 - \frac{\sqrt{6}}{2}$,$f''(x) > 0$ (Local Minima).
For $x = -1 + \frac{\sqrt{6}}{2}$,$f''(x) < 0$ (Local Maxima).
Thus,the local maximum occurs at $x = \frac{\sqrt{6}-2}{2}$.
Substituting $x = \frac{\sqrt{6}-2}{2}$ into $f(x)$,we calculate $M = f\left(\frac{\sqrt{6}-2}{2}\right) = 12\sqrt{6} - \frac{33}{2}$.

(C) Given the function $f(x)=2 x^3+(2 p-7) x^2+3(2 p-9) x-6$.
First,find the derivative $f'(x) = 6x^2 + 2(2p-7)x + 3(2p-9)$.
For the function to have a local maximum at $x < 0$ and a local minimum at $x > 0$,the quadratic equation $f'(x) = 0$ must have two distinct real roots,one negative and one positive.
Let the roots be $\alpha$ and $\beta$ such that $\alpha < 0 < \beta$.
For a quadratic $ax^2 + bx + c = 0$ to have roots of opposite signs,the product of the roots must be negative,i.e.,$\frac{c}{a} < 0$.
Here,$a = 6$ and $c = 3(2p-9)$.
Thus,$\frac{3(2p-9)}{6} < 0$,which simplifies to $\frac{2p-9}{2} < 0$.
This implies $2p - 9 < 0$,or $p < \frac{9}{2}$.
Therefore,the set of all values of $p$ is $\left(-\infty, \frac{9}{2}\right)$.

(D) For a function to have an extreme point,its derivative must be zero at that point.
$f'(x) = x^2 + ax + 2b$
$g'(x) = x^2 + 2bx + a$
Let $x_0$ be the common extreme point. Then $f'(x_0) = 0$ and $g'(x_0) = 0$.
$x_0^2 + ax_0 + 2b = 0$ ---$(1)$
$x_0^2 + 2bx_0 + a = 0$ ---$(2)$
Subtracting equation $(2)$ from $(1)$:
$(a - 2b)x_0 + (2b - a) = 0$
$(a - 2b)x_0 - (a - 2b) = 0$
$(a - 2b)(x_0 - 1) = 0$
Since $a \neq 2b$,we must have $x_0 - 1 = 0$,which implies $x_0 = 1$.
Substituting $x_0 = 1$ into equation $(1)$:
$1^2 + a(1) + 2b = 0$
$1 + a + 2b = 0$
$a + 2b = -1$
We need to find the value of $a + 2b + 7$.
Substituting $a + 2b = -1$ into the expression:
$-1 + 7 = 6$.

(A) Given the total length of the wire is $\ell_1 + \ell_2 = 20$.
The side of the square formed by $\ell_1$ is $s = \frac{\ell_1}{4}$,so the area $A_1 = (\frac{\ell_1}{4})^2 = \frac{\ell_1^2}{16}$.
The radius of the circle formed by $\ell_2$ is $r = \frac{\ell_2}{2\pi}$,so the area $A_2 = \pi(\frac{\ell_2}{2\pi})^2 = \frac{\ell_2^2}{4\pi}$.
Let $S = 2A_1 + 3A_2 = 2(\frac{\ell_1^2}{16}) + 3(\frac{\ell_2^2}{4\pi}) = \frac{\ell_1^2}{8} + \frac{3\ell_2^2}{4\pi}$.
Substitute $\ell_2 = 20 - \ell_1$,then $S = \frac{\ell_1^2}{8} + \frac{3(20 - \ell_1)^2}{4\pi}$.
To find the minimum,differentiate with respect to $\ell_1$ and set to zero: $\frac{dS}{d\ell_1} = \frac{2\ell_1}{8} + \frac{6(20 - \ell_1)(-1)}{4\pi} = 0$.
$\frac{\ell_1}{4} = \frac{6(20 - \ell_1)}{4\pi} = \frac{6\ell_2}{4\pi}$.
$\frac{\pi \ell_1}{4} = \frac{6\ell_2}{4} \Rightarrow \frac{\pi \ell_1}{\ell_2} = 6$.

(A) Let $g(x) = x^2 - x + 1 = (x - \frac{1}{2})^2 + \frac{3}{4}$.
For $x \in [-1, 2]$,the range of $g(x)$ is $[\frac{3}{4}, 3]$.
Since $g(x) \ge \frac{3}{4}$,we have $|g(x)| = g(x)$.
Thus,$f(x) = g(x) + [g(x)]$.
To minimize $f(x)$,we consider the values of $g(x)$ in its range $[\frac{3}{4}, 3]$.
If $\frac{3}{4} \le g(x) < 1$,then $[g(x)] = 0$,so $f(x) = g(x) + 0 = g(x)$. The minimum value in this sub-interval is $\frac{3}{4}$.
If $1 \le g(x) < 2$,then $[g(x)] = 1$,so $f(x) = g(x) + 1$. The minimum value is $1 + 1 = 2$.
If $2 \le g(x) \le 3$,then $[g(x)] = 2$,so $f(x) = g(x) + 2$. The minimum value is $2 + 2 = 4$.
Comparing these,the absolute minimum value is $\frac{3}{4}$.

(A) Given $f(x) = |x^2 - 5x + 6| - 3x + 2$. The roots of $x^2 - 5x + 6 = 0$ are $x = 2$ and $x = 3$.
For $x \in [-1, 2]$,$x^2 - 5x + 6 \ge 0$,so $f(x) = x^2 - 5x + 6 - 3x + 2 = x^2 - 8x + 8$.
For $x \in [2, 3]$,$x^2 - 5x + 6 \le 0$,so $f(x) = -(x^2 - 5x + 6) - 3x + 2 = -x^2 + 2x - 4$.
Now,check critical points and endpoints:
$1$. For $x \in [-1, 2]$,$f'(x) = 2x - 8$. Setting $f'(x) = 0$ gives $x = 4$,which is outside the interval. Thus,check endpoints: $f(-1) = (-1)^2 - 8(-1) + 8 = 1 + 8 + 8 = 17$ and $f(2) = 2^2 - 8(2) + 8 = 4 - 16 + 8 = -4$.
$2$. For $x \in [2, 3]$,$f'(x) = -2x + 2$. Setting $f'(x) = 0$ gives $x = 1$,which is outside the interval. Thus,check endpoints: $f(2) = -4$ and $f(3) = -(3)^2 + 2(3) - 4 = -9 + 6 - 4 = -7$.
The absolute maximum value is $17$ and the absolute minimum value is $-7$.
The sum of the absolute maximum and minimum values is $17 + (-7) = 10$.

(C) Let $f(x) = x^5-20x^3+50x+2$.
To find the number of points where the curve crosses the $x$-axis,we analyze the local maxima and minima by finding the derivative:
$f'(x) = 5x^4-60x^2+50 = 5(x^4-12x^2+10)$.
Setting $f'(x) = 0$,we get $x^4-12x^2+10 = 0$.
Using the quadratic formula for $x^2$:
$x^2 = \frac{12 \pm \sqrt{144-40}}{2} = 6 \pm \sqrt{26} \approx 6 \pm 5.1$.
So,$x^2 \approx 11.1$ or $x^2 \approx 0.9$.
This gives critical points at $x \approx \pm 3.3$ and $x \approx \pm 0.95$.
Evaluating the function at these points:
$f(-3.3) \approx -100 < 0$
$f(-0.95) \approx -28 < 0$
$f(0.95) \approx 32 > 0$
$f(3.3) \approx 104 > 0$
Also,$f(-4) < 0$,$f(-2) > 0$,$f(0) = 2$,$f(2) = -14$,$f(4) > 0$.
By the Intermediate Value Theorem,the function changes sign between $(-4, -2)$,$(-2, 0)$,$(0, 2)$,and $(2, 4)$.
Thus,there are $5$ points where the curve crosses the $x$-axis.

(B) Let $f(x) = \frac{x^3}{x^4+147}$.
To find the maximum value,we calculate the derivative $f'(x)$:
$f'(x) = \frac{(x^4+147)(3x^2) - (x^3)(4x^3)}{(x^4+147)^2}$
$f'(x) = \frac{3x^6 + 441x^2 - 4x^6}{(x^4+147)^2} = \frac{441x^2 - x^6}{(x^4+147)^2} = \frac{x^2(441 - x^4)}{(x^4+147)^2}$.
Setting $f'(x) = 0$,we get $x^2 = 0$ or $x^4 = 441$,which implies $x^2 = 21$ (since $x > 0$),so $x = \sqrt{21} \approx 4.58$.
Since $f(x)$ increases for $x < \sqrt{21}$ and decreases for $x > \sqrt{21}$,the maximum term in the sequence $a_n$ occurs at $n$ values closest to $\sqrt{21}$.
We compare $a_4$ and $a_5$:
$a_4 = \frac{4^3}{4^4+147} = \frac{64}{256+147} = \frac{64}{403} \approx 0.1588$.
$a_5 = \frac{5^3}{5^4+147} = \frac{125}{625+147} = \frac{125}{772} \approx 0.1619$.
Since $a_5 > a_4$,the greatest term is at $n = 5$.

(C) Let the side of the square cut from each corner be $x\,cm$.
The dimensions of the resulting box will be length $= (30-2x)\,cm$,width $= (30-2x)\,cm$,and height $= x\,cm$.
The volume $V$ of the box is given by $V = x(30-2x)^2$.
To find the maximum volume,we differentiate $V$ with respect to $x$:
$V = x(900 - 120x + 4x^2) = 4x^3 - 120x^2 + 900x$.
$\frac{dV}{dx} = 12x^2 - 240x + 900$.
Setting $\frac{dV}{dx} = 0$ for critical points:
$12(x^2 - 20x + 75) = 0$
$12(x-5)(x-15) = 0$.
Thus,$x = 5$ or $x = 15$. Since $x=15$ would result in a side length of $0$,we take $x = 5\,cm$.
The surface area $S$ of the open box is the area of the original square minus the area of the four cut-out squares:
$S = (30)^2 - 4x^2 = 900 - 4(5)^2 = 900 - 100 = 800\,cm^2$.

(C) Given the inequality: $(x \ln x) f'(x) + (\ln x + 1) f(x) \geq 1$.
This can be written as $\frac{d}{dx} (x \ln x \cdot f(x)) \geq 1$.
Let $g(x) = x \ln x \cdot f(x) - x$. Then $g'(x) = \frac{d}{dx} (x \ln x \cdot f(x)) - 1 \geq 0$.
Thus,$g(x)$ is a non-decreasing function on $[2,4]$.
Calculating values at endpoints:
$g(2) = 2 \ln 2 \cdot f(2) - 2 = 2 \ln 2 \cdot \frac{1}{2} - 2 = \ln 2 - 2$.
$g(4) = 4 \ln 4 \cdot f(4) - 4 = 4 \ln 4 \cdot \frac{1}{4} - 4 = \ln 4 - 4 = 2 \ln 2 - 4$.
Since $g(x)$ is non-decreasing,$g(2) \leq g(x) \leq g(4)$ for $x \in [2,4]$.
$\ln 2 - 2 \leq x \ln x \cdot f(x) - x \leq 2 \ln 2 - 4$.
Adding $x$ and dividing by $x \ln x$ gives:
$\frac{\ln 2 - 2}{x \ln x} + \frac{1}{\ln x} \leq f(x) \leq \frac{2 \ln 2 - 4}{x \ln x} + \frac{1}{\ln x}$.
For $x \in [2,4]$,the upper bound is $\leq \frac{2 \ln 2 - 4}{2 \ln 2} + \frac{1}{\ln 2} = 1 - \frac{2}{\ln 2} + \frac{1}{\ln 2} = 1 - \frac{1}{\ln 2} < 1$. Thus,$(A)$ is true.
For $x \in [2,4]$,the lower bound is $\geq \frac{\ln 2 - 2}{4 \ln 4} + \frac{1}{\ln 4} = \frac{\ln 2 - 2}{8 \ln 2} + \frac{1}{2 \ln 2} = \frac{\ln 2 - 2 + 4}{8 \ln 2} = \frac{\ln 2 + 2}{8 \ln 2} = \frac{1}{8} + \frac{1}{4 \ln 2} > \frac{1}{8}$. Thus,$(B)$ is true.

(C) Let $y=\left(\frac{\sqrt{3 e}}{2 \sin x}\right)^{\sin ^2 x}$.
Taking natural logarithm on both sides: $\ln y = \sin^2 x \cdot \ln \left(\frac{\sqrt{3 e}}{2 \sin x}\right)$.
Differentiating with respect to $x$: $\frac{1}{y} \frac{dy}{dx} = 2 \sin x \cos x \ln \left(\frac{\sqrt{3 e}}{2 \sin x}\right) + \sin^2 x \cdot \frac{2 \sin x}{\sqrt{3 e}} \cdot \frac{\sqrt{3 e}}{2} \cdot (-\csc x \cot x)$.
Simplifying the derivative: $\frac{1}{y} \frac{dy}{dx} = \sin x \cos x \left[ 2 \ln \left(\frac{\sqrt{3 e}}{2 \sin x}\right) - 1 \right]$.
Setting $\frac{dy}{dx} = 0$,we get $2 \ln \left(\frac{\sqrt{3 e}}{2 \sin x}\right) = 1$,which implies $\ln \left(\frac{3 e}{4 \sin^2 x}\right) = 1$.
Thus,$\frac{3 e}{4 \sin^2 x} = e$,so $\sin^2 x = \frac{3}{4}$.
Since $x \in (0, \frac{\pi}{2})$,$\sin x = \frac{\sqrt{3}}{2}$.
The local maximum value is $f(x) = \left(\frac{\sqrt{3 e}}{2 \cdot \frac{\sqrt{3}}{2}}\right)^{3/4} = (\sqrt{e})^{3/4} = e^{3/8}$.
Given $e^{3/8} = \frac{k}{e}$,we have $k = e^{1 + 3/8} = e^{11/8}$.
Then $k^8 = (e^{11/8})^8 = e^{11}$.
Substituting these values: $\left(\frac{k}{e}\right)^8 + \frac{k^8}{e^5} + k^8 = (e^{3/8})^8 + \frac{e^{11}}{e^5} + e^{11} = e^3 + e^6 + e^{11}$.

(A) Let $f(x) = x - 2 \sin x \cos x + \frac{1}{3} \sin 3x = x - \sin 2x + \frac{1}{3} \sin 3x$.
To find the maximum,we differentiate $f(x)$ with respect to $x$:
$f'(x) = 1 - 2 \cos 2x + \cos 3x$.
Setting $f'(x) = 0$:
$1 - 2(2 \cos^2 x - 1) + (4 \cos^3 x - 3 \cos x) = 0$
$1 - 4 \cos^2 x + 2 + 4 \cos^3 x - 3 \cos x = 0$
$4 \cos^3 x - 4 \cos^2 x - 3 \cos x + 3 = 0$
$4 \cos^2 x (\cos x - 1) - 3 (\cos x - 1) = 0$
$(4 \cos^2 x - 3)(\cos x - 1) = 0$.
This gives $\cos x = 1$ or $\cos x = \pm \frac{\sqrt{3}}{2}$.
For $x \in [0, \pi]$,$x = 0, \frac{\pi}{6}, \frac{5\pi}{6}, \pi$.
Evaluating $f(x)$ at these points:
$f(0) = 0 - 0 + 0 = 0$.
$f(\pi) = \pi - 0 + 0 = \pi$.
$f(\frac{\pi}{6}) = \frac{\pi}{6} - \sin(\frac{\pi}{3}) + \frac{1}{3} \sin(\frac{\pi}{2}) = \frac{\pi}{6} - \frac{\sqrt{3}}{2} + \frac{1}{3} = \frac{\pi + 2 - 3\sqrt{3}}{6}$.
$f(\frac{5\pi}{6}) = \frac{5\pi}{6} - \sin(\frac{5\pi}{3}) + \frac{1}{3} \sin(\frac{5\pi}{2}) = \frac{5\pi}{6} - (-\frac{\sqrt{3}}{2}) + \frac{1}{3}(1) = \frac{5\pi + 3\sqrt{3} + 2}{6}$.
Comparing the values,the maximum value is $\frac{5\pi + 2 + 3\sqrt{3}}{6}$.

(C) Given function is $f(x) = 2x + 3x^{\frac{2}{3}}$.
First,find the derivative $f'(x)$:
$f'(x) = 2 + 3 \cdot \frac{2}{3} x^{\frac{2}{3} - 1} = 2 + 2x^{-\frac{1}{3}} = 2 + \frac{2}{x^{\frac{1}{3}}} = 2 \left( \frac{x^{\frac{1}{3}} + 1}{x^{\frac{1}{3}}} \right)$.
Critical points occur where $f'(x) = 0$ or $f'(x)$ is undefined.
$f'(x) = 0 \implies x^{\frac{1}{3}} + 1 = 0 \implies x = -1$.
$f'(x)$ is undefined at $x = 0$.
Now,check the sign of $f'(x)$ around these points:
For $x < -1$,$f'(x) > 0$.
For $-1 < x < 0$,$f'(x) < 0$.
For $x > 0$,$f'(x) > 0$.
Since $f'(x)$ changes from positive to negative at $x = -1$,there is a local maximum at $x = -1$.
Since $f'(x)$ changes from negative to positive at $x = 0$,there is a local minimum at $x = 0$.
Thus,the function has exactly one point of local maxima and exactly one point of local minima.

(D) The vertices of the triangle are $(0,0)$,$(x, y)$,and $(-x, y)$.
The area of the triangle is given by the determinant formula:
$\text{Area} = \frac{1}{2} |0(y - y) + x(y - 0) + (-x)(0 - y)|$
$\text{Area} = \frac{1}{2} |xy + xy| = |xy|$
Given the curve $y = -2x^2 + 54$,we substitute $y$ into the area expression:
$A(x) = |x(-2x^2 + 54)| = |-2x^3 + 54x|$
Since $y > 0$,we have $-2x^2 + 54 > 0$,which implies $x^2 < 27$,so $x \in (-\sqrt{27}, \sqrt{27})$.
For $x > 0$,$A(x) = -2x^3 + 54x$.
To find the maximum area,we differentiate $A(x)$ with respect to $x$:
$\frac{dA}{dx} = -6x^2 + 54$
Setting $\frac{dA}{dx} = 0$:
$-6x^2 + 54 = 0 \Rightarrow x^2 = 9 \Rightarrow x = 3$ (since $x > 0$).
Now,calculate the maximum area at $x = 3$:
$A(3) = |3(-2(3)^2 + 54)| = |3(-18 + 54)| = |3(36)| = 108$.
Thus,the maximum area of the triangle is $108$.

(C) Given $f(x)=(x+3)^2(x-2)^3$.
To find the critical points,we differentiate $f(x)$ with respect to $x$:
$f^{\prime}(x) = \frac{d}{dx}[(x+3)^2(x-2)^3]$
$f^{\prime}(x) = 2(x+3)(x-2)^3 + 3(x-2)^2(x+3)^2$
$f^{\prime}(x) = (x+3)(x-2)^2 [2(x-2) + 3(x+3)]$
$f^{\prime}(x) = (x+3)(x-2)^2 [2x - 4 + 3x + 9]$
$f^{\prime}(x) = (x+3)(x-2)^2 (5x + 5)$
$f^{\prime}(x) = 5(x+3)(x-2)^2 (x+1)$
Setting $f^{\prime}(x) = 0$,we get critical points $x = -3, -1, 2$.
Now,we evaluate $f(x)$ at the critical points and the endpoints of the interval $[-4, 4]$:
$f(-4) = (-4+3)^2(-4-2)^3 = (-1)^2(-6)^3 = 1 \times (-216) = -216$
$f(-3) = (-3+3)^2(-3-2)^3 = 0$
$f(-1) = (-1+3)^2(-1-2)^3 = (2)^2(-3)^3 = 4 \times (-27) = -108$
$f(2) = (2+3)^2(2-2)^3 = 0$
$f(4) = (4+3)^2(4-2)^3 = (7)^2(2)^3 = 49 \times 8 = 392$
Comparing these values,the maximum value $M = 392$ and the minimum value $m = -216$.
Therefore,$M - m = 392 - (-216) = 392 + 216 = 608$.

(B) By the Fundamental Theorem of Calculus,$f'(x) = (e^x-1)^{11}(2x-1)^5(x-2)^7(x-3)^{12}(2x-10)^{61}$.
To find critical points,set $f'(x) = 0$,which gives $x = 0, \frac{1}{2}, 2, 3, 5$.
We analyze the sign of $f'(x)$ around these points:
- At $x=0$: $f'(x)$ changes from $+$ to $-$,so $f(x)$ has a local maximum at $x=0$.
- At $x=\frac{1}{2}$: $f'(x)$ changes from $-$ to $+$,so $f(x)$ has a local minimum at $x=\frac{1}{2}$.
- At $x=2$: $f'(x)$ changes from $+$ to $-$,so $f(x)$ has a local maximum at $x=2$.
- At $x=3$: $f'(x)$ does not change sign (power is $12$),so it is a point of inflection.
- At $x=5$: $f'(x)$ changes from $-$ to $+$,so $f(x)$ has a local minimum at $x=5$.
Local maxima occur at $x=0$ and $x=2$. Thus,$p = 0^2 + 2^2 = 4$.
Local minima occur at $x=\frac{1}{2}$ and $x=5$. Thus,$q = \frac{1}{2} + 5 = \frac{11}{2}$.
Finally,$p^2 + 2q = 4^2 + 2(\frac{11}{2}) = 16 + 11 = 27$.

(D) Let $y = \frac{2x^2 - 3x + 8}{2x^2 + 3x + 8}$.
Rearranging the terms,we get $y(2x^2 + 3x + 8) = 2x^2 - 3x + 8$.
$x^2(2y - 2) + x(3y + 3) + 8y - 8 = 0$.
For $x$ to be real,the discriminant $D \geq 0$.
$D = (3y + 3)^2 - 4(2y - 2)(8y - 8) \geq 0$.
$9(y + 1)^2 - 4 \cdot 2(y - 1) \cdot 8(y - 1) \geq 0$.
$9(y^2 + 2y + 1) - 64(y^2 - 2y + 1) \geq 0$.
$9y^2 + 18y + 9 - 64y^2 + 128y - 64 \geq 0$.
$-55y^2 + 146y - 55 \geq 0$.
$55y^2 - 146y + 55 \leq 0$.
Solving the quadratic equation $55y^2 - 146y + 55 = 0$ using the quadratic formula $y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:
$y = \frac{146 \pm \sqrt{21316 - 12100}}{110} = \frac{146 \pm \sqrt{9216}}{110} = \frac{146 \pm 96}{110}$.
$y_1 = \frac{242}{110} = \frac{11}{5}$ and $y_2 = \frac{50}{110} = \frac{5}{11}$.
Thus,the range is $[\frac{5}{11}, \frac{11}{5}]$.
The maximum value is $\frac{11}{5}$ and the minimum value is $\frac{5}{11}$.
Sum $= \frac{11}{5} + \frac{5}{11} = \frac{121 + 25}{55} = \frac{146}{55}$.
Here $m = 146$ and $n = 55$. Since $\gcd(146, 55) = 1$,$m + n = 146 + 55 = 201$.

(B) Given the function $f(x)=3 \sqrt{x-2}+\sqrt{4-x}$.
The domain of the function is determined by $x-2 \geq 0$ and $4-x \geq 0$,which implies $x \in [2, 4]$.
To find the range,we use the Cauchy-Schwarz inequality or trigonometric substitution. Let $u = \sqrt{x-2}$ and $v = \sqrt{4-x}$. Then $u^2 + v^2 = (x-2) + (4-x) = 2$.
The function is $f(x) = 3u + v$. By the Cauchy-Schwarz inequality,$(3u + v)^2 \leq (3^2 + 1^2)(u^2 + v^2) = (9+1)(2) = 20$.
Thus,$f(x)^2 \leq 20$,so $f(x) \leq \sqrt{20} = \beta$.
At the boundaries of the domain:
If $x=2$,$f(2) = 3(0) + \sqrt{2} = \sqrt{2}$.
If $x=4$,$f(4) = 3\sqrt{2} + 0 = 3\sqrt{2} = \sqrt{18}$.
Since $\sqrt{2} < \sqrt{18} < \sqrt{20}$,the minimum value $\alpha = \sqrt{2}$.
Therefore,$\alpha^2 + 2\beta^2 = (\sqrt{2})^2 + 2(\sqrt{20})^2 = 2 + 2(20) = 2 + 40 = 42$.

(A) Let $\theta$ be the angle that the side $AB$ of the rectangle $ABCD$ makes with the side $PQ$ of the rectangle $PQRS$.
From the geometry of the figure,the sides of the rectangle $PQRS$ are $a = 4 \cos \theta + 2 \sin \theta$ and $b = 4 \sin \theta + 2 \cos \theta$.
The area of the rectangle $PQRS$ is $A = a \times b = (4 \cos \theta + 2 \sin \theta)(4 \sin \theta + 2 \cos \theta)$.
Expanding this,we get $A = 16 \sin \theta \cos \theta + 8 \cos^2 \theta + 8 \sin^2 \theta + 4 \sin \theta \cos \theta = 8(\sin^2 \theta + \cos^2 \theta) + 20 \sin \theta \cos \theta$.
Using trigonometric identities,$A = 8 + 10 \sin 2\theta$.
The area is maximum when $\sin 2\theta = 1$,which occurs at $\theta = 45^{\circ}$.
At $\theta = 45^{\circ}$,the sides are $a = 4(\frac{1}{\sqrt{2}}) + 2(\frac{1}{\sqrt{2}}) = \frac{6}{\sqrt{2}} = 3\sqrt{2}$ and $b = 4(\frac{1}{\sqrt{2}}) + 2(\frac{1}{\sqrt{2}}) = 3\sqrt{2}$.
Then $(a+b)^2 = (3\sqrt{2} + 3\sqrt{2})^2 = (6\sqrt{2})^2 = 36 \times 2 = 72$.

(C) Let $f(x) = (\frac{1}{x})^{2x}$.
Taking the natural logarithm on both sides,we get $\ln(f(x)) = 2x \ln(\frac{1}{x}) = -2x \ln(x)$.
To find the critical points,we differentiate with respect to $x$:
$\frac{f'(x)}{f(x)} = -2(1 \cdot \ln(x) + x \cdot \frac{1}{x}) = -2(\ln(x) + 1)$.
Setting $f'(x) = 0$,we get $\ln(x) = -1$,which implies $x = \frac{1}{e}$.
Since $f'(x) > 0$ for $x < \frac{1}{e}$ and $f'(x) < 0$ for $x > \frac{1}{e}$,the function attains its maximum value at $x = \frac{1}{e}$.
Thus,$f(x) \leq f(\frac{1}{e})$ for all $x > 0$.
$f(\frac{1}{e}) = (\frac{1}{1/e})^{2(1/e)} = e^{2/e}$.
For any $x$,$(\frac{1}{x})^{2x} \leq e^{2/e}$.
Taking $x = \frac{1}{\pi}$,we have $(\frac{1}{1/\pi})^{2(1/\pi)} \leq e^{2/e}$.
$\pi^{2/\pi} \leq e^{2/e}$.
Raising both sides to the power of $\frac{\pi e}{2}$,we get $(\pi^{2/\pi})^{\pi e/2} \leq (e^{2/e})^{\pi e/2}$.
$\pi^e \leq e^\pi$.
Since $\pi \neq e$,we have $e^\pi > \pi^e$.

(A) Given function: $f(x) = (x-2)^{2/3}(2x+1)$.
To find the critical points,we calculate the derivative $f'(x)$ using the product rule:
$f'(x) = \frac{d}{dx}[(x-2)^{2/3}] \cdot (2x+1) + (x-2)^{2/3} \cdot \frac{d}{dx}[2x+1]$
$f'(x) = \frac{2}{3}(x-2)^{-1/3}(2x+1) + 2(x-2)^{2/3}$
Factor out $2(x-2)^{-1/3}$:
$f'(x) = 2(x-2)^{-1/3} [\frac{1}{3}(2x+1) + (x-2)]$
$f'(x) = \frac{2}{3(x-2)^{1/3}} [2x + 1 + 3x - 6]$
$f'(x) = \frac{2(5x-5)}{3(x-2)^{1/3}} = \frac{10(x-1)}{3(x-2)^{1/3}}$
Critical points occur where $f'(x) = 0$ or $f'(x)$ is undefined.
$f'(x) = 0 \implies x-1 = 0 \implies x = 1$.
$f'(x)$ is undefined when the denominator is zero,i.e.,$x-2 = 0 \implies x = 2$.
Thus,the critical points are $x=1$ and $x=2$.
There are $2$ critical points.

(B) Given $f(x)=4 \cos ^3 x+3 \sqrt{3} \cos ^2 x-10$ for $x \in(0, 2 \pi)$.
First,find the derivative $f^{\prime}(x)$:
$f^{\prime}(x) = 4(3 \cos ^2 x)(-\sin x) + 3 \sqrt{3}(2 \cos x)(-\sin x)$
$f^{\prime}(x) = -12 \cos ^2 x \sin x - 6 \sqrt{3} \cos x \sin x$
$f^{\prime}(x) = -6 \sin x \cos x (2 \cos x + \sqrt{3})$
$f^{\prime}(x) = -3 \sin(2x) (2 \cos x + \sqrt{3})$
To find critical points,set $f^{\prime}(x) = 0$:
$-3 \sin(2x) = 0 \Rightarrow 2x = 0, \pi, 2\pi, 3\pi, 4\pi \Rightarrow x = 0, \frac{\pi}{2}, \pi, \frac{3\pi}{2}, 2\pi$
$2 \cos x + \sqrt{3} = 0 \Rightarrow \cos x = -\frac{\sqrt{3}}{2} \Rightarrow x = \frac{5\pi}{6}, \frac{7\pi}{6}$
Critical points in $(0, 2\pi)$ are $\frac{\pi}{2}, \frac{5\pi}{6}, \pi, \frac{7\pi}{6}, \frac{3\pi}{2}$.
Analyzing the sign of $f^{\prime}(x)$:
For $x \in (0, \frac{\pi}{2})$,$f^{\prime}(x) < 0$.
For $x \in (\frac{\pi}{2}, \frac{5\pi}{6})$,$f^{\prime}(x) > 0$.
For $x \in (\frac{5\pi}{6}, \pi)$,$f^{\prime}(x) < 0$.
For $x \in (\pi, \frac{7\pi}{6})$,$f^{\prime}(x) > 0$.
For $x \in (\frac{7\pi}{6}, \frac{3\pi}{2})$,$f^{\prime}(x) < 0$.
For $x \in (\frac{3\pi}{2}, 2\pi)$,$f^{\prime}(x) > 0$.
Local maxima occur where $f^{\prime}(x)$ changes from positive to negative. This happens at $x = \frac{5\pi}{6}$ and $x = \frac{3\pi}{2}$.
Thus,there are $2$ points of local maxima.

(A) Given $f(x)=2x^3-9ax^2+12a^2x+1$.
Find the derivative: $f'(x)=6x^2-18ax+12a^2$.
For local extrema,set $f'(x)=0$,so $6(x^2-3ax+2a^2)=0$,which factors to $6(x-a)(x-2a)=0$.
The roots are $x=a$ and $x=2a$.
Given the local maximum at $x=\alpha$ and local minimum at $x=\alpha^2$,we have two cases:
Case $1$: $\alpha=a$ and $\alpha^2=2a$. Then $a^2=2a \Rightarrow a(a-2)=0$. Since $a>0$,$a=2$.
If $a=2$,the roots are $\alpha=2$ and $\alpha^2=4$. The equation is $(x-2)(x-4)=x^2-6x+8=0$.
Case $2$: $\alpha=2a$ and $\alpha^2=a$. Then $(2a)^2=a \Rightarrow 4a^2-a=0 \Rightarrow a(4a-1)=0$. Since $a>0$,$a=1/4$.
If $a=1/4$,the roots are $\alpha=1/2$ and $\alpha^2=1/4$. The equation is $(x-1/2)(x-1/4)=x^2-(3/4)x+1/8=0$,or $8x^2-6x+1=0$.
Checking the second derivative $f''(x)=12x-18a$:
For $a=2$,$f''(2)=24-36=-12 < 0$ (max) and $f''(4)=48-36=12 > 0$ (min). This matches.
For $a=1/4$,$f''(1/2)=6-18(1/4)=1.5 > 0$ (min) and $f''(1/4)=3-18(1/4)=-1.5 < 0$ (max). This contradicts the problem statement.
Thus,the correct equation is $x^2-6x+8=0$.

(A) Let the rectangle have vertices at $(x, b)$,$(24, b)$,$(24, -b)$,and $(x, -b)$.
Since the vertex $(x, b)$ lies on the parabola $y^2=2x$,we have $b^2=2x$,which implies $x = \frac{b^2}{2}$.
The width of the rectangle is $(24 - x) = (24 - \frac{b^2}{2})$ and the height is $2b$.
The area $A$ of the rectangle is given by $A = 2b(24 - \frac{b^2}{2}) = 48b - b^3$.
To find the maximum area,we differentiate $A$ with respect to $b$:
$\frac{dA}{db} = 48 - 3b^2$.
Setting $\frac{dA}{db} = 0$,we get $3b^2 = 48$,so $b^2 = 16$,which gives $b = 4$ (since $b > 0$).
The maximum area is $A = 48(4) - (4)^3 = 192 - 64 = 128$.

(C) First,solve the inequality $\frac{x^2+x+2}{x^2+5x+6} < 0$.
Since the numerator $x^2+x+2$ has a discriminant $D = 1^2 - 4(1)(2) = -7 < 0$,it is always positive for all real $x$.
Thus,the inequality reduces to $\frac{1}{(x+2)(x+3)} < 0$,which implies $(x+2)(x+3) < 0$.
This gives $x \in (-3, -2)$.
Next,consider the function $f(x) = 1 + x\lambda^2 - x^3$.
Find the derivative: $f'(x) = \lambda^2 - 3x^2$.
Set $f'(x) = 0$ to find critical points: $3x^2 = \lambda^2 \Rightarrow x = \pm \frac{\lambda}{\sqrt{3}}$.
Using the second derivative test: $f''(x) = -6x$.
For local minimum,$f''(x) > 0$,so $-6x > 0 \Rightarrow x < 0$.
Thus,the point of local minimum is $x = -\frac{\lambda}{\sqrt{3}}$.
Given that this point must satisfy $x \in (-3, -2)$,we have:
$-3 < -\frac{\lambda}{\sqrt{3}} < -2$
Multiplying by $-1$ reverses the inequality:
$2 < \frac{\lambda}{\sqrt{3}} < 3$
$2\sqrt{3} < \lambda < 3\sqrt{3}$.
So,$\alpha = 2\sqrt{3}$ and $\beta = 3\sqrt{3}$.
Then $\alpha^2 + \beta^2 = (2\sqrt{3})^2 + (3\sqrt{3})^2 = 12 + 27 = 39$.

(B) Given $f(x) = (p^2 - 6p + 8)(\sin^2 2x - \cos^2 2x) + 2(2 - p)x + 7$.
Using $\cos 4x = \cos^2 2x - \sin^2 2x$,we have $f(x) = -(p^2 - 6p + 8)\cos 4x + 2(2 - p)x + 7$.
For $f(x)$ to have no critical points,$f'(x) \neq 0$ for all $x \in \mathbb{R}$.
$f'(x) = 4(p^2 - 6p + 8)\sin 4x + 2(2 - p)$.
Setting $f'(x) = 0$,we get $4(p - 4)(p - 2)\sin 4x = 2(p - 2)$.
If $p = 2$,$f'(x) = 0$ for all $x$,so $p \neq 2$.
For $p \neq 2$,$\sin 4x = \frac{2(p - 2)}{4(p - 4)(p - 2)} = \frac{1}{2(p - 4)}$.
For no critical points,the equation $\sin 4x = \frac{1}{2(p - 4)}$ must have no solution.
This happens if $\left| \frac{1}{2(p - 4)} \right| > 1$.
$|2(p - 4)| < 1 \implies -1 < 2p - 8 < 1 \implies 7 < 2p < 9 \implies p \in (3.5, 4.5)$.
Thus,$a = 3.5 = \frac{7}{2}$ and $b = 4.5 = \frac{9}{2}$.
$16ab = 16 \times \frac{7}{2} \times \frac{9}{2} = 4 \times 7 \times 9 = 252$.

(B, A, A) $1.$ For $k \leq 0$,let $g(x) = ke^x - x$. Since $k \leq 0$,$g'(x) = ke^x - 1 < 0$ for all $x \in R$. Thus,$g(x)$ is a strictly decreasing function. As $x \to -\infty$,$g(x) \to \infty$,and as $x \to \infty$,$g(x) \to -\infty$. By the Intermediate Value Theorem,$g(x)=0$ has exactly one root. Thus,the line $y=x$ meets $y=ke^x$ at one point. Correct option is $(B)$.
$2.$ Let $f(x) = ke^x - x$. For $k>0$,$f'(x) = ke^x - 1$. Setting $f'(x)=0$ gives $e^x = 1/k$,so $x = -\ln k$. The minimum value is $f(-\ln k) = k(1/k) - (-\ln k) = 1 + \ln k$. For exactly one root,the minimum value must be $0$. Thus,$1 + \ln k = 0 \Rightarrow \ln k = -1 \Rightarrow k = 1/e$. Correct option is $(A)$.
$3.$ For two distinct roots,the minimum value must be negative,i.e.,$1 + \ln k < 0$. This implies $\ln k < -1$,so $k < 1/e$. Since $k>0$,the set of values is $(0, 1/e)$. Correct option is $(A)$.

(C) To find the local maxima and local minima,we analyze the function $f(x)$ in its given intervals:
$1$. For $-3 < x \leq -1$,$f(x) = (2+x)^3$. The derivative is $f'(x) = 3(2+x)^2$. Since $f'(x) \geq 0$,the function is increasing on $(-3, -1]$. At $x = -1$,$f(-1) = (2-1)^3 = 1$. Since the function is increasing up to $x = -1$,this point acts as a local maximum.
$2$. For $-1 < x < 2$,$f(x) = x^{2/3}$. The derivative is $f'(x) = \frac{2}{3}x^{-1/3} = \frac{2}{3\sqrt[3]{x}}$.
$3$. At $x = 0$,$f'(x)$ is undefined. For $x < 0$,$f'(x) < 0$ (decreasing),and for $x > 0$,$f'(x) > 0$ (increasing). Thus,$x = 0$ is a local minimum with $f(0) = 0$.
$4$. Comparing the values,we have a local maximum at $x = -1$ and a local minimum at $x = 0$.
Therefore,the total number of local maxima and local minima is $2$.

(C) Let $f(x) = 4 \alpha x^2 + \frac{1}{x}$ for $x > 0$.
To find the minimum value of $f(x)$,we differentiate with respect to $x$:
$f'(x) = 8 \alpha x - \frac{1}{x^2}$.
Setting $f'(x) = 0$ for critical points:
$8 \alpha x = \frac{1}{x^2}$ $\Rightarrow x^3 = \frac{1}{8 \alpha}$ $\Rightarrow x = \frac{1}{2 \alpha^{1/3}}$.
Substituting this value into $f(x)$ to find the minimum:
$f\left(\frac{1}{2 \alpha^{1/3}}\right) = 4 \alpha \left(\frac{1}{4 \alpha^{2/3}}\right) + \frac{1}{1/(2 \alpha^{1/3})} = \alpha^{1/3} + 2 \alpha^{1/3} = 3 \alpha^{1/3}$.
Given $f(x) \geq 1$ for all $x > 0$,the minimum value must be at least $1$:
$3 \alpha^{1/3} \geq 1$ $\Rightarrow \alpha^{1/3} \geq \frac{1}{3}$ $\Rightarrow \alpha \geq \frac{1}{27}$.
Thus,the least value of $\alpha$ is $\frac{1}{27}$.

(A) Let $S$ be the set of all twice differentiable functions $f: R \rightarrow R$ such that $\frac{d^2 f}{dx^2} > 0$ for all $x \in (-1, 1)$.
This implies that the graph of $f$ is strictly concave upward (convex) on the interval $(-1, 1)$.
Let $\phi(x) = f(x) - x$. Then $\phi''(x) = f''(x) - 0 = f''(x) > 0$ for all $x \in (-1, 1)$.
Since $\phi''(x) > 0$,the function $\phi(x)$ is strictly convex.
$A$ strictly convex function can intersect the $x$-axis at most at two points.
Therefore,the equation $\phi(x) = 0$,which is equivalent to $f(x) = x$,can have at most two solutions in the interval $(-1, 1)$.
Thus,$X_f \leq 2$ for every $f \in S$,which makes statement $(B)$ true.
By choosing appropriate convex functions,we can construct examples where $X_f = 0$ (e.g.,$f(x) = x^2 + 2$),$X_f = 1$ (e.g.,$f(x) = x^2 + x + 0.1$),and $X_f = 2$ (e.g.,$f(x) = x^2 + 0.1$).
Since $X_f$ can be $0, 1,$ or $2$,statement $(A)$ is true,statement $(C)$ is true,and statement $(D)$ is false.
Therefore,the correct statements are $(A)$,$(B)$,and $(C)$.

(A) Given $f(x) = x \cos \frac{1}{x}$ for $x \geq 1$.
First,find the derivative $f^{\prime}(x)$:
$f^{\prime}(x) = \cos \frac{1}{x} + x \left( -\sin \frac{1}{x} \right) \left( -\frac{1}{x^2} \right) = \cos \frac{1}{x} + \frac{1}{x} \sin \frac{1}{x}$.
Now,evaluate $\lim _{x \rightarrow \infty} f^{\prime}(x)$:
$\lim _{x \rightarrow \infty} \left( \cos \frac{1}{x} + \frac{1}{x} \sin \frac{1}{x} \right) = \cos(0) + 0 \cdot \sin(0) = 1 + 0 = 1$. Thus,statement $(B)$ is correct.
Next,find the second derivative $f^{\prime \prime}(x)$:
$f^{\prime \prime}(x) = -\sin \frac{1}{x} \left( -\frac{1}{x^2} \right) + \left( -\frac{1}{x^2} \right) \sin \frac{1}{x} + \frac{1}{x} \cos \frac{1}{x} \left( -\frac{1}{x^2} \right) = \frac{1}{x^2} \sin \frac{1}{x} - \frac{1}{x^2} \sin \frac{1}{x} - \frac{1}{x^3} \cos \frac{1}{x} = -\frac{1}{x^3} \cos \frac{1}{x}$.
For $x \in [1, \infty)$,$\frac{1}{x} \in (0, 1]$. Since $\cos \theta > 0$ for $\theta \in (0, 1]$,$f^{\prime \prime}(x) = -\frac{1}{x^3} \cos \frac{1}{x} < 0$. Thus,$f^{\prime}(x)$ is strictly decreasing,so statement $(D)$ is correct.
By the Mean Value Theorem $(LMVT)$ on $[x, x+2]$,there exists $c \in (x, x+2)$ such that $\frac{f(x+2)-f(x)}{2} = f^{\prime}(c)$.
Since $f^{\prime}(x)$ is strictly decreasing and $\lim _{x \rightarrow \infty} f^{\prime}(x) = 1$,we have $f^{\prime}(x) > 1$ for all $x \in [1, \infty)$.
Therefore,$f^{\prime}(c) > 1$,which implies $\frac{f(x+2)-f(x)}{2} > 1$,or $f(x+2)-f(x) > 2$. Thus,statement $(C)$ is correct and $(A)$ is incorrect.
The correct combination is $(B, C, D)$.

(B) Given the set $A = \{x \mid x^2+20 \leq 9x\}$.
Solving the inequality $x^2-9x+20 \leq 0$:
$(x-4)(x-5) \leq 0$,which implies $x \in [4, 5]$.
Now,consider the function $f(x) = 2x^3-15x^2+36x-48$.
Find the derivative $f'(x) = 6x^2-30x+36 = 6(x^2-5x+6) = 6(x-2)(x-3)$.
For $x \in [4, 5]$,$f'(x) > 0$ because both $(x-2)$ and $(x-3)$ are positive in this interval.
Since $f'(x) > 0$ for all $x \in [4, 5]$,the function $f(x)$ is strictly increasing on the interval $[4, 5]$.
Therefore,the maximum value occurs at the right endpoint $x = 5$.
$f(5) = 2(5)^3 - 15(5)^2 + 36(5) - 48$
$f(5) = 2(125) - 15(25) + 180 - 48$
$f(5) = 250 - 375 + 180 - 48 = 7$.

(B) Let $p(x) = ax^4 + bx^3 + cx^2 + dx + e$.
Since $\lim_{x \rightarrow 0} (1 + \frac{p(x)}{x^2}) = 2$,we have $\lim_{x \rightarrow 0} \frac{p(x)}{x^2} = 1$. This implies $e = 0$ and $d = 0$,and $c = 1$.
Thus,$p(x) = ax^4 + bx^3 + x^2$.
The derivative is $p'(x) = 4ax^3 + 3bx^2 + 2x$.
Given extrema at $x=1$ and $x=2$,$p'(1) = 0$ and $p'(2) = 0$.
$p'(1) = 4a + 3b + 2 = 0 \implies 4a + 3b = -2$ (Equation $1$).
$p'(2) = 4a(8) + 3b(4) + 2(2) = 32a + 12b + 4 = 0 \implies 8a + 3b = -1$ (Equation $2$).
Subtracting Equation $1$ from Equation $2$: $(8a - 4a) = -1 - (-2) \implies 4a = 1 \implies a = \frac{1}{4}$.
Substituting $a = \frac{1}{4}$ into Equation $1$: $4(\frac{1}{4}) + 3b = -2 \implies 1 + 3b = -2 \implies 3b = -3 \implies b = -1$.
So,$p(x) = \frac{1}{4}x^4 - x^3 + x^2$.
Calculating $p(2)$: $p(2) = \frac{1}{4}(16) - (8) + (4) = 4 - 8 + 4 = 0$.