Maxima and Minima Questions in English

(A) Given $xy = 1$ and $x > 0$,we have $y = \frac{1}{x}$.
Let $f(x) = x + y = x + \frac{1}{x}$.
To find the minimum value,we differentiate $f(x)$ with respect to $x$:
$f'(x) = 1 - \frac{1}{x^2}$.
Setting $f'(x) = 0$ for critical points:
$1 - \frac{1}{x^2} = 0 \Rightarrow x^2 = 1$.
Since $x > 0$,we take $x = 1$.
Now,find the second derivative:
$f''(x) = \frac{2}{x^3}$.
At $x = 1$,$f''(1) = \frac{2}{1^3} = 2 > 0$.
Since the second derivative is positive,$x = 1$ is a point of local minima.
The minimum value is $f(1) = 1 + \frac{1}{1} = 2$.

(A) Let $y = f(x) = x^5 - 5x^4 + 5x^3 - 10$.
First,find the derivative: $\frac{dy}{dx} = 5x^4 - 20x^3 + 15x^2 = 5x^2(x^2 - 4x + 3) = 5x^2(x - 3)(x - 1)$.
Setting $\frac{dy}{dx} = 0$,we get critical points at $x = 0, 1, 3$.
Now,find the second derivative: $\frac{d^2y}{dx^2} = 20x^3 - 60x^2 + 30x = 10x(2x^2 - 6x + 3)$.
For $x = 0$: $\frac{d^2y}{dx^2} = 0$. Checking the third derivative: $\frac{d^3y}{dx^3} = 60x^2 - 120x + 30$. At $x = 0$,$\frac{d^3y}{dx^3} = 30 \neq 0$. Thus,$x = 0$ is a point of inflection,not a local extremum.
For $x = 1$: $\frac{d^2y}{dx^2} = 10(1)(2 - 6 + 3) = -10 < 0$. Thus,$x = 1$ is a local maximum.
$f(1) = 1 - 5 + 5 - 10 = -9$.
For $x = 3$: $\frac{d^2y}{dx^2} = 10(3)(2(9) - 6(3) + 3) = 30(18 - 18 + 3) = 90 > 0$. Thus,$x = 3$ is a local minimum.
$f(3) = 3^5 - 5(3^4) + 5(3^3) - 10 = 243 - 405 + 135 - 10 = -37$.
Therefore,the minimum value is $-37$ and the maximum value is $-9$.

(B) Let the two parts be $x$ and $y$ such that $x + y = 20$.
We want to maximize the product $z = x \cdot y^3$.
Substituting $x = 20 - y$ into the equation,we get $z = (20 - y)y^3 = 20y^3 - y^4$.
To find the critical points,we differentiate $z$ with respect to $y$: $\frac{dz}{dy} = 60y^2 - 4y^3$.
Setting $\frac{dz}{dy} = 0$,we get $4y^2(15 - y) = 0$,which gives $y = 0$ or $y = 15$.
Now,we find the second derivative: $\frac{d^2z}{dy^2} = 120y - 12y^2$.
At $y = 15$,$\frac{d^2z}{dy^2} = 120(15) - 12(15^2) = 1800 - 2700 = -900 < 0$.
Since the second derivative is negative,$y = 15$ is the point of maxima.
If $y = 15$,then $x = 20 - 15 = 5$.
Thus,the two parts are $(5, 15)$.

(C) Let $f(x) = {x^3} - 18{x^2} + 96x$.
First,find the derivative: $f'(x) = 3{x^2} - 36x + 96$.
Set $f'(x) = 0$ to find critical points: $3(x^2 - 12x + 32) = 0 \Rightarrow 3(x - 4)(x - 8) = 0$.
Thus,the critical points are $x = 4$ and $x = 8$.
Now,find the second derivative: $f''(x) = 6x - 36$.
At $x = 4$,$f''(4) = 6(4) - 36 = -12 < 0$,so $f(x)$ has a local maximum at $x = 4$.
The maximum value is $f(4) = (4)^3 - 18(4)^2 + 96(4) = 64 - 288 + 384 = 160$.
At $x = 8$,$f''(8) = 6(8) - 36 = 12 > 0$,so $f(x)$ has a local minimum at $x = 8$.
The minimum value is $f(8) = (8)^3 - 18(8)^2 + 96(8) = 512 - 1152 + 768 = 128$.
Therefore,the maximum and minimum values are $160$ and $128$ respectively.

(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 maximum,we find the derivative $f'(x)$ and set it to $0$:
$f'(x) = \cos x + \cos 2x = 0$.
Using the identity $\cos 2x = 2 \cos^2 x - 1$,we get:
$2 \cos^2 x + \cos x - 1 = 0$.
Let $u = \cos x$,then $2u^2 + u - 1 = 0$,which factors as $(2u - 1)(u + 1) = 0$.
So,$\cos x = \frac{1}{2}$ or $\cos x = -1$.
For $\cos x = \frac{1}{2}$,$x = \frac{\pi}{3}$ (in the first quadrant).
For $\cos x = -1$,$x = \pi$,but at $x = \pi$,$f(x) = 0$,which is the minimum.
Checking the second derivative $f''(x) = -\sin x - 2\sin 2x$:
At $x = \frac{\pi}{3}$,$f''(\frac{\pi}{3}) = -\sin(\frac{\pi}{3}) - 2\sin(\frac{2\pi}{3}) = -\frac{\sqrt{3}}{2} - 2(\frac{\sqrt{3}}{2}) = -\frac{3\sqrt{3}}{2} < 0$.
Since the second derivative is negative,the function has a maximum at $x = \frac{\pi}{3}$.

(B) Let $f(x) = \frac{x}{1 + x \tan x}$. To find the maxima of $f(x)$,we can find the minima of its reciprocal $g(x) = \frac{1}{f(x)} = \frac{1 + x \tan x}{x} = \frac{1}{x} + \tan x$.
Now,differentiate $g(x)$ with respect to $x$:
$g'(x) = -\frac{1}{x^2} + \sec^2 x$.
For critical points,set $g'(x) = 0$:
$-\frac{1}{x^2} + \sec^2 x = 0 \implies \sec^2 x = \frac{1}{x^2} \implies \cos^2 x = x^2 \implies x = \cos x$ (considering the domain where $x > 0$).
Now,find the second derivative $g''(x)$:
$g''(x) = \frac{2}{x^3} + 2 \sec x (\sec x \tan x) = \frac{2}{x^3} + 2 \sec^2 x \tan x$.
At $x = \cos x$,$g''(x) = \frac{2}{\cos^3 x} + 2 \sec^2 x \tan x = 2 \sec^3 x + 2 \sec^2 x \tan x = 2 \sec^2 x (\sec x + \tan x)$.
Since $x = \cos x$ lies in the first quadrant,$\sec x > 0$ and $\tan x > 0$,so $g''(x) > 0$.
Thus,$g(x)$ has a minimum at $x = \cos x$,which implies $f(x)$ has a maximum at $x = \cos x$.

(B) Let $y = \frac{x^2 - x + 1}{x^2 + x + 1}$.
To find the extrema,we differentiate $y$ with respect to $x$:
$\frac{dy}{dx} = \frac{(x^2 + x + 1)(2x - 1) - (x^2 - x + 1)(2x + 1)}{(x^2 + x + 1)^2}$
$= \frac{(2x^3 + 2x^2 + 2x - x^2 - x - 1) - (2x^3 - 2x^2 + 2x + x^2 - x + 1)}{(x^2 + x + 1)^2}$
$= \frac{(2x^3 + x^2 + x - 1) - (2x^3 - x^2 + x + 1)}{(x^2 + x + 1)^2}$
$= \frac{2x^2 - 2}{(x^2 + x + 1)^2}$.
Setting $\frac{dy}{dx} = 0$,we get $2x^2 - 2 = 0$,which implies $x^2 = 1$,so $x = 1$ or $x = -1$.
For $x = 1$,$y = \frac{1 - 1 + 1}{1 + 1 + 1} = \frac{1}{3}$.
For $x = -1$,$y = \frac{1 + 1 + 1}{1 - 1 + 1} = \frac{3}{1} = 3$.
Thus,the greatest value is $3$ and the least value is $\frac{1}{3}$.

(D) Let $f(x) = \frac{\log x}{x}$.
Find the derivative: $f'(x) = \frac{x \cdot \frac{1}{x} - \log x \cdot 1}{x^2} = \frac{1 - \log x}{x^2}$.
Set $f'(x) = 0$ to find critical points: $1 - \log x = 0 \Rightarrow \log x = 1 \Rightarrow x = e$.
Since $e \approx 2.718$,$x = e$ lies in the interval $[2, \infty)$.
Evaluate the function at the critical point and the boundary:
$f(e) = \frac{\log e}{e} = \frac{1}{e} \approx 0.367$.
$f(2) = \frac{\log 2}{2} \approx \frac{0.693}{2} = 0.346$.
As $x \to \infty$,$\frac{\log x}{x} \to 0$ (by $L$'Hopital's rule).
Since $f(2) \approx 0.346$ and the limit as $x \to \infty$ is $0$,and the function decreases for $x > e$,the function values approach $0$ but never reach it in the interval $[2, \infty)$.
Therefore,the minimum value does not exist.

(B) Let the function be $f(x) = x + \frac{1}{x}$ for $x > 0$.
To find the minimum value,we find the derivative $f'(x) = 1 - \frac{1}{x^2}$.
Setting $f'(x) = 0$,we get $1 - \frac{1}{x^2} = 0$,which implies $x^2 = 1$. Since $x > 0$,we have $x = 1$.
Now,we check the second derivative $f''(x) = \frac{2}{x^3}$.
At $x = 1$,$f''(1) = \frac{2}{1^3} = 2 > 0$.
Since the second derivative is positive,the function $f(x)$ has a local minimum at $x = 1$.

(B) Let the numerator be $x$. Then the denominator is $x^2 + 16$.
The fraction is $f(x) = \frac{x}{x^2 + 16}$.
To find the extrema,we find the derivative $f'(x)$:
$f'(x) = \frac{(x^2 + 16)(1) - x(2x)}{(x^2 + 16)^2} = \frac{16 - x^2}{(x^2 + 16)^2}$.
Setting $f'(x) = 0$ gives $16 - x^2 = 0$,so $x = 4$ or $x = -4$.
We evaluate $f(x)$ at these critical points:
For $x = 4$,$f(4) = \frac{4}{16 + 16} = \frac{4}{32} = \frac{1}{8}$.
For $x = -4$,$f(-4) = \frac{-4}{16 + 16} = \frac{-4}{32} = -\frac{1}{8}$.
Comparing the values,the least value is $-1/8$.

(B) Let the number be $x$. We want to maximize the difference between the number and its cube,so we define the function $f(x) = x - x^3$.
To find the maximum,we take the first derivative: $f'(x) = 1 - 3x^2$.
Setting $f'(x) = 0$,we get $1 - 3x^2 = 0$,which implies $x^2 = 1/3$,so $x = \pm 1/\sqrt{3}$.
Now,we check the second derivative: $f''(x) = -6x$.
For $x = 1/\sqrt{3}$,$f''(1/\sqrt{3}) = -6/\sqrt{3} < 0$,which indicates a local maximum.
For $x = -1/\sqrt{3}$,$f''(-1/\sqrt{3}) = 6/\sqrt{3} > 0$,which indicates a local minimum.
Thus,the number that most exceeds its cube is $1/\sqrt{3}$.

(C) Given function: $f(x) = \frac{x}{4 + x + x^2}$.
To find the critical points,we differentiate $f(x)$ with respect to $x$ using the quotient rule:
$f'(x) = \frac{(4 + x + x^2)(1) - x(1 + 2x)}{(4 + x + x^2)^2}$
$f'(x) = \frac{4 + x + x^2 - x - 2x^2}{(4 + x + x^2)^2} = \frac{4 - x^2}{(4 + x + x^2)^2}$.
Setting $f'(x) = 0$ gives $4 - x^2 = 0$,which implies $x = 2$ or $x = -2$.
Since both $x = 2$ and $x = -2$ lie outside the given interval $[-1, 1]$,the maximum and minimum values must occur at the endpoints of the interval.
Evaluating at the endpoints:
$f(-1) = \frac{-1}{4 - 1 + 1} = \frac{-1}{4} = -0.25$.
$f(1) = \frac{1}{4 + 1 + 1} = \frac{1}{6} \approx 0.1667$.
Comparing these values,the maximum value is $1/6$.

(A) Let the radius of the sphere be $R$,so the diameter is $2R$.
Let the height of the cone be $y$ and the radius of the base of the cone be $x$.
From the geometry of the sphere,the distance from the center of the sphere to the base of the cone is $|y - R|$.
Using the Pythagorean theorem in the triangle formed by the radius of the sphere,the radius of the cone,and the distance from the center: $x^2 + (y - R)^2 = R^2$.
$x^2 = R^2 - (y - R)^2 = R^2 - (y^2 - 2Ry + R^2) = 2Ry - y^2$.
The volume of the cone is $V = \frac{1}{3}\pi x^2 y = \frac{1}{3}\pi (2Ry - y^2)y = \frac{1}{3}\pi (2Ry^2 - y^3)$.
To find the maximum volume,differentiate $V$ with respect to $y$: $\frac{dV}{dy} = \frac{1}{3}\pi (4Ry - 3y^2)$.
Setting $\frac{dV}{dy} = 0$,we get $y(4R - 3y) = 0$. Since $y \neq 0$,we have $y = \frac{4}{3}R$.
The second derivative $\frac{d^2V}{dy^2} = \frac{1}{3}\pi (4R - 6y)$. At $y = \frac{4}{3}R$,$\frac{d^2V}{dy^2} = \frac{1}{3}\pi (4R - 8R) = -\frac{4}{3}\pi R < 0$,confirming a maximum.
The ratio of the height of the cone $(y)$ to the diameter of the sphere $(2R)$ is $\frac{y}{2R} = \frac{4/3 R}{2R} = \frac{2}{3}$.

(B) Let the radius of the sphere be $R$ and the diameter be $AE = 2R$.
Let the radius of the cone be $x$ and its height be $y$.
In the right-angled triangle $BDE$,by the property of chords,$BD^2 = AD \times DE$,where $AD = y$ and $DE = 2R - y$.
So,$x^2 = y(2R - y)$.
The volume of the cone is $V = \frac{1}{3}\pi x^2 y = \frac{1}{3}\pi y(2R - y)y = \frac{1}{3}\pi (2Ry^2 - y^3)$.
To find the maximum volume,differentiate $V$ with respect to $y$:
$\frac{dV}{dy} = \frac{1}{3}\pi (4Ry - 3y^2)$.
Set $\frac{dV}{dy} = 0$:
$\frac{1}{3}\pi y(4R - 3y) = 0$.
Since $y \neq 0$,we have $y = \frac{4}{3}R$.
Using the second derivative test,$\frac{d^2V}{dy^2} = \frac{1}{3}\pi (4R - 6y)$.
At $y = \frac{4}{3}R$,$\frac{d^2V}{dy^2} = \frac{1}{3}\pi (4R - 8R) = -\frac{4}{3}\pi R < 0$,which confirms the volume is maximum.
The ratio of the height of the cone to the radius of the sphere is $\frac{y}{R} = \frac{4}{3}$.

(C) Given the function $f(x) = x + \sin x$.
First,we find the derivative: $f'(x) = 1 + \cos x$.
Setting $f'(x) = 0$,we get $1 + \cos x = 0$,which implies $\cos x = -1$.
This occurs at $x = (2n+1)\pi$ for any integer $n$.
Now,we check the second derivative: $f''(x) = -\sin x$.
At $x = \pi$,$f''(\pi) = -\sin(\pi) = 0$.
Since the second derivative is zero,we check the third derivative: $f'''(x) = -\cos x$.
At $x = \pi$,$f'''(\pi) = -\cos(\pi) = -(-1) = 1 \neq 0$.
Since the first non-zero derivative at the critical point is of an odd order (the third derivative),the function $f(x)$ has a point of inflection at $x = \pi$ and does not have a local maximum or local minimum.

(D) Given the function $f(x) = ax + \frac{b}{x}$.
To find the critical points,we differentiate $f(x)$ with respect to $x$:
$f'(x) = a - \frac{b}{x^2}$.
Setting $f'(x) = 0$ for critical points:
$a - \frac{b}{x^2} = 0 \Rightarrow a = \frac{b}{x^2} \Rightarrow x^2 = \frac{b}{a} \Rightarrow x = \sqrt{\frac{b}{a}}$ (since $x > 0$).
Now,we find the second derivative to check for minima:
$f''(x) = \frac{d}{dx}(a - bx^{-2}) = 0 - b(-2)x^{-3} = \frac{2b}{x^3}$.
Since $b > 0$ and $x > 0$,$f''(x) = \frac{2b}{x^3} > 0$ for all $x > 0$.
Since $f''(x) > 0$ at $x = \sqrt{\frac{b}{a}}$,the function attains its least value at $x = \sqrt{\frac{b}{a}}$.

(B) Given $xy = c^2$,we can write $y = \frac{c^2}{x}$.
Let $f(x) = ax + by = ax + \frac{bc^2}{x}$.
To find the minimum value,differentiate $f(x)$ with respect to $x$:
$f'(x) = a - \frac{bc^2}{x^2}$.
Set $f'(x) = 0$ for critical points:
$a - \frac{bc^2}{x^2} = 0 \implies ax^2 = bc^2 \implies x^2 = \frac{bc^2}{a}$.
Since $x$ must be positive for a minimum value in this context (assuming $a, b, c > 0$),we take $x = c\sqrt{\frac{b}{a}}$.
Substituting $x$ back into $f(x)$:
$f\left(c\sqrt{\frac{b}{a}}\right) = a\left(c\sqrt{\frac{b}{a}}\right) + \frac{bc^2}{c\sqrt{\frac{b}{a}}} = ac\sqrt{\frac{b}{a}} + bc\sqrt{\frac{a}{b}} = c\sqrt{ab} + c\sqrt{ab} = 2c\sqrt{ab}$.

(C) Given the equation ${a^2}{x^4} + {b^2}{y^4} = {c^6}$.
Using the $AM$-$GM$ inequality,we know that for positive numbers $u$ and $v$,$\frac{u+v}{2} \ge \sqrt{uv}$.
Let $u = {a^2}{x^4}$ and $v = {b^2}{y^4}$.
Then $\frac{{{a^2}{x^4} + {b^2}{y^4}}}{2} \ge \sqrt{{a^2}{x^4} \cdot {b^2}{y^4}}$.
Substituting the given equation: $\frac{{{c^6}}}{2} \ge \sqrt{{a^2}{b^2}{x^4}{y^4}}$.
$\frac{{{c^6}}}{2} \ge ab{x^2}{y^2}$.
${x^2}{y^2} \le \frac{{{c^6}}}{{2ab}}$.
Taking the square root on both sides,we get $xy \le \sqrt{\frac{{{c^6}}}{{2ab}}}$.
Therefore,the maximum value of $xy$ is $\frac{{{c^3}}}{{\sqrt {2ab} }}$.

(A) Given function is $f(x) = 2x^3 - 15x^2 + 36x + 4$.
First,find the first derivative $f'(x)$:
$f'(x) = \frac{d}{dx}(2x^3 - 15x^2 + 36x + 4) = 6x^2 - 30x + 36$.
To find the critical points,set $f'(x) = 0$:
$6(x^2 - 5x + 6) = 0$
$6(x - 2)(x - 3) = 0$
So,the critical points are $x = 2$ and $x = 3$.
Now,find the second derivative $f''(x)$ to check for maxima or minima:
$f''(x) = \frac{d}{dx}(6x^2 - 30x + 36) = 12x - 30$.
Evaluate $f''(x)$ at the critical points:
For $x = 2$: $f''(2) = 12(2) - 30 = 24 - 30 = -6$.
Since $f''(2) < 0$,the function has a local maximum at $x = 2$.
For $x = 3$: $f''(3) = 12(3) - 30 = 36 - 30 = 6$.
Since $f''(3) > 0$,the function has a local minimum at $x = 3$.
Therefore,the function is maximum at $x = 2$.

(B) Let the function be $f(x) = -x^3 + 3x^2 + 9x - 27$.
The slope of the curve at any point $x$ is given by the derivative $f'(x) = -3x^2 + 6x + 9$.
To find the maximum slope,let $g(x) = f'(x) = -3x^2 + 6x + 9$.
We find the derivative of $g(x)$ with respect to $x$: $g'(x) = -6x + 6$.
Setting $g'(x) = 0$ for critical points gives $-6x + 6 = 0$,which implies $x = 1$.
To verify the maximum,we find the second derivative: $g''(x) = -6$. Since $g''(1) = -6 < 0$,the function $g(x)$ attains its maximum value at $x = 1$.
Substituting $x = 1$ into $g(x)$,we get the maximum slope: $g(1) = -3(1)^2 + 6(1) + 9 = -3 + 6 + 9 = 12$.

(B) Given function: $f(x) = 2x^3 - 3x^2 - 12x + 4$
Step $1$: Find the first derivative $f'(x)$.
$f'(x) = \frac{d}{dx}(2x^3 - 3x^2 - 12x + 4) = 6x^2 - 6x - 12$
Step $2$: Find critical points by setting $f'(x) = 0$.
$6(x^2 - x - 2) = 0$
$6(x - 2)(x + 1) = 0$
So,the critical points are $x = 2$ and $x = -1$.
Step $3$: Find the second derivative $f''(x)$ to check for maxima/minima.
$f''(x) = \frac{d}{dx}(6x^2 - 6x - 12) = 12x - 6$
Step $4$: Evaluate $f''(x)$ at critical points.
For $x = 2$: $f''(2) = 12(2) - 6 = 24 - 6 = 18 > 0$. Since $f''(2) > 0$,the function has a local minimum at $x = 2$.
For $x = -1$: $f''(-1) = 12(-1) - 6 = -12 - 6 = -18 < 0$. Since $f''(-1) < 0$,the function has a local maximum at $x = -1$.
Conclusion: The function has one maximum and one minimum.

(D) Given the function $f(x) = x + \frac{1}{x}$ for $x > 0$.
First,we find the derivative of the function with respect to $x$:
$f'(x) = \frac{d}{dx}(x + x^{-1}) = 1 - x^{-2} = 1 - \frac{1}{x^2}$.
To find the critical points,we set $f'(x) = 0$:
$1 - \frac{1}{x^2} = 0 \implies x^2 = 1 \implies x = \pm 1$.
Since the domain is $x > 0$,we only consider $x = 1$.
Now,we check the second derivative to determine the nature of the critical point:
$f''(x) = \frac{d}{dx}(1 - x^{-2}) = 2x^{-3} = \frac{2}{x^3}$.
At $x = 1$,$f''(1) = \frac{2}{1^3} = 2 > 0$.
Since $f''(1) > 0$,the function has a local minimum at $x = 1$,and the minimum value is $f(1) = 1 + \frac{1}{1} = 2$.
As $x \to 0^+$,$f(x) \to \infty$,and as $x \to \infty$,$f(x) \to \infty$.
Therefore,the function does not have a greatest value (maximum) on the interval $(0, \infty)$.
Thus,the correct option is $(d)$.

(D) Let $r$ be the radius and $s$ be the arc length of the sector. The perimeter $p$ is given by $p = 2r + s$,so $s = p - 2r$.
The area $A$ of the sector is given by $A = \frac{1}{2}rs$.
Substituting $s = p - 2r$ into the area formula,we get $A = \frac{1}{2}r(p - 2r) = \frac{1}{2}pr - r^2$.
To find the maximum area,we differentiate $A$ with respect to $r$ and set the derivative to zero:
$\frac{dA}{dr} = \frac{1}{2}p - 2r = 0$.
Solving for $r$,we get $2r = \frac{p}{2}$,which implies $r = \frac{p}{4}$.
Since $\frac{d^2A}{dr^2} = -2 < 0$,the area is maximum at $r = \frac{p}{4}$.

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

(C) Given $f(x) = \int_{-10}^x (t^4 - 4)e^{-4t} dt$.
By the Fundamental Theorem of Calculus,$f'(x) = (x^4 - 4)e^{-4x}$.
To find the critical points,set $f'(x) = 0$:
$(x^4 - 4)e^{-4x} = 0$.
Since $e^{-4x} \neq 0$ for all $x$,we have $x^4 - 4 = 0$,which implies $x^2 = 2$ or $x^2 = -2$.
In the real domain,$x = \sqrt{2}$ and $x = -\sqrt{2}$.
Both values $\sqrt{2} \approx 1.414$ and $-\sqrt{2} \approx -1.414$ lie within the interval $(-4, 4)$.
Since $f'(x)$ changes sign at $x = \sqrt{2}$ and $x = -\sqrt{2}$,the function has two extrema in the given interval.

(A) Let $f(x) = x^2 \log x$.
To find the critical points,we find the derivative $f'(x)$:
$f'(x) = \frac{d}{dx}(x^2) \cdot \log x + x^2 \cdot \frac{d}{dx}(\log x) = 2x \log x + x^2 \cdot \frac{1}{x} = 2x \log x + x = x(2 \log x + 1)$.
Setting $f'(x) = 0$,we get $x(2 \log x + 1) = 0$.
Since $x \in [1, e]$,$x \neq 0$. Thus,$2 \log x + 1 = 0 \implies \log x = -\frac{1}{2} \implies x = e^{-1/2} = \frac{1}{\sqrt{e}}$.
Since $\frac{1}{\sqrt{e}} \approx 0.606$,which is not in the interval $[1, e]$,there are no critical points in the given interval.
Therefore,the maximum value must occur at the endpoints $x = 1$ or $x = e$.
$f(1) = 1^2 \cdot \log(1) = 1 \cdot 0 = 0$.
$f(e) = e^2 \cdot \log(e) = e^2 \cdot 1 = e^2$.
Comparing the values $0$ and $e^2$,the greatest value is $e^2$.

(C) Let $f(x) = y = x^{-x}$.
Taking the natural logarithm on both sides,we get $\ln y = -x \ln x$.
Differentiating both sides with respect to $x$,we have $\frac{1}{y} \frac{dy}{dx} = -[x \cdot \frac{1}{x} + \ln x \cdot 1] = -(1 + \ln x)$.
Thus,$\frac{dy}{dx} = -x^{-x}(1 + \ln x)$.
For critical points,set $\frac{dy}{dx} = 0$,which implies $1 + \ln x = 0$.
$\ln x = -1$,so $x = e^{-1} = \frac{1}{e}$.
To verify the maximum,we check the second derivative or the sign change of the first derivative. Since $f'(x) > 0$ for $x < 1/e$ and $f'(x) < 0$ for $x > 1/e$,the function attains a maximum at $x = 1/e$.

(B) Given $ab = 2a + 3b$. Since $a > 0$ and $b > 0$,we can express $b$ in terms of $a$:
$b(a - 3) = 2a \implies b = \frac{2a}{a - 3}$.
Since $b > 0$ and $a > 0$,we must have $a - 3 > 0$,so $a > 3$.
Let $z = ab = a \left( \frac{2a}{a - 3} \right) = \frac{2a^2}{a - 3}$.
To find the minimum,we differentiate $z$ with respect to $a$:
$\frac{dz}{da} = \frac{(a - 3)(4a) - 2a^2(1)}{(a - 3)^2} = \frac{4a^2 - 12a - 2a^2}{(a - 3)^2} = \frac{2a^2 - 12a}{(a - 3)^2}$.
Setting $\frac{dz}{da} = 0$ gives $2a(a - 6) = 0$. Since $a > 3$,we have $a = 6$.
At $a = 6$,$b = \frac{2(6)}{6 - 3} = \frac{12}{3} = 4$.
The value of $ab = 6 \times 4 = 24$.
Using the second derivative test or checking values around $a=6$,we confirm this is the minimum value.
Thus,the minimum value of $ab$ is $24$.

(D) Let the lengths of the sides $PQ$ and $PR$ be $a$ and $b$ respectively.
The area $\Delta$ of the triangle is given by the formula:
$\Delta = \frac{1}{2} ab \sin \theta$,where $\theta$ is the angle between the sides $PQ$ and $PR$.
Since $a$ and $b$ are constant lengths,the area $\Delta$ is directly proportional to $\sin \theta$.
The area is maximized when $\sin \theta$ attains its maximum value.
We know that the maximum value of $\sin \theta$ is $1$,which occurs when $\theta = \frac{\pi}{2}$.
Therefore,the angle between the sides that gives the maximum area is $\frac{\pi}{2}$.

(A) Given the function $y = a(1 - \cos x)$.
First,find the derivative $y'$ with respect to $x$:
$y' = \frac{d}{dx}[a(1 - \cos x)] = a(0 - (-\sin x)) = a \sin x$.
To find the critical points,set $y' = 0$:
$a \sin x = 0 \Rightarrow \sin x = 0 \Rightarrow x = n\pi$ for any integer $n$.
Within the standard interval $[0, 2\pi]$,the critical points are $x = 0$ and $x = \pi$.
Now,find the second derivative $y''$:
$y'' = \frac{d}{dx}(a \sin x) = a \cos x$.
Evaluate $y''$ at the critical points:
For $x = 0$,$y''(0) = a \cos(0) = a$.
For $x = \pi$,$y''(\pi) = a \cos(\pi) = -a$.
Assuming $a > 0$,the function has a local maximum when $y'' < 0$,which occurs at $x = \pi$.

(A) Let $f(x) = x^2 + \frac{250}{x}$.
To find the minimum value,we first find the derivative $f'(x)$:
$f'(x) = 2x - \frac{250}{x^2}$.
Set $f'(x) = 0$ to find the critical points:
$2x - \frac{250}{x^2} = 0 \implies 2x^3 = 250 \implies x^3 = 125 \implies x = 5$.
Now,we check the second derivative $f''(x)$ to confirm the nature of the extremum:
$f''(x) = 2 + \frac{500}{x^3}$.
At $x = 5$,$f''(5) = 2 + \frac{500}{125} = 2 + 4 = 6$.
Since $f''(5) > 0$,the function has a local minimum at $x = 5$.
The minimum value is $f(5) = 5^2 + \frac{250}{5} = 25 + 50 = 75$.

(A) Let $f(x) = x^2 + \frac{1}{1 + x^2}$.
To find the critical points,we differentiate $f(x)$ with respect to $x$:
$f'(x) = 2x - \frac{1}{(1 + x^2)^2} \cdot (2x) = 2x \left( 1 - \frac{1}{(1 + x^2)^2} \right)$.
Setting $f'(x) = 0$,we get $2x = 0$ or $1 - \frac{1}{(1 + x^2)^2} = 0$.
From $2x = 0$,we find $x = 0$.
From $1 - \frac{1}{(1 + x^2)^2} = 0$,we get $(1 + x^2)^2 = 1$,which implies $1 + x^2 = 1$ (since $1 + x^2 > 0$),so $x^2 = 0$,which again gives $x = 0$.
Evaluating the function at $x = 0$,$f(0) = 0^2 + \frac{1}{1 + 0^2} = 1$.
Since $x^2 \ge 0$,let $u = x^2$ where $u \ge 0$. Then $g(u) = u + \frac{1}{1+u} = \frac{u^2 + u + 1}{1+u} = \frac{u^2 - 1 + u + 2}{1+u} = u - 1 + \frac{2}{1+u}$.
At $x=0$,$u=0$,$f(0)=1$. For $x \neq 0$,$x^2 > 0$,the function value increases. Thus,the minimum value is at $x = 0$.

(A) Given the equation $x - 2y = 4,$ we can express $x$ in terms of $y$ as $x = 2y + 4$ $(i).$
Let the product be $P = xy.$
Substituting $(i)$ into the expression for $P,$ we get $P = y(2y + 4) = 2y^2 + 4y.$
To find the minimum value,we differentiate $P$ with respect to $y$ and set it to zero:
$\frac{dP}{dy} = 4y + 4 = 0 \Rightarrow y = -1.$
Now,we find the second derivative to check for a minimum:
$\frac{d^2P}{dy^2} = 4,$ which is greater than $0,$ confirming that $y = -1$ gives a minimum value.
Substituting $y = -1$ into $(i),$ we get $x = 2(-1) + 4 = 2.$
Therefore,the minimum value of $xy$ is $P = (2)(-1) = -2.$

(A) Let $f(x, y) = 2x + 3y$ subject to the constraint $xy = 6.$
Since $xy = 6,$ we have $y = \frac{6}{x}.$
Substituting this into the function,we get $f(x) = 2x + 3\left(\frac{6}{x}\right) = 2x + \frac{18}{x}.$
To find the minimum,we find the derivative $f'(x) = 2 - \frac{18}{x^2}.$
Setting $f'(x) = 0,$ we get $2 = \frac{18}{x^2},$ which implies $x^2 = 9,$ so $x = \pm 3.$
Since $x$ and $y$ must be positive for a minimum value in this context (as $xy=6$),we take $x = 3.$
Then $y = \frac{6}{3} = 2.$
The minimum value is $f(3) = 2(3) + 3(2) = 6 + 6 = 12.$

(B) Given the function $f(x) = 3x^4 - 8x^3 + 12x^2 - 48x + 25$.
First,find the derivative $f'(x)$:
$f'(x) = 12x^3 - 24x^2 + 24x - 48$
$f'(x) = 12(x^3 - 2x^2 + 2x - 4)$
$f'(x) = 12[x^2(x - 2) + 2(x - 2)] = 12(x - 2)(x^2 + 2)$.
To find critical points,set $f'(x) = 0$:
Since $x^2 + 2 = 0$ has no real roots,the only critical point is $x = 2$.
Now,evaluate $f(x)$ at the critical point $x = 2$ and the endpoints of the interval $[0, 3]$:
$f(0) = 3(0)^4 - 8(0)^3 + 12(0)^2 - 48(0) + 25 = 25$
$f(2) = 3(2)^4 - 8(2)^3 + 12(2)^2 - 48(2) + 25 = 3(16) - 8(8) + 12(4) - 96 + 25 = 48 - 64 + 48 - 96 + 25 = -39$
$f(3) = 3(3)^4 - 8(3)^3 + 12(3)^2 - 48(3) + 25 = 3(81) - 8(27) + 12(9) - 144 + 25 = 243 - 216 + 108 - 144 + 25 = 16$.
The minimum value is $\min\{25, -39, 16\} = -39$.

(B) Let $y = x^{1/x}$.
Taking the natural logarithm on both sides,we get $\ln y = \frac{1}{x} \ln x$.
Differentiating both sides with respect to $x$ using the product rule:
$\frac{1}{y} \frac{dy}{dx} = \left(-\frac{1}{x^2}\right) \ln x + \left(\frac{1}{x}\right) \left(\frac{1}{x}\right) = \frac{1 - \ln x}{x^2}$.
Thus,$\frac{dy}{dx} = x^{1/x} \left(\frac{1 - \ln x}{x^2}\right)$.
For critical points,set $\frac{dy}{dx} = 0$,which implies $1 - \ln x = 0$,so $\ln x = 1$,which gives $x = e$.
Since the derivative changes from positive to negative at $x = e$,the function attains a maximum at $x = e$.
The maximum value is $y(e) = e^{1/e}$.

(D) Let a point on the curve be $P(h, k)$.
Since the point lies on the curve ${x^2} = 2y$,we have ${h^2} = 2k$ ... $(i)$.
The distance $D$ between the point $(0, 5)$ and $P(h, k)$ is given by $D = \sqrt{(h - 0)^2 + (k - 5)^2} = \sqrt{h^2 + (k - 5)^2}$.
Substituting ${h^2} = 2k$ from $(i)$,we get $D = \sqrt{2k + (k - 5)^2} = \sqrt{2k + k^2 - 10k + 25} = \sqrt{k^2 - 8k + 25}$.
To minimize $D$,we minimize $f(k) = k^2 - 8k + 25$.
Taking the derivative,$f'(k) = 2k - 8$.
Setting $f'(k) = 0$,we get $2k = 8$,which implies $k = 4$.
For $k = 4$,${h^2} = 2(4) = 8$,so $h = \pm 2\sqrt{2}$.
The points on the curve closest to $(0, 5)$ are $(2\sqrt{2}, 4)$ and $(-2\sqrt{2}, 4)$.
Since these points are not listed in options $A$,$B$,or $C$,the correct option is $D$.

(D) Given function is $f(a) = (2a^2 - 3) + 3(3 - a) + 4$.
Simplifying the expression:
$f(a) = 2a^2 - 3 + 9 - 3a + 4$
$f(a) = 2a^2 - 3a + 10$.
To find the minimum value,we find the derivative $f'(a)$:
$f'(a) = \frac{d}{da}(2a^2 - 3a + 10) = 4a - 3$.
Setting $f'(a) = 0$ for critical points:
$4a - 3 = 0 \Rightarrow a = \frac{3}{4}$.
Since the second derivative $f''(a) = 4 > 0$,the function has a local minimum at $a = \frac{3}{4}$.
The minimum value is:
$f\left(\frac{3}{4}\right) = 2\left(\frac{3}{4}\right)^2 - 3\left(\frac{3}{4}\right) + 10$
$f\left(\frac{3}{4}\right) = 2\left(\frac{9}{16}\right) - \frac{9}{4} + 10$
$f\left(\frac{3}{4}\right) = \frac{9}{8} - \frac{18}{8} + \frac{80}{8} = \frac{71}{8}$.

(B) The given function is a geometric series with first term $a = 1$,common ratio $r = 2x^2$,and $n = 11$ terms.
$f(x) = \sum_{k=0}^{10} (2x^2)^k = \frac{(2x^2)^{11} - 1}{2x^2 - 1}$.
Alternatively,we can differentiate the polynomial directly:
$f(x) = 1 + 2x^2 + 4x^4 + 8x^6 + \dots + 1024x^{20}$.
$f'(x) = 4x + 16x^3 + 48x^5 + \dots + 20480x^{19}$.
$f'(x) = 4x(1 + 4x^2 + 12x^4 + \dots + 5120x^{18})$.
Setting $f'(x) = 0$,we see that $x = 0$ is the only real critical point because the term in the bracket is always positive for all real $x$.
Calculating the second derivative at $x = 0$:
$f''(x) = 4 + 48x^2 + 240x^4 + \dots$
$f''(0) = 4 > 0$.
Since $f'(0) = 0$ and $f''(0) > 0$,the function has a local minimum at $x = 0$. Since it is a polynomial of even degree with a positive leading coefficient,this is the global minimum. Thus,$f(x)$ has exactly one minimum.

(B) Consider the function $f(x) = x - \log_e(1 + x)$ for $x \in (0, 1)$.
Taking the derivative,we get $f'(x) = 1 - \frac{1}{1 + x} = \frac{1 + x - 1}{1 + x} = \frac{x}{1 + x}$.
Since $x > 0$,$f'(x) > 0$ for all $x \in (0, 1)$.
This implies that $f(x)$ is a strictly increasing function on $(0, 1)$.
Since $f(0) = 0 - \log_e(1) = 0$,and $f(x)$ is strictly increasing,$f(x) > f(0)$ for all $x > 0$.
Therefore,$x - \log_e(1 + x) > 0$,which means $\log_e(1 + x) < x$ for all $x \in (0, 1)$.
Thus,option $(b)$ is correct.

(B) Let $r$ be the radius and $h$ be the height of the cylinder inscribed in a sphere of radius $R$. From the geometry of the figure,we have the relation:
${r^2} + {\left( {\frac{h}{2}} \right)^2} = {R^2}$
${h^2} = 4({R^2} - {r^2}) \implies h = 2\sqrt {{R^2} - {r^2}}$
The volume $V$ of the cylinder is given by:
$V = \pi {r^2}h = 2\pi {r^2}\sqrt {{R^2} - {r^2}}$
To find the maximum volume,we differentiate $V$ with respect to $r$:
$\frac{{dV}}{{dr}} = 2\pi \left[ {2r\sqrt {{R^2} - {r^2}} + {r^2} \cdot \frac{1}{{2\sqrt {{R^2} - {r^2}} }} \cdot ( - 2r)} \right]$
$\frac{{dV}}{{dr}} = 2\pi \left[ {2r\sqrt {{R^2} - {r^2}} - \frac{{{r^3}}}{{\sqrt {{R^2} - {r^2}} }}} \right]$
Setting $\frac{{dV}}{{dr}} = 0$ for critical points:
$2r\sqrt {{R^2} - {r^2}} = \frac{{{r^3}}}{{\sqrt {{R^2} - {r^2}} }}$
$2({R^2} - {r^2}) = {r^2}$
$2{R^2} - 2{r^2} = {r^2}$
$3{r^2} = 2{R^2}$
${r^2} = \frac{2}{3}{R^2}$
$r = \sqrt {\frac{2}{3}} R$
Thus,the radius of the cylinder of maximum volume is $\sqrt {\frac{2}{3}} R$.

(D) Given $f(x) = \int\limits_0^x \frac{\cos t}{t} dt$ for $x > 0$.
By the Fundamental Theorem of Calculus,$f'(x) = \frac{\cos x}{x}$.
For critical points,set $f'(x) = 0$,which implies $\cos x = 0$ (since $x > 0$).
Thus,$x = (2n + 1)\frac{\pi}{2}$ for $n \in \mathbb{Z}$.
Now,$f''(x) = \frac{-x \sin x - \cos x}{x^2}$.
At critical points $x_n = (2n + 1)\frac{\pi}{2}$,we have $\cos x_n = 0$ and $\sin x_n = (-1)^n$.
So,$f''(x_n) = \frac{-x_n \sin x_n}{x_n^2} = \frac{-\sin x_n}{x_n} = \frac{-(-1)^n}{(2n + 1)\frac{\pi}{2}} = \frac{2(-1)^{n+1}}{(2n + 1)\pi}$.
For a maximum,we need $f''(x_n) < 0$. This occurs when $(-1)^{n+1} < 0$,which means $n+1$ is odd,so $n$ is even $(n = 0, 2, 4, \dots)$.
For a minimum,we need $f''(x_n) > 0$. This occurs when $(-1)^{n+1} > 0$,which means $n+1$ is even,so $n$ is odd $(n = 1, 3, 5, \dots)$.
Since none of the provided options match these conditions correctly,the answer is $(d)$.

(C) The total number of batches is $\frac{N}{x}$.
The time taken per batch is $(\alpha + \beta x^2)$ seconds.
Therefore,the total processing time $T$ is given by $T = \frac{N}{x}(\alpha + \beta x^2) = N(\frac{\alpha}{x} + \beta x)$.
For fast processing,$T$ must be minimized.
Taking the derivative with respect to $x$: $\frac{dT}{dx} = N(-\frac{\alpha}{x^2} + \beta)$.
Setting $\frac{dT}{dx} = 0$ for critical points: $-\frac{\alpha}{x^2} + \beta = 0 \implies x^2 = \frac{\alpha}{\beta} \implies x = \sqrt{\frac{\alpha}{\beta}}$.
Checking the second derivative: $\frac{d^2T}{dx^2} = N(\frac{2\alpha}{x^3})$. Since $x, \alpha, \beta > 0$,$\frac{d^2T}{dx^2} > 0$,which confirms that $T$ is minimum at $x = \sqrt{\frac{\alpha}{\beta}}$.

(D) Let $f(x) = {x^{25}}{(1 - x)^{75}}$.
To find the maximum value,we differentiate $f(x)$ with respect to $x$:
$f'(x) = {x^{25}} \cdot 75{(1 - x)^{74}} \cdot (-1) + 25{x^{24}} \cdot {(1 - x)^{75}}$
$f'(x) = 25{x^{24}}{(1 - x)^{74}} [ -3x + (1 - x) ]$
$f'(x) = 25{x^{24}}{(1 - x)^{74}} (1 - 4x)$
For critical points,set $f'(x) = 0$:
$25{x^{24}}{(1 - x)^{74}} (1 - 4x) = 0$
This gives $x = 0, x = 1,$ or $x = 1/4$.
Since $f(0) = 0$ and $f(1) = 0$,and $f(x) > 0$ for $x \in (0, 1)$,the maximum must occur at the critical point $x = 1/4$.
Thus,the function takes its maximum value at $x = 1/4$.

(B) Given $f(x) = \int_{-1}^x {t({e^t} - 1)(t - 1){(t - 2)}^3{(t - 3)}^5} dt$.
By the Fundamental Theorem of Calculus,$f'(x) = x({e^x} - 1)(x - 1){(x - 2)}^3{(x - 3)}^5$.
For local extrema,we set $f'(x) = 0$,which gives critical points $x = 0, 1, 2, 3$.
We examine the sign change of $f'(x)$ around these points:
$1$. At $x = 0$: $f'(x)$ changes from negative to negative (no extremum).
$2$. At $x = 1$: $f'(x)$ changes from negative to positive (local minimum).
$3$. At $x = 2$: $f'(x)$ changes from positive to negative (local maximum).
$4$. At $x = 3$: $f'(x)$ changes from negative to positive (local minimum).
Thus,the function has local minima at $x = 1$ and $x = 3$. Since $1$ is an option,the correct answer is $1$.

(C) Given the function $f(x) = 2x^3 - 9ax^2 + 12a^2x + 1$.
Find the first derivative: $f'(x) = 6x^2 - 18ax + 12a^2$.
Set $f'(x) = 0$ to find critical points: $6(x^2 - 3ax + 2a^2) = 0 \Rightarrow 6(x - a)(x - 2a) = 0$.
The critical points are $x = a$ and $x = 2a$.
Find the second derivative: $f''(x) = 12x - 18a$.
At $x = a$,$f''(a) = 12a - 18a = -6a < 0$ (since $a > 0$),so $x = a$ is a point of local maximum. Thus,$p = a$.
At $x = 2a$,$f''(2a) = 12(2a) - 18a = 6a > 0$,so $x = 2a$ is a point of local minimum. Thus,$q = 2a$.
Given the condition $p^2 = q$,we have $a^2 = 2a$.
Since $a > 0$,we can divide by $a$ to get $a = 2$.

(A) Let the sum be $S = x + \frac{1}{x}$.
By the Arithmetic Mean-Geometric Mean $(AM-GM)$ inequality,for positive real numbers $x$ and $\frac{1}{x}$:
$\frac{x + \frac{1}{x}}{2} \ge \sqrt{x \cdot \frac{1}{x}}$
$\frac{x + \frac{1}{x}}{2} \ge \sqrt{1}$
$x + \frac{1}{x} \ge 2$.
The minimum value of the sum is $2$,which occurs when $x = \frac{1}{x}$,i.e.,$x^2 = 1$. Since $x$ is a positive real number,$x = 1$.

(B) Let $A = x + y$. Since $xy = 1$,we have $y = \frac{1}{x}$.
Substituting this into the expression for $A$,we get $A(x) = x + \frac{1}{x}$ for $x > 0$.
To find the minimum value,we find the derivative: $\frac{dA}{dx} = 1 - \frac{1}{x^2}$.
Setting $\frac{dA}{dx} = 0$,we get $1 = \frac{1}{x^2}$,which implies $x^2 = 1$. Since $x > 0$,we have $x = 1$.
Now,we check the second derivative: $\frac{d^2A}{dx^2} = \frac{2}{x^3}$.
At $x = 1$,$\frac{d^2A}{dx^2} = \frac{2}{1^3} = 2 > 0$.
Since the second derivative is positive,$x = 1$ is a point of local minima.
The minimum value is $A(1) = 1 + \frac{1}{1} = 2$.

(A) Given function is $f(x) = x^2 \log x$ on the interval $[1, e]$.
First,find the derivative $f'(x)$:
$f'(x) = x^2 \cdot \frac{1}{x} + (\log x) \cdot 2x = x + 2x \log x = x(1 + 2 \log x)$.
For $x \in [1, e]$,$x$ is always positive and $\log x \ge 0$.
Since $1 + 2 \log x > 0$ for all $x \in [1, e]$,$f'(x) > 0$.
Thus,$f(x)$ is a strictly increasing function on the interval $[1, e]$.
The maximum value occurs at the right endpoint of the interval,which is $x = e$.
$f(e) = e^2 \log e = e^2 \cdot 1 = e^2$.

Given $f(x) = x^4(12\ln x - 7)$.
$f'(x) = 4x^3(12\ln x - 7) + x^4(\frac{12}{x}) = 48x^3\ln x - 28x^3 + 12x^3 = 48x^3\ln x - 16x^3 = 16x^3(3\ln x - 1)$.
$f''(x) = 48x^2(3\ln x - 1) + 16x^3(\frac{3}{x}) = 144x^2\ln x - 48x^2 + 48x^2 = 144x^2\ln x$.
$(A)$ Point of inflection occurs where $f''(x) = 0$,so $144x^2\ln x = 0 \implies \ln x = 0 \implies x = 1$. Then $f(1) = 1^4(12\ln 1 - 7) = -7$. So $(a, b) = (1, -7)$. $a - b = 1 - (-7) = 8$. Thus $(A) \to S$.
$(B)$ Local minimum occurs where $f'(x) = 0$ and $f''(x) > 0$. $16x^3(3\ln x - 1) = 0 \implies \ln x = 1/3 \implies x = e^{1/3}$. Here $t = 1/3$. So $12t = 12(1/3) = 4$. Thus $(B) \to R$.
$(C)$ Concave down where $f''(x) < 0$,i.e.,$144x^2\ln x < 0 \implies \ln x < 0 \implies 0 < x < 1$. So $(d, e) = (0, 1)$. $d + 3e = 0 + 3(1) = 3$. Thus $(C) \to P$.
$(D)$ Concave up where $f''(x) > 0$,i.e.,$144x^2\ln x > 0 \implies \ln x > 0 \implies x > 1$. So $p = 1$. Thus $(D) \to Q$.