Maxima and Minima Questions in English

(B) Given,$f(x)=a \log |x|+b x^2+x$.
The derivative is $f^{\prime}(x) = \frac{a}{x} + 2bx + 1$.
Since the function has extreme values at $x=-1$ and $x=2$,we have $f^{\prime}(-1)=0$ and $f^{\prime}(2)=0$.
For $x=-1$: $f^{\prime}(-1) = \frac{a}{-1} + 2b(-1) + 1 = 0 \Rightarrow -a - 2b + 1 = 0 \Rightarrow a + 2b = 1$ $(i)$.
For $x=2$: $f^{\prime}(2) = \frac{a}{2} + 2b(2) + 1 = 0 \Rightarrow \frac{a}{2} + 4b + 1 = 0 \Rightarrow a + 8b = -2$ (ii).
Subtracting $(i)$ from (ii): $(a + 8b) - (a + 2b) = -2 - 1 \Rightarrow 6b = -3 \Rightarrow b = -\frac{1}{2}$.
Substituting $b = -\frac{1}{2}$ into $(i)$: $a + 2(-\frac{1}{2}) = 1 \Rightarrow a - 1 = 1 \Rightarrow a = 2$.
Thus,the ordered pair $(a, b) = \left(2, -\frac{1}{2}\right)$.

(C) Given,$f(x) = 2x^3 - 9ax^2 + 12a^2x + 1$.
To find the critical points,we calculate the derivative: $f'(x) = 6x^2 - 18ax + 12a^2$.
Setting $f'(x) = 0$,we get $6(x^2 - 3ax + 2a^2) = 0$,which factors as $6(x - a)(x - 2a) = 0$.
Thus,the critical points are $x = a$ and $x = 2a$.
We find the second derivative: $f''(x) = 12x - 18a$.
For $x = a$,$f''(a) = 12a - 18a = -6a$. Since $a > 0$,$f''(a) < 0$,so $f(x)$ has a maximum at $p = a$.
For $x = 2a$,$f''(2a) = 12(2a) - 18a = 6a$. Since $a > 0$,$f''(2a) > 0$,so $f(x)$ has a minimum at $q = 2a$.
Given the condition $p^2 = q$,we substitute the values: $a^2 = 2a$.
Since $a > 0$,we can divide by $a$ to get $a = 2$.

(C) Given the function $f(x) = x^2 e^{2x}$ on the interval $[-2, 2]$.
First,we find the critical points by setting the derivative $f'(x) = 0$.
$f'(x) = 2x e^{2x} + x^2 (2 e^{2x}) = 2x e^{2x} (1 + x)$.
Setting $f'(x) = 0$,we get $x = 0$ or $x = -1$.
Now,we evaluate the function at the critical points and the endpoints of the interval $[-2, 2]$:
$f(-2) = (-2)^2 e^{2(-2)} = 4 e^{-4}$.
$f(-1) = (-1)^2 e^{2(-1)} = 1 e^{-2} = e^{-2}$.
$f(0) = (0)^2 e^{2(0)} = 0$.
$f(2) = (2)^2 e^{2(2)} = 4 e^4$.
Comparing these values,the global maximum $p = 4 e^4$ and the global minimum $q = 0$.
Finally,we calculate $p e^{-4} + q e^4 = (4 e^4) e^{-4} + (0) e^4 = 4 e^0 + 0 = 4(1) = 4$.

(A) Let $h$ be the height and $r$ be the radius of the cylinder inscribed in a sphere of radius $R = 3$ units.
From the geometry of the inscribed cylinder,we have the relation $r^2 + (h/2)^2 = R^2$.
Substituting $R = 3$,we get $r^2 + h^2/4 = 9$,which implies $r^2 = 9 - h^2/4$.
The volume $V$ of the cylinder is given by $V = \pi r^2 h$.
Substituting $r^2$,we get $V = \pi (9 - h^2/4) h = \pi (9h - h^3/4)$.
To find the maximum volume,we differentiate $V$ with respect to $h$:
$\frac{dV}{dh} = \pi (9 - 3h^2/4)$.
Setting $\frac{dV}{dh} = 0$,we get $9 - 3h^2/4 = 0$,which implies $3h^2/4 = 9$,so $h^2 = 12$ and $h = 2\sqrt{3}$ (since height must be positive).
Then $r^2 = 9 - (12/4) = 9 - 3 = 6$.
The maximum volume is $V = \pi (6) (2\sqrt{3}) = 12\sqrt{3}\pi$ cubic units.

(A) Given the function $f(x) = 2x^3 - 3x^2 - 12x + 5$.
To find the critical points,we calculate the derivative $f'(x) = 6x^2 - 6x - 12$.
Setting $f'(x) = 0$,we get $6(x^2 - x - 2) = 0$,which factors as $6(x + 1)(x - 2) = 0$.
Thus,the critical points are $x = -1$ and $x = 2$,both of which lie within the interval $[-2, 4]$.
Now,we evaluate the function at the critical points and the endpoints of the interval:
$f(-2) = 2(-8) - 3(4) - 12(-2) + 5 = -16 - 12 + 24 + 5 = 1$.
$f(-1) = 2(-1) - 3(1) - 12(-1) + 5 = -2 - 3 + 12 + 5 = 12$.
$f(2) = 2(8) - 3(4) - 12(2) + 5 = 16 - 12 - 24 + 5 = -15$.
$f(4) = 2(64) - 3(16) - 12(4) + 5 = 128 - 48 - 48 + 5 = 37$.
Comparing these values,the absolute maximum is $37$,which occurs at $x = 4$.
Therefore,the correct option is $A$.

(B) Let $v$ be the speed of the motor boat relative to the water,where $v > C$. The speed of the boat relative to the land,while moving against the water flow,is $(v - C)$.
If the distance to be covered is $S$,the time taken is $t = \frac{S}{v - C}$.
The rate of petrol consumption per hour is proportional to the cube of the velocity relative to the water,i.e.,$k v^3$,where $k$ is a constant.
The total petrol consumed $f(v)$ is given by:
$f(v) = (k v^3) \times \left( \frac{S}{v - C} \right) = k S \frac{v^3}{v - C}$.
To find the minimum consumption,we differentiate $f(v)$ with respect to $v$ and set it to zero:
$f'(v) = k S \left[ \frac{(v - C)(3v^2) - v^3(1)}{(v - C)^2} \right] = 0$.
This implies $3v^2(v - C) - v^3 = 0$,which simplifies to $3v^3 - 3v^2C - v^3 = 0$.
$2v^3 - 3v^2C = 0$.
Since $v \neq 0$,we have $2v - 3C = 0$,which gives $v = \frac{3C}{2}$.

(B) Given the function $f(x) = 3x^4 - 2x^3 - 6x^2 + 6x + 4$.
First,find the derivative $f'(x)$:
$f'(x) = 12x^3 - 6x^2 - 12x + 6$
Factorize the derivative:
$f'(x) = 6(2x^3 - x^2 - 2x + 1) = 6[x^2(2x - 1) - 1(2x - 1)] = 6(x^2 - 1)(2x - 1) = 6(x - 1)(x + 1)(2x - 1)$.
Set $f'(x) = 0$ to find critical points:
$6(x - 1)(x + 1)(2x - 1) = 0$.
The critical points are $x = 1, x = -1, x = 1/2$.
Since we are considering the interval $(0, 2)$,we only consider $x = 1$ and $x = 1/2$ (as $-1$ is outside the interval).
Evaluate $f(x)$ at these points:
$f(1) = 3(1)^4 - 2(1)^3 - 6(1)^2 + 6(1) + 4 = 3 - 2 - 6 + 6 + 4 = 5$.
$f(1/2) = 3(1/16) - 2(1/8) - 6(1/4) + 6(1/2) + 4 = 3/16 - 4/16 - 24/16 + 48/16 + 64/16 = 87/16$.
Comparing the values,the maximum value is $87/16$ and the minimum value is $5$.
The sum of the maximum and minimum values is $87/16 + 5 = (87 + 80) / 16 = 167/16$.

(B) Given that,$s = t^5 - 40t^3 + 30t^2 + 80t - 250$.
Velocity $v = \frac{ds}{dt} = 5t^4 - 120t^2 + 60t + 80$.
Acceleration $a = \frac{dv}{dt} = 20t^3 - 240t + 60$.
Let $f(t) = 20t^3 - 240t + 60$.
To find the minimum acceleration,we find the critical points by setting $f'(t) = 0$:
$f'(t) = 60t^2 - 240 = 0 \Rightarrow t^2 = 4 \Rightarrow t = \pm 2$.
Now,check the second derivative $f''(t) = 120t$.
At $t = 2$,$f''(2) = 120(2) = 240 > 0$,so $f(t)$ has a minimum at $t = 2$.
Minimum acceleration $a_{\min} = f(2) = 20(2)^3 - 240(2) + 60 = 160 - 480 + 60 = -260$.

(C) To find the local extrema of the function $f(x) = 2x^6 - 3$,we first find its derivative.
$f'(x) = \frac{d}{dx}(2x^6 - 3) = 12x^5$.
Setting the derivative equal to zero to find critical points:
$12x^5 = 0 \implies x = 0$.
Now,we use the first derivative test or second derivative test to check the nature of the point $x = 0$.
Using the first derivative test:
For $x < 0$,$f'(x) = 12x^5 < 0$ (function is decreasing).
For $x > 0$,$f'(x) = 12x^5 > 0$ (function is increasing).
Since the derivative changes sign from negative to positive at $x = 0$,the function has a local minimum at $x = 0$.
The local minimum value is $f(0) = 2(0)^6 - 3 = -3$.
Thus,$-3$ is the local minimum value of $f$.

(C) To find the greatest value of $f(x) = (x+1)^{1/3} - (x-1)^{1/3}$ on $[0, 1]$,we find the derivative $f'(x)$.
$f'(x) = \frac{1}{3}(x+1)^{-2/3} - \frac{1}{3}(x-1)^{-2/3} = \frac{1}{3} \left[ \frac{1}{(x+1)^{2/3}} - \frac{1}{(x-1)^{2/3}} \right]$.
Setting $f'(x) = 0$,we get $(x-1)^{2/3} = (x+1)^{2/3}$,which implies $|x-1| = |x+1|$.
Squaring both sides,$(x-1)^2 = (x+1)^2$,so $x^2 - 2x + 1 = x^2 + 2x + 1$,which gives $4x = 0$,so $x = 0$.
Now,we evaluate $f(x)$ at the critical point $x=0$ and the endpoints $x=0$ and $x=1$.
At $x=0$,$f(0) = (0+1)^{1/3} - (0-1)^{1/3} = 1 - (-1) = 2$.
At $x=1$,$f(1) = (1+1)^{1/3} - (1-1)^{1/3} = 2^{1/3} - 0 = 2^{1/3}$.
Comparing $f(0) = 2$ and $f(1) = 2^{1/3} \approx 1.26$,the greatest value is $2$.

(A) The area $A$ of a triangle with vertices $(0,0)$,$(x_1, y_1)$,and $(x_2, y_2)$ is given by $A = \frac{1}{2} |x_1 y_2 - x_2 y_1|$.
Here,the vertices are $(0,0)$,$(x, \cos x)$,and $(\sin^3 x, 0)$.
Thus,$A(x) = \frac{1}{2} |x(0) - (\sin^3 x)(\cos x)| = \frac{1}{2} \sin^3 x \cos x$ for $x \in (0, \frac{\pi}{2})$.
To find the maximum,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 $\sin x \neq 0$,we have $3 \cos^2 x = \sin^2 x$,which implies $\tan^2 x = 3$,so $\tan x = \sqrt{3}$ (since $x \in (0, \frac{\pi}{2})$).
Thus,$x = \frac{\pi}{3}$.
At $x = \frac{\pi}{3}$,$\sin x = \frac{\sqrt{3}}{2}$ and $\cos x = \frac{1}{2}$.
Substituting these values into $A(x)$:
$A = \frac{1}{2} (\frac{\sqrt{3}}{2})^3 (\frac{1}{2}) = \frac{1}{2} (\frac{3 \sqrt{3}}{8}) (\frac{1}{2}) = \frac{3 \sqrt{3}}{32}$.

(C) Given that the volume of the cylindrical vessel is $V$ and it has no lid on the top.
Let $r$ be the radius and $h$ be the height of the cylinder.
The volume of the cylinder is given by $V = \pi r^2 h$,which implies $h = \frac{V}{\pi r^2}$.
The surface area $S$ of the metal sheet used (including the base but no lid) is given by:
$S = 2\pi rh + \pi r^2$
Substituting $h = \frac{V}{\pi r^2}$ into the expression for $S$:
$S(r) = 2\pi r \left(\frac{V}{\pi r^2}\right) + \pi r^2 = \frac{2V}{r} + \pi r^2$
To find the minimum surface area,we differentiate $S(r)$ with respect to $r$ and set it to zero:
$S'(r) = -\frac{2V}{r^2} + 2\pi r = 0$
$2\pi r = \frac{2V}{r^2} \Rightarrow r^3 = \frac{V}{\pi} \Rightarrow r = \sqrt[3]{\frac{V}{\pi}}$
Now,substitute $r$ back into the expression for $h$:
$h = \frac{V}{\pi r^2} = \frac{V}{\pi (V/\pi)^{2/3}} = \frac{V}{\pi} \cdot \left(\frac{\pi}{V}\right)^{2/3} = \left(\frac{V}{\pi}\right)^{1/3} = \sqrt[3]{\frac{V}{\pi}}$
Thus,the radius and height for minimum surface area are $r = \sqrt[3]{\frac{V}{\pi}}$ and $h = \sqrt[3]{\frac{V}{\pi}}$.

(B) Let $R(x)$ be the revenue function and $C(x)$ be the cost function.
Revenue $R(x) = x \times \left(5 - \frac{x}{100}\right) = 5x - \frac{x^2}{100}$.
Profit function $P(x) = R(x) - C(x) = \left(5x - \frac{x^2}{100}\right) - \left(\frac{x}{5} + 500\right)$.
$P(x) = 5x - \frac{x^2}{100} - \frac{x}{5} - 500 = 4.8x - \frac{x^2}{100} - 500$.
To find the maximum profit,we find the derivative $P'(x)$ and set it to $0$.
$P'(x) = 4.8 - \frac{2x}{100} = 4.8 - \frac{x}{50}$.
Setting $P'(x) = 0$,we get $4.8 = \frac{x}{50}$,which implies $x = 4.8 \times 50 = 240$.
To verify,we check the second derivative $P''(x) = -\frac{1}{50}$.
Since $P''(x) < 0$,the profit is maximum at $x = 240$.

(B) Given the function $f(x)=x^3+a x^2+b x+c$.
To find the extreme values,we differentiate $f(x)$ with respect to $x$:
$f^{\prime}(x)=3 x^2+2 a x+b$.
For extreme values,we set $f^{\prime}(x)=0$:
$3 x^2+2 a x+b=0$.
The roots of this quadratic equation are given by the quadratic formula:
$x = \frac{-2 a \pm \sqrt{(2 a)^2 - 4(3)(b)}}{2(3)} = \frac{-2 a \pm \sqrt{4 a^2 - 12 b}}{6} = \frac{-2 a \pm 2 \sqrt{a^2 - 3 b}}{6} = \frac{-a \pm \sqrt{a^2 - 3 b}}{3}$.
Given the condition $a^2 \leq 3 b$,the term $a^2 - 3 b \leq 0$.
If $a^2 - 3 b < 0$,the roots are imaginary,meaning $f^{\prime}(x)$ is never zero for any real $x$.
If $a^2 = 3 b$,then $f^{\prime}(x) = 3(x + \frac{a}{3})^2$,which is always $\geq 0$,so the function is monotonically increasing and has no local extrema.
Thus,the function has no extreme value.

(D) Let $y = \frac{\log x}{x}$.
Find the derivative with respect to $x$ using the quotient rule:
$\frac{dy}{dx} = \frac{x \cdot \frac{1}{x} - \log x \cdot 1}{x^2} = \frac{1 - \log x}{x^2}$.
For critical points,set $\frac{dy}{dx} = 0$:
$1 - \log x = 0 \Rightarrow \log x = 1 \Rightarrow x = e$.
To verify the maximum,find the second derivative:
$\frac{d^2y}{dx^2} = \frac{x^2(-\frac{1}{x}) - (1 - \log x)(2x)}{x^4} = \frac{-x - 2x + 2x \log x}{x^4} = \frac{2 \log x - 3}{x^3}$.
At $x = e$,$\frac{d^2y}{dx^2} = \frac{2(1) - 3}{e^3} = -\frac{1}{e^3} < 0$,so $x = e$ is a point of local maxima.
The maximum value is $y(e) = \frac{\log e}{e} = \frac{1}{e} = e^{-1}$.

(A) Let $f(x) = \frac{x^2-x+1}{x^2+x+1}$ ... $(i)$
On differentiating with respect to $x$,we get
$f'(x) = \frac{(x^2+x+1)(2x-1) - (x^2-x+1)(2x+1)}{(x^2+x+1)^2}$
For maximum or minimum,put $f'(x) = 0$
$(x^2+x+1)(2x-1) - (x^2-x+1)(2x+1) = 0$
$(2x^3 - x^2 + 2x^2 - x + 2x - 1) - (2x^3 + x^2 - 2x^2 - x + 2x + 1) = 0$
$(2x^3 + x^2 + x - 1) - (2x^3 - x^2 + x + 1) = 0$
$2x^2 - 2 = 0 \Rightarrow x^2 = 1 \Rightarrow x = \pm 1$
Now,$f'(x) = \frac{2x^2-2}{(x^2+x+1)^2}$
Again differentiating,we get $f''(x) = \frac{(x^2+x+1)^2(4x) - (2x^2-2) \cdot 2(x^2+x+1)(2x+1)}{(x^2+x+1)^4}$
At $x = 1$,$f''(1) = \frac{(1+1+1)^2(4) - 0}{(1+1+1)^4} = \frac{9 \times 4}{81} = \frac{36}{81} > 0$
Therefore,the function has a minimum at $x = 1$.
Putting $x = 1$ in equation $(i)$,we get
$f(1) = \frac{1^2-1+1}{1^2+1+1} = \frac{1}{3}$
$\therefore$ The minimum value is $\frac{1}{3}$.

(C) Given the equation of motion: $s = 490t - 4.9t^2$.
To find the maximum height,we differentiate $s$ with respect to $t$ and set the derivative to zero:
$\frac{ds}{dt} = 490 - 9.8t$.
Setting $\frac{ds}{dt} = 0$ for maximum height:
$490 - 9.8t = 0$
$9.8t = 490$
$t = \frac{490}{9.8} = 50 \text{ seconds}$.
Now,substitute $t = 50$ into the original equation to find the maximum height $s$:
$s = 490(50) - 4.9(50)^2$
$s = 24500 - 4.9(2500)$
$s = 24500 - 12250$
$s = 12250$.

(A) Given the function $f(x) = x e^{-x}$.
To find the critical points,we calculate the first derivative $f'(x)$ using the product rule:
$f'(x) = (1)e^{-x} + x(-e^{-x}) = e^{-x}(1 - x)$.
Setting $f'(x) = 0$ for maxima or minima:
$e^{-x}(1 - x) = 0$.
Since $e^{-x} \neq 0$ for any $x \in R$,we have $1 - x = 0$,which gives $x = 1$.
Now,we find the second derivative $f''(x)$ to check for maxima:
$f''(x) = \frac{d}{dx}[e^{-x} - x e^{-x}] = -e^{-x} - [e^{-x} - x e^{-x}] = e^{-x}(x - 2)$.
Evaluating at $x = 1$:
$f''(1) = e^{-1}(1 - 2) = -e^{-1} < 0$.
Since the second derivative is negative at $x = 1$,the function attains a maximum value at $x = 1$.

(D) Given,$x+y=60$.
Using the $AM \geq GM$ inequality for the terms $x, \frac{y}{3}, \frac{y}{3}, \frac{y}{3}$:
$\frac{x + \frac{y}{3} + \frac{y}{3} + \frac{y}{3}}{4} \geq \sqrt[4]{x \cdot \frac{y}{3} \cdot \frac{y}{3} \cdot \frac{y}{3}}$
$\frac{x+y}{4} \geq \sqrt[4]{\frac{x y^3}{27}}$
Substituting $x+y=60$:
$\frac{60}{4} \geq \sqrt[4]{\frac{x y^3}{27}}$
$15 \geq \sqrt[4]{\frac{x y^3}{27}}$
Raising both sides to the power of $4$:
$(15)^4 \geq \frac{x y^3}{27}$
$x y^3 \leq 27 \times (15)^4$
$x y^3 \leq 3^3 \times (3 \times 5)^4$
$x y^3 \leq 3^3 \times 3^4 \times 5^4$
$x y^3 \leq 3^7 \times 5^4$
Since $\frac{(45)^4}{3} = \frac{(3^2 \times 5)^4}{3} = \frac{3^8 \times 5^4}{3} = 3^7 \times 5^4$,the maximum value is $\frac{(45)^4}{3}$.

(D) Given $x + 2y = 10$,where $x, y$ are positive integers.
Possible pairs $(x, y)$ are:
$(8, 1) \implies x^2 y^3 = 8^2 \times 1^3 = 64$
$(6, 2) \implies x^2 y^3 = 6^2 \times 2^3 = 36 \times 8 = 288$
$(4, 3) \implies x^2 y^3 = 4^2 \times 3^3 = 16 \times 27 = 432$
$(2, 4) \implies x^2 y^3 = 2^2 \times 4^3 = 4 \times 64 = 256$
The maximum value of $x^2 y^3$ is $432$,which occurs when $x = 4$ and $y = 3$.
We need to find $x^2 + 2y^3 = 4^2 + 2(3^3) = 16 + 2(27) = 16 + 54 = 70$.

(D) Given the function $y = 2x^3 - 8x^2 + 10x - 4$.
To find the slope of the tangent,we differentiate $y$ with respect to $x$:
$\frac{dy}{dx} = 6x^2 - 16x + 10$.
Since the tangent is parallel to the $X$-axis,its slope must be zero:
$\frac{dy}{dx} = 0 \implies 6x^2 - 16x + 10 = 0$.
Dividing by $2$,we get $3x^2 - 8x + 5 = 0$.
Factoring the quadratic equation:
$3x^2 - 3x - 5x + 5 = 0 \implies 3x(x - 1) - 5(x - 1) = 0$.
$(3x - 5)(x - 1) = 0$.
This gives $x = 1$ or $x = \frac{5}{3}$.
Since the point $(a, b)$ satisfies $a \in (1, 2)$,we discard $x = 1$ and accept $a = \frac{5}{3}$.

(D) Let $R = 3$ be the radius of the sphere. Let $h$ be the height of the cone and $r$ be the radius of the base of the cone.
Let $x$ be the distance from the center of the sphere to the base of the cone. Then $h = R + x = 3 + x$.
In the right triangle formed by the radius of the sphere,the radius of the cone base,and the distance $x$,we have $r^2 + x^2 = R^2 = 3^2 = 9$,so $r^2 = 9 - x^2$.
The volume of the cone is $V = \frac{1}{3} \pi r^2 h = \frac{1}{3} \pi (9 - x^2)(3 + x)$.
$V = \frac{\pi}{3} (27 + 9x - 3x^2 - x^3)$.
To maximize $V$,we find $\frac{dV}{dx} = \frac{\pi}{3} (9 - 6x - 3x^2) = 0$.
$x^2 + 2x - 3 = 0 \Rightarrow (x + 3)(x - 1) = 0$.
Since $x$ is a distance,$x = 1$ (as $x \neq -3$).
For $x = 1$,$r^2 = 9 - 1^2 = 8$,so $r = \sqrt{8} = 2\sqrt{2}$.
The height $h = 3 + 1 = 4$.
The semi-vertical angle $\theta$ satisfies $\tan \theta = \frac{r}{h} = \frac{2\sqrt{2}}{4} = \frac{1}{\sqrt{2}}$.
Thus,$\theta = \tan ^{-1}\left(\frac{1}{\sqrt{2}}\right)$.

(C) Establish Relationships:
In $\triangle ABC$,we are given $\angle B=90^{\circ}$,making it a right-angled triangle where $b$ is the hypotenuse and $a, c$ are the legs. From the Pythagorean theorem,$c=\sqrt{b^2-a^2}$. We are given that $a+b=k$ (a constant),so $b=k-a$. The area $A$ is:
$A=\frac{1}{2}ac=\frac{1}{2}a\sqrt{b^2-a^2}$
Express Area in Terms of a Single Variable:
Substitute $b=k-a$ into the area equation:
$A=\frac{1}{2}a\sqrt{(k-a)^2-a^2}=\frac{1}{2}a\sqrt{k^2-2ka+a^2-a^2}=\frac{1}{2}a\sqrt{k^2-2ka}$
To maximize $A$,we maximize $f(a)=4A^2=a^2(k^2-2ka)=k^2a^2-2ka^3$.
Differentiate and Solve for Extremum:
Find the derivative of $f(a)$ with respect to $a$ and set it to zero:
$f'(a)=2k^2a-6ka^2=0 \Longrightarrow 2ka(k-3a)=0$
Since $a, k \neq 0$,we have $a=\frac{k}{3}$.
Then $b=k-\frac{k}{3}=\frac{2k}{3}$.
Determine Angle $C$:
In the right triangle,$\cos C=\frac{a}{b}$.
$\cos C=\frac{a}{b}=\frac{k/3}{2k/3}=\frac{1}{2}$
Since $\cos C=\frac{1}{2}$,then $C=\frac{\pi}{3}$.

(A) For a function $f(x)$ to have no extreme value,its derivative $f'(x)$ must not change sign. This means $f'(x)$ must be either always non-negative or always non-positive.
Given $f(x) = x^3 + px^2 + qx + r$.
The derivative is $f'(x) = 3x^2 + 2px + q$.
For $f'(x)$ to have no sign change,the quadratic equation $3x^2 + 2px + q = 0$ must have no real roots or a repeated root.
This implies the discriminant $D \le 0$.
The discriminant $D = (2p)^2 - 4(3)(q) = 4p^2 - 12q$.
Setting $D \le 0$,we get $4p^2 - 12q \le 0$.
Dividing by $4$,we get $p^2 - 3q \le 0$,or $p^2 \le 3q$.
Given the options,the condition $p^2 < 3q$ is the correct choice.

(A) Given $f(x) = \frac{x^2+2x+2}{x+1} = \frac{(x+1)^2+1}{x+1} = (x+1) + \frac{1}{x+1}$.
Let $u = x+1$. Then $f(u) = u + \frac{1}{u}$.
To find local extrema,we find the derivative: $f'(u) = 1 - \frac{1}{u^2}$.
Setting $f'(u) = 0$,we get $u^2 = 1$,so $u = 1$ or $u = -1$.
For $u = 1$,$x+1 = 1 \implies x = 0$. $f(0) = 0 + \frac{1}{1} = 2$. This is the local minimum value $m = 2$ at $\beta = 0$.
For $u = -1$,$x+1 = -1 \implies x = -2$. $f(-2) = -2 + \frac{1}{-1} = -2 - 1 = -3$. This is the local maximum value $l = -3$ at $\alpha = -2$.
Thus,$l = -3, m = 2, \alpha = -2, \beta = 0$.
The value of $\frac{l+m}{\alpha+\beta} = \frac{-3+2}{-2+0} = \frac{-1}{-2} = \frac{1}{2}$.
Wait,checking the options,let us re-evaluate. If $f(x) = x+1 + \frac{1}{x+1}$,for $x > -1$,$u > 0$,$f(u) \ge 2$ ($AM$-$GM$). For $x < -1$,$u < 0$,$f(u) \le -2$. The local maximum is $-2$ at $x = -2$ and local minimum is $2$ at $x = 0$. The expression $\frac{l+m}{\alpha+\beta} = \frac{-2+2}{-2+0} = 0$.

(B) Given $xy = 4$,we can express $x$ as $x = \frac{4}{y}$.
Let $f(y) = \sqrt{x} + \frac{y^2}{2} = \sqrt{\frac{4}{y}} + \frac{y^2}{2} = 2y^{-1/2} + \frac{y^2}{2}$.
To find the minimum,we differentiate $f(y)$ with respect to $y$:
$f'(y) = 2(-\frac{1}{2})y^{-3/2} + \frac{1}{2}(2y) = -y^{-3/2} + y$.
Setting $f'(y) = 0$ gives $y = y^{-3/2}$,which implies $y^{5/2} = 1$,so $y = 1$.
For $y = 1$,$x = \frac{4}{1} = 4$.
The value of the expression at $y = 1$ is $f(1) = \sqrt{4} + \frac{1^2}{2} = 2 + 0.5 = 2.5 = \frac{5}{2}$.
Since $f''(y) = \frac{3}{2}y^{-5/2} + 1 > 0$ for $y > 0$,the function has a local minimum at $y = 1$.

(C) Given $f(x) = \frac{ax + b}{x^2 - 5x + 4}$. Since the local maximum exists at $(2, -1)$,we have $f(2) = -1$.
Substituting $x = 2$ into the function: $f(2) = \frac{2a + b}{(2 - 1)(2 - 4)} = \frac{2a + b}{(1)(-2)} = \frac{2a + b}{-2} = -1$.
This implies $2a + b = 2$.
Also,for a local maximum at $x = 2$,the derivative $f'(2) = 0$.
Using the quotient rule $f'(x) = \frac{a(x^2 - 5x + 4) - (ax + b)(2x - 5)}{(x^2 - 5x + 4)^2}$.
Setting $f'(2) = 0$: $a(4 - 10 + 4) - (2a + b)(4 - 5) = 0$.
$a(-2) - (2a + b)(-1) = 0 \implies -2a + 2a + b = 0 \implies b = 0$.
Substituting $b = 0$ into $2a + b = 2$,we get $2a = 2$,so $a = 1$.
Thus,$a + b = 1 + 0 = 1$.

(B) Given the function $f(x) = 2x^3 - 9ax^2 + 12a^2x + 1$.
First,find the derivative $f'(x) = 6x^2 - 18ax + 12a^2$.
Set $f'(x) = 0$ to find critical points: $6(x^2 - 3ax + 2a^2) = 0$.
Factoring the quadratic gives $6(x - a)(x - 2a) = 0$,so 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 local maximum. Thus,$p = a$.
At $x = 2a$,$f''(2a) = 12(2a) - 18a = 6a > 0$ (since $a > 0$),so $x = 2a$ is a local minimum. Thus,$q = 2a$.
Given the condition $p^2 = q$,we substitute the values: $a^2 = 2a$.
Since $a > 0$,we can divide by $a$ to get $a = 2$.

(B) Let $f(x) = 4+3x-7x^2$.
Taking the derivative,$f'(x) = 3-14x$. Setting $f'(x) = 0$,we get $x = \frac{3}{14} = \alpha$.
Since $f''(x) = -14 < 0$,$f(x)$ attains its maximum at $x = \alpha = \frac{3}{14}$.
The maximum value $M = f(\frac{3}{14}) = 4 + 3(\frac{3}{14}) - 7(\frac{3}{14})^2 = 4 + \frac{9}{14} - \frac{63}{196} = 4 + \frac{126-63}{196} = 4 + \frac{63}{196} = 4 + \frac{9}{28} = \frac{112+9}{28} = \frac{121}{28}$.
Now,let $g(x) = 5x^2-2x+1$.
Taking the derivative,$g'(x) = 10x-2$. Setting $g'(x) = 0$,we get $x = \frac{1}{5} = \beta$.
Since $g''(x) = 10 > 0$,$g(x)$ attains its minimum at $x = \beta = \frac{1}{5}$.
The minimum value $m = g(\frac{1}{5}) = 5(\frac{1}{5})^2 - 2(\frac{1}{5}) + 1 = \frac{1}{5} - \frac{2}{5} + 1 = \frac{4}{5}$.
Finally,calculate $\frac{28(M-\alpha)}{5(m+\beta)} = \frac{28(\frac{121}{28} - \frac{3}{14})}{5(\frac{4}{5} + \frac{1}{5})} = \frac{28(\frac{121-6}{28})}{5(1)} = \frac{115}{5} = 23$.

(C) Given $x+y=24$,so $y=24-x$.
Let $P = x^3 y^5 = x^3(24-x)^5$.
To find the maximum,we differentiate $P$ with respect to $x$:
$\frac{dP}{dx} = 3x^2(24-x)^5 + x^3 \cdot 5(24-x)^4(-1) = 0$.
$x^2(24-x)^4 [3(24-x) - 5x] = 0$.
$x^2(24-x)^4 [72 - 3x - 5x] = 0$.
$x^2(24-x)^4 (72 - 8x) = 0$.
Since $x$ and $y$ are positive integers,$x \neq 0$ and $x \neq 24$.
Thus,$72 - 8x = 0 \Rightarrow x = 9$.
Then $y = 24 - 9 = 15$.
Finally,$x^2 + y^2 = 9^2 + 15^2 = 81 + 225 = 306$.

(C) Let $y = \frac{x^2-x+1}{x^2+x+1}$.
Then $y(x^2+x+1) = x^2-x+1$,which implies $(y-1)x^2 + (y+1)x + (y-1) = 0$.
For $x$ to be real,the discriminant $D \ge 0$.
$D = (y+1)^2 - 4(y-1)^2 \ge 0$.
$(y+1-2(y-1))(y+1+2(y-1)) \ge 0$.
$(-y+3)(3y-1) \ge 0$.
$(y-3)(3y-1) \le 0$.
Thus,$\frac{1}{3} \le y \le 3$.
The maximum value is $3$ and the minimum value is $\frac{1}{3}$.
The sum of the maximum and minimum values is $3 + \frac{1}{3} = \frac{10}{3}$.

(A) Given the function $f(x)=2x^3+9x^2+12x+1$ on the interval $[-3,0]$.
First,find the derivative $f'(x)$:
$f'(x)=6x^2+18x+12=6(x^2+3x+2)=6(x+1)(x+2)$.
Setting $f'(x)=0$,we get critical points $x=-1$ and $x=-2$.
Now,evaluate $f(x)$ at the critical points and the endpoints of the interval $[-3,0]$:
$f(-3)=2(-27)+9(9)+12(-3)+1 = -54+81-36+1 = -8$.
$f(-2)=2(-8)+9(4)+12(-2)+1 = -16+36-24+1 = -3$.
$f(-1)=2(-1)+9(1)+12(-1)+1 = -2+9-12+1 = -4$.
$f(0)=2(0)+9(0)+12(0)+1 = 1$.
Comparing these values,the absolute minimum $m = -8$ and the absolute maximum $M = 1$.
Therefore,$m+M = -8+1 = -7$.

(D) Given the function $f(x) = \frac{x}{5} + \frac{5}{x}$.
First,find the derivative $f'(x) = \frac{1}{5} - \frac{5}{x^2}$.
Set $f'(x) = 0$ to find critical points: $\frac{1}{5} = \frac{5}{x^2} \Rightarrow x^2 = 25 \Rightarrow x = \pm 5$.
Find the second derivative $f''(x) = \frac{10}{x^3}$.
For $x = 5$,$f''(5) = \frac{10}{125} > 0$,so $x = 5$ is a local minimum.
For $x = -5$,$f''(-5) = \frac{10}{-125} < 0$,so $x = -5$ is a local maximum.
Thus,$a = -5$.
Now,calculate $\sqrt{a^2 + 2a - 6} = \sqrt{(-5)^2 + 2(-5) - 6} = \sqrt{25 - 10 - 6} = \sqrt{9} = 3$.

(D) Given $2x + 3y = 50$. We want to maximize $P = x^2 y^3$.
From the constraint,$y = \frac{50 - 2x}{3}$.
Substituting $y$ in $P$,we get $P = x^2 \left(\frac{50 - 2x}{3}\right)^3 = \frac{1}{27} x^2 (50 - 2x)^3$.
To find the maximum,we differentiate $P$ with respect to $x$:
$\frac{dP}{dx} = \frac{1}{27} [x^2 \cdot 3(50 - 2x)^2(-2) + 2x(50 - 2x)^3] = 0$.
$\Rightarrow 2x(50 - 2x)^2 [-3x + 50 - 2x] = 0$.
$\Rightarrow 2x(50 - 2x)^2 (50 - 5x) = 0$.
Since $x, y$ are positive integers,$x \neq 0$ and $x \neq 25$. Thus,$50 - 5x = 0 \Rightarrow x = 10$.
For $x = 10$,$y = \frac{50 - 2(10)}{3} = \frac{30}{3} = 10$.
Thus,$\alpha = 10$ and $\beta = 10$.
Finally,$\frac{\alpha}{2} + \frac{\beta}{5} = \frac{10}{2} + \frac{10}{5} = 5 + 2 = 7$.

(A) Given function: $f(x) = x^3 - 2x^2 + x - 3$ on the interval $[0, 2]$.
First,find the derivative: $f'(x) = 3x^2 - 4x + 1$.
Set $f'(x) = 0$ to find critical points:
$3x^2 - 3x - x + 1 = 0$
$3x(x - 1) - 1(x - 1) = 0$
$(3x - 1)(x - 1) = 0$
So,$x = \frac{1}{3}$ and $x = 1$.
Both critical points lie within the interval $[0, 2]$.
Now,evaluate the function at the critical points and the endpoints:
$f(0) = 0^3 - 2(0)^2 + 0 - 3 = -3$
$f\left(\frac{1}{3}\right) = \left(\frac{1}{3}\right)^3 - 2\left(\frac{1}{3}\right)^2 + \frac{1}{3} - 3 = \frac{1}{27} - \frac{2}{9} + \frac{1}{3} - 3 = \frac{1 - 6 + 9 - 81}{27} = \frac{-77}{27}$
$f(1) = 1^3 - 2(1)^2 + 1 - 3 = 1 - 2 + 1 - 3 = -3$
$f(2) = 2^3 - 2(2)^2 + 2 - 3 = 8 - 8 + 2 - 3 = -1$
Comparing these values: $\{-3, \frac{-77}{27}, -3, -1\}$.
The absolute maximum value $M = -1$.
The absolute minimum value $m = -3$.
Therefore,$M + m = -1 + (-3) = -4$.

(B) Given function: $f(x) = x e^{-x}$.
To find the maximum value,we first find the derivative $f'(x)$:
$f'(x) = 1 \cdot e^{-x} + x \cdot (-e^{-x}) = e^{-x}(1-x)$.
Setting $f'(x) = 0$ gives $e^{-x}(1-x) = 0$. Since $e^{-x} \neq 0$ for all $x \in R$,we have $1-x = 0$,which implies $x = 1$.
Now,we find the second derivative $f''(x)$ to check for maxima:
$f''(x) = \frac{d}{dx}[e^{-x} - x e^{-x}] = -e^{-x} - (e^{-x} - x e^{-x}) = e^{-x}(x-2)$.
At $x = 1$,$f''(1) = e^{-1}(1-2) = -e^{-1} = -\frac{1}{e} < 0$.
Since $f''(1) < 0$,the function attains a local maximum at $x = 1$.
Thus,$\alpha = 1$.
The maximum value $\beta = f(1) = 1 \cdot e^{-1} = \frac{1}{e}$.
Therefore,$(\alpha, \beta) = \left(1, \frac{1}{e}\right)$.

(B) Let $r$ be the radius and $h$ be the height of the cylinder inscribed in a sphere of radius $R = 12$.
From the geometry of the sphere,we have the relation $R^2 = r^2 + (h/2)^2$,which gives $12^2 = r^2 + \frac{h^2}{4}$.
Thus,$r^2 = 144 - \frac{h^2}{4}$.
The volume $V$ of the cylinder is given by $V = \pi r^2 h$.
Substituting $r^2$,we get $V = \pi (144 - \frac{h^2}{4}) h = 144 \pi h - \frac{\pi}{4} h^3$.
To find the maximum volume,we differentiate $V$ with respect to $h$: $\frac{dV}{dh} = 144 \pi - \frac{3 \pi}{4} h^2$.
Setting $\frac{dV}{dh} = 0$,we get $144 \pi = \frac{3 \pi}{4} h^2$,which implies $h^2 = 144 \times \frac{4}{3} = 192$.
So,$h = \sqrt{192} = 8 \sqrt{3}$.
Now,$r^2 = 144 - \frac{192}{4} = 144 - 48 = 96$.
The maximum volume is $V = \pi r^2 h = \pi \times 96 \times 8 \sqrt{3} = 768 \sqrt{3} \pi$ cubic units.

(C) Given that $f(x)$ is continuous in $[a, b]$ and $f(c - \delta) < f(c)$ and $f(c + \delta) < f(c)$ for every $\delta > 0$,it follows by the definition of a local maximum that $f(x)$ has a local maximum at $x = c$.
Now,consider the condition $(f(\alpha - \delta) - f(\alpha))(f(\alpha + \delta) - f(\alpha)) < 0$ for all $\alpha \in (a, b)$ where $\alpha \neq c$.
This inequality implies that the two factors $(f(\alpha - \delta) - f(\alpha))$ and $(f(\alpha + \delta) - f(\alpha))$ must have opposite signs.
Case $1$: $f(\alpha - \delta) - f(\alpha) > 0$ and $f(\alpha + \delta) - f(\alpha) < 0$. This implies $f(\alpha - \delta) > f(\alpha)$ and $f(\alpha + \delta) < f(\alpha)$. This indicates that the function is decreasing in the neighborhood of $\alpha$,so $\alpha$ is a point of inflection (neither a local maximum nor a local minimum).
Case $2$: $f(\alpha - \delta) - f(\alpha) < 0$ and $f(\alpha + \delta) - f(\alpha) > 0$. This implies $f(\alpha - \delta) < f(\alpha)$ and $f(\alpha + \delta) > f(\alpha)$. This indicates that the function is increasing in the neighborhood of $\alpha$,so $\alpha$ is also a point of inflection.
Therefore,$f(x)$ has only one local maximum at $c$ and no local extrema at $\alpha$.

(B) Given function is $f(x) = -(x-2)^3(x+2)^2$.
To find the local maximum,we first find the derivative $f'(x)$:
$f'(x) = -[3(x-2)^2(x+2)^2 + (x-2)^3 \cdot 2(x+2)]$
$f'(x) = -(x-2)^2(x+2) [3(x+2) + 2(x-2)]$
$f'(x) = -(x-2)^2(x+2) [3x + 6 + 2x - 4]$
$f'(x) = -(x-2)^2(x+2)(5x + 2)$.
Setting $f'(x) = 0$,we get critical points $x = 2, x = -2, x = -\frac{2}{5}$.
Using the first derivative test:
For $x < -2$,$f'(x) < 0$.
For $-2 < x < -\frac{2}{5}$,$f'(x) > 0$.
For $-\frac{2}{5} < x < 2$,$f'(x) < 0$.
For $x > 2$,$f'(x) < 0$.
Since $f'(x)$ changes from positive to negative at $x = -\frac{2}{5}$,it is a point of local maximum.
The local maximum value is $f(-\frac{2}{5}) = -(-\frac{2}{5}-2)^3(-\frac{2}{5}+2)^2$.
$f(-\frac{2}{5}) = -(-\frac{12}{5})^3(\frac{8}{5})^2 = -(-\frac{1728}{125}) \cdot \frac{64}{25} = \frac{1728 \cdot 64}{3125} = \frac{12^3 \cdot 8^2}{5^5}$.

(B) Given,$f(x)=\frac{6 x^2-18 x+21}{6 x^2-18 x+17}$.
We can rewrite the function as $f(x)=1+\frac{4}{6 x^2-18 x+17}$.
Let $g(x)=6 x^2-18 x+17$. To find the maximum value of $f(x)$,we need to find the minimum value of $g(x)$.
$g'(x)=12 x-18$. Setting $g'(x)=0$,we get $x=\frac{3}{2}$.
The minimum value of $g(x)$ is $g(\frac{3}{2}) = 6(\frac{9}{4}) - 18(\frac{3}{2}) + 17 = \frac{27}{2} - 27 + 17 = \frac{27-54+34}{2} = \frac{7}{2}$.
Thus,the maximum value of $f(x)$ is $m = 1 + \frac{4}{7/2} = 1 + \frac{8}{7} = \frac{15}{7}$.
As $x \to \pm \infty$,$f(x) \to 1$. Since $g(x) > 0$ for all $x$,$f(x) > 1$ for all $x \in R$. Thus,$n = 1$.
Finally,$14 m - 7 n = 14(\frac{15}{7}) - 7(1) = 30 - 7 = 23$.

(D) Let the radius of the cone be $R = \sqrt{3}$ and the semi-vertical angle be $\alpha = \frac{\pi}{3}$.
The height of the cone is $H = R \cot(\alpha) = \sqrt{3} \cot(\frac{\pi}{3}) = \sqrt{3} \times \frac{1}{\sqrt{3}} = 1$.
Let the radius of the inscribed cylinder be $r$ and its height be $h$.
By similar triangles,$\frac{R-r}{h} = \frac{R}{H} = \frac{\sqrt{3}}{1} = \sqrt{3}$.
Thus,$h = \frac{R-r}{\sqrt{3}} = \frac{\sqrt{3}-r}{\sqrt{3}} = 1 - \frac{r}{\sqrt{3}}$.
The volume of the cylinder is $V = \pi r^2 h = \pi r^2 (1 - \frac{r}{\sqrt{3}}) = \pi (r^2 - \frac{r^3}{\sqrt{3}})$.
To maximize $V$,we find $\frac{dV}{dr} = \pi (2r - \frac{3r^2}{\sqrt{3}}) = \pi (2r - \sqrt{3}r^2) = 0$.
Since $r \neq 0$,we have $2 - \sqrt{3}r = 0$,so $r = \frac{2}{\sqrt{3}}$.
The height of the cylinder is $h = 1 - \frac{1}{\sqrt{3}} (\frac{2}{\sqrt{3}}) = 1 - \frac{2}{3} = \frac{1}{3}$.

(B) Given the function $f(x) = 2x^3 - 3x^2 - 36x + 9$ on the interval $[-3, 3]$.
First,find the derivative $f'(x) = 6x^2 - 6x - 36$.
Set $f'(x) = 0$ to find critical points:
$6(x^2 - x - 6) = 0 \Rightarrow 6(x - 3)(x + 2) = 0$.
The critical points are $x = 3$ and $x = -2$.
Both points lie within the interval $[-3, 3]$.
Now,evaluate $f(x)$ at the critical points and the endpoints:
$f(-3) = 2(-3)^3 - 3(-3)^2 - 36(-3) + 9 = -54 - 27 + 108 + 9 = 36$.
$f(-2) = 2(-2)^3 - 3(-2)^2 - 36(-2) + 9 = -16 - 12 + 72 + 9 = 53$.
$f(3) = 2(3)^3 - 3(3)^2 - 36(3) + 9 = 54 - 27 - 108 + 9 = -72$.
The absolute maximum value is the largest of these values,which is $53$.

(A) Let $f(x) = -x^2 + 6x + 12$.
To find the extreme value,we find the derivative $f'(x) = -2x + 6$.
Setting $f'(x) = 0$,we get $-2x + 6 = 0$,which implies $x = 3$.
Thus,$\alpha = 3$.
The extreme value $\beta$ is $f(\alpha) = f(3) = -(3)^2 + 6(3) + 12 = -9 + 18 + 12 = 21$.
Since $\alpha = 3$,we can write $\beta = 21 = 7 \times 3 = 7 \alpha$.
Therefore,$\beta = 7 \alpha$.

(C) Given the function $f(x)=2 x^3-9 x^2+12 x+1$ on the interval $[0,2]$.
First,find the critical points by calculating the derivative $f'(x)$ and setting it to $0$:
$f'(x) = 6x^2 - 18x + 12$
Setting $f'(x) = 0$ gives:
$6(x^2 - 3x + 2) = 0$
$6(x-1)(x-2) = 0$
Thus,the critical points are $x=1$ and $x=2$.
Now,evaluate the function $f(x)$ at the critical points and the endpoints of the interval $[0,2]$:
At $x=0$: $f(0) = 2(0)^3 - 9(0)^2 + 12(0) + 1 = 1$
At $x=1$: $f(1) = 2(1)^3 - 9(1)^2 + 12(1) + 1 = 2 - 9 + 12 + 1 = 6$
At $x=2$: $f(2) = 2(2)^3 - 9(2)^2 + 12(2) + 1 = 16 - 36 + 24 + 1 = 5$
Comparing the values ${1, 6, 5}$,the absolute maximum value is $6$.