Maxima and Minima Questions in English

(C) Given $f(x) = 2\sqrt{2} \sin x - \tan x$ on $[0, \pi/3]$.
First,find the derivative $f'(x) = 2\sqrt{2} \cos x - \sec^2 x$.
Set $f'(x) = 0$ to find critical points:
$2\sqrt{2} \cos x = \frac{1}{\cos^2 x} \implies \cos^3 x = \frac{1}{2\sqrt{2}} = \left(\frac{1}{\sqrt{2}}\right)^3$.
Thus,$\cos x = \frac{1}{\sqrt{2}}$,which gives $x = \pi/4$.
Now,evaluate $f(x)$ at critical points and endpoints:
$f(0) = 2\sqrt{2}(0) - 0 = 0$.
$f(\pi/4) = 2\sqrt{2}(\frac{1}{\sqrt{2}}) - 1 = 2 - 1 = 1$.
$f(\pi/3) = 2\sqrt{2}(\frac{\sqrt{3}}{2}) - \sqrt{3} = \sqrt{6} - \sqrt{3} \approx 2.45 - 1.73 = 0.72$.
Comparing values: $m = 0$ and $M = 1$.
Therefore,$m + M = 0 + 1 = 1$.

(B) To find the turning points,we need to find the points where the derivative $f'(x) = 0$.
Given $f(x) = 2 \cos x - \sin 2x$.
Differentiating with respect to $x$:
$f'(x) = -2 \sin x - 2 \cos 2x$.
Setting $f'(x) = 0$:
$-2 \sin x - 2 \cos 2x = 0$
$\sin x + \cos 2x = 0$.
Using the identity $\cos 2x = 1 - 2 \sin^2 x$:
$\sin x + 1 - 2 \sin^2 x = 0$
$2 \sin^2 x - \sin x - 1 = 0$.
Let $u = \sin x$,then $2u^2 - u - 1 = 0$.
$(2u + 1)(u - 1) = 0$.
So,$\sin x = 1$ or $\sin x = -1/2$.
For $\sin x = 1$ in $[-\pi, \pi]$,$x = \pi/2$.
For $\sin x = -1/2$ in $[-\pi, \pi]$,$x = -\pi/6$ and $x = -5\pi/6$.
Thus,there are $3$ turning points in the given interval.

(C) To find the absolute maximum and minimum values of $f(x) = 2x^3 - 15x^2 + 36x - 30$ on $[-1, 4]$,we first find the critical points by setting $f'(x) = 0$.
$f'(x) = 6x^2 - 30x + 36 = 6(x^2 - 5x + 6) = 6(x - 2)(x - 3)$.
The critical points are $x = 2$ and $x = 3$,both of which lie in $[-1, 4]$.
Now,we evaluate $f(x)$ at the critical points and the endpoints of the interval:
$f(-1) = 2(-1)^3 - 15(-1)^2 + 36(-1) - 30 = -2 - 15 - 36 - 30 = -83$.
$f(2) = 2(8) - 15(4) + 36(2) - 30 = 16 - 60 + 72 - 30 = -2$.
$f(3) = 2(27) - 15(9) + 36(3) - 30 = 54 - 135 + 108 - 30 = -3$.
$f(4) = 2(64) - 15(16) + 36(4) - 30 = 128 - 240 + 144 - 30 = 2$.
The absolute maximum value is $2$ and the absolute minimum value is $-83$.
The difference is $2 - (-83) = 85$.

(A) Let $y = \frac{x^2-x+1}{x^2+x+1}$.
To find the extrema,we differentiate $y$ with respect to $x$:
$\frac{dy}{dx} = \frac{(2x-1)(x^2+x+1) - (2x+1)(x^2-x+1)}{(x^2+x+1)^2} = 0$.
Simplifying the numerator:
$(2x^3 + 2x^2 + 2x - x^2 - x - 1) - (2x^3 - 2x^2 + 2x + x^2 - x + 1) = 0$.
$(2x^3 + x^2 + x - 1) - (2x^3 - x^2 + x + 1) = 0$.
$2x^2 - 2 = 0 \Rightarrow x^2 = 1 \Rightarrow x = \pm 1$.
For $x = -1$,$y = \frac{(-1)^2 - (-1) + 1}{(-1)^2 + (-1) + 1} = \frac{1+1+1}{1-1+1} = \frac{3}{1} = 3$.
For $x = 1$,$y = \frac{1^2 - 1 + 1}{1^2 + 1 + 1} = \frac{1}{3}$.
Thus,the maximum value $\alpha = 3$ and the minimum value $\beta = \frac{1}{3}$.
Therefore,$\alpha + \beta = 3 + \frac{1}{3} = \frac{10}{3}$.

(B) Let the length of the rectangular portion be $x$ and the radius of the semicircular ends be $r$. The total perimeter of the track is given by $P = 2x + 2\pi r = 500 \ ft$.
From this,we get $x + \pi r = 250$,so $x = 250 - \pi r$.
The area of the rectangular portion is $A = x \times (2r) = (250 - \pi r)(2r) = 500r - 2\pi r^2$.
To maximize the area,we find the derivative with respect to $r$ and set it to zero:
$\frac{dA}{dr} = 500 - 4\pi r = 0 \Rightarrow r = \frac{125}{\pi}$.
Now,substitute $r$ back into the expression for $x$:
$x = 250 - \pi \left(\frac{125}{\pi}\right) = 250 - 125 = 125 \ ft$.
Thus,the length of the rectangular portion for maximum area is $125 \ ft$.

(C) Given $f(x) = px^3 + qx^2 + rx + t$.
The derivative is $f'(x) = 3px^2 + 2qx + r$ ... $(i)$
Since $f(x)$ has local extrema at $x = -2$ and $x = 2$,$f'(x) = k(x+2)(x-2) = k(x^2 - 4)$ ... (ii)
Comparing $(i)$ and (ii),we get $3p = k$,$2q = 0$,and $r = -4k$.
Thus,$q = 0$ and $r = -4(3p) = -12p$.
The second derivative is $f''(x) = 6px + 2q = 6px$.
For local minimum at $x = -2$,$f''(-2) = -12p > 0$,which implies $p < 0$.
Given $p$ is a root of $9x^2 - 1 = 0$,so $x = \pm \frac{1}{3}$.
Since $p < 0$,we have $p = -\frac{1}{3}$.
Then $r = -12(-\frac{1}{3}) = 4$.
Therefore,$p + q + r = -\frac{1}{3} + 0 + 4 = \frac{11}{3}$.

(D) Let $l$ and $b$ be the length and width of the rectangle.
Given that the perimeter is $48 \text{ cm}$.
$2(l + b) = 48 \Rightarrow l + b = 24 \Rightarrow b = 24 - l$ ... $(i)$
When the rectangle is rotated about side $b$,the radius of the cylinder is $r = l$ and the height is $h = b$.
The volume of the cylinder is $V = \pi r^2 h = \pi l^2 b$.
Substituting $b$ from $(i)$: $V = \pi l^2(24 - l) = 24\pi l^2 - \pi l^3$.
To find the maximum volume,differentiate with respect to $l$:
$\frac{dV}{dl} = 48\pi l - 3\pi l^2$.
Set $\frac{dV}{dl} = 0$ for critical points:
$3\pi l(16 - l) = 0 \Rightarrow l = 0$ or $l = 16$.
Since $l$ must be positive,$l = 16$.
Check the second derivative: $\frac{d^2V}{dl^2} = 48\pi - 6\pi l$.
At $l = 16$,$\frac{d^2V}{dl^2} = 48\pi - 96\pi = -48\pi < 0$.
Since the second derivative is negative,the volume is maximum at $l = 16$.
Then $b = 24 - 16 = 8$.
Thus,the dimensions of the rectangle are $8 \text{ cm}$ and $16 \text{ cm}$.

(A) Given the function $f(x) = \frac{4}{3}x^3 - 4x$ on the interval $[0, 2]$.
First,find the critical points by setting the derivative to zero:
$f'(x) = 4x^2 - 4 = 0 \Rightarrow x^2 = 1 \Rightarrow x = \pm 1$.
Since we are considering the interval $[0, 2]$,we only consider $x = 1$ as a critical point within the interval.
Now,evaluate the function at the critical point and the endpoints of the interval:
$f(0) = \frac{4}{3}(0)^3 - 4(0) = 0$.
$f(1) = \frac{4}{3}(1)^3 - 4(1) = \frac{4}{3} - 4 = -\frac{8}{3}$.
$f(2) = \frac{4}{3}(2)^3 - 4(2) = \frac{32}{3} - 8 = \frac{32 - 24}{3} = \frac{8}{3}$.
Comparing these values,the global minimum is $-\frac{8}{3}$ and the global maximum is $\frac{8}{3}$.
The sum of the global minimum and global maximum values is $\frac{8}{3} + (-\frac{8}{3}) = 0$.

(C) Given that $A(1,15), B(3,-12), C(6,12)$ are three consecutive turning points of a continuous curve $y=f(x)$ and the curve intersects the $x$-axis at $x=\alpha$ and $x=\beta$.
From the given graph,it is clear that:
$1 < \alpha < 3$ and $3 < \beta < 6$.
We want to find the range for $|\beta-\alpha|$.
Since $1 < \alpha < 3$,we have $-3 < -\alpha < -1$.
Also,$3 < \beta < 6$.
Adding these inequalities,we get:
$3 - 3 < \beta - \alpha < 6 - 1$
$0 < \beta - \alpha < 5$.
Therefore,$|\beta-\alpha| < 5$.

(B) Given $f(x) = x^2 + \frac{54}{x}$.
First,find the derivative: $f'(x) = 2x - \frac{54}{x^2}$.
For $x \in (-\infty, 0)$,$x^2 > 0$ and $x < 0$,so $2x < 0$ and $-\frac{54}{x^2} < 0$. Thus,$f'(x) < 0$,which means $f(x)$ is decreasing in $(-\infty, 0)$.
For critical points,set $f'(x) = 0$: $2x - \frac{54}{x^2} = 0 \Rightarrow 2x^3 = 54 \Rightarrow x^3 = 27 \Rightarrow x = 3$.
Since the only critical point $x = 3$ lies in $(0, \infty)$,there is no critical point in $(-\infty, 0)$.
Therefore,$f(x)$ has neither maximum nor minimum in $(-\infty, 0)$.

(B) Using the $AM$-$GM$ inequality for two positive terms $a^2 x^4$ and $b^2 y^4$:
$\frac{a^2 x^4 + b^2 y^4}{2} \ge \sqrt{(a^2 x^4)(b^2 y^4)}$
Given $a^2 x^4 + b^2 y^4 = c^6$,we substitute:
$\frac{c^6}{2} \ge \sqrt{a^2 b^2 x^4 y^4}$
$\frac{c^6}{2} \ge ab x^2 y^2$
Squaring both sides:
$\frac{c^{12}}{4} \ge a^2 b^2 x^4 y^4$
$x^4 y^4 \le \frac{c^{12}}{4 a^2 b^2}$
Thus,the maximum value is $\frac{c^{12}}{4 a^2 b^2}$.

(B) Given $f(x) = 3x + \frac{12}{x}$.
First,find the derivative: $f'(x) = 3 - \frac{12}{x^2}$.
For critical points,set $f'(x) = 0$:
$3 - \frac{12}{x^2} = 0 \implies x^2 = 4 \implies x = 2, -2$.
Now,find the second derivative: $f''(x) = \frac{24}{x^3}$.
At $x = 2$,$f''(2) = \frac{24}{8} = 3 > 0$ (Local minimum).
At $x = -2$,$f''(-2) = \frac{24}{-8} = -3 < 0$ (Local maximum).
The local maximum value $M = f(-2) = 3(-2) + \frac{12}{-2} = -6 - 6 = -12$.
We need to evaluate $\lim_{x \rightarrow M} f(x) = \lim_{x \rightarrow -12} (3x + \frac{12}{x})$.
Substituting $x = -12$: $3(-12) + \frac{12}{-12} = -36 - 1 = -37$.

(A) Given $f(x) = \frac{4}{\sin x} + \frac{1}{1-\sin x}$.
To find the extreme value,we differentiate $f(x)$ with respect to $x$:
$f'(x) = -\frac{4 \cos x}{\sin^2 x} + \frac{\cos x}{(1-\sin x)^2}$.
Setting $f'(x) = 0$ for critical points:
$-\frac{4 \cos x}{\sin^2 x} + \frac{\cos x}{(1-\sin x)^2} = 0$.
Since $x \in (0, \pi/2)$,$\cos x \neq 0$,so we can divide by $\cos x$:
$-\frac{4}{\sin^2 x} + \frac{1}{(1-\sin x)^2} = 0$.
$\frac{1}{(1-\sin x)^2} = \frac{4}{\sin^2 x}$.
Taking the square root on both sides:
$\frac{1}{1-\sin x} = \frac{2}{\sin x}$ (since $\sin x > 0$ and $1-\sin x > 0$ in the given interval).
$\sin x = 2 - 2 \sin x$.
$3 \sin x = 2 \Rightarrow \sin x = \frac{2}{3}$.
Now,substitute $\sin x = \frac{2}{3}$ into $f(x)$:
$f(x) = \frac{4}{2/3} + \frac{1}{1-2/3} = 4 \times \frac{3}{2} + \frac{1}{1/3} = 6 + 3 = 9$.

(A) Let the height of the cone be $h$ and the radius of its base be $r$. The sphere has radius $R$. From the geometry of the cone inscribed in a sphere,we have the relation $r^2 = R^2 - (h - R)^2 = 2hR - h^2$.
The volume of the cone is $V = \frac{1}{3} \pi r^2 h = \frac{1}{3} \pi (2hR - h^2) h = \frac{1}{3} \pi (2Rh^2 - h^3)$.
To find the maximum volume,we differentiate $V$ with respect to $h$ and set it to zero:
$\frac{dV}{dh} = \frac{1}{3} \pi (4Rh - 3h^2) = 0$.
Since $h \neq 0$,we have $4R - 3h = 0$,which gives $h = \frac{4R}{3}$.
Thus,$k = \frac{4}{3}$.
The volume of the cone at $h = \frac{4R}{3}$ is $V_{cone} = \frac{1}{3} \pi (2R(\frac{4R}{3})^2 - (\frac{4R}{3})^3) = \frac{1}{3} \pi (\frac{32R^3}{9} - \frac{64R^3}{27}) = \frac{1}{3} \pi (\frac{96R^3 - 64R^3}{27}) = \frac{32}{81} \pi R^3$.
The volume of the sphere is $V_{sphere} = \frac{4}{3} \pi R^3$.
The ratio of the volume of the cone to the volume of the sphere is $\frac{\frac{32}{81} \pi R^3}{\frac{4}{3} \pi R^3} = \frac{32}{81} \times \frac{3}{4} = \frac{8}{27}$.

(C) Given function is $f(x) = \frac{x}{4 + x + x^2}$.
To find the maximum value,we differentiate $f(x)$ with respect to $x$ using the quotient rule:
$f'(x) = \frac{(4 + x + x^2)(1) - x(1 + 2x)}{(4 + x + x^2)^2}$
$f'(x) = \frac{4 + x + x^2 - x - 2x^2}{(4 + x + x^2)^2} = \frac{4 - x^2}{(4 + x + x^2)^2}$.
For critical points,set $f'(x) = 0$,which implies $4 - x^2 = 0$,so $x = \pm 2$.
Since $x = \pm 2$ are not in the interval $[-1, 1]$,the function has no critical points in the given interval.
Since $4 - x^2 > 0$ for all $x \in [-1, 1]$,$f'(x) > 0$,meaning $f(x)$ is strictly increasing on $[-1, 1]$.
The maximum value occurs at the right endpoint $x = 1$:
$f(1) = \frac{1}{4 + 1 + 1^2} = \frac{1}{6}$.

(D) Given function: $f(x) = x + \frac{4}{x + 2}$ for $x > -2$.
To find the critical points,we calculate the derivative $f'(x)$ and set it to $0$:
$f'(x) = 1 - \frac{4}{(x + 2)^2}$
Setting $f'(x) = 0$:
$1 = \frac{4}{(x + 2)^2} \implies (x + 2)^2 = 4$
Since $x > -2$,we have $x + 2 = 2$,which gives $x = 0$.
Now,we check the second derivative $f''(x) = \frac{8}{(x + 2)^3}$.
At $x = 0$,$f''(0) = \frac{8}{2^3} = 1 > 0$.
Since $f''(0) > 0$,the function has a local minimum at $x = 0$.
The minimum value is $f(0) = 0 + \frac{4}{0 + 2} = \frac{4}{2} = 2$.

(D) Given the function $f(x) = ax^3 + bx^2 + cx + d$.
To find the extreme values,we first find the derivative of $f(x)$ with respect to $x$:
$f'(x) = 3ax^2 + 2bx + c$.
$A$ function has no extreme values if its derivative $f'(x)$ does not change sign,which means $f'(x) = 0$ has no real roots or has equal roots such that the sign does not change.
For the quadratic equation $3ax^2 + 2bx + c = 0$ to have no real roots,its discriminant $D$ must be less than $0$.
$D = (2b)^2 - 4(3a)(c) < 0$.
$4b^2 - 12ac < 0$.
Dividing by $4$,we get $b^2 - 3ac < 0$,which implies $b^2 < 3ac$.

(A) Let the height of the closed cylinder be $h$ and the base radius be $r$.
The volume $V$ of the cylinder is given by $V = \pi r^2 h$,which implies $h = \frac{V}{\pi r^2}$.
The total surface area $S$ of the closed cylinder is $S = 2 \pi r h + 2 \pi r^2$.
Substituting the value of $h$ in the surface area formula: $S = 2 \pi r (\frac{V}{\pi r^2}) + 2 \pi r^2 = \frac{2V}{r} + 2 \pi r^2$.
To find the minimum surface area,we differentiate $S$ with respect to $r$ and set it to zero: $\frac{dS}{dr} = -\frac{2V}{r^2} + 4 \pi r = 0$.
This gives $4 \pi r = \frac{2V}{r^2}$,so $V = 2 \pi r^3$.
Substituting $V = \pi r^2 h$ into this equation: $\pi r^2 h = 2 \pi r^3$.
Dividing both sides by $\pi r^2$,we get $h = 2r$,which means the ratio $h : r = 2 : 1$.

(A) The coordinates of the particles are $P(t, t^3 - 16t - 3)$ and $Q(t + 1, t^3 - 6t - 6)$.
The distance $PQ$ between the two particles is given by the distance formula:
$PQ = \sqrt{((t + 1) - t)^2 + ((t^3 - 6t - 6) - (t^3 - 16t - 3))^2}$
$PQ = \sqrt{(1)^2 + (t^3 - 6t - 6 - t^3 + 16t + 3)^2}$
$PQ = \sqrt{1 + (10t - 3)^2}$
To find the minimum distance,we minimize the square of the distance,$y = PQ^2 = 1 + (10t - 3)^2$.
For the minimum value,we set the derivative with respect to $t$ to zero:
$\frac{dy}{dt} = 2(10t - 3) \times 10 = 20(10t - 3) = 0$
This gives $10t - 3 = 0$,or $t = \frac{3}{10}$.
Substituting $t = \frac{3}{10}$ into the expression for $PQ$:
$PQ_{min} = \sqrt{1 + (10(\frac{3}{10}) - 3)^2} = \sqrt{1 + (3 - 3)^2} = \sqrt{1 + 0} = 1$.
Thus,the minimum distance is $1$.

(A) Given the function $y = \frac{b^2}{a-x} + \frac{a^2}{x}$.
To find the minimum,we differentiate with respect to $x$:
$\frac{dy}{dx} = \frac{b^2}{(a-x)^2} - \frac{a^2}{x^2}$.
Setting $\frac{dy}{dx} = 0$,we get $\frac{b^2}{(a-x)^2} = \frac{a^2}{x^2}$,which implies $\frac{b}{a-x} = \pm \frac{a}{x}$.
Since $0 < x < a$ and $a, b > 0$,we take the positive root: $\frac{b}{a-x} = \frac{a}{x} \Rightarrow bx = a^2 - ax \Rightarrow x(a+b) = a^2 \Rightarrow x = \frac{a^2}{a+b}$.
Using the second derivative test: $\frac{d^2y}{dx^2} = \frac{2b^2}{(a-x)^3} + \frac{2a^2}{x^3}$.
Since $x = \frac{a^2}{a+b}$ lies in $(0, a)$,both terms are positive,so $\frac{d^2y}{dx^2} > 0$,confirming a minimum.
Substituting $x = \frac{a^2}{a+b}$ into $y$:
$y_{\min} = \frac{b^2}{a - \frac{a^2}{a+b}} + \frac{a^2}{\frac{a^2}{a+b}} = \frac{b^2}{\frac{a^2+ab-a^2}{a+b}} + (a+b) = \frac{b^2(a+b)}{ab} + (a+b) = \frac{b(a+b)}{a} + (a+b) = (a+b)(\frac{b}{a} + 1) = (a+b)(\frac{a+b}{a}) = \frac{(a+b)^2}{a}$.

(A) Let the right-angled triangle be $ABC$ with hypotenuse $AC = h$. Let $\angle A = \theta$.
Then the sides are $AB = h \cos \theta$ and $BC = h \sin \theta$.
The area $A$ of the triangle is given by $A = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} (h \cos \theta)(h \sin \theta)$.
$A = \frac{h^2}{2} \sin \theta \cos \theta = \frac{h^2}{4} (2 \sin \theta \cos \theta) = \frac{h^2}{4} \sin 2\theta$.
For the area to be maximum,$\sin 2\theta$ must be maximum,which is $1$ when $2\theta = 90^{\circ}$ or $\theta = 45^{\circ}$.
Thus,the maximum area is $\frac{h^2}{4} \times 1 = \frac{h^2}{4}$.

(D) Given the function $f(x)=a \sin x+\frac{1}{3} \sin 3 x$.
To find the extremum,we first find the derivative $f'(x)$:
$f'(x) = \frac{d}{dx}(a \sin x + \frac{1}{3} \sin 3x) = a \cos x + \frac{1}{3} \cdot 3 \cos 3x = a \cos x + \cos 3x$.
Since the function has an extremum at $x = \frac{\pi}{3}$,we must have $f'(\frac{\pi}{3}) = 0$.
Substituting $x = \frac{\pi}{3}$ into the derivative:
$a \cos(\frac{\pi}{3}) + \cos(3 \cdot \frac{\pi}{3}) = 0$.
$a(\frac{1}{2}) + \cos(\pi) = 0$.
Since $\cos(\pi) = -1$,we have:
$\frac{a}{2} - 1 = 0$.
$\frac{a}{2} = 1 \implies a = 2$.

(D) Let the two given sides of the triangle be $a$ and $b$,and let $\theta$ be the angle between them.
The area $A$ of the triangle is given by the formula $A = \frac{1}{2} ab \sin \theta$.
Since the sides $a$ and $b$ are fixed,the area $A$ depends only on $\sin \theta$.
For the area to be maximum,$\sin \theta$ must be at its maximum value.
The maximum value of $\sin \theta$ is $1$,which occurs when $\theta = 90^{\circ}$ or $\frac{\pi}{2} \text{ radians}$.
Therefore,the angle between the given sides for the maximum area is $90^{\circ}$.

(B) The velocity $V$ of the particle is the rate of change of displacement with respect to time:
$V = \frac{dS}{dt} = \frac{d}{dt}(6t - \frac{t^3}{2}) = 6 - \frac{3}{2}t^2$
To find the maximum velocity,we analyze the derivative of velocity with respect to time:
$\frac{dV}{dt} = \frac{d}{dt}(6 - \frac{3}{2}t^2) = -3t$
Setting $\frac{dV}{dt} = 0$ gives $t = 0$.
Since $\frac{d^2V}{dt^2} = -3 < 0$,the velocity is maximum at $t = 0$.
Substituting $t = 0$ into the velocity equation:
$V_{max} = 6 - \frac{3}{2}(0)^2 = 6$.

(B) Given $f(x)=x^2+\frac{1}{x^2}$ and $g(x)=x-\frac{1}{x}$.
We can write $f(x)$ in terms of $g(x)$ as $f(x)=\left(x-\frac{1}{x}\right)^2+2$.
Let $t=x-\frac{1}{x}$. Then $f(x)=t^2+2$ and $g(x)=t$.
We define $h(t)=\frac{f(x)}{g(x)}=\frac{t^2+2}{t}=t+\frac{2}{t}$.
To find the local minimum,we find the derivative $h'(t)=1-\frac{2}{t^2}$.
Setting $h'(t)=0$,we get $1-\frac{2}{t^2}=0$,which implies $t^2=2$,so $t=\pm \sqrt{2}$.
Using the second derivative test,$h''(t)=\frac{4}{t^3}$.
For $t=\sqrt{2}$,$h''(\sqrt{2})=\frac{4}{(\sqrt{2})^3} > 0$,so $t=\sqrt{2}$ is a point of local minimum.
The local minimum value is $h(\sqrt{2})=\sqrt{2}+\frac{2}{\sqrt{2}}=\sqrt{2}+\sqrt{2}=2\sqrt{2}$.

(D) Let $r$ be the radius and $\theta$ be the sectorial angle in radians. The perimeter $P$ of the sector is given by $P = 2r + r\theta = r(2 + \theta)$.
Since the perimeter is constant,let $P = k$,where $k$ is a constant.
Thus,$r(2 + \theta) = k$,which implies $r = \frac{k}{2 + \theta}$.
The area $A$ of the sector is given by $A = \frac{1}{2} r^2 \theta$.
Substituting the value of $r$,we get $A = \frac{1}{2} \left(\frac{k}{2 + \theta}\right)^2 \theta = \frac{k^2}{2} \cdot \frac{\theta}{(2 + \theta)^2}$.
To find the maximum area,we differentiate $A$ with respect to $\theta$:
$\frac{dA}{d\theta} = \frac{k^2}{2} \left[ \frac{(2 + \theta)^2(1) - \theta(2)(2 + \theta)}{(2 + \theta)^4} \right] = \frac{k^2}{2} \cdot \frac{(2 + \theta) - 2\theta}{(2 + \theta)^3} = \frac{k^2}{2} \cdot \frac{2 - \theta}{(2 + \theta)^3}$.
Setting $\frac{dA}{d\theta} = 0$,we get $2 - \theta = 0$,which implies $\theta = 2$.
For $\theta = 2$,the second derivative $\frac{d^2A}{d\theta^2}$ is negative,confirming that the area is maximum at $\theta = 2^c$.

(B) Given the function $f(x)=2 x^3-9 a x^2+12 a^2 x+1$ with $a>0$.
First,find the derivative $f^{\prime}(x)$:
$f^{\prime}(x)=6 x^2-18 a x+12 a^2$
$f^{\prime}(x)=6(x^2-3 a x+2 a^2)=6(x-a)(x-2 a)$
For local maxima or minima,set $f^{\prime}(x)=0$:
$6(x-a)(x-2 a)=0 \Rightarrow x=a, 2 a$
Using the second derivative test,$f^{\prime\prime}(x)=12 x-18 a$:
At $x=a$,$f^{\prime\prime}(a)=12 a-18 a=-6 a < 0$ (since $a>0$),so $x=a$ is a point of local maxima.
At $x=2 a$,$f^{\prime\prime}(2 a)=24 a-18 a=6 a > 0$ (since $a>0$),so $x=2 a$ is a point of local minima.
Thus,$p=a$ and $q=2 a$.
Given the condition $p^2=q$,we have:
$a^2=2 a$
$a^2-2 a=0$
$a(a-2)=0$
Since $a>0$,we get $a=2$.

(A) Let the point on the curve be $P(x, y) = (x, x^2-4)$.
The square of the distance $D$ from the origin $(0,0)$ is $S = x^2 + y^2 = x^2 + (x^2-4)^2$.
$S = x^2 + x^4 - 8x^2 + 16 = x^4 - 7x^2 + 16$.
To find the minimum,differentiate $S$ with respect to $x$ and set it to zero:
$\frac{dS}{dx} = 4x^3 - 14x = 0$.
$2x(2x^2 - 7) = 0$.
This gives $x = 0$ or $x^2 = \frac{7}{2}$.
If $x = 0$,$S = 16$. If $x^2 = \frac{7}{2}$,$S = (\frac{7}{2})^2 - 7(\frac{7}{2}) + 16 = \frac{49}{4} - \frac{49}{2} + 16 = 16 - \frac{49}{4} = \frac{64-49}{4} = \frac{15}{4}$.
The minimum distance is $\sqrt{S} = \sqrt{\frac{15}{4}} = \frac{\sqrt{15}}{2}$.

(A) Let the radius of the base be $R$ and the height be $H$. The surface area $A$ of an open cylinder is given by $A = 2\pi RH + \pi R^2$.
From this,we can express $H$ as $H = \frac{A - \pi R^2}{2\pi R}$.
The volume $V$ is given by $V = \pi R^2 H = \pi R^2 \left( \frac{A - \pi R^2}{2\pi R} \right) = \frac{R}{2}(A - \pi R^2) = \frac{AR}{2} - \frac{\pi R^3}{2}$.
To find the maximum volume,we differentiate $V$ with respect to $R$: $\frac{dV}{dR} = \frac{A}{2} - \frac{3\pi R^2}{2}$.
Setting $\frac{dV}{dR} = 0$,we get $A = 3\pi R^2$.
Checking the second derivative,$\frac{d^2V}{dR^2} = -3\pi R$,which is negative for $R > 0$,confirming a maximum.
Substituting $A = 3\pi R^2$ into the surface area formula: $3\pi R^2 = 2\pi RH + \pi R^2$.
This simplifies to $2\pi R^2 = 2\pi RH$,which implies $R = H$.
Thus,the radius is equal to the height of the cylinder.

(C) Given $x+y=k$,where $x>0$ and $y>0$.
We can write $y=k-x$.
Let $f(x) = x^2+y^2 = x^2+(k-x)^2$.
Expanding the expression: $f(x) = x^2+k^2-2kx+x^2 = 2x^2-2kx+k^2$.
To find the minimum,we differentiate $f(x)$ with respect to $x$ and set it to zero:
$f'(x) = 4x-2k = 0 \Rightarrow x = \frac{k}{2}$.
Now,check the second derivative:
$f''(x) = 4 > 0$.
Since the second derivative is positive,the function has a minimum at $x = \frac{k}{2}$.
Substituting $x = \frac{k}{2}$ into $y = k-x$,we get $y = k - \frac{k}{2} = \frac{k}{2}$.
Thus,$x=y=\frac{k}{2}$ gives the minimum value of $x^2+y^2$.

(B) Let the length and breadth of the rectangle be $x$ and $y$ respectively. The rectangle is inscribed in a circle of radius $R = 10 \text{ cm}$.
The diagonal of the rectangle is equal to the diameter of the circle,so $x^2 + y^2 = (2R)^2 = (20)^2 = 400$.
The area of the rectangle is $A = xy$. To maximize the area,we maximize $A^2 = x^2y^2$.
Let $u = x^2$ and $v = y^2$,then $u + v = 400$. We want to maximize $uv$.
By the $AM$-$GM$ inequality,$\frac{u+v}{2} \geq \sqrt{uv}$,so $\sqrt{uv} \leq \frac{400}{2} = 200$.
Thus,$uv \leq (200)^2 = 40000$.
Alternatively,for a rectangle inscribed in a circle,the area is maximum when it is a square.
If it is a square with side $a$,then $a^2 + a^2 = (20)^2 \Rightarrow 2a^2 = 400 \Rightarrow a^2 = 200$.
The area of the square is $a^2 = 200 \text{ cm}^2$.

(C) Given $f(x)=\cos x-1+\frac{x^2}{2}-\frac{x^3}{3}$.
First,find the first derivative $f'(x)$:
$f'(x) = -\sin x + x - x^2$.
Next,find the second derivative $f''(x)$:
$f''(x) = -\cos x + 1 - 2x$.
Evaluate at $x=0$:
$f'(0) = -\sin(0) + 0 - 0^2 = 0$.
$f''(0) = -\cos(0) + 1 - 2(0) = -1 + 1 = 0$.
Since $f'(0)=0$ and $f''(0)=0$,we check the third derivative $f'''(x)$:
$f'''(x) = \sin x - 2$.
Evaluate at $x=0$:
$f'''(0) = \sin(0) - 2 = -2$.
Since the first non-zero derivative at $x=0$ is of odd order (the third derivative),$x=0$ is a point of inflection and not an extremum point.
Therefore,$f(x)$ has no extremum value at $x=0$.

(B) Given,$a > 0$.
$f(x) = 2 x^3 - 9 a x^2 + 12 a^2 x + 1$.
Find the derivative: $f'(x) = 6 x^2 - 18 a x + 12 a^2$.
Factor the derivative: $f'(x) = 6 (x^2 - 3 a x + 2 a^2) = 6 (x - 2 a) (x - a)$.
Setting $f'(x) = 0$ gives critical points $x = a$ and $x = 2 a$.
Find the second derivative: $f''(x) = 6 (2 x - 3 a) = 12 x - 18 a$.
Check for local maxima and minima:
At $x = a$,$f''(a) = 6 (2 a - 3 a) = -6 a < 0$ (since $a > 0$),so $\alpha = a$ is the point of local maximum.
At $x = 2 a$,$f''(2 a) = 6 (4 a - 3 a) = 6 a > 0$ (since $a > 0$),so $\beta = 2 a$ is the point of local minimum.
Given the condition $2 \alpha + \beta = 8$,substitute $\alpha = a$ and $\beta = 2 a$:
$2(a) + 2 a = 8$.
$4 a = 8$.
$a = 2$.

(B) Given: $xy = 6$
$\implies y = \frac{6}{x}$
Let $f(x) = 2x + 3y = 2x + 3 \left( \frac{6}{x} \right) = 2x + \frac{18}{x}$
To find the minimum value,differentiate $f(x)$ with respect to $x$:
$f'(x) = 2 - \frac{18}{x^2}$
Set $f'(x) = 0$ for critical points:
$2 - \frac{18}{x^2} = 0 \implies x^2 = 9 \implies x = 3$ (assuming $x > 0$ for positive values)
Now,find the second derivative:
$f''(x) = \frac{36}{x^3}$
At $x = 3$,$f''(3) = \frac{36}{27} > 0$,which confirms a local minimum.
The minimum value is $f(3) = 2(3) + \frac{18}{3} = 6 + 6 = 12$.

(C) Given $f(x) = 2x^3 - 9ax^2 + 12a^2x + 1$.
Differentiating $f(x)$ with respect to $x$,we get $f'(x) = 6x^2 - 18ax + 12a^2$.
For maximum or minimum,we set $f'(x) = 0$:
$6(x^2 - 3ax + 2a^2) = 0$
$6(x - a)(x - 2a) = 0$
Thus,the critical points are $x = a$ and $x = 2a$.
Now,we find the second derivative $f''(x) = 12x - 18a$.
At $x = a$,$f''(a) = 12a - 18a = -6a$. Since $a > 0$,$f''(a) < 0$,so $f(x)$ has a maximum at $p = a$.
At $x = 2a$,$f''(2a) = 12(2a) - 18a = 6a$. 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:
$a^2 = 2a$
$a^2 - 2a = 0$
$a(a - 2) = 0$
Since $a > 0$,we must have $a = 2$.

(B) Let $f(x) = x^4-x^2-2x+5$.
Find the derivative: $f'(x) = 4x^3-2x-2$.
Set $f'(x) = 0$ to find critical points: $4x^3-2x-2 = 0 \Rightarrow 2x^3-x-1 = 0$.
Factoring the cubic equation: $2x^3-2x^2+2x^2-2x+x-1 = 0 \Rightarrow 2x^2(x-1) + 2x(x-1) + 1(x-1) = 0 \Rightarrow (x-1)(2x^2+2x+1) = 0$.
The real root is $x = 1$. The quadratic factor $2x^2+2x+1$ has a negative discriminant $(D = 4-8 = -4)$,so it yields no real roots.
Check the second derivative: $f''(x) = 12x^2-2$.
At $x = 1$,$f''(1) = 12(1)^2-2 = 10 > 0$.
Since $f''(1) > 0$,the function has a local minimum at $x = 1$.
As $x \to \pm \infty$,$f(x) \to \infty$,so the local minimum is the absolute minimum.
Calculating the value: $f(1) = (1)^4-(1)^2-2(1)+5 = 1-1-2+5 = 3$.
Thus,the absolute minimum value is $3$.

(B) Given,the perimeter of the circular sector is $P = 60 \ m$.
Let the radius be $r \ m$ and the arc length be $l \ m$.
The perimeter of a circular sector is given by $P = l + 2r$.
Substituting the given perimeter,we have $60 = l + 2r$,which implies $l = 60 - 2r$.
The area $A$ of a circular sector is given by $A = \frac{1}{2}lr$.
Substituting $l = 60 - 2r$ into the area formula,we get:
$A = \frac{1}{2}(60 - 2r)r = 30r - r^2$.
To maximize the area,we find the derivative of $A$ with respect to $r$ and set it to zero:
$\frac{dA}{dr} = \frac{d}{dr}(30r - r^2) = 30 - 2r$.
Setting $\frac{dA}{dr} = 0$ gives $30 - 2r = 0$,so $2r = 30$,which means $r = 15 \ m$.
To confirm this is a maximum,we check the second derivative: $\frac{d^2A}{dr^2} = -2$,which is less than $0$,confirming that $r = 15 \ m$ provides the maximum area.

(D) Given $y = x(\log x)^2$.
To find the critical points,we differentiate $y$ with respect to $x$:
$\frac{dy}{dx} = x \cdot (2 \log x \cdot \frac{1}{x}) + (\log x)^2 \cdot 1 = 2 \log x + (\log x)^2$.
Setting $\frac{dy}{dx} = 0$:
$\log x (2 + \log x) = 0$.
This gives $\log x = 0$ or $\log x = -2$.
Thus,$x = e^0 = 1$ or $x = e^{-2}$.
Now,we use the second derivative test:
$\frac{d^2y}{dx^2} = \frac{2}{x} + 2 \log x \cdot \frac{1}{x} = \frac{2 + 2 \log x}{x}$.
At $x = 1$,$\frac{d^2y}{dx^2} = \frac{2 + 0}{1} = 2 > 0$,so $x = 1$ is a point of local minima.
At $x = e^{-2}$,$\frac{d^2y}{dx^2} = \frac{2 + 2(-2)}{e^{-2}} = \frac{-2}{e^{-2}} < 0$,so $x = e^{-2}$ is a point of local maxima.
The maximum value is $y(e^{-2}) = e^{-2}(\log e^{-2})^2 = e^{-2}(-2)^2 = 4e^{-2}$.

(C) Given the function $f(x) = x^3 - 6x^2 - 12x - 3$.
First,find the first derivative: $f'(x) = 3x^2 - 12x - 12$.
Next,find the second derivative: $f''(x) = 6x - 12$.
Evaluate the second derivative at $x = 2$: $f''(2) = 6(2) - 12 = 12 - 12 = 0$.
Since $f'(2) = 3(2)^2 - 12(2) - 12 = 12 - 24 - 12 = -24 \neq 0$,$x=2$ is not a critical point.
However,checking the options provided,the question implies evaluating the nature of the point where $f''(x)=0$.
Since $f''(x)$ changes sign at $x=2$ (for $x < 2$,$f''(x) < 0$ and for $x > 2$,$f''(x) > 0$),$x=2$ is a point of inflection.

(B) Let $f(x) = x^{40} - x^{20}$ on the interval $[0, 1]$.
To find the absolute maximum,we first find the critical points by setting the derivative $f'(x) = 0$.
$f'(x) = 40x^{39} - 20x^{19} = 20x^{19}(2x^{20} - 1)$.
Setting $f'(x) = 0$,we get $x = 0$ or $2x^{20} = 1$,which implies $x^{20} = \frac{1}{2}$,so $x = (\frac{1}{2})^{1/20}$.
Now,evaluate $f(x)$ at the critical point and the endpoints $x=0$ and $x=1$:
$f(0) = 0^{40} - 0^{20} = 0$.
$f(1) = 1^{40} - 1^{20} = 1 - 1 = 0$.
$f((\frac{1}{2})^{1/20}) = ((\frac{1}{2})^{1/20})^{40} - ((\frac{1}{2})^{1/20})^{20} = (\frac{1}{2})^2 - (\frac{1}{2})^1 = \frac{1}{4} - \frac{1}{2} = -\frac{1}{4}$.
Comparing the values $f(0)=0$,$f(1)=0$,and $f((\frac{1}{2})^{1/20}) = -\frac{1}{4}$,the absolute maximum value is $0$.

(B) Let the sides of the rectangle be $x$ and $y$.
Given,the perimeter is $20$ units.
$2(x + y) = 20 \Rightarrow x + y = 10 \Rightarrow y = 10 - x$.
The area $A$ of the rectangle is given by $A = xy$.
Substituting $y$,we get $A = x(10 - x) = 10x - x^2$.
To find the maximum area,we differentiate $A$ with respect to $x$:
$\frac{dA}{dx} = 10 - 2x$.
Setting $\frac{dA}{dx} = 0$,we get $10 - 2x = 0$,which implies $x = 5$.
Now,we check the second derivative: $\frac{d^2A}{dx^2} = -2$.
Since $\frac{d^2A}{dx^2} < 0$,the area is maximum at $x = 5$.
When $x = 5$,$y = 10 - 5 = 5$.
Thus,the maximum area is $A = 5 \times 5 = 25$ sq units.

(A) Given $f(x) = x^5 - 5x^4 + 5x^3 - 10$.
First,find the derivative $f'(x)$:
$f'(x) = 5x^4 - 20x^3 + 15x^2$.
For local maxima and minima,set $f'(x) = 0$:
$5x^2(x^2 - 4x + 3) = 0
\Rightarrow 5x^2(x - 1)(x - 3) = 0$.
The critical points are $x = 0, 1, 3$.
Now,find the second derivative $f''(x)$:
$f''(x) = 20x^3 - 60x^2 + 30x$.
Check the nature of the critical points:
For $x = 1$: $f''(1) = 20(1)^3 - 60(1)^2 + 30(1) = 20 - 60 + 30 = -10 < 0$. Thus,$x = 1$ is a point of local maxima. So,$a = 1$.
For $x = 3$: $f''(3) = 20(27) - 60(9) + 30(3) = 540 - 540 + 90 = 90 > 0$. Thus,$x = 3$ is a point of local minima. So,$b = 3$.
For $x = 0$: $f''(0) = 0$,which is a point of inflection.
Finally,calculate $2a + b$:
$2a + b = 2(1) + 3 = 5$.

(C) Given function: $f(x) = a \sin(x) + \frac{1}{3} \sin(3x)$.
First,find the derivative $f'(x)$:
$f'(x) = a \cos(x) + \frac{1}{3} \cos(3x) \cdot 3 = a \cos(x) + \cos(3x)$.
Since the function attains its maximum at $x = \frac{\pi}{3}$,the first derivative must be zero at this point:
$f'\left(\frac{\pi}{3}\right) = 0$.
Substituting $x = \frac{\pi}{3}$ into the derivative:
$a \cos\left(\frac{\pi}{3}\right) + \cos\left(3 \cdot \frac{\pi}{3}\right) = 0$.
$a \cdot \frac{1}{2} + \cos(\pi) = 0$.
Since $\cos(\pi) = -1$,we have:
$\frac{a}{2} - 1 = 0$.
$\frac{a}{2} = 1$.
$a = 2$.

(B) Given,$y = \frac{ax - b}{(x - 1)(x - 4)} . . . . . . (i)$ has a turning point $P(2, -1)$.
Since point $P$ lies on the curve,it must satisfy equation $(i)$:
$-1 = \frac{2a - b}{(2 - 1)(2 - 4)} = \frac{2a - b}{-2}$
$2a - b = 2 . . . . . . (ii)$
At a turning point,the derivative $\frac{dy}{dx} = 0$.
From equation $(i)$,$y(x^2 - 5x + 4) = ax - b$.
Differentiating with respect to $x$ using the product rule:
$y(2x - 5) + (x^2 - 5x + 4) \frac{dy}{dx} = a$.
At $x = 2$ and $y = -1$,$\frac{dy}{dx} = 0$:
$-1(2(2) - 5) + (2^2 - 5(2) + 4)(0) = a$
$-1(4 - 5) = a$
$a = 1$.
Substituting $a = 1$ into equation $(ii)$:
$2(1) - b = 2$
$b = 0$.
Thus,$a = 1$ and $b = 0$.

(B) Let $R = 22 \ cm$ be the radius of the sphere and $h$ be the height of the cylinder inscribed in it. Let $r$ be the radius of the cylinder.
From the geometry of the sphere,we have $r^2 + (h/2)^2 = R^2$,which implies $r^2 = R^2 - h^2/4$.
The curved surface area $A$ of the cylinder is given by $A = 2\pi rh = 2\pi h \sqrt{R^2 - h^2/4} = \pi h \sqrt{4R^2 - h^2}$.
To maximize $A$,we maximize $A^2 = \pi^2 h^2 (4R^2 - h^2) = \pi^2 (4R^2h^2 - h^4)$.
Let $f(h) = 4R^2h^2 - h^4$. Differentiating with respect to $h$,we get $f'(h) = 8R^2h - 4h^3$.
Setting $f'(h) = 0$,we get $4h(2R^2 - h^2) = 0$. Since $h > 0$,we have $h^2 = 2R^2$,so $h = R\sqrt{2}$.
Given $R = 22 \ cm$,the height $h = 22\sqrt{2} \ cm$.

(A) Given function is $f(x) = -x + \sin 2x$ on the interval $[-\frac{\pi}{2}, \frac{\pi}{2}]$.
First,find the critical points by setting $f'(x) = 0$.
$f'(x) = -1 + 2 \cos 2x = 0$
$\cos 2x = \frac{1}{2}$
Since $x \in [-\frac{\pi}{2}, \frac{\pi}{2}]$,$2x \in [-\pi, \pi]$.
Thus,$2x = \pm \frac{\pi}{3}$,which gives $x = \pm \frac{\pi}{6}$.
Now,evaluate $f(x)$ at the critical points and the endpoints:
$f(-\frac{\pi}{2}) = -(-\frac{\pi}{2}) + \sin(-\pi) = \frac{\pi}{2} + 0 = \frac{\pi}{2} \approx 1.57$
$f(-\frac{\pi}{6}) = -(-\frac{\pi}{6}) + \sin(-\frac{\pi}{3}) = \frac{\pi}{6} - \frac{\sqrt{3}}{2} \approx 0.52 - 0.866 = -0.346$
$f(\frac{\pi}{6}) = -(\frac{\pi}{6}) + \sin(\frac{\pi}{3}) = -\frac{\pi}{6} + \frac{\sqrt{3}}{2} \approx -0.52 + 0.866 = 0.346$
$f(\frac{\pi}{2}) = -(\frac{\pi}{2}) + \sin(\pi) = -\frac{\pi}{2} + 0 = -\frac{\pi}{2} \approx -1.57$
Comparing these values,the greatest value is $M = \frac{\pi}{2}$ and the least value is $m = -\frac{\pi}{2}$.
The difference is $M - m = \frac{\pi}{2} - (-\frac{\pi}{2}) = \pi$.

(C) Let the two positive numbers be $x$ and $y$. Given that $x + y = t$,so $y = t - x$.
We want to minimize the sum of their squares,$S = x^2 + y^2$.
Substituting $y = t - x$,we get $S(x) = x^2 + (t - x)^2 = x^2 + t^2 - 2tx + x^2 = 2x^2 - 2tx + t^2$.
To find the minimum,we take the derivative with respect to $x$: $\frac{dS}{dx} = 4x - 2t$.
Setting $\frac{dS}{dx} = 0$,we get $4x = 2t$,which implies $x = \frac{t}{2}$.
Since $\frac{d^2S}{dx^2} = 4 > 0$,the function has a minimum at $x = \frac{t}{2}$.
Then $y = t - \frac{t}{2} = \frac{t}{2}$.
Thus,the two numbers are $\frac{t}{2}$ and $\frac{t}{2}$.

(B) To find the absolute minimum of $f(x) = x^{40} - x^{20}$ on the interval $[0, 1]$,we first find the critical points by setting the derivative $f'(x) = 0$.
$f'(x) = 40x^{39} - 20x^{19} = 20x^{19}(2x^{20} - 1)$.
Setting $f'(x) = 0$,we get $x = 0$ or $2x^{20} = 1$,which implies $x^{20} = 1/2$,so $x = (1/2)^{1/20}$.
Since $(1/2)^{1/20}$ is in the interval $(0, 1)$,we evaluate $f(x)$ at the critical point and the endpoints $x = 0$ and $x = 1$.
$f(0) = 0^{40} - 0^{20} = 0$.
$f(1) = 1^{40} - 1^{20} = 0$.
$f((1/2)^{1/20}) = ((1/2)^{1/20})^{40} - ((1/2)^{1/20})^{20} = (1/2)^2 - (1/2)^1 = 1/4 - 1/2 = -1/4$.
Comparing the values $0, 0,$ and $-1/4$,the absolute minimum value is $-1/4$.