Maxima and Minima Questions in English

(C) For a cubic function $f(x) = ax^3 + bx^2 + cx + d$ to have no extreme values,its derivative $f'(x) = 3ax^2 + 2bx + c$ must not change sign.
This means $f'(x)$ must be either always $\ge 0$ or always $\le 0$ for all $x \in R$.
This implies that the quadratic expression $g(x) = 3ax^2 + 2bx + c$ does not change sign.
For a quadratic $g(x) = Ax^2 + Bx + C$ to maintain a constant sign,its discriminant $D = B^2 - 4AC$ must be $\le 0$.
Here,$D = (2b)^2 - 4(3a)(c) = 4b^2 - 12ac \le 0$,which implies $b^2 \le 3ac$.
If $f'(x)$ does not change sign,then $f'(x_1)$ and $f'(x_2)$ must have the same sign for any $x_1, x_2 \in R$.
Specifically,$f'(1) = 3a + 2b + c$ and $f'(0) = c$ must have the same sign or be zero.
Therefore,their product must be non-negative: $(3a + 2b + c)c \ge 0$.

(A) The volume of a right circular cylinder is given by $V = \pi r^2 h$.
Given the relation $r^2 + h = 6$,we can express $h$ as $h = 6 - r^2$.
Substituting this into the volume formula: $V = \pi r^2 (6 - r^2) = \pi (6r^2 - r^4)$.
To find the maximum volume,we differentiate $V$ with respect to $r$: $\frac{dV}{dr} = \pi (12r - 4r^3)$.
Setting $\frac{dV}{dr} = 0$,we get $12r - 4r^3 = 0$,which implies $4r(3 - r^2) = 0$.
Since $r > 0$,we have $r^2 = 3$,so $r = \sqrt{3}$.
Substituting $r^2 = 3$ into the relation $h = 6 - r^2$,we get $h = 6 - 3 = 3$.
Therefore,the ratio $\frac{r}{h} = \frac{\sqrt{3}}{3} = \frac{1}{\sqrt{3}}$.

(D) For $x \in (0, 1)$,$\ln x < 0$,so $x \ln x < 0$.
Thus,$f(x) = |x \ln x| = -x \ln x$.
To find the critical points,we calculate the derivative: $f'(x) = -(1 \cdot \ln x + x \cdot \frac{1}{x}) = -(1 + \ln x)$.
Setting $f'(x) = 0$,we get $1 + \ln x = 0$,which implies $\ln x = -1$,so $x = e^{-1} = \frac{1}{e}$.
Since $f'(x) > 0$ for $x < \frac{1}{e}$ and $f'(x) < 0$ for $x > \frac{1}{e}$,the function has a local maximum at $x = \frac{1}{e}$.
The maximum value is $f(\frac{1}{e}) = -(\frac{1}{e}) \ln(\frac{1}{e}) = -(\frac{1}{e})(-1) = \frac{1}{e} = e^{-1}$.

(C) Given $f(x) = \frac{x^2 - 2}{\sqrt{1 + x^2}}$.
To find the critical points,we calculate the derivative $f'(x)$ using the quotient rule:
$f'(x) = \frac{\sqrt{1+x^2}(2x) - (x^2-2)\frac{1}{2\sqrt{1+x^2}}(2x)}{1+x^2}$
$f'(x) = \frac{2x(1+x^2) - x(x^2-2)}{(1+x^2)^{3/2}}$
$f'(x) = \frac{2x + 2x^3 - x^3 + 2x}{(1+x^2)^{3/2}} = \frac{x^3 + 4x}{(1+x^2)^{3/2}} = \frac{x(x^2 + 4)}{(1+x^2)^{3/2}}$.
Since $x^2 + 4 > 0$ and $(1+x^2)^{3/2} > 0$ for all $x \in \mathbb{R}$,the sign of $f'(x)$ depends only on $x$.
For $x < 0$,$f'(x) < 0$ (function is decreasing).
For $x > 0$,$f'(x) > 0$ (function is increasing).
At $x = 0$,$f'(x) = 0$ and the sign changes from negative to positive,which indicates that $x = 0$ is a point of local minima.
Thus,the function has exactly one point of minima.

(B) Given $f''(x) > 0$,$f'(x)$ is a strictly increasing function.
$g'(x) = 2f'(2x^3 - 3x^2) \cdot (6x^2 - 6x) + f'(6x^2 - 4x^3 - 3) \cdot (12x - 12x^2)$
$g'(x) = 12(x^2 - x) f'(2x^3 - 3x^2) - 12(x^2 - x) f'(6x^2 - 4x^3 - 3)$
$g'(x) = 12x(x - 1) [f'(2x^3 - 3x^2) - f'(6x^2 - 4x^3 - 3)]$
For $g'(x) > 0$,we analyze the sign of $12x(x - 1)$ and the difference $[f'(2x^3 - 3x^2) - f'(6x^2 - 4x^3 - 3)]$.
Since $f'(x)$ is increasing,$f'(A) > f'(B) \iff A > B$.
Let $A = 2x^3 - 3x^2$ and $B = 6x^2 - 4x^3 - 3$.
$A - B = (2x^3 - 3x^2) - (6x^2 - 4x^3 - 3) = 6x^3 - 9x^2 + 3 = 3(2x^3 - 3x^2 + 1) = 3(x - 1)^2(2x + 1)$.
Thus,$f'(A) - f'(B) > 0$ when $3(x - 1)^2(2x + 1) > 0$,which implies $x > -\frac{1}{2}$ (excluding $x = 1$).
Now,$g'(x) > 0$ when $12x(x - 1)$ and $(f'(A) - f'(B))$ have the same sign.
Case $I$: $x(x - 1) > 0$ and $x > -\frac{1}{2}$. This occurs when $x \in (1, \infty)$.
Case $II$: $x(x - 1) < 0$ and $x < -\frac{1}{2}$. This occurs when $x \in (-\frac{1}{2}, 0)$.
Combining these,$g'(x) > 0$ for $x \in (-\frac{1}{2}, 0) \cup (1, \infty)$.

(A) Given the function $f(x) = x + \sin x$.
To find the local maxima,we first 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 sign of $f'(x)$ around these critical points. Since $\cos x \geq -1$ for all $x$,$f'(x) = 1 + \cos x \geq 0$ for all $x$.
Because $f'(x)$ does not change sign from positive to negative at any point $x = (2n + 1)\pi$,the function is strictly increasing and does not have any local maxima.
Therefore,the number of points of local maxima is $0$.

(C) Let $AF = x$. Since $FM \parallel ED$ and $FE \parallel AD$,the parallelogram is $AMFE$ (or $AFED$ based on the diagram). Let $AF = x$. Since $\Delta AFE \sim \Delta ABC$,we have $\frac{FE}{BC} = \frac{AF}{AB} = \frac{AE}{AC}$.
Let $AF = x$. Then $FE = \frac{x}{b} \cdot c$. The height of the parallelogram with respect to base $AF$ is $h = x \sin A$.
The area of the parallelogram is $Area = \text{base} \times \text{height} = FE \times (x \sin A) = (\frac{cx}{b}) \times (x \sin A) = \frac{c \sin A}{b} x^2$. However,looking at the diagram,the base is $AD$ and height is $FM$.
Let $AF = x$. Then $FE = \frac{x}{b} c$. The height of the parallelogram from $FE$ to $AB$ is $x \sin A$.
Area $A(x) = FE \cdot (x \sin A) = (\frac{c}{b} x) \cdot (x \sin A) = \frac{c \sin A}{b} x^2$. This is not correct as $x$ is limited.
Correct approach: Let $AF = x$. Then $FE = \frac{c}{b}(b-x)$. The height of the parallelogram is $h = x \sin A$.
Area $A(x) = FE \cdot h = \frac{c}{b}(b-x) \cdot x \sin A = \frac{c \sin A}{b} (bx - x^2)$.
To maximize,$\frac{dA}{dx} = \frac{c \sin A}{b} (b - 2x) = 0 \Rightarrow x = \frac{b}{2}$.
Maximum Area $= \frac{c \sin A}{b} (b(\frac{b}{2}) - (\frac{b}{2})^2) = \frac{c \sin A}{b} (\frac{b^2}{4}) = \frac{bc}{4} \sin A$.

(B) The given curve is $y = e^x + e^{-x}$.
We want to find the shortest distance from the origin $(0,0)$ to this curve.
Let $f(x) = e^x + e^{-x}$. By the $AM$-$GM$ inequality,$e^x + e^{-x} \ge 2\sqrt{e^x \cdot e^{-x}} = 2\sqrt{1} = 2$.
The minimum value of $y$ is $2$,which occurs at $x = 0$.
Thus,the point on the curve closest to the origin is $(0,2)$.
The distance between $(0,0)$ and $(0,2)$ is $\sqrt{(0-0)^2 + (2-0)^2} = \sqrt{4} = 2$.

(D) Given the function $f(x) = x^3 - 12x + \lambda$.
To find the points of local maxima and minima,we calculate the first derivative:
$f'(x) = 3x^2 - 12$.
Setting $f'(x) = 0$ gives $3(x^2 - 4) = 0$,which implies $x = 2$ or $x = -2$.
Now,we check the second derivative:
$f''(x) = 6x$.
At $x = -2$,$f''(-2) = 6(-2) = -12 < 0$,which indicates a local maximum at $x = -2$.
At $x = 2$,$f''(2) = 6(2) = 12 > 0$,which indicates a local minimum at $x = 2$.
Since the existence of local maxima and minima for a cubic polynomial $f(x) = ax^3 + bx^2 + cx + d$ depends only on the derivative $f'(x) = 3ax^2 + 2bx + c$,and here $f'(x) = 3x^2 - 12$ is independent of $\lambda$,the function $f(x)$ will have a point of local maxima for all values of $\lambda$.
Given $\lambda \in [0, 20]$,the integral values of $\lambda$ are $\{0, 1, 2, \dots, 20\}$.
The total number of such values is $20 - 0 + 1 = 21$.

(D) Given the function $f(x) = -1 + \frac{2}{2^{x^2} + 1}$.
To find the greatest value of $f(x)$,we need to maximize the term $\frac{2}{2^{x^2} + 1}$.
$A$ fraction is maximized when its denominator is minimized.
The denominator $2^{x^2} + 1$ is minimized when $2^{x^2}$ is minimized.
Since $x^2 \ge 0$ for all real $x$,the minimum value of $x^2$ is $0$,which occurs at $x = 0$.
At $x = 0$,$2^{x^2} = 2^0 = 1$.
Thus,the minimum value of the denominator is $1 + 1 = 2$.
The maximum value of the fraction is $\frac{2}{2} = 1$.
Therefore,the greatest value of $f(x)$ is $f(0) = -1 + 1 = 0$.

(A) Given the function $y = a \ln |x + 1| + b(x + 1)^2 + x$.
Since the function has an extremum at $x = 0$,the value of the function at $x = 0$ is $4$.
Substituting $x = 0$ and $y = 4$ into the equation:
$4 = a \ln |0 + 1| + b(0 + 1)^2 + 0$
$4 = a \ln(1) + b(1)^2 + 0$
Since $\ln(1) = 0$,we get $4 = 0 + b + 0$,which implies $b = 4$.
Now,find the derivative $\frac{dy}{dx}$:
$\frac{dy}{dx} = \frac{a}{x + 1} + 2b(x + 1) + 1$.
At an extremum point,the derivative is zero. Thus,at $x = 0$,$\frac{dy}{dx} = 0$:
$\frac{a}{0 + 1} + 2b(0 + 1) + 1 = 0$
$a + 2b + 1 = 0$.
Substitute $b = 4$ into the equation:
$a + 2(4) + 1 = 0$
$a + 8 + 1 = 0$
$a + 9 = 0 \Rightarrow a = -9$.
Therefore,$(a, b) = (-9, 4)$.

(B) For $f(x)$ to have a local minimum at $x = 1$,the function must satisfy $f(1) \le f(1 - h)$ and $f(1) \le f(1 + h)$ for small $h > 0$.
First,we evaluate the left-hand limit at $x = 1$:
$f(1^-) = \lim_{x \to 1^-} (3 - x) = 3 - 1 = 2$.
Next,we evaluate the function value at $x = 1$:
$f(1) = 1^2 + \log_e b = 1 + \log_e b$.
For $x > 1$,$f(x) = x^2 + \log_e b$. Since $f'(x) = 2x > 0$ for $x > 1$,the function is strictly increasing for $x > 1$. Thus,$f(x) \ge f(1)$ for all $x > 1$ in the neighborhood of $1$.
For $x < 1$,$f(x) = 3 - x$. Since $f'(x) = -1 < 0$ for $x < 1$,the function is strictly decreasing for $x < 1$. Thus,$f(x) > f(1^-) = 2$ for all $x < 1$ in the neighborhood of $1$.
For $x = 1$ to be a local minimum,we require $f(1) \le f(x)$ for all $x$ in a neighborhood of $1$. Specifically,we need $f(1) \le f(1^-)$:
$1 + \log_e b \le 2$
$\log_e b \le 1$
$b \le e^1 = e$.
Since the logarithm $\log_e b$ is defined only for $b > 0$,the set of values for $b$ is $(0, e]$.

(A) Let $h$ be the height of the cone inscribed in a sphere of radius $R$. Let the center of the sphere be $O$. The distance from the center $O$ to the base of the cone is $|h-R|$.
By the Pythagorean theorem,the radius $r$ of the base of the cone satisfies $r^2 + (h-R)^2 = R^2$,so $r^2 = R^2 - (h^2 - 2hR + R^2) = 2hR - h^2$.
The volume $V$ of the cone is $V = \frac{1}{3} \pi r^2 h = \frac{1}{3} \pi (2hR - h^2) h = \frac{1}{3} \pi (2h^2R - h^3)$.
To maximize $V$,we find the derivative with respect to $h$: $\frac{dV}{dh} = \frac{1}{3} \pi (4hR - 3h^2)$.
Setting $\frac{dV}{dh} = 0$ gives $h(4R - 3h) = 0$. Since $h \neq 0$,we have $h = \frac{4R}{3}$.
The diameter of the sphere is $D = 2R$.
The ratio of the altitude of the cone to the diameter of the sphere is $\frac{h}{D} = \frac{4R/3}{2R} = \frac{4}{6} = \frac{2}{3}$.

(D) To determine the nature of the function at $x = 2$,we find the first and second derivatives of $f(x) = x^3 - 6x^2 + 12x - 3$.
Step $1$: Find the first derivative $f'(x)$.
$f'(x) = \frac{d}{dx}(x^3 - 6x^2 + 12x - 3) = 3x^2 - 12x + 12$.
Step $2$: Find the critical points by setting $f'(x) = 0$.
$3x^2 - 12x + 12 = 0 \implies 3(x^2 - 4x + 4) = 0 \implies 3(x - 2)^2 = 0$.
This gives $x = 2$ as the critical point.
Step $3$: Find the second derivative $f''(x)$.
$f''(x) = \frac{d}{dx}(3x^2 - 12x + 12) = 6x - 12$.
Step $4$: Evaluate $f''(x)$ at $x = 2$.
$f''(2) = 6(2) - 12 = 12 - 12 = 0$.
Since $f''(2) = 0$,the second derivative test is inconclusive. We examine the sign of $f'(x)$ around $x = 2$.
$f'(x) = 3(x - 2)^2$. Since $(x - 2)^2$ is always non-negative for all real $x$,$f'(x) \geq 0$ for all $x$.
Because the derivative does not change sign as $x$ passes through $2$,$x = 2$ is a point of inflection,not a local maximum or minimum.

(B) Let $y = (\frac{1}{x})^{2x^2} = x^{-2x^2}$.
Taking the natural logarithm on both sides,we get $\ln y = -2x^2 \ln x$.
Differentiating with respect to $x$,we have $\frac{1}{y} \frac{dy}{dx} = -4x \ln x - 2x^2 (\frac{1}{x}) = -4x \ln x - 2x = -2x(2 \ln x + 1)$.
Thus,$\frac{dy}{dx} = -2x(2 \ln x + 1) y$.
For critical points,set $\frac{dy}{dx} = 0$. Since $x > 0$ and $y > 0$,we have $2 \ln x + 1 = 0$,which implies $\ln x = -\frac{1}{2}$,so $x = e^{-1/2} = \frac{1}{\sqrt{e}}$.
At $x = \frac{1}{\sqrt{e}}$,the function attains its maximum value.
Substituting $x = e^{-1/2}$ into the original expression:
$y_{max} = (\frac{1}{e^{-1/2}})^{2(e^{-1/2})^2} = (e^{1/2})^{2(e^{-1})} = e^{(1/2) \times (2/e)} = e^{1/e} = \sqrt[e]{e}$.

(C) To find the maximum value,we first find the derivative $f'(x)$ and set it to $0$.
$f(x) = (7-x)^4 (2+x)^5$
Using the product rule and chain rule:
$f'(x) = 4(7-x)^3(-1)(2+x)^5 + 5(7-x)^4(2+x)^4$
$f'(x) = (7-x)^3(2+x)^4 [-4(2+x) + 5(7-x)]$
$f'(x) = (7-x)^3(2+x)^4 [-8 - 4x + 35 - 5x]$
$f'(x) = (7-x)^3(2+x)^4 [27 - 9x]$
Setting $f'(x) = 0$,we get critical points $x = 7, x = -2, x = 3$.
For $x \in (-2, 7)$,the function is positive. At $x = 3$,the derivative changes from positive to negative,indicating a local maximum.
The maximum value is $f(3) = (7-3)^4 (2+3)^5 = 4^4 \times 5^5$.

(D) Let the square have side length $2 \ cm$. $A$ cut is made from one corner to a point on an adjacent side,creating a right-angled triangle and a quadrilateral (trapezium).
Let the cut start from a corner and end at a distance $x$ from the corner on one side and $y$ from the corner on the other side. However,based on the figure,the cut is a straight line segment of length $L$ connecting a point at distance $x$ from the corner on one side to the corner itself on the other side.
Let the triangle have sides $x$,$2$,and $L$,where $L = \sqrt{x^2 + 2^2}$.
The perimeter of the triangle is $P_1 = x + 2 + L$.
The remaining figure is a quadrilateral with sides $2$,$2$,$(2-x)$,and $L$.
The perimeter of the quadrilateral is $P_2 = 2 + 2 + (2-x) + L = 6 - x + L$.
The sum of the perimeters is $S = P_1 + P_2 = (x + 2 + L) + (6 - x + L) = 8 + 2L$.
To maximize $S$,we must maximize $L$. The maximum length of a straight line segment $L$ that can be cut from a corner of a square of side $2$ is the diagonal of the square.
Thus,$L = \sqrt{2^2 + 2^2} = \sqrt{8} = 2\sqrt{2}$.
The maximum sum of the perimeters is $S_{max} = 8 + 2(2\sqrt{2}) = 8 + 4\sqrt{2} \ cm$.

(B) Given $f(x) = \int_{x}^{x^2} (t - 1) \, dt$.
Using the Leibniz integral rule,we differentiate $f(x)$ with respect to $x$:
$f'(x) = (x^2 - 1) \cdot \frac{d}{dx}(x^2) - (x - 1) \cdot \frac{d}{dx}(x)$
$f'(x) = (x^2 - 1)(2x) - (x - 1)(1)$
$f'(x) = 2x^3 - 2x - x + 1 = 2x^3 - 3x + 1$.
We can factorize $f'(x)$ by observing that $x=1$ is a root:
$f'(x) = (x - 1)(2x^2 + 2x - 1)$.
For $x \in [1, 3]$,$f'(x) \ge 0$,so $f(x)$ is an increasing function.
The maximum value occurs at the right endpoint,$x = 3$.
$f(3) = \int_{3}^{9} (t - 1) \, dt = \left[ \frac{t^2}{2} - t \right]_{3}^{9}$
$f(3) = \left( \frac{81}{2} - 9 \right) - \left( \frac{9}{2} - 3 \right) = \frac{72}{2} - 6 = 36 - 6 = 30$.

(C) Let $f(x) = x^2 - x \sin x - \cos x$.
Then,the derivative is $f'(x) = 2x - (\sin x + x \cos x) - (-\sin x) = 2x - x \cos x = x(2 - \cos x)$.
Since $-1 \le \cos x \le 1$,we have $2 - \cos x > 0$ for all real $x$.
Thus,$f'(x) = 0$ only at $x = 0$.
For $x < 0$,$f'(x) < 0$,so $f(x)$ is strictly decreasing.
For $x > 0$,$f'(x) > 0$,so $f(x)$ is strictly increasing.
Therefore,$x = 0$ is the point of global minimum.
The minimum value is $f(0) = 0^2 - 0 \sin(0) - \cos(0) = -1$.
As $x \to \infty$,$f(x) \to \infty$,and as $x \to -\infty$,$f(x) \to \infty$.
Since $f(x)$ is continuous,it crosses the $x$-axis exactly once in $(-\infty, 0)$ and exactly once in $(0, \infty)$.
Thus,there are exactly $2$ points where $f(x) = 0$.

(D) Let $f(x) = x^4 e^{-x^2}$.
To find the maximum value,we find the derivative $f'(x)$:
$f'(x) = 4x^3 e^{-x^2} + x^4 e^{-x^2}(-2x)$
$f'(x) = 2x^3 e^{-x^2}(2 - x^2)$
Setting $f'(x) = 0$,we get $x = 0$ or $x^2 = 2$,which means $x = 0, \pm \sqrt{2}$.
Analyzing the sign of $f'(x)$:
For $x < -\sqrt{2}$,$f'(x) > 0$.
For $-\sqrt{2} < x < 0$,$f'(x) < 0$.
For $0 < x < \sqrt{2}$,$f'(x) > 0$.
For $x > \sqrt{2}$,$f'(x) < 0$.
Thus,$f(x)$ has local maxima at $x = \pm \sqrt{2}$.
The maximum value is $f(\pm \sqrt{2}) = (\pm \sqrt{2})^4 e^{-(\pm \sqrt{2})^2} = 4 e^{-2}$.

(B) Let $g(x) = 4x^3 - 12x^2 + 11x - 3$.
Then $g'(x) = 12x^2 - 24x + 11$.
We can write $g'(x) = 12(x^2 - 2x) + 11 = 12(x^2 - 2x + 1 - 1) + 11 = 12(x - 1)^2 - 1$.
For $x \in [2, 3]$,$(x - 1) \in [1, 2]$,so $(x - 1)^2 \in [1, 4]$.
Thus,$g'(x) \geq 12(1) - 1 = 11 > 0$ for all $x \in [2, 3]$.
Since $g'(x) > 0$,$g(x)$ is a strictly increasing function on $[2, 3]$.
Consequently,$f(x) = \log_{10}(g(x))$ is also strictly increasing on $[2, 3]$.
The global maximum value occurs at the right endpoint $x = 3$.
$f(3) = \log_{10}(4(3)^3 - 12(3)^2 + 11(3) - 3) = \log_{10}(4(27) - 12(9) + 33 - 3) = \log_{10}(108 - 108 + 30) = \log_{10}(30)$.
$f(3) = \log_{10}(10 \times 3) = \log_{10}10 + \log_{10}3 = 1 + \log_{10}3$.

(A) Let $R = a/2$ 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.
From the geometry,if $x$ is the distance from the center of the sphere to the base of the cone,then $h = R + x = a/2 + x$.
The radius of the base of the cone is given by $r^2 = R^2 - x^2 = (a/2)^2 - x^2$.
The volume of the cone is $V = \frac{1}{3} \pi r^2 h = \frac{1}{3} \pi ((a/2)^2 - x^2)(a/2 + x)$.
Expanding this,$V = \frac{\pi}{3} (a^2/4 - x^2)(a/2 + x) = \frac{\pi}{3} (a^3/8 + a^2x/4 - ax^2/2 - x^3)$.
To maximize volume,find $dV/dx = 0$:
$dV/dx = \frac{\pi}{3} (a^2/4 - ax - 3x^2) = 0$.
$3x^2 + ax - a^2/4 = 0$.
Using the quadratic formula,$x = \frac{-a \pm \sqrt{a^2 - 4(3)(-a^2/4)}}{2(3)} = \frac{-a \pm \sqrt{a^2 + 3a^2}}{6} = \frac{-a \pm 2a}{6}$.
Since $x$ must be positive,$x = a/6$.
Therefore,the height $h = a/2 + a/6 = (3a + a)/6 = 4a/6 = (2/3)a$.

(A) Let $R = 3 \, cm$ be the radius of the sphere.
Let $h$ be the height and $b$ be the base radius of the inscribed cone.
From the geometry of the sphere,the relationship between $h, b,$ and $R$ is given by $(h-R)^2 + b^2 = R^2$.
Thus,$b^2 = R^2 - (h-R)^2 = 2hR - h^2$.
The volume of the cone is $V = \frac{1}{3} \pi b^2 h = \frac{1}{3} \pi (2hR - h^2) h = \frac{\pi}{3} (2Rh^2 - h^3)$.
To maximize volume,we find $\frac{dV}{dh} = \frac{\pi}{3} (4Rh - 3h^2) = 0$.
This gives $h(4R - 3h) = 0$. Since $h \neq 0$,we have $h = \frac{4R}{3} = \frac{4(3)}{3} = 4 \, cm$.
Now,$b^2 = 2(4)(3) - (4)^2 = 24 - 16 = 8$,so $b = \sqrt{8} = 2\sqrt{2} \, cm$.
The slant height $l = \sqrt{h^2 + b^2} = \sqrt{4^2 + (2\sqrt{2})^2} = \sqrt{16 + 8} = \sqrt{24} = 2\sqrt{6} \, cm$.
The curved surface area is $A = \pi b l = \pi (2\sqrt{2}) (2\sqrt{6}) = 4\pi \sqrt{12} = 4\pi (2\sqrt{3}) = 8\sqrt{3} \pi \, cm^2$.

(D) Since $f(x)$ is a polynomial of degree $4$,let $f(x) = Ax^4 + Bx^3 + Cx^2 + Dx + E$.
Given $\lim_{x \to 0} \left( \frac{f(x)}{x^2} + 1 \right) = 3$,we have $\lim_{x \to 0} \left( \frac{Ax^4 + Bx^3 + Cx^2 + Dx + E}{x^2} + 1 \right) = 3$.
This implies $\lim_{x \to 0} \left( Ax^2 + Bx + C + \frac{D}{x} + \frac{E}{x^2} + 1 \right) = 3$.
For the limit to exist and be finite,we must have $D = 0$ and $E = 0$. Thus,$C + 1 = 3$,which gives $C = 2$.
Now,$f(x) = Ax^4 + Bx^3 + 2x^2$. The derivative is $f'(x) = 4Ax^3 + 3Bx^2 + 4x$.
Since $f(x)$ has extreme values at $x = 1$ and $x = 2$,$f'(1) = 0$ and $f'(2) = 0$.
$f'(1) = 4A + 3B + 4 = 0 \implies 4A + 3B = -4$ (Equation $1$).
$f'(2) = 4A(8) + 3B(4) + 4(2) = 32A + 12B + 8 = 0 \implies 8A + 3B = -2$ (Equation $2$).
Subtracting Equation $1$ from Equation $2$: $(8A + 3B) - (4A + 3B) = -2 - (-4) \implies 4A = 2 \implies A = \frac{1}{2}$.
Substituting $A = \frac{1}{2}$ into Equation $1$: $4(\frac{1}{2}) + 3B = -4 \implies 2 + 3B = -4 \implies 3B = -6 \implies B = -2$.
Thus,$f(x) = \frac{1}{2}x^4 - 2x^3 + 2x^2$.
Finally,$f(-1) = \frac{1}{2}(-1)^4 - 2(-1)^3 + 2(-1)^2 = \frac{1}{2} + 2 + 2 = \frac{9}{2}$.

(A) Given function is $f(x) = 2x^3 - 9x^2 + 12x + 5$.
To find local maxima and minima,we find the derivative $f'(x)$:
$f'(x) = 6x^2 - 18x + 12$.
Setting $f'(x) = 0$ for critical points:
$6(x^2 - 3x + 2) = 0 \Rightarrow 6(x - 1)(x - 2) = 0$.
So,the critical points are $x = 1$ and $x = 2$.
Now,we check the second derivative $f''(x) = 12x - 18$.
At $x = 1$,$f''(1) = 12(1) - 18 = -6 < 0$,so $x = 1$ is a local maximum.
$M = f(1) = 2(1)^3 - 9(1)^2 + 12(1) + 5 = 2 - 9 + 12 + 5 = 10$.
At $x = 2$,$f''(2) = 12(2) - 18 = 6 > 0$,so $x = 2$ is a local minimum.
$m = f(2) = 2(2)^3 - 9(2)^2 + 12(2) + 5 = 16 - 36 + 24 + 5 = 9$.
Therefore,$M - m = 10 - 9 = 1$.

(A) Let $f(x) = \frac{(1 + x)^{3/5}}{1 + x^{3/5}}$ for $x \in [0, 1]$.
Taking the derivative $f'(x)$:
$f'(x) = \frac{(1 + x^{3/5}) \cdot \frac{3}{5}(1 + x)^{-2/5} - (1 + x)^{3/5} \cdot \frac{3}{5}x^{-2/5}}{(1 + x^{3/5})^2}$
$f'(x) = \frac{3}{5} \left[ \frac{1 + x^{3/5}}{(1 + x)^{2/5}} - \frac{(1 + x)^{3/5}}{x^{2/5}} \right]$
$f'(x) = \frac{3}{5} \left[ \frac{x^{2/5}(1 + x^{3/5}) - (1 + x)^{3/5}(1 + x)^{2/5}}{x^{2/5}(1 + x)^{2/5}} \right]$
$f'(x) = \frac{3}{5} \left[ \frac{x^{2/5} + x - (1 + x)}{x^{2/5}(1 + x)^{2/5}} \right] = \frac{3}{5} \left[ \frac{x^{2/5} - 1}{x^{2/5}(1 + x)^{2/5}} \right]$
Since $x \in [0, 1]$,$x^{2/5} \le 1$,so $f'(x) \le 0$. The function is monotonically decreasing.
Thus,the maximum value $K = f(0) = \frac{(1+0)^{0.6}}{1+0^{0.6}} = 1$.
The minimum value $k = f(1) = \frac{(1+1)^{0.6}}{1+1^{0.6}} = \frac{2^{0.6}}{2} = 2^{0.6 - 1} = 2^{-0.4}$.
Therefore,the ordered pair $(k, K)$ is $(2^{-0.4}, 1)$.

(C) Let the radius of the sphere be $R = \sqrt{3}$. Let the height of the cylinder be $h$ and its radius be $r$.
In the right-angled triangle formed by the radius of the sphere,the radius of the cylinder,and half the height of the cylinder,we have:
$R^2 = r^2 + (h/2)^2$
$(\sqrt{3})^2 = r^2 + \frac{h^2}{4}$
$3 = r^2 + \frac{h^2}{4} \Rightarrow r^2 = 3 - \frac{h^2}{4}$
The volume of the cylinder is $V = \pi r^2 h = \pi (3 - \frac{h^2}{4})h = 3\pi h - \frac{\pi h^3}{4}$.
To maximize the volume,we find the derivative with respect to $h$ and set it to zero:
$\frac{dV}{dh} = 3\pi - \frac{3\pi h^2}{4} = 0$
$3\pi = \frac{3\pi h^2}{4} \Rightarrow h^2 = 4 \Rightarrow h = 2$.
Substituting $h = 2$ into the volume formula:
$V = \pi (3 - \frac{2^2}{4})(2) = \pi (3 - 1)(2) = 4\pi$.

(D) Let the base be $b$ and the hypotenuse be $h$.
Then the altitude (perpendicular) is $\sqrt{h^2 - b^2}$.
The area $A$ of the triangle is given by $A = \frac{1}{2} \times \text{base} \times \text{altitude} = \frac{1}{2} b \sqrt{h^2 - b^2}$.
To maximize the area,we differentiate $A$ with respect to $b$:
$\frac{dA}{db} = \frac{1}{2} \left[ \sqrt{h^2 - b^2} + b \cdot \frac{-2b}{2\sqrt{h^2 - b^2}} \right] = \frac{1}{2} \left[ \frac{h^2 - b^2 - b^2}{\sqrt{h^2 - b^2}} \right] = \frac{h^2 - 2b^2}{2\sqrt{h^2 - b^2}}$.
Setting $\frac{dA}{db} = 0$,we get $h^2 - 2b^2 = 0$,which implies $b^2 = \frac{h^2}{2}$,or $b = \frac{h}{\sqrt{2}}$.
Substituting this value into the area formula:
$A_{max} = \frac{1}{2} \cdot \frac{h}{\sqrt{2}} \cdot \sqrt{h^2 - \frac{h^2}{2}} = \frac{1}{2} \cdot \frac{h}{\sqrt{2}} \cdot \frac{h}{\sqrt{2}} = \frac{h^2}{4}$.

(C) Let the cost function be $C(v) = av + \frac{b}{v}$.
Given that at $v = 30 \text{ km/h}$,$C = 75$,so $30a + \frac{b}{30} = 75 \implies 900a + b = 2250 \quad (i)$.
Given that at $v = 40 \text{ km/h}$,$C = 65$,so $40a + \frac{b}{40} = 65 \implies 1600a + b = 2600 \quad (ii)$.
Subtracting $(i)$ from $(ii)$,we get $700a = 350$,which implies $a = 0.5$.
Substituting $a = 0.5$ into $(i)$,we get $900(0.5) + b = 2250 \implies 450 + b = 2250 \implies b = 1800$.
To find the most economical speed,we minimize $C(v)$ by setting $\frac{dC}{dv} = 0$.
$\frac{dC}{dv} = a - \frac{b}{v^2} = 0 \implies v^2 = \frac{b}{a}$.
$v^2 = \frac{1800}{0.5} = 3600 \implies v = 60 \text{ km/h}$.

(D) Let the slant height be $l = 3 \, m$,radius be $r$,and height be $h$.
We know that $l^2 = r^2 + h^2$,so $r^2 = l^2 - h^2 = 9 - h^2$.
The volume of the cone is $V = \frac{1}{3} \pi r^2 h$.
Substituting $r^2$,we get $V(h) = \frac{1}{3} \pi (9 - h^2) h = \frac{1}{3} \pi (9h - h^3)$.
To find the maximum volume,differentiate $V$ with respect to $h$:
$\frac{dV}{dh} = \frac{1}{3} \pi (9 - 3h^2)$.
Setting $\frac{dV}{dh} = 0$,we get $9 - 3h^2 = 0$,which implies $h^2 = 3$,so $h = \sqrt{3}$.
Now,calculate the maximum volume:
$V = \frac{1}{3} \pi (9 - 3) \sqrt{3} = \frac{1}{3} \pi (6) \sqrt{3} = 2\sqrt{3}\pi \, m^3$.

(A) Let $P(x, y)$ be a point on the curve $y = \sqrt{x}$. Then $P$ can be represented as $(t^2, t)$ for $t > 0$.
The squared distance $D^2$ between $P(t^2, t)$ and $\left( \frac{3}{2}, 0 \right)$ is given by:
$f(t) = (t^2 - \frac{3}{2})^2 + (t - 0)^2 = t^4 - 3t^2 + \frac{9}{4} + t^2 = t^4 - 2t^2 + \frac{9}{4}$.
To find the minimum distance,we differentiate $f(t)$ with respect to $t$ and set it to zero:
$f'(t) = 4t^3 - 4t = 4t(t^2 - 1) = 0$.
Since $t > 0$,we have $t^2 = 1$,which gives $t = 1$.
For $t = 1$,the point $P$ is $(1^2, 1) = (1, 1)$.
The shortest distance is the distance between $(1, 1)$ and $\left( \frac{3}{2}, 0 \right)$:
$d = \sqrt{(\frac{3}{2} - 1)^2 + (0 - 1)^2} = \sqrt{(\frac{1}{2})^2 + (-1)^2} = \sqrt{\frac{1}{4} + 1} = \sqrt{\frac{5}{4}} = \frac{\sqrt{5}}{2}$.

(C) Let the point on the curve be $P(x, x^{3/2} + 7)$. The soldier is at $A(1/2, 7)$.
The squared distance $D^2 = (x - 1/2)^2 + (x^{3/2} + 7 - 7)^2 = (x - 1/2)^2 + x^3$.
Let $f(x) = (x - 1/2)^2 + x^3$. For minimum distance,$f'(x) = 0$.
$f'(x) = 2(x - 1/2) + 3x^2 = 3x^2 + 2x - 1 = 0$.
$(3x - 1)(x + 1) = 0$. Since $x \geq 0$,we have $x = 1/3$.
The point $P$ is $(1/3, (1/3)^{3/2} + 7) = (1/3, 7 + \frac{1}{3\sqrt{3}})$.
The distance $AD = \sqrt{(1/3 - 1/2)^2 + (7 + \frac{1}{3\sqrt{3}} - 7)^2} = \sqrt{(-1/6)^2 + (\frac{1}{3\sqrt{3}})^2}$.
$AD = \sqrt{\frac{1}{36} + \frac{1}{27}} = \sqrt{\frac{3 + 4}{108}} = \sqrt{\frac{7}{108}} = \frac{1}{6}\sqrt{\frac{7}{3}}$.

(C) Let the vertices of the rectangle on the parabola be $(\alpha, 12 - \alpha^2)$ and $(-\alpha, 12 - \alpha^2)$,where $\alpha > 0$.
The base of the rectangle lies on the $x-$axis,so the length of the base is $2\alpha$ and the height is $12 - \alpha^2$.
The area $A$ of the rectangle is given by $A = \text{length} \times \text{height} = 2\alpha(12 - \alpha^2) = 24\alpha - 2\alpha^3$.
To find the maximum area,we differentiate $A$ with respect to $\alpha$:
$\frac{dA}{d\alpha} = 24 - 6\alpha^2$.
Setting $\frac{dA}{d\alpha} = 0$,we get $24 - 6\alpha^2 = 0$,which implies $\alpha^2 = 4$,so $\alpha = 2$ (since $\alpha > 0$).
To verify this is a maximum,we check the second derivative: $\frac{d^2A}{d\alpha^2} = -12\alpha$. At $\alpha = 2$,$\frac{d^2A}{d\alpha^2} = -24 < 0$,so it is a maximum.
The maximum area is $A = 2(2)(12 - 2^2) = 4(12 - 4) = 4(8) = 32$ sq. units.

(C) Given $f(x) = x^3 - 3(a - 2)x^2 + 3ax + 7$.
The derivative is $f'(x) = 3x^2 - 6(a - 2)x + 3a$.
Since $f(x)$ is increasing in $(0, 1]$ and decreasing in $[1, 5)$,the function must have a local maximum at $x = 1$.
Thus,$f'(1) = 0$.
$f'(1) = 3(1)^2 - 6(a - 2)(1) + 3a = 3 - 6a + 12 + 3a = 15 - 3a = 0$.
Solving for $a$,we get $a = 5$.
Substituting $a = 5$ into $f(x)$,we get $f(x) = x^3 - 3(5 - 2)x^2 + 3(5)x + 7 = x^3 - 9x^2 + 15x + 7$.
We need to solve $\frac{f(x) - 14}{(x - 1)^2} = 0$,which implies $f(x) - 14 = 0$ for $x \neq 1$.
$x^3 - 9x^2 + 15x + 7 - 14 = x^3 - 9x^2 + 15x - 7 = 0$.
Since $x = 1$ is a root of $f(x) - 14 = 0$,we can divide by $(x - 1)^2$:
$x^3 - 9x^2 + 15x - 7 = (x - 1)^2(x - 7) = 0$.
Thus,the root of the equation $\frac{f(x) - 14}{(x - 1)^2} = 0$ is $x = 7$.

(A) Given the function $f(x) = 9x^4 + 12x^3 - 36x^2 + 25$.
To find the critical points,we find the first derivative $f'(x)$:
$f'(x) = 36x^3 + 36x^2 - 72x$
Setting $f'(x) = 0$:
$36x(x^2 + x - 2) = 0$
$36x(x - 1)(x + 2) = 0$
The critical points are $x = -2, 0, 1$.
Now,we use the second derivative test $f''(x) = 108x^2 + 72x - 72$:
For $x = -2$: $f''(-2) = 108(4) + 72(-2) - 72 = 432 - 144 - 72 = 216 > 0$ (Local Minima).
For $x = 0$: $f''(0) = -72 < 0$ (Local Maxima).
For $x = 1$: $f''(1) = 108 + 72 - 72 = 108 > 0$ (Local Minima).
Thus,the set of local minimum points $S_1 = \{-2, 1\}$ and the set of local maximum points $S_2 = \{0\}$.

(C) Let the radius of the sphere be $R=3$. Let the height of the cylinder be $h$ and its radius be $r$.
From the geometry of the sphere and cylinder,we have $r^2 + (h/2)^2 = R^2 = 3^2 = 9$.
Thus,$r^2 = 9 - \frac{h^2}{4}$.
The volume of the cylinder is $V = \pi r^2 h = \pi (9 - \frac{h^2}{4}) h = \pi (9h - \frac{h^3}{4})$.
To maximize the volume,we find the derivative with respect to $h$ and set it to zero:
$\frac{dV}{dh} = \pi (9 - \frac{3h^2}{4}) = 0$.
$9 = \frac{3h^2}{4} \Rightarrow h^2 = 12 \Rightarrow h = \sqrt{12} = 2\sqrt{3}$.
Thus,the height of the cylinder of maximum volume is $2\sqrt{3}$.

(C) Since $f(x)$ is a polynomial of degree $4$ and has local extreme points at $x = -1, 0, 1$,its derivative $f'(x)$ must be a polynomial of degree $3$ with roots at $-1, 0, 1$.
Thus,$f'(x) = \lambda(x + 1)(x)(x - 1) = \lambda(x^3 - x)$,where $\lambda \neq 0$.
Integrating $f'(x)$,we get $f(x) = \lambda(\frac{x^4}{4} - \frac{x^2}{2}) + \mu$,where $\mu$ is the constant of integration.
We are given the set $S = \{x \in R; f(x) = f(0)\}$.
Substituting $f(0) = \mu$ into the equation $f(x) = f(0)$,we get $\lambda(\frac{x^4}{4} - \frac{x^2}{2}) + \mu = \mu$.
This simplifies to $\lambda(\frac{x^4}{4} - \frac{x^2}{2}) = 0$.
Since $\lambda \neq 0$,we have $\frac{x^4}{4} - \frac{x^2}{2} = 0$,which implies $x^2(\frac{x^2}{4} - \frac{1}{2}) = 0$.
This gives $x^2 = 0$ or $x^2 = 2$.
Thus,the roots are $x = 0, 0, \sqrt{2}, -\sqrt{2}$.
The distinct elements in the set $S$ are $0, \sqrt{2}, -\sqrt{2}$.
Here,$0$ is a rational number,and $\sqrt{2}, -\sqrt{2}$ are irrational numbers.
Therefore,the set $S$ contains two irrational numbers and one rational number.

(D) Given $f(x) = x\sqrt{kx - x^2}$.
For $f(x)$ to be defined,$kx - x^2 \geq 0$,which implies $x(k - x) \geq 0$. For $x \in [0, 3]$,this requires $k \geq 3$.
The derivative is $f'(x) = \sqrt{kx - x^2} + x \cdot \frac{k - 2x}{2\sqrt{kx - x^2}} = \frac{2(kx - x^2) + kx - 2x^2}{2\sqrt{kx - x^2}} = \frac{3kx - 4x^2}{2\sqrt{kx - x^2}}$.
For $f(x)$ to be increasing on $[0, 3]$,we need $f'(x) \geq 0$ for all $x \in (0, 3)$.
This implies $3kx - 4x^2 \geq 0$,or $x(3k - 4x) \geq 0$.
Since $x > 0$,we require $3k - 4x \geq 0$,or $k \geq \frac{4x}{3}$.
For this to hold for all $x \in [0, 3]$,$k$ must be at least the maximum value of $\frac{4x}{3}$ on $[0, 3]$,which is $\frac{4(3)}{3} = 4$.
Thus,$m = 4$.
Now,for $k = 4$,$f(x) = x\sqrt{4x - x^2}$.
To find the maximum value $M$ on $[0, 3]$,we check the derivative $f'(x) = \frac{12x - 4x^2}{2\sqrt{4x - x^2}} = \frac{2x(3 - x)}{\sqrt{4x - x^2}}$.
The critical point is $x = 3$. Since $f(0) = 0$ and $f(3) = 3\sqrt{4(3) - 3^2} = 3\sqrt{12 - 9} = 3\sqrt{3}$,the maximum value $M = 3\sqrt{3}$.
Therefore,$(m, M) = (4, 3\sqrt{3})$.

(B) Given $\lim _{x \rightarrow 0}\left(2+\frac{f(x)}{x^{3}}\right)=4$,we have $\lim _{x \rightarrow 0} \frac{f(x)}{x^{3}}=2$. Since $f(x)$ is a polynomial of degree $5$,let $f(x) = ax^5 + bx^4 + cx^3 + dx^2 + ex + g$. For the limit to exist and be $2$,we must have $g=e=d=0$ and $c=2$. Thus,$f(x) = ax^5 + bx^4 + 2x^3$.
$f'(x) = 5ax^4 + 4bx^3 + 6x^2$.
Since $x=\pm 1$ are critical points,$f'(1) = 5a + 4b + 6 = 0$ and $f'(-1) = 5a - 4b + 6 = 0$.
Adding these gives $10a + 12 = 0 \Rightarrow a = -6/5$. Subtracting gives $8b = 0 \Rightarrow b = 0$.
So,$f(x) = 2x^3 - \frac{6}{5}x^5$.
$f'(x) = 6x^2 - 6x^4 = 6x^2(1-x^2) = 6x^2(1-x)(1+x)$.
For $x < -1$,$f'(x) < 0$. For $-1 < x < 0$,$f'(x) > 0$. For $0 < x < 1$,$f'(x) > 0$. For $x > 1$,$f'(x) < 0$.
At $x = -1$,$f'(x)$ changes from negative to positive,so it is a point of local minima.
At $x = 1$,$f'(x)$ changes from positive to negative,so it is a point of local maxima.
Option $B$ states $x=1$ is a point of minima and $x=-1$ is a point of maxima,which is false.

(B) Since $f(x)$ is a polynomial of degree $3$,$f^{\prime \prime}(x)$ is a linear function. Given $f^{\prime}(x)$ has a critical point at $x=1$,$f^{\prime \prime}(1)=0$. Thus,$f^{\prime \prime}(x) = \lambda(x-1)$.
Integrating $f^{\prime \prime}(x)$,we get $f^{\prime}(x) = \frac{\lambda x^2}{2} - \lambda x + C$.
Since $f(x)$ has a critical point at $x=-1$,$f^{\prime}(-1) = 0$. Substituting $x=-1$ gives $\frac{\lambda}{2} + \lambda + C = 0$,so $C = -\frac{3\lambda}{2}$.
Thus,$f^{\prime}(x) = \frac{\lambda x^2}{2} - \lambda x - \frac{3\lambda}{2}$.
Integrating $f^{\prime}(x)$,we get $f(x) = \frac{\lambda x^3}{6} - \frac{\lambda x^2}{2} - \frac{3\lambda x}{2} + d$.
Using $f(1) = -6$: $\frac{\lambda}{6} - \frac{\lambda}{2} - \frac{3\lambda}{2} + d = -6 \Rightarrow -\frac{11\lambda}{6} + d = -6 \Rightarrow -11\lambda + 6d = -36 \dots (i)$.
Using $f(-1) = 10$: $-\frac{\lambda}{6} - \frac{\lambda}{2} + \frac{3\lambda}{2} + d = 10 \Rightarrow \frac{5\lambda}{6} + d = 10 \Rightarrow 5\lambda + 6d = 60 \dots (ii)$.
Subtracting $(i)$ from $(ii)$: $16\lambda = 96 \Rightarrow \lambda = 6$.
Substituting $\lambda = 6$ into $(ii)$: $5(6) + 6d = 60 \Rightarrow 30 + 6d = 60 \Rightarrow d = 5$.
So,$f(x) = x^3 - 3x^2 - 9x + 5$.
$f^{\prime}(x) = 3x^2 - 6x - 9 = 3(x^2 - 2x - 3) = 3(x-3)(x+1)$.
Setting $f^{\prime}(x) = 0$ gives critical points $x=3$ and $x=-1$.
$f^{\prime \prime}(x) = 6x - 6$. At $x=3$,$f^{\prime \prime}(3) = 18 - 6 = 12 > 0$,so $f(x)$ has a local minima at $x=3$.

(A) Given $F(x) = \int_{1}^{x} t^{2} g(t) dt$. By the Fundamental Theorem of Calculus,$F'(x) = x^{2} g(x) = x^{2} \int_{1}^{x} f(u) du$.
At $x=1$,$F'(1) = 1^{2} \int_{1}^{1} f(u) du = 0$. Thus,$x=1$ is a critical point.
Now,find the second derivative: $F''(x) = \frac{d}{dx} [x^{2} g(x)] = x^{2} g'(x) + 2x g(x)$.
Since $g(t) = \int_{1}^{t} f(u) du$,$g'(t) = f(t)$.
So,$F''(x) = x^{2} f(x) + 2x \int_{1}^{x} f(u) du$.
Evaluating at $x=1$: $F''(1) = 1^{2} f(1) + 2(1) \int_{1}^{1} f(u) du = f(1) + 0 = 3$.
Since $F'(1) = 0$ and $F''(1) = 3 > 0$,by the Second Derivative Test,$x=1$ is a point of local minima.

(N/A) The function is $f(x) = x^{2}$.
Since $x^{2} \geq 0$ for all $x \in R$,the minimum value of the function is $0$,which occurs at $x = 0$.
As $x \to \infty$ or $x \to -\infty$,$f(x) = x^{2} \to \infty$.
Therefore,the function $f(x) = x^{2}$ has no maximum value in $R$.

(N/A) From the graph of the given function,note that:
$f(x) \geq 0$ for all $x \in R$ and $f(x)=0$ if $x=0.$
Therefore,the function $f$ has a minimum value of $0$ and the point of minimum value of $f$ is $x=0.$
Also,the graph clearly shows that $f$ has no maximum value in $R$ and hence no point of maximum value in $R.$

(NONE) The given function $f(x) = x$ is a strictly increasing function in the open interval $(0, 1)$.
From the graph of the function,it appears that the minimum value should be at a point closest to $0$ on its right,and the maximum value should be at a point closest to $1$ on its left.
However,such points do not exist in the open interval $(0, 1)$.
For any point $x_0 \in (0, 1)$,we can always find a smaller point $\frac{x_0}{2} \in (0, 1)$ such that $\frac{x_0}{2} < x_0$. Thus,there is no minimum value.
Similarly,for any point $x_1 \in (0, 1)$,we can always find a larger point $\frac{x_1 + 1}{2} \in (0, 1)$ such that $\frac{x_1 + 1}{2} > x_1$. Thus,there is no maximum value.
Therefore,the function $f(x) = x$ has neither a maximum nor a minimum value in the interval $(0, 1)$.

(A) We have $f(x) = x^3 - 3x + 3$.
Taking the derivative,we get $f'(x) = 3x^2 - 3 = 3(x - 1)(x + 1)$.
Setting $f'(x) = 0$,we find critical points at $x = 1$ and $x = -1$.
To determine the nature of these points,we use the first derivative test:
For $x = 1$:
- For values close to $1$ and to the right,$f'(x) > 0$.
- For values close to $1$ and to the left,$f'(x) < 0$.
Since the sign changes from negative to positive,$x = 1$ is a point of local minima,and the local minimum value is $f(1) = 1 - 3 + 3 = 1$.
For $x = -1$:
- For values close to $-1$ and to the left,$f'(x) > 0$.
- For values close to $-1$ and to the right,$f'(x) < 0$.
Since the sign changes from positive to negative,$x = -1$ is a point of local maxima,and the local maximum value is $f(-1) = -1 + 3 + 3 = 5$.

(D) Given function is $f(x) = 2x^3 - 6x^2 + 6x + 5$.
First,we find the derivative $f'(x)$:
$f'(x) = \frac{d}{dx}(2x^3 - 6x^2 + 6x + 5) = 6x^2 - 12x + 6$.
Setting $f'(x) = 0$ to find critical points:
$6(x^2 - 2x + 1) = 0$
$6(x - 1)^2 = 0$
This gives $x = 1$ as the only critical point.
Now,we examine the sign of $f'(x)$ around $x = 1$:
For $x < 1$,let $x = 0$,then $f'(0) = 6(0-1)^2 = 6 > 0$.
For $x > 1$,let $x = 2$,then $f'(2) = 6(2-1)^2 = 6 > 0$.
Since $f'(x)$ does not change sign as $x$ passes through $1$ (it remains positive),$x = 1$ is neither a point of local maxima nor a point of local minima.
Therefore,the function has no local maxima and no local minima.

(A) The given function is $f(x) = 3 + |x|$.
Note that the function is not differentiable at $x = 0$,so the second derivative test cannot be applied.
We use the first derivative test. The critical point is $x = 0$.
For $x < 0$,$f(x) = 3 - x$,so $f'(x) = -1 < 0$.
For $x > 0$,$f(x) = 3 + x$,so $f'(x) = 1 > 0$.
Since the derivative changes sign from negative to positive at $x = 0$,$x = 0$ is a point of local minima.
The local minimum value is $f(0) = 3 + |0| = 3$.

(N/A) Given the function $f(x) = 3x^4 + 4x^3 - 12x^2 + 12$.
First,find the derivative $f'(x) = 12x^3 + 12x^2 - 24x$.
Factorizing the derivative,we get $f'(x) = 12x(x^2 + x - 2) = 12x(x - 1)(x + 2)$.
Setting $f'(x) = 0$,we find the critical points $x = 0, x = 1, x = -2$.
Now,find the second derivative $f''(x) = 36x^2 + 24x - 24$.
Evaluating the second derivative at the critical points:
$f''(0) = -24 < 0$,so $x = 0$ is a point of local maxima. The local maximum value is $f(0) = 12$.
$f''(1) = 36(1)^2 + 24(1) - 24 = 36 > 0$,so $x = 1$ is a point of local minima. The local minimum value is $f(1) = 3(1)^4 + 4(1)^3 - 12(1)^2 + 12 = 3 + 4 - 12 + 12 = 7$.
$f''(-2) = 36(-2)^2 + 24(-2) - 24 = 144 - 48 - 24 = 72 > 0$,so $x = -2$ is a point of local minima. The local minimum value is $f(-2) = 3(-2)^4 + 4(-2)^3 - 12(-2)^2 + 12 = 48 - 32 - 48 + 12 = -20$.

(NONE) Given the function $f(x) = 2x^3 - 6x^2 + 6x + 5$.
First,we find the first derivative: $f'(x) = 6x^2 - 12x + 6 = 6(x^2 - 2x + 1) = 6(x - 1)^2$.
Setting $f'(x) = 0$,we get $6(x - 1)^2 = 0$,which implies $x = 1$.
Now,we examine the sign of $f'(x)$ around $x = 1$.
For $x < 1$,$(x - 1)^2 > 0$,so $f'(x) > 0$.
For $x > 1$,$(x - 1)^2 > 0$,so $f'(x) > 0$.
Since the sign of $f'(x)$ does not change as $x$ passes through $1$ (it remains positive on both sides),the function $f(x)$ is strictly increasing at $x = 1$.
Therefore,$x = 1$ is neither a point of local maxima nor a point of local minima. It is a point of inflexion.

(A) Let the two positive numbers be $x$ and $y$. Given that $x + y = 15$,so $y = 15 - x$.
Let $S$ be the sum of the squares of the numbers,so $S = x^2 + y^2$.
Substituting $y = 15 - x$,we get $S(x) = x^2 + (15 - x)^2 = x^2 + 225 - 30x + x^2 = 2x^2 - 30x + 225$.
To find the minimum,we find the first derivative $S'(x) = 4x - 30$.
Setting $S'(x) = 0$,we get $4x = 30$,which implies $x = \frac{30}{4} = 7.5$.
Now,find the second derivative $S''(x) = 4$.
Since $S''(7.5) = 4 > 0$,the function $S(x)$ has a local minimum at $x = 7.5$.
Thus,the two numbers are $x = 7.5$ and $y = 15 - 7.5 = 7.5$.