Maxima and Minima Questions in English

(B) Given the function $f(x)=a x^3+b x^2-\frac{18}{5} x+\frac{19}{10}$.
Taking the derivative,$f^{\prime}(x)=3 a x^2+2 b x-\frac{18}{5}$.
Since $f(x)$ has a maximum at $x=-3$,$f^{\prime}(-3)=0$:
$3 a(-3)^2+2 b(-3)-\frac{18}{5}=0 \Rightarrow 27 a-6 b=\frac{18}{5} \Rightarrow 9 a-2 b=\frac{6}{5} \cdots (1)$.
Since $f(x)$ has a minimum at $x=2$,$f^{\prime}(2)=0$:
$3 a(2)^2+2 b(2)-\frac{18}{5}=0 \Rightarrow 12 a+4 b=\frac{18}{5} \Rightarrow 6 a+2 b=\frac{9}{5} \cdots (2)$.
Adding $(1)$ and $(2)$:
$(9 a-2 b)+(6 a+2 b)=\frac{6}{5}+\frac{9}{5} \Rightarrow 15 a=\frac{15}{5} \Rightarrow 15 a=3 \Rightarrow a=\frac{1}{5}$.
Substituting $a=\frac{1}{5}$ into $(2)$:
$6(\frac{1}{5})+2 b=\frac{9}{5} \Rightarrow 2 b=\frac{9}{5}-\frac{6}{5}=\frac{3}{5} \Rightarrow b=\frac{3}{10}$.
Thus,$f(x)=\frac{1}{5} x^3+\frac{3}{10} x^2-\frac{18}{5} x+\frac{19}{10}$.
Calculating $f(1)$:
$f(1)=\frac{1}{5}(1)^3+\frac{3}{10}(1)^2-\frac{18}{5}(1)+\frac{19}{10} = \frac{2}{10}+\frac{3}{10}-\frac{36}{10}+\frac{19}{10} = \frac{2+3-36+19}{10} = \frac{-12}{10} = \frac{-6}{5}$.
Therefore,$f(1)=\frac{-6}{5}$.

(A) Let $y = \frac{x^2+x+1}{x^2-x+1}$.
To find the critical points,we differentiate $y$ with respect to $x$ using the quotient rule:
$\frac{dy}{dx} = \frac{(x^2-x+1)(2x+1) - (x^2+x+1)(2x-1)}{(x^2-x+1)^2}$
$= \frac{(2x^3 - 2x^2 + 2x + x^2 - x + 1) - (2x^3 + 2x^2 + 2x - x^2 - x - 1)}{(x^2-x+1)^2}$
$= \frac{(2x^3 - x^2 + x + 1) - (2x^3 + x^2 + x - 1)}{(x^2-x+1)^2}$
$= \frac{-2x^2 + 2}{(x^2-x+1)^2} = \frac{2(1-x^2)}{(x^2-x+1)^2}$.
Setting $\frac{dy}{dx} = 0$,we get $1-x^2 = 0$,which implies $x = 1$ or $x = -1$.
For $x = 1$,$y = \frac{1+1+1}{1-1+1} = 3$ (Maximum value).
For $x = -1$,$y = \frac{1-1+1}{1+1+1} = \frac{1}{3}$ (Minimum value).
Thus,the minimum value $\frac{1}{3}$ occurs at $l = -1$ and the maximum value $3$ occurs at $m = 1$.
Therefore,$l+m = -1 + 1 = 0$.

(A) Given,$f(x) = x^3 - 4x^2 + 4x + 3$ on the interval $[-1, 3]$.
First,find the critical points by setting $f'(x) = 0$.
$f'(x) = 3x^2 - 8x + 4 = 0$.
$(3x - 2)(x - 2) = 0$,which gives $x = \frac{2}{3}$ and $x = 2$.
Both points lie within the interval $[-1, 3]$.
Now,evaluate $f(x)$ at the critical points and the endpoints:
$f(-1) = (-1)^3 - 4(-1)^2 + 4(-1) + 3 = -1 - 4 - 4 + 3 = -6$.
$f(\frac{2}{3}) = (\frac{2}{3})^3 - 4(\frac{2}{3})^2 + 4(\frac{2}{3}) + 3 = \frac{8}{27} - \frac{16}{9} + \frac{8}{3} + 3 = \frac{8 - 48 + 72 + 81}{27} = \frac{113}{27} \approx 4.18$.
$f(2) = (2)^3 - 4(2)^2 + 4(2) + 3 = 8 - 16 + 8 + 3 = 3$.
$f(3) = (3)^3 - 4(3)^2 + 4(3) + 3 = 27 - 36 + 12 + 3 = 6$.
Comparing these values: $\{-6, 4.18, 3, 6\}$,the minimum value is $-6$ at $x = -1$.

(B) Given the function $f(x) = 2x^2 - \ln|x|$ for $x \geq 1$.
Since $x \geq 1$,we have $|x| = x$,so $f(x) = 2x^2 - \ln x$.
Find the derivative: $f'(x) = 4x - \frac{1}{x}$.
For critical points,set $f'(x) = 0$:
$4x - \frac{1}{x} = 0 \implies 4x^2 = 1 \implies x^2 = \frac{1}{4} \implies x = \pm \frac{1}{2}$.
Since the domain is $x \geq 1$,there are no critical points in the interval $(1, \infty)$.
Check the behavior of the function: $f'(x) = \frac{4x^2 - 1}{x}$. For $x > 1$,$4x^2 - 1 > 3$,so $f'(x) > 0$.
Since $f'(x) > 0$ for all $x \geq 1$,the function is strictly increasing on $[1, \infty)$.
Therefore,the minimum value occurs at the boundary $x = 1$.
$f(1) = 2(1)^2 - \ln(1) = 2 - 0 = 2$.

(A) Let $P(x) = ax^3 + bx^2 + cx + d$.
Then $P'(x) = 3ax^2 + 2bx + c$.
Since $P(x)$ has an extreme value at $x=1$,we have $P'(1) = 0$,which implies $3a + 2b + c = 0$ ... $(i)$.
Given $\lim_{x \rightarrow 0} \left(\frac{P(x)+4}{x^2} + 2\right) = 6$,we have $\lim_{x \rightarrow 0} \frac{P(x)+4}{x^2} = 4$.
Substituting $P(x)$,we get $\lim_{x \rightarrow 0} \frac{ax^3 + bx^2 + cx + d + 4}{x^2} = 4$.
For the limit to be finite,the coefficients of $x^{-2}$ and $x^{-1}$ must be zero.
Thus,$d+4 = 0 \Rightarrow d = -4$ and $c = 0$.
Then $\lim_{x \rightarrow 0} (ax + b) = 4 \Rightarrow b = 4$.
Substituting $b=4$ and $c=0$ into equation $(i)$,we get $3a + 2(4) + 0 = 0 \Rightarrow 3a = -8 \Rightarrow a = -\frac{8}{3}$.
So,$P(x) = -\frac{8}{3}x^3 + 4x^2 - 4$.
The derivative is $P'(x) = -8x^2 + 8x$.
Evaluating at $x = \frac{1}{2}$,we get $P'(\frac{1}{2}) = -8(\frac{1}{4}) + 8(\frac{1}{2}) = -2 + 4 = 2$.

(B) The given function is $f(x) = (x - 3)^{2018}(x - 2)^{2019}$.
Applying the product rule,$f'(x) = 2018(x - 3)^{2017}(x - 2)^{2019} + 2019(x - 3)^{2018}(x - 2)^{2018}$.
Factoring out common terms,$f'(x) = (x - 3)^{2017}(x - 2)^{2018} \{2018(x - 2) + 2019(x - 3)\}$.
Simplifying the expression inside the braces,$2018x - 4036 + 2019x - 6057 = 4037x - 10093$.
Thus,$f'(x) = (x - 3)^{2017}(x - 2)^{2018} \cdot 4037 \left( x - \frac{10093}{4037} \right)$.
The critical points are $x = 3, 2, \frac{10093}{4037}$.
Analyzing the sign of $f'(x)$ around $x = \alpha = \frac{10093}{4037}$,the derivative changes from positive to negative,indicating a local maximum.
At $x = \alpha = \frac{10093}{4037}$,we have $x - 3 = \frac{10093 - 12111}{4037} = -\frac{2018}{4037}$ and $x - 2 = \frac{10093 - 8074}{4037} = \frac{2019}{4037}$.
Thus,$f(\alpha) = \left( -\frac{2018}{4037} \right)^{2018} \left( \frac{2019}{4037} \right)^{2019} = \left( \frac{2018}{4037} \right)^{2018} \left( \frac{2019}{4037} \right)^{2019}$.
Therefore,$2\alpha + 3f(\alpha) = 2 \left( \frac{10093}{4037} \right) + 3 \left( \frac{2018}{4037} \right)^{2018} \left( \frac{2019}{4037} \right)^{2019} = \frac{20186}{4037} + 3 \left( \frac{2018}{4037} \right)^{2018} \left( \frac{2019}{4037} \right)^{2019}$.
Note: The provided option $B$ has a minus sign,but based on the calculation,the result is positive.

(A) Let the radius of the semi-circle be $R = 2$. Let the rectangle have length $2x$ and breadth $y$ inscribed in the semi-circle.
By the Pythagorean theorem,$x^2 + y^2 = R^2 = 2^2 = 4$.
Thus,$x = \sqrt{4 - y^2}$.
The area of the rectangle is $A = (2x) \times y = 2y \sqrt{4 - y^2}$.
To maximize $A$,we maximize $A^2 = 4y^2(4 - y^2) = 16y^2 - 4y^4$.
Let $f(y) = 16y^2 - 4y^4$. Differentiating with respect to $y$,we get $f'(y) = 32y - 16y^3$.
Setting $f'(y) = 0$,we have $16y(2 - y^2) = 0$.
Since $y > 0$,$y^2 = 2$,so $y = \sqrt{2}$.
Then $x = \sqrt{4 - 2} = \sqrt{2}$.
The sides of the rectangle are $2x = 2\sqrt{2}$ and $y = \sqrt{2}$.
The smaller side is $\sqrt{2}$.

(B) Let the height of the solid cylinder be $h$ and its radius be $r$.
The volume $V$ is given by $V = \pi r^2 h$,which implies $h = \frac{V}{\pi r^2}$.
The total surface area $S$ is given by $S = 2\pi rh + 2\pi r^2$.
Substituting $h$ in terms of $V$ and $r$: $S = 2\pi r \left(\frac{V}{\pi r^2}\right) + 2\pi r^2 = \frac{2V}{r} + 2\pi r^2$.
To find the minimum surface area,differentiate $S$ with respect to $r$: $\frac{dS}{dr} = -\frac{2V}{r^2} + 4\pi r$.
Setting $\frac{dS}{dr} = 0$,we get $4\pi r = \frac{2V}{r^2}$,which implies $V = 2\pi r^3$.
Substituting $V = \pi r^2 h$ into this equation: $\pi r^2 h = 2\pi r^3$,which simplifies to $h = 2r$.
Since $2r$ is the diameter,the surface area is least when the height is equal to the diameter.

(C) An extreme point $c$ of a differentiable function $f$ satisfies $f'(c) = 0$.
For statement $Y$: Let $g(x) = rf(x)$. Then $g'(x) = rf'(x)$. At $x=c$,$g'(c) = rf'(c) = r(0) = 0$. Thus,$c$ is an extreme point of $rf$.
For statement $M$: Let $h(x) = r + f(x)$. Then $h'(x) = f'(x)$. At $x=c$,$h'(c) = f'(c) = 0$. Thus,$c$ is an extreme point of $r+f$.
Therefore,both statements are true.

(B) Given the function $f(x) = 2x^3 - 15x^2 + 36x - 8$.
To find the turning points,we calculate the derivative $f'(x)$ and set it to $0$:
$f'(x) = 6x^2 - 30x + 36 = 0$.
Dividing by $6$,we get $x^2 - 5x + 6 = 0$.
Factoring the quadratic equation: $(x - 2)(x - 3) = 0$.
Thus,the $x$-coordinates of the turning points are $x = 2$ and $x = 3$.
For $x = 2$,$f(2) = 2(8) - 15(4) + 36(2) - 8 = 16 - 60 + 72 - 8 = 20$.
For $x = 3$,$f(3) = 2(27) - 15(9) + 36(3) - 8 = 54 - 135 + 108 - 8 = 19$.
Given $\alpha < \gamma$,we have $\alpha = 2$ and $\gamma = 3$.
Correspondingly,$\beta = 20$ and $\delta = 19$.
Now,calculate $\alpha - \gamma - \beta + \delta = 2 - 3 - 20 + 19 = -2$.

(B) Let $R$ be the radius of the sphere and $r$ and $h$ be the radius and height of the cylinder,respectively.
From the geometry of the sphere and inscribed cylinder,we have the relation: $R^2 = r^2 + (h/2)^2$.
Thus,$r^2 = R^2 - h^2/4$.
The volume $V$ of the cylinder is given by $V = \pi r^2 h = \pi (R^2 - h^2/4) h = \pi R^2 h - \pi h^3/4$.
To find the maximum volume,we differentiate $V$ with respect to $h$: $\frac{dV}{dh} = \pi R^2 - \frac{3\pi h^2}{4}$.
Setting $\frac{dV}{dh} = 0$,we get $\pi R^2 = \frac{3\pi h^2}{4}$,which implies $h^2 = \frac{4R^2}{3}$,so $h = \frac{2R}{\sqrt{3}}$.
Given $R = 3 \ cm$,we have $h = \frac{2 \times 3}{\sqrt{3}} = 2 \sqrt{3} \ cm$.
Since $\frac{d^2V}{dh^2} = -\frac{6\pi h}{4} < 0$ for $h > 0$,the volume is maximum at $h = 2 \sqrt{3} \ cm$.

(B) Given the function $f(x) = x^2 e^{-2x}$ for $x > 0$.
To find the maximum value,we first find the derivative $f'(x)$:
$f'(x) = \frac{d}{dx}(x^2) \cdot e^{-2x} + x^2 \cdot \frac{d}{dx}(e^{-2x})$
$f'(x) = 2x e^{-2x} + x^2(-2)e^{-2x}$
$f'(x) = 2x e^{-2x}(1 - x)$
Setting $f'(x) = 0$ for critical points:
$2x e^{-2x}(1 - x) = 0$
Since $x > 0$ and $e^{-2x} \neq 0$,we have $1 - x = 0$,which gives $x = 1$.
Using the first derivative test:
For $0 < x < 1$,$f'(x) > 0$ (function is increasing).
For $x > 1$,$f'(x) < 0$ (function is decreasing).
Thus,$f(x)$ has a local maximum at $x = 1$.
The maximum value is $f(1) = (1)^2 e^{-2(1)} = e^{-2} = \frac{1}{e^2}$.

(A) Given,$y=x^3-3 x^2+5$.
On differentiating both sides with respect to $x$,we get $\frac{dy}{dx} = 3x^2 - 6x$.
For local maxima or local minima,we set $\frac{dy}{dx} = 0$.
$3x^2 - 6x = 0 \Rightarrow 3x(x-2) = 0 \Rightarrow x = 0$ or $x = 2$.
Now,differentiating $\frac{dy}{dx}$ with respect to $x$,we get $\frac{d^2y}{dx^2} = 6x - 6$.
At $x = 0$,$\frac{d^2y}{dx^2} = 6(0) - 6 = -6 < 0$.
Since the second derivative is negative at $x = 0$,$x = 0$ is a point of local maxima.
At $x = 2$,$\frac{d^2y}{dx^2} = 6(2) - 6 = 6 > 0$.
Since the second derivative is positive at $x = 2$,$x = 2$ is a point of local minima.
Therefore,the local maximum is attained at $x = 0$.

(C) Let the height of the cone be $h$ and the radius of the cone be $r$. Given,the radius of the sphere is $R$.
In $\triangle OPB$,by the Pythagorean theorem,we have:
$R^2 = r^2 + (h - R)^2$
$\Rightarrow r^2 = R^2 - (h - R)^2 = R^2 - (h^2 - 2Rh + R^2) = 2Rh - h^2$.
The volume $V$ of the cone is given by:
$V = \frac{1}{3} \pi r^2 h = \frac{1}{3} \pi (2Rh - h^2) h = \frac{\pi}{3} (2Rh^2 - h^3)$.
To find the maximum volume,differentiate $V$ with respect to $h$:
$\frac{dV}{dh} = \frac{\pi}{3} (4Rh - 3h^2)$.
Setting $\frac{dV}{dh} = 0$ for critical points:
$\frac{\pi}{3} h(4R - 3h) = 0$.
Since $h \neq 0$,we have $h = \frac{4R}{3}$.
Checking the second derivative:
$\frac{d^2V}{dh^2} = \frac{\pi}{3} (4R - 6h)$.
At $h = \frac{4R}{3}$,$\frac{d^2V}{dh^2} = \frac{\pi}{3} (4R - 6(\frac{4R}{3})) = \frac{\pi}{3} (4R - 8R) = -\frac{4\pi R}{3} < 0$.
Since the second derivative is negative,the volume is maximum at $h = \frac{4R}{3}$.

(B) Given the function $f(x) = x^3 + ax^2 + bx + c$.
To find the extreme values,we differentiate $f(x)$ with respect to $x$:
$f'(x) = 3x^2 + 2ax + b$.
For extreme values,we set $f'(x) = 0$:
$3x^2 + 2ax + b = 0$.
The discriminant of this quadratic equation is $D = (2a)^2 - 4(3)(b) = 4a^2 - 12b = 4(a^2 - 3b)$.
Given the condition $a^2 \leq 3b$,it follows that $a^2 - 3b \leq 0$,which implies $D \leq 0$.
If $D < 0$,the quadratic equation $f'(x) = 0$ has no real roots,meaning $f'(x)$ does not change sign and the function is strictly monotonic.
If $D = 0$,$f'(x) = 3(x + a/3)^2$,which is always $\geq 0$,so the function is monotonically increasing and has a point of inflection,not an extreme value.
Therefore,the function has no extreme value.

(B) Given the function $f(x) = (x-1)^2 + 3$ defined on the interval $x \in [-3, 1]$.
To find the critical points,we differentiate $f(x)$ with respect to $x$:
$f'(x) = 2(x-1)$.
Setting $f'(x) = 0$,we get $2(x-1) = 0$,which implies $x = 1$.
Since $x = 1$ is an endpoint of the interval $[-3, 1]$,we evaluate the function at the critical point and the boundaries:
At $x = 1$,$f(1) = (1-1)^2 + 3 = 3$.
At $x = -3$,$f(-3) = (-3-1)^2 + 3 = (-4)^2 + 3 = 16 + 3 = 19$.
Comparing the values,the minimum value $m = 3$ and the maximum value $M = 19$.
Thus,the ordered pair $(m, M)$ is $(3, 19)$.

(A) Given,$f(x)=x e^{-x}$.
First,find the derivative $f^{\prime}(x)$ using the product rule:
$f^{\prime}(x) = (1)e^{-x} + x(-e^{-x}) = e^{-x}(1-x)$.
For critical points,set $f^{\prime}(x) = 0$:
$e^{-x}(1-x) = 0 \Rightarrow x = 1$.
Next,find the second derivative $f^{\prime \prime}(x)$:
$f^{\prime \prime}(x) = -e^{-x}(1-x) + e^{-x}(-1) = e^{-x}(x-1-1) = e^{-x}(x-2)$.
Evaluate at $x=1$:
$f^{\prime \prime}(1) = e^{-1}(1-2) = -e^{-1} = -\frac{1}{e} < 0$.
Since $f^{\prime}(1) = 0$ and $f^{\prime \prime}(1) < 0$,the function $f(x)$ has a local maximum at $x=1$.
Therefore,both $(A)$ and $(R)$ are true and $(R)$ is the correct explanation for $(A)$.

(C) Let $y = 2x^2 + x - 1$.
To find the minimum value,we find the derivative $y' = 4x + 1$.
Setting $y' = 0$ for critical points,we get $4x + 1 = 0$,which implies $x = -\frac{1}{4}$.
The second derivative is $y'' = 4$,which is positive $(> 0)$,confirming that the function has a minimum at $x = -\frac{1}{4}$.
Substituting $x = -\frac{1}{4}$ into the original expression:
$y = 2(-\frac{1}{4})^2 + (-\frac{1}{4}) - 1$
$y = 2(\frac{1}{16}) - \frac{1}{4} - 1$
$y = \frac{1}{8} - \frac{2}{8} - \frac{8}{8} = -\frac{9}{8}$.
Thus,the minimum value is $-\frac{9}{8}$.

(A) Let the two numbers be $x$ and $y$.
Given that $x + y = 20$,so $y = 20 - x$.
Let the product be $P = x^2 y^3$.
Substituting $y$,we get $P(x) = x^2(20 - x)^3$.
To find the maximum,differentiate $P$ with respect to $x$:
$\frac{dP}{dx} = 2x(20 - x)^3 + x^2 \cdot 3(20 - x)^2(-1) = x(20 - x)^2 [2(20 - x) - 3x] = x(20 - x)^2 (40 - 5x)$.
Setting $\frac{dP}{dx} = 0$,we get $x = 0$,$x = 20$,or $x = 8$.
Since $x$ and $y$ must be positive,we test $x = 8$.
If $x = 8$,then $y = 20 - 8 = 12$.
Thus,the numbers are $8$ and $12$.

(C) Let $M = xy$.
Given $x + y = 7$,we can write $y = 7 - x$.
Substituting this into the expression for $M$,we get $M = x(7 - x) = 7x - x^2$.
To find the maximum value,we differentiate $M$ with respect to $x$: $\frac{dM}{dx} = 7 - 2x$.
Setting $\frac{dM}{dx} = 0$ for critical points,we get $7 - 2x = 0$,which implies $x = \frac{7}{2}$.
Now,we find the second derivative: $\frac{d^2M}{dx^2} = -2$.
Since $\frac{d^2M}{dx^2} < 0$,the function $M$ has a maximum at $x = \frac{7}{2}$.
The maximum value is $M = \frac{7}{2}(7 - \frac{7}{2}) = \frac{7}{2} \times \frac{7}{2} = \frac{49}{4}$.

(B) Let $f(x) = x \log x + x - 3$.
$f'(x) = x \cdot \frac{1}{x} + \log x + 1 = 1 + \log x + 1 = \log x + 2$.
For $x \in (1, 3)$,$\log x > 0$,so $f'(x) = \log x + 2 > 2 > 0$.
Thus,$f(x)$ is strictly increasing on $(1, 3)$.
$f(1) = 1 \cdot \log(1) + 1 - 3 = 0 + 1 - 3 = -2$.
$f(3) = 3 \log 3 + 3 - 3 = 3 \log 3 \approx 3(1.098) = 3.294 > 0$.
Since $f(1) < 0$ and $f(3) > 0$ and $f(x)$ is continuous and strictly increasing,by the Intermediate Value Theorem,there exists exactly one root in $(1, 3)$.

(C) Given the height function: $x(t) = 100t - \frac{25}{2}t^2$.
To find the maximum height,we differentiate $x$ with respect to $t$:
$\frac{dx}{dt} = 100 - 25t$.
For maximum height,set the first derivative to zero:
$100 - 25t = 0 \implies t = 4 \text{ s}$.
Now,check the second derivative to confirm it is a maximum:
$\frac{d^2x}{dt^2} = -25$.
Since $\frac{d^2x}{dt^2} < 0$,the function attains a maximum at $t = 4 \text{ s}$.
Substitute $t = 4$ into the original equation:
$x_{\text{max}} = 100(4) - \frac{25}{2}(4)^2 = 400 - \frac{25}{2}(16) = 400 - 200 = 200 \text{ m}$.

(B, C) Given the acceleration $f = \frac{dv}{dt} = 6 - \sqrt{1.2t}$.
At $t = 0$,$v = 0$.
The velocity is maximum when the acceleration $f = 0$.
Setting $f = 0$,we get $6 - \sqrt{1.2t} = 0 \Rightarrow \sqrt{1.2t} = 6 \Rightarrow 1.2t = 36 \Rightarrow t = \frac{36}{1.2} = 30 \text{ sec}$.
Thus,the time $T = 30 \text{ sec}$.
To find the maximum velocity $v$,we integrate the acceleration with respect to time from $t = 0$ to $t = 30$:
$v = \int_{0}^{30} (6 - \sqrt{1.2} \cdot t^{1/2}) dt$
$v = [6t - \sqrt{1.2} \cdot \frac{t^{3/2}}{3/2}]_{0}^{30}$
$v = 6(30) - \frac{2}{3} \sqrt{1.2} \cdot (30)^{3/2}$
$v = 180 - \frac{2}{3} \sqrt{1.2} \cdot 30 \sqrt{30} = 180 - 20 \sqrt{36} = 180 - 20(6) = 180 - 120 = 60 \text{ ft/sec}$.
Therefore,the maximum velocity is $v = 60 \text{ ft/sec}$ at $T = 30 \text{ sec}$.

(A) Given acceleration $v'(t) = a - kt^2$. Since the particle starts from rest,$v(0) = 0$.
Integrating with respect to $t$,we get $v(t) = \int (a - kt^2) dt = at - \frac{k}{3}t^3 + C$.
Using $v(0) = 0$,we find $C = 0$,so $v(t) = at - \frac{k}{3}t^3$.
To find the maximum velocity,we set the acceleration to zero: $a - kt^2 = 0$,which gives $t^2 = \frac{a}{k}$,or $t = \sqrt{\frac{a}{k}}$ (since $t > 0$).
Substituting this value of $t$ into the velocity equation:
$v_{\max} = a \sqrt{\frac{a}{k}} - \frac{k}{3} \left(\sqrt{\frac{a}{k}}\right)^3$
$v_{\max} = a \sqrt{\frac{a}{k}} - \frac{k}{3} \cdot \frac{a}{k} \sqrt{\frac{a}{k}}$
$v_{\max} = a \sqrt{\frac{a}{k}} - \frac{a}{3} \sqrt{\frac{a}{k}} = \frac{2a}{3} \sqrt{\frac{a}{k}} = \frac{2}{3} \sqrt{\frac{a^3}{k}}$.

(C) Given $f^{\prime}(x)=(x-1)^{2}(4-x).$
To find the critical points,we set $f^{\prime}(x)=0.$
$(x-1)^{2}(4-x)=0 \implies x=1, 4.$
We analyze the sign of $f^{\prime}(x)$ in the intervals $(-\infty, 1), (1, 4),$ and $(4, \infty).$
For $x \in (-\infty, 1),$ $f^{\prime}(x) > 0.$
For $x \in (1, 4),$ $f^{\prime}(x) > 0.$
For $x \in (4, \infty),$ $f^{\prime}(x) < 0.$
Since $f^{\prime}(x) > 0$ for $x \in (-\infty, 4),$ the function is increasing in $(-\infty, 4).$
Since $f^{\prime}(x) < 0$ for $x \in (4, \infty),$ the function is decreasing in $(4, \infty).$
At $x=4,$ $f^{\prime}(x)=0,$ so $x=4$ is a critical point.
Thus,option $C$ is correct.

(C) Given function is $f(x) = e^{(x^4 - x^3 + x^2)}$.
To find the minimum value,we find the derivative $f'(x)$:
$f'(x) = e^{(x^4 - x^3 + x^2)} \cdot \frac{d}{dx}(x^4 - x^3 + x^2)$
$f'(x) = e^{(x^4 - x^3 + x^2)} \cdot (4x^3 - 3x^2 + 2x)$
$f'(x) = e^{(x^4 - x^3 + x^2)} \cdot x(4x^2 - 3x + 2)$.
For the quadratic factor $4x^2 - 3x + 2$,the discriminant $D = (-3)^2 - 4(4)(2) = 9 - 32 = -23 < 0$.
Since the leading coefficient $4 > 0$ and $D < 0$,the expression $4x^2 - 3x + 2$ is always positive for all real $x$.
Thus,the sign of $f'(x)$ depends only on $x$.
For $x < 0$,$f'(x) < 0$,so the function is decreasing.
For $x > 0$,$f'(x) > 0$,so the function is increasing.
Therefore,the function attains its minimum value at $x = 0$.
The minimum value is $f(0) = e^{(0^4 - 0^3 + 0^2)} = e^0 = 1$.

(B) Since $p(x)$ has a local maximum at $x=1$ and a local minimum at $x=3$,its derivative $p^{\prime}(x)$ must have roots at $x=1$ and $x=3$. Thus,$p^{\prime}(x) = a(x-1)(x-3)$ for some constant $a \neq 0$.
Integrating $p^{\prime}(x)$,we get $p(x) = a(\frac{x^3}{3} - 2x^2 + 3x) + b$.
Using $p(1) = 6$: $a(\frac{1}{3} - 2 + 3) + b = 6 \implies \frac{4a}{3} + b = 6 \implies 4a + 3b = 18$.
Using $p(3) = 2$: $a(\frac{27}{3} - 2(9) + 3(3)) + b = 2 \implies a(9 - 18 + 9) + b = 2 \implies b = 2$.
Substituting $b = 2$ into $4a + 3b = 18$,we get $4a + 6 = 18 \implies 4a = 12 \implies a = 3$.
Thus,$p^{\prime}(x) = 3(x-1)(x-3)$.
Evaluating at $x=0$,$p^{\prime}(0) = 3(0-1)(0-3) = 3(-1)(-3) = 9$.

(C) To find the local maxima and minima,we first find the derivative of the function $f(x) = 2x^3 - 3x^2 - 12x + 4$.
$f'(x) = 6x^2 - 6x - 12$.
Setting $f'(x) = 0$ for critical points:
$6(x^2 - x - 2) = 0 \Rightarrow 6(x - 2)(x + 1) = 0$.
Thus,the critical points are $x = 2$ and $x = -1$.
Now,we find the second derivative $f''(x) = 12x - 6$.
For $x = -1$,$f''(-1) = 12(-1) - 6 = -18 < 0$,so $x = -1$ is a point of local maxima.
For $x = 2$,$f''(2) = 12(2) - 6 = 18 > 0$,so $x = 2$ is a point of local minima.
Therefore,the function has one local maximum and one local minimum.

(C) The function is defined as $f(x) = |x^2 - 1|$.
We can analyze the behavior of the function by looking at its graph or by examining its critical points.
$1$. The function $f(x) = |x^2 - 1|$ is non-negative for all $x \in R$.
$2$. At $x = 1$ and $x = -1$,$f(x) = |1^2 - 1| = 0$. Since $f(x) \ge 0$ for all $x$,the points $x = 1$ and $x = -1$ are points of local minima where the minimum value is $0$.
$3$. At $x = 0$,$f(0) = |0^2 - 1| = |-1| = 1$. For values of $x$ in a small neighborhood around $0$ (e.g.,$x \in (-1, 1)$),$f(x) = |x^2 - 1| = 1 - x^2$. Since $1 - x^2 < 1$ for all $x \neq 0$ in this neighborhood,$x = 0$ is a point of local maxima.
Thus,$f$ has local minima at $x = \pm 1$ and a local maxima at $x = 0$.

(B) The function is $f(x) = x(x - 1)(x - 2) \dots (x - 100)$.
This is a polynomial of degree $101$.
The roots of the function are $x = 0, 1, 2, \dots, 100$.
Since the leading coefficient is positive and the degree is odd,the graph starts from $-\infty$ and goes to $+\infty$.
Between any two consecutive roots $x = k$ and $x = k+1$,there must be at least one local extremum (by Rolle's Theorem).
There are $100$ intervals of the form $(k, k+1)$ for $k = 0, 1, \dots, 99$.
In each interval,there is exactly one local extremum.
Since the function is positive in $(0, 1)$,negative in $(1, 2)$,positive in $(2, 3)$,and so on,the local extrema alternate between local maxima and local minima.
The intervals are $(0, 1), (1, 2), (2, 3), \dots, (99, 100)$.
Local maxima occur in intervals $(0, 1), (2, 3), \dots, (98, 99)$.
The number of such intervals is $50$ (i.e.,$k = 0, 2, \dots, 98$).
Thus,there are $50$ local maxima.

(C) Given the function $f(x) = e^{\sin x} + e^{\cos x}$.
To find the critical points,we set the derivative $f'(x) = 0$:
$f'(x) = e^{\sin x} \cdot \cos x + e^{\cos x} \cdot (-\sin x) = 0$
$e^{\sin x} \cos x = e^{\cos x} \sin x$
$\frac{e^{\sin x}}{e^{\cos x}} = \frac{\sin x}{\cos x}$
$e^{\sin x - \cos x} = \tan x$
Since $f(x)$ is a periodic function with period $2\pi$,the solutions for $x$ occur periodically.
At $x = \frac{\pi}{4}$,$e^{\sin(\pi/4) - \cos(\pi/4)} = e^0 = 1$ and $\tan(\pi/4) = 1$. Thus,$x = \frac{\pi}{4}$ is a critical point.
Due to the periodicity of $\sin x$ and $\cos x$,the function $f(x)$ repeats its values every $2\pi$.
Therefore,the global maximum occurs at $x = 2n\pi + \frac{\pi}{4}$ for all integers $n$.
Since there are infinitely many such points,the global maximum exists at infinitely many points.

(B) Given $f(x) = 3x^{2/3} - x^2$.
To find the extrema,we differentiate $f(x)$ with respect to $x$:
$f'(x) = 3 \cdot \frac{2}{3} x^{-1/3} - 2x = 2x^{-1/3} - 2x = 2 \left( \frac{1 - x^{4/3}}{x^{1/3}} \right)$.
Setting $f'(x) = 0$,we get $1 - x^{4/3} = 0$,which implies $x^{4/3} = 1$,so $x = 1$ or $x = -1$.
Also,$f'(x)$ is undefined at $x = 0$.
We analyze the sign of $f'(x)$ around the critical points $x = -1, 0, 1$:
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 positive to negative,so $f$ has a local maximum at $x = -1$.
At $x = 0$,$f'(x)$ changes from negative to positive,so $f$ has a local minimum at $x = 0$.
At $x = 1$,$f'(x)$ changes from positive to negative,so $f$ has a local maximum at $x = 1$.
Thus,$f$ is maximum at two points $x = 1$ and $x = -1$.

(C) Given $f(x)=(x-2)^{17}(x+5)^{24}$.
To find the critical points,we find the derivative $f'(x)$:
$f'(x) = 17(x-2)^{16}(x+5)^{24} + 24(x-2)^{17}(x+5)^{23}$
$f'(x) = (x-2)^{16}(x+5)^{23} [17(x+5) + 24(x-2)]$
$f'(x) = (x-2)^{16}(x+5)^{23} [17x + 85 + 24x - 48]$
$f'(x) = (x-2)^{16}(x+5)^{23} (41x + 37)$
The critical points are $x=2, x=-5, x=-\frac{37}{41}$.
Now,examine the sign of $f'(x)$ around $x=2$:
Since $(x-2)^{16}$ is always non-negative,the sign of $f'(x)$ depends on $(x+5)^{23}(41x+37)$.
For $x$ slightly less than $2$,$(x+5)^{23} > 0$ and $(41x+37) > 0$,so $f'(x) > 0$.
For $x$ slightly greater than $2$,$(x+5)^{23} > 0$ and $(41x+37) > 0$,so $f'(x) > 0$.
Since the sign of $f'(x)$ does not change as $x$ passes through $2$,$f(x)$ has neither a maximum nor a minimum at $x=2$.

(C) Given $f(x) = e^{\sin x} + e^{\cos x}$.
To find the maximum value,we use the Arithmetic Mean-Geometric Mean ($AM$-$GM$) inequality or derivative analysis.
Let $u = \sin x$ and $v = \cos x$. We know that $u^2 + v^2 = 1$.
Using the property that for $f(x) = e^{\sin x} + e^{\cos x}$,the function attains its maximum when $\sin x = \cos x$.
Setting $\sin x = \cos x$,we get $\tan x = 1$,which implies $x = \frac{\pi}{4} + n\pi$.
For $x = \frac{\pi}{4}$,$\sin x = \frac{1}{\sqrt{2}}$ and $\cos x = \frac{1}{\sqrt{2}}$.
Substituting these values into $f(x)$:
$f\left(\frac{\pi}{4}\right) = e^{1/\sqrt{2}} + e^{1/\sqrt{2}} = 2e^{1/\sqrt{2}}$.
Thus,the maximum value is $2e^{1/\sqrt{2}}$.

(C) Let the radius of the circular sector be $r$ and the arc length be $\ell$.
Given that the total length of the wire is $20 \ m$,the perimeter of the sector is $2r + \ell = 20$.
Thus,$\ell = 20 - 2r$.
The area $A$ of a circular sector is given by $A = \frac{1}{2} r \ell$.
Substituting the value of $\ell$,we get $A = \frac{1}{2} r (20 - 2r) = 10r - r^2$.
To find the maximum area,we differentiate $A$ with respect to $r$ and set it to zero:
$\frac{dA}{dr} = 10 - 2r = 0$.
Solving for $r$,we get $2r = 10$,which implies $r = 5 \ m$.
To verify that this is a maximum,we check the second derivative: $\frac{d^2A}{dr^2} = -2$,which is less than $0$,confirming that the area is maximum at $r = 5 \ m$.

(C) The function is defined as $f(x) = |x^{2} - 1|$.
We can analyze the behavior of the function by looking at its graph or by testing points.
At $x = \pm 1$,$f(x) = |(\pm 1)^{2} - 1| = |1 - 1| = 0$. Since the absolute value function is always non-negative,$f(x) \geq 0$ for all $x \in R$. Thus,$f(x) = 0$ at $x = \pm 1$ represents the absolute minimum value of the function,which is also a local minimum.
At $x = 0$,$f(0) = |0^{2} - 1| = |-1| = 1$. For values of $x$ close to $0$,say $x = 0.1$ or $x = -0.1$,$f(0.1) = |(0.1)^{2} - 1| = |0.01 - 1| = 0.99$. Since $f(0) = 1 > 0.99$,$x = 0$ is a point of local maximum.
Therefore,$f$ has local minima at $x = \pm 1$ and a local maximum at $x = 0$.

(B) Given $f(x) = \tan^{-1} x - \frac{1}{2} \ln x$ on the interval $x \in \left[\frac{1}{\sqrt{3}}, \sqrt{3}\right]$.
First,find the derivative $f'(x)$:
$f'(x) = \frac{1}{1+x^2} - \frac{1}{2x} = \frac{2x - (1+x^2)}{2x(1+x^2)} = \frac{-(x^2 - 2x + 1)}{2x(1+x^2)} = \frac{-(x-1)^2}{2x(1+x^2)}$.
Since $-(x-1)^2 \leq 0$ and $2x(1+x^2) > 0$ for $x \in \left[\frac{1}{\sqrt{3}}, \sqrt{3}\right]$,we have $f'(x) \leq 0$ for all $x$ in the interval.
Thus,$f(x)$ is a strictly decreasing function on the given interval.
The maximum value occurs at the left endpoint $x = \frac{1}{\sqrt{3}}$:
$f_{\max} = f\left(\frac{1}{\sqrt{3}}\right) = \tan^{-1}\left(\frac{1}{\sqrt{3}}\right) - \frac{1}{2} \ln\left(\frac{1}{\sqrt{3}}\right) = \frac{\pi}{6} - \frac{1}{2} \ln(3^{-1/2}) = \frac{\pi}{6} + \frac{1}{4} \ln 3$.
The minimum value occurs at the right endpoint $x = \sqrt{3}$:
$f_{\min} = f(\sqrt{3}) = \tan^{-1}(\sqrt{3}) - \frac{1}{2} \ln(\sqrt{3}) = \frac{\pi}{3} - \frac{1}{2} \ln(3^{1/2}) = \frac{\pi}{3} - \frac{1}{4} \ln 3$.

(C) Given the function $f(x) = 1 - \sqrt{x^2}$.
Since $\sqrt{x^2} = |x|$,the function can be rewritten as $f(x) = 1 - |x|$.
At $x = 0$,$f(0) = 1 - |0| = 1$.
For any $x \neq 0$,$|x| > 0$,so $f(x) = 1 - |x| < 1$.
Since $f(x) \leq f(0)$ for all $x$ in the domain,the function attains its maximum value at $x = 0$.
Therefore,$f$ has a maxima at $x = 0$.

(A) Given the function $f(x)=2x^{3}-9ax^{2}+12a^{2}x+1$.
First,find the derivative $f'(x)$:
$f'(x) = 6x^{2}-18ax+12a^{2}$.
Set $f'(x)=0$ to find critical points:
$6(x^{2}-3ax+2a^{2})=0 \Rightarrow 6(x-a)(x-2a)=0$.
So,the critical points are $x=a$ and $x=2a$.
Now,find the second derivative $f''(x)$:
$f''(x) = 12x-18a$.
Check the nature of the critical points:
For $x=a$: $f''(a) = 12a-18a = -6a$. Since $a>0$,$f''(a) < 0$,so $f(x)$ has a local maximum at $p=a$.
For $x=2a$: $f''(2a) = 12(2a)-18a = 6a$. Since $a>0$,$f''(2a) > 0$,so $f(x)$ has a local minimum at $q=2a$.
Given the condition $p^{2}=q$,substitute the values:
$a^{2} = 2a$.
Since $a>0$,we can divide by $a$:
$a = 2$.

(B, C) Let $S_1$ be the displacement of the first particle and $S_2$ be the displacement of the second particle at time $t$.
For the first particle: $S_1 = u t$.
For the second particle: $S_2 = \frac{1}{2} f t^2$.
The distance between them is $D(t) = |S_1 - S_2| = |u t - \frac{1}{2} f t^2|$.
To find the greatest distance,we differentiate $D(t)$ with respect to $t$ and set it to zero:
$\frac{dD}{dt} = u - f t = 0$.
This gives $t = \frac{u}{f}$.
At $t = \frac{u}{f}$,the distance is $D = u(\frac{u}{f}) - \frac{1}{2} f(\frac{u}{f})^2 = \frac{u^2}{f} - \frac{u^2}{2f} = \frac{u^2}{2f}$.
Thus,the particles are at the greatest distance at $t = \frac{u}{f}$ and the maximum distance is $\frac{u^2}{2f}$.
Therefore,options $B$ and $C$ are correct.

(C) Given $f(x) = \exp\left(x^{\frac{1}{x}}\right)$. Let $g(x) = x^{\frac{1}{x}}$. Since $\exp(u)$ is an increasing function,the minimum of $f(x)$ occurs where $g(x)$ is minimum.
Taking the natural logarithm of $g(x)$,we have $\ln(g(x)) = \frac{\ln x}{x}$.
Differentiating with respect to $x$: $\frac{g'(x)}{g(x)} = \frac{x(\frac{1}{x}) - \ln x(1)}{x^2} = \frac{1 - \ln x}{x^2}$.
Setting $g'(x) = 0$ gives $1 - \ln x = 0$,so $x = e \approx 2.718$.
For $x < e$,$g'(x) > 0$ (increasing),and for $x > e$,$g'(x) < 0$ (decreasing).
Since $e \in [2, 5]$,the function $g(x)$ increases on $[2, e]$ and decreases on $[e, 5]$.
The minimum value of $g(x)$ on $[2, 5]$ must occur at the endpoints $x = 2$ or $x = 5$.
Comparing $g(2) = 2^{1/2} = \sqrt{2} \approx 1.414$ and $g(5) = 5^{1/5} \approx 1.3797$.
Since $g(5) < g(2)$,the minimum value of $g(x)$ is $5^{1/5}$.
Thus,the minimum value of $f(x)$ is $\exp\left(5^{\frac{1}{5}}\right)$.

(B) Given,$f(x) = 2|x-1| + |x-2|$.
We analyze the function in different intervals:
Case $1$: If $x < 1$,then $f(x) = -2(x-1) - (x-2) = -2x + 2 - x + 2 = -3x + 4$. Since the slope is $-3$,the function is decreasing in this interval.
Case $2$: If $1 \leq x < 2$,then $f(x) = 2(x-1) - (x-2) = 2x - 2 - x + 2 = x$. Since the slope is $1$,the function is increasing in this interval.
Case $3$: If $x \geq 2$,then $f(x) = 2(x-1) + (x-2) = 2x - 2 + x - 2 = 3x - 4$. Since the slope is $3$,the function is increasing in this interval.
Comparing these,the function decreases until $x=1$ and then starts increasing.
Therefore,the minimum value occurs at $x=1$.
$f(1) = 2|1-1| + |1-2| = 2(0) + |-1| = 0 + 1 = 1$.

(C) Given,$f(x)=\sin x+2 \cos ^{2} x$ for $x \in \left[\frac{\pi}{4}, \frac{3 \pi}{4}\right]$.
$f'(x) = \cos x - 4 \cos x \sin x = \cos x(1 - 2 \sin 2x)$ is incorrect. Let us differentiate correctly:
$f'(x) = \cos x + 2(2 \cos x)(-\sin x) = \cos x - 2 \sin 2x$.
Setting $f'(x) = 0$,we have $\cos x(1 - 4 \sin x) = 0$.
This gives $\cos x = 0$ or $\sin x = \frac{1}{4}$.
For $x \in \left[\frac{\pi}{4}, \frac{3 \pi}{4}\right]$,$\cos x = 0 \Rightarrow x = \frac{\pi}{2}$.
Also,$\sin x = \frac{1}{4}$ has no solution in this interval since $\sin x \geq \sin \frac{\pi}{4} = \frac{1}{\sqrt{2}} \approx 0.707 > 0.25$.
Checking the critical point and endpoints:
$f\left(\frac{\pi}{4}\right) = \sin \frac{\pi}{4} + 2 \cos^2 \frac{\pi}{4} = \frac{1}{\sqrt{2}} + 2(\frac{1}{2}) = \frac{1}{\sqrt{2}} + 1 \approx 1.707$.
$f\left(\frac{\pi}{2}\right) = \sin \frac{\pi}{2} + 2 \cos^2 \frac{\pi}{2} = 1 + 0 = 1$.
$f\left(\frac{3 \pi}{4}\right) = \sin \frac{3 \pi}{4} + 2 \cos^2 \frac{3 \pi}{4} = \frac{1}{\sqrt{2}} + 2(\frac{1}{2}) = \frac{1}{\sqrt{2}} + 1 \approx 1.707$.
Thus,$f(x)$ attains its minimum at $x = \frac{\pi}{2}$.