Maxima and Minima Questions in English

(B) Given,$f(x) = x^{25}(1-x)^{75}$.
To find the maximum value,we calculate the derivative $f'(x)$ using the product rule:
$f'(x) = 25x^{24}(1-x)^{75} - 75x^{25}(1-x)^{74}$.
Factor out the common terms:
$f'(x) = 25x^{24}(1-x)^{74} [ (1-x) - 3x ]$.
$f'(x) = 25x^{24}(1-x)^{74} (1-4x)$.
Setting $f'(x) = 0$ gives critical points $x = 0, 1, 1/4$.
For $x \in (0, 1/4)$,$f'(x) > 0$ (function is increasing).
For $x \in (1/4, 1)$,$f'(x) < 0$ (function is decreasing).
Since the derivative changes sign from positive to negative at $x = 1/4$,the function attains its maximum value at $x = 1/4$.

(C) Given,$f(x)=2 x^{3}-9 a x^{2}+12 a^{2} x+1$. The critical points are found by setting $f^{\prime}(x)=0$.
$f^{\prime}(x) = 6x^{2} - 18ax + 12a^{2} = 6(x^{2} - 3ax + 2a^{2}) = 6(x-a)(x-2a) = 0$.
So,the critical points are $x=a$ and $x=2a$.
We find the second derivative $f^{\prime \prime}(x) = 12x - 18a$.
For a maximum at $p$,$f^{\prime \prime}(p) < 0 \Rightarrow 12p - 18a < 0 \Rightarrow p < \frac{3}{2}a$.
For a minimum at $q$,$f^{\prime \prime}(q) > 0 \Rightarrow 12q - 18a > 0 \Rightarrow q > \frac{3}{2}a$.
Comparing the critical points $a$ and $2a$ with these conditions,we get $p=a$ and $q=2a$.
Given $p^{2}=q$,we substitute the values: $a^{2} = 2a$.
$a^{2} - 2a = 0 \Rightarrow a(a-2) = 0$.
Thus,$a=0$ or $a=2$.
If $a=0$,$f(x)=2x^{3}+1$,which is a strictly increasing function and has no local maxima or minima.
Therefore,$a=2$ is the only valid solution.

(C) Given $f(x)=x^3-3(a-2)x^2+3ax+7$.
Since $f(x)$ is increasing in $(0,1]$ and decreasing in $[1,5)$,$f(x)$ must have a local maximum at $x=1$.
Thus,$f'(1)=0$.
$f'(x)=3x^2-6(a-2)x+3a$.
Substituting $x=1$: $3(1)^2-6(a-2)(1)+3a=0$.
$3-6a+12+3a=0 \Rightarrow -3a+15=0 \Rightarrow a=5$.
Now,substitute $a=5$ into $f(x)$: $f(x)=x^3-3(5-2)x^2+3(5)x+7 = x^3-9x^2+15x+7$.
We need to solve $\frac{f(x)-14}{(x-1)^2}=0$.
$f(x)-14 = x^3-9x^2+15x+7-14 = x^3-9x^2+15x-7$.
By synthetic division or polynomial division,$x^3-9x^2+15x-7 = (x-1)^2(x-7)$.
So,$\frac{(x-1)^2(x-7)}{(x-1)^2} = 0 \Rightarrow x-7=0 \Rightarrow x=7$.

(A) Given that $x+y=60$,so $x=60-y$.
Let $f(y) = x y^3 = (60-y) y^3 = 60 y^3 - y^4$.
To find the maximum,we differentiate $f(y)$ with respect to $y$:
$f'(y) = \frac{d}{dy}(60 y^3 - y^4) = 180 y^2 - 4 y^3$.
Setting $f'(y) = 0$ for critical points:
$4 y^2(45 - y) = 0$.
Since $y$ is a positive number,$y=45$.
Now,we check the second derivative:
$f''(y) = 360 y - 12 y^2$.
At $y=45$,$f''(45) = 360(45) - 12(45)^2 = 45(360 - 540) = 45(-180) < 0$.
Since the second derivative is negative,the function is maximum at $y=45$.
Then $x = 60 - 45 = 15$.
Thus,the numbers are $x=15$ and $y=45$.

(B) Let $y = \frac{1-x+x^{2}}{1+x+x^{2}}$.
We can rewrite the expression as $y = \frac{(x^{2}+x+1) - 2x}{x^{2}+x+1} = 1 - \frac{2x}{x^{2}+x+1}$.
To find the minimum value of $y$,we need to maximize the term $\frac{2x}{x^{2}+x+1}$.
Let $f(x) = \frac{x}{x^{2}+x+1}$.
Taking the derivative with respect to $x$ and setting it to $0$:
$f'(x) = \frac{(x^{2}+x+1)(1) - x(2x+1)}{(x^{2}+x+1)^{2}} = \frac{x^{2}+x+1-2x^{2}-x}{(x^{2}+x+1)^{2}} = \frac{1-x^{2}}{(x^{2}+x+1)^{2}}$.
Setting $f'(x) = 0$ gives $1-x^{2} = 0$,so $x = 1$ or $x = -1$.
For $x = 1$,$f(1) = \frac{1}{1+1+1} = \frac{1}{3}$.
For $x = -1$,$f(-1) = \frac{-1}{1-1+1} = -1$.
Since we want to minimize $y = 1 - 2f(x)$,we choose the maximum value of $f(x)$,which is $1/3$.
Thus,the minimum value of $y$ is $1 - 2(1/3) = 1 - 2/3 = 1/3$.

(A) Given the function $y=a \log x+b x^2+x$.
Find the derivative with respect to $x$: $\frac{d y}{d x}=\frac{a}{x}+2 b x+1$.
Since the function has extreme values at $x=-1$ and $x=2$,the derivative must be zero at these points.
At $x=-1$: $\frac{a}{-1}+2 b(-1)+1=0 \Rightarrow -a-2 b+1=0 \Rightarrow a+2 b=1$ (Equation $i$).
At $x=2$: $\frac{a}{2}+2 b(2)+1=0 \Rightarrow \frac{a}{2}+4 b+1=0 \Rightarrow a+8 b=-2$ (Equation $ii$).
Subtracting Equation $i$ from Equation $ii$: $(a+8 b)-(a+2 b)=-2-1 \Rightarrow 6 b=-3 \Rightarrow b=-\frac{1}{2}$.
Substitute $b=-\frac{1}{2}$ into Equation $i$: $a+2(-\frac{1}{2})=1 \Rightarrow a-1=1 \Rightarrow a=2$.
Therefore,$a+b=2+(-\frac{1}{2})=\frac{3}{2}$.

(C) Let $f(x) = \frac{1-x+x^2}{1+x+x^2}$.
Using the quotient rule,$f'(x) = \frac{(1+x+x^2)(-1+2x) - (1-x+x^2)(1+2x)}{(1+x+x^2)^2}$.
Expanding the numerator: $(1+x+x^2)(-1+2x) = -1+2x-x+2x^2-x^2+2x^3 = 2x^3+x^2+x-1$.
$(1-x+x^2)(1+2x) = 1+2x-x-2x^2+x^2+2x^3 = 2x^3-x^2+x+1$.
Subtracting these: $(2x^3+x^2+x-1) - (2x^3-x^2+x+1) = 2x^2-2$.
Thus,$f'(x) = \frac{2x^2-2}{(1+x+x^2)^2}$.
Setting $f'(x) = 0$ gives $2x^2-2 = 0$,so $x^2 = 1$,which means $x = 1$ or $x = -1$.
Evaluating $f(x)$ at these points:
$f(1) = \frac{1-1+1}{1+1+1} = \frac{1}{3}$.
$f(-1) = \frac{1-(-1)+(-1)^2}{1+(-1)+(-1)^2} = \frac{1+1+1}{1-1+1} = \frac{3}{1} = 3$.
Therefore,the minimum value of $f(x)$ is $\frac{1}{3}$.

(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 the function changes behavior at $x = 1$,we have $f'(1) = 0$.
$3(1)^2 - 6(a - 2)(1) + 3a = 0$
$3 - 6a + 12 + 3a = 0$
$-3a + 15 = 0 \Rightarrow a = 5$.
Substituting $a = 5$ into $f(x)$,we get $f(x) = x^3 - 3(5 - 2)x^2 + 3(5)x + 7 = x^3 - 9x^2 + 15x + 7$.
Now,solve $\frac{f(x) - 14}{(x - 1)^2} = 0$ for $x \neq 1$.
$f(x) - 14 = 0 \Rightarrow x^3 - 9x^2 + 15x + 7 - 14 = 0$.
$x^3 - 9x^2 + 15x - 7 = 0$.
Since $x = 1$ is a root,we divide by $(x - 1)^2$:
$(x - 1)^2(x - 7) = 0$.
The roots are $x = 1$ and $x = 7$.
Since $x \neq 1$,the root is $x = 7$.

(D) Let $f(\theta) = a \sec \theta - b \tan \theta$.
Then $f'(\theta) = a \sec \theta \tan \theta - b \sec^2 \theta = \sec \theta (a \tan \theta - b \sec \theta)$.
Setting $f'(\theta) = 0$,we get $a \tan \theta - b \sec \theta = 0$ (since $\sec \theta \neq 0$ for $0 < \theta < \frac{\pi}{2}$).
This implies $a \sin \theta = b$,or $\sin \theta = \frac{b}{a}$.
Since $\sin \theta = \frac{b}{a}$,we have $\cos \theta = \sqrt{1 - \frac{b^2}{a^2}} = \frac{\sqrt{a^2 - b^2}}{a}$.
Thus,$\sec \theta = \frac{a}{\sqrt{a^2 - b^2}}$ and $\tan \theta = \frac{b}{\sqrt{a^2 - b^2}}$.
The second derivative $f''(\theta) = a \sec \theta (1 + 2 \tan^2 \theta) > 0$,confirming a minimum at $\sin \theta = \frac{b}{a}$.
The minimum value is $f(\theta) = a \left( \frac{a}{\sqrt{a^2 - b^2}} \right) - b \left( \frac{b}{\sqrt{a^2 - b^2}} \right) = \frac{a^2 - b^2}{\sqrt{a^2 - b^2}} = \sqrt{a^2 - b^2}$.

(B) $f(x) = \bar{a} \cdot (\bar{b} \times \bar{c}) = \begin{vmatrix} x & -2 & 3 \\ -2 & x & -1 \\ 7 & -2 & x \end{vmatrix}$
Expanding the determinant:
$f(x) = x(x^2 - 2) + 2(-2x + 7) + 3(4 - 7x)$
$f(x) = x^3 - 2x - 4x + 14 + 12 - 21x$
$f(x) = x^3 - 27x + 26$
To find local minima,we find $f'(x) = 0$:
$f'(x) = 3x^2 - 27 = 0$
$3(x^2 - 9) = 0 \Rightarrow x = \pm 3$
Using the second derivative test:
$f''(x) = 6x$
$f''(3) = 18 > 0$,so $x_0 = 3$ is the point of local minima.
At $x = 3$:
$\bar{a} = 3 \hat{i} - 2 \hat{j} + 3 \hat{k}$
$\bar{b} = -2 \hat{i} + 3 \hat{j} - \hat{k}$
$\bar{a} \cdot \bar{b} = (3)(-2) + (-2)(3) + (3)(-1)$
$\bar{a} \cdot \bar{b} = -6 - 6 - 3 = -15$

(B) The function $f(x)$ is defined as the scalar triple product of vectors $\overline{a}, \overline{b}, \text{ and } \overline{c}$:
$f(x) = \begin{vmatrix} x & -2 & 3 \\ -2 & x & -1 \\ 7 & -2 & x \end{vmatrix}$
Expanding the determinant along the first row:
$f(x) = x(x^2 - 2) - (-2)(-2x + 7) + 3(4 - 7x)$
$f(x) = x^3 - 2x + 4x - 14 + 12 - 21x$
$f(x) = x^3 - 19x - 2$ (Note: Correcting the expansion: $x(x^2-2) + 2(-2x+7) + 3(4-7x) = x^3 - 2x - 4x + 14 + 12 - 21x = x^3 - 27x + 26$)
To find the local minima,we find the critical points by setting $f'(x) = 0$:
$f'(x) = 3x^2 - 27 = 0$
$3(x^2 - 9) = 0 \Rightarrow x = \pm 3$
Using the second derivative test,$f''(x) = 6x$:
At $x = 3$,$f''(3) = 18 > 0$,so $x_0 = 3$ is the point of local minima.
At $x = 3$,$\overline{a} = 3 \hat{i} - 2 \hat{j} + 3 \hat{k}$ and $\overline{b} = -2 \hat{i} + 3 \hat{j} - \hat{k}$.
Calculating the dot product $\overline{a} \cdot \overline{b}$:
$\overline{a} \cdot \overline{b} = (3)(-2) + (-2)(3) + (3)(-1) = -6 - 6 - 3 = -15$.

(D) To find the absolute maximum value of $f(x) = \sin x + \cos x$ on the interval $[0, \pi]$,we first find the critical points by setting the derivative to zero.
$f'(x) = \cos x - \sin x$.
Setting $f'(x) = 0$,we get $\cos x = \sin x$,which implies $\tan x = 1$.
For $x \in [0, \pi]$,the solution is $x = \pi/4$.
Now,we evaluate $f(x)$ at the critical point and the endpoints of the interval:
$1$. At $x = 0$: $f(0) = \sin(0) + \cos(0) = 0 + 1 = 1$.
$2$. At $x = \pi/4$: $f(\pi/4) = \sin(\pi/4) + \cos(\pi/4) = 1/\sqrt{2} + 1/\sqrt{2} = 2/\sqrt{2} = \sqrt{2}$.
$3$. At $x = \pi$: $f(\pi) = \sin(\pi) + \cos(\pi) = 0 - 1 = -1$.
Comparing these values,the absolute maximum value is $\sqrt{2}$.

(B) To find the maximum value of $f(x) = [x(x-1) + 1]^{\frac{1}{3}}$ on the interval $[0, 1]$,we first find the derivative $f'(x)$.
Let $g(x) = x(x-1) + 1 = x^2 - x + 1$.
Then $f(x) = [g(x)]^{\frac{1}{3}}$.
$f'(x) = \frac{1}{3} [g(x)]^{-\frac{2}{3}} \cdot g'(x) = \frac{1}{3} [x^2 - x + 1]^{-\frac{2}{3}} (2x - 1)$.
Setting $f'(x) = 0$,we get $2x - 1 = 0$,which implies $x = \frac{1}{2}$.
Now,we evaluate $f(x)$ at the critical point and the endpoints:
$f(0) = [0(0-1) + 1]^{\frac{1}{3}} = 1^{\frac{1}{3}} = 1$.
$f(1) = [1(1-1) + 1]^{\frac{1}{3}} = 1^{\frac{1}{3}} = 1$.
$f(\frac{1}{2}) = [\frac{1}{2}(\frac{1}{2}-1) + 1]^{\frac{1}{3}} = [-\frac{1}{4} + 1]^{\frac{1}{3}} = (\frac{3}{4})^{\frac{1}{3}}$.
Comparing the values $1$,$1$,and $(\frac{3}{4})^{\frac{1}{3}}$,the maximum value is $1$.

(A) To find the local maximum value of $f(x) = x + \frac{1}{x}$,we first find the derivative $f'(x) = 1 - \frac{1}{x^2}$.
Setting $f'(x) = 0$,we get $1 = \frac{1}{x^2}$,which implies $x^2 = 1$,so $x = 1$ or $x = -1$.
We use the second derivative test: $f''(x) = \frac{2}{x^3}$.
For $x = 1$,$f''(1) = 2 > 0$,so $x = 1$ is a local minimum.
For $x = -1$,$f''(-1) = \frac{2}{(-1)^3} = -2 < 0$,so $x = -1$ is a local maximum.
The local maximum value is $f(-1) = -1 + \frac{1}{-1} = -1 - 1 = -2$.

(A) Let the point on the curve $x^2 = 2y$ be $(x, y)$. Since $x^2 = 2y$,we have $y = \frac{x^2}{2}$.
The distance $D$ between $(x, y)$ and $(0, 5)$ is given by $D^2 = (x - 0)^2 + (y - 5)^2$.
Substituting $y = \frac{x^2}{2}$,we get $f(x) = D^2 = x^2 + (\frac{x^2}{2} - 5)^2$.
$f(x) = x^2 + \frac{x^4}{4} - 5x^2 + 25 = \frac{x^4}{4} - 4x^2 + 25$.
To find the minimum distance,we differentiate $f(x)$ with respect to $x$ and set it to zero:
$f'(x) = x^3 - 8x = 0$.
$x(x^2 - 8) = 0$,which gives $x = 0$ or $x^2 = 8$ (i.e.,$x = \pm 2\sqrt{2}$).
For $x = 0$,$y = 0$. The distance squared is $(0-0)^2 + (0-5)^2 = 25$.
For $x^2 = 8$,$y = \frac{8}{2} = 4$. The distance squared is $8 + (4-5)^2 = 8 + 1 = 9$.
Since $9 < 25$,the points $(\pm 2\sqrt{2}, 4)$ are the nearest points.
Comparing with the given options,$(2\sqrt{2}, 4)$ is the correct choice.

(A) Let $x$ and $y$ be two positive numbers.
Given that their sum is constant,$x + y = a$.
Let $z = x^3 + y^3$.
Substituting $y = a - x$,we get $z = x^3 + (a - x)^3$.
Differentiating with respect to $x$: $\frac{dz}{dx} = 3x^2 - 3(a - x)^2 = 3(x^2 - (a^2 - 2ax + x^2)) = 3(2ax - a^2) = 3a(2x - a)$.
Setting $\frac{dz}{dx} = 0$,we find $2x - a = 0$,which implies $x = \frac{a}{2}$.
Since $y = a - x$,we get $y = a - \frac{a}{2} = \frac{a}{2}$.
Checking the second derivative: $\frac{d^2z}{dx^2} = 6a$. Since $a > 0$,$\frac{d^2z}{dx^2} > 0$,confirming a minimum at $x = \frac{a}{2}$.
Thus,$x = y$,meaning the numbers are equal.

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

(B) Let the radius of the sector be $r$ and the arc length be $s$. The perimeter of the sector is given by $P = 2r + s = 20 \text{ cm}$.
Thus,$s = 20 - 2r$.
The area $A$ of the sector is given by $A = \frac{1}{2} s r$.
Substituting the value of $s$,we get $A = \frac{1}{2} (20 - 2r) r = 10r - r^2$.
To find the maximum area,we differentiate $A$ with respect to $r$ and set it to zero:
$\frac{dA}{dr} = 10 - 2r = 0 \Rightarrow r = 5 \text{ cm}$.
Substituting $r = 5$ into the area equation:
$A = 10(5) - (5)^2 = 50 - 25 = 25 \text{ cm}^2$.

(C) Given the displacement equation: $s = 22t - 11t^{2}$.
To find the maximum height, we differentiate $s$ with respect to $t$ to find the velocity $v = \frac{ds}{dt}$.
$\frac{ds}{dt} = 22 - 22t$.
At the maximum height, the velocity is zero, so $22 - 22t = 0$, which gives $t = 1 \text{ s}$.
Substituting $t = 1$ into the displacement equation:
$s = 22(1) - 11(1)^{2} = 22 - 11 = 11 \text{ units}$.
Since the ball starts from $s=0$ and reaches a maximum height of $11 \text{ units}$, the total distance travelled is $11 \text{ units}$.

(C) Given the displacement from the top of the tower is $s=48t-16t^{2}$.
The velocity is given by $v = \frac{ds}{dt} = 48-32t$.
At the greatest height,the velocity $v=0$.
$48-32t=0 \Rightarrow t = \frac{48}{32} = 1.5 \ s$.
The maximum displacement from the top of the tower is $s = 48(1.5) - 16(1.5)^{2} = 72 - 36 = 36 \ m$.
The total height above the ground is the height of the tower plus the maximum displacement: $H = 64 + 36 = 100 \ m$.

(B) Given the height function: $x = 80t - 16t^2$.
To find the time at which the stone reaches maximum height,we differentiate $x$ with respect to $t$:
$\frac{dx}{dt} = 80 - 32t$.
At maximum height,the velocity of the stone is zero,so $\frac{dx}{dt} = 0$.
Setting the derivative to zero: $80 - 32t = 0$.
$32t = 80$.
$t = \frac{80}{32} = 2.5 \text{ s}$.

(B) Given the function $f(x)=x^3-6x^2+12x-3$. \\ First,find the first derivative: $f'(x)=3x^2-12x+12$. \\ Set $f'(x)=0$ to find critical points: $3(x^2-4x+4)=0 \Rightarrow 3(x-2)^2=0 \Rightarrow x=2$. \\ Next,find the second derivative: $f''(x)=6x-12$. \\ Evaluate the second derivative at $x=2$: $f''(2)=6(2)-12=0$. \\ Since $f''(2)=0$,we check the third derivative: $f'''(x)=6$. \\ Since $f'''(2)=6 \neq 0$,the concavity changes at $x=2$. \\ Therefore,$x=2$ is a point of inflexion.

(B) $\because$ Slant height of the cone,$L = 6$ units.
Let the radius be $r$ and height be $h$.
Volume $(V) = \frac{1}{3} \pi r^2 h$.
Since $L^2 = r^2 + h^2$,we have $r^2 = L^2 - h^2 = 36 - h^2$.
Substituting this into the volume formula: $V = \frac{1}{3} \pi (36 - h^2) h = \frac{1}{3} \pi (36h - h^3)$.
To find the maximum volume,differentiate $V$ with respect to $h$: $\frac{dV}{dh} = \frac{1}{3} \pi (36 - 3h^2)$.
Setting $\frac{dV}{dh} = 0$,we get $36 - 3h^2 = 0 \Rightarrow h^2 = 12 \Rightarrow h = 2 \sqrt{3}$ units.
Checking the second derivative: $\frac{d^2 V}{dh^2} = \frac{1}{3} \pi (-6h) = -2 \pi h$.
At $h = 2 \sqrt{3}$,$\frac{d^2 V}{dh^2} = -4 \sqrt{3} \pi < 0$,so the volume is maximum.
Maximum volume $V = \frac{1}{3} \pi (36 - (2 \sqrt{3})^2) (2 \sqrt{3}) = \frac{1}{3} \pi (36 - 12) (2 \sqrt{3}) = \frac{1}{3} \pi (24) (2 \sqrt{3}) = 16 \sqrt{3} \pi$ cu units.

(C) Let $P(x, y)$ be the position of the jet and the soldier be at $A(3, 2)$.
The distance $AP$ is given by $AP = \sqrt{(x - 3)^2 + (y - 2)^2}$.
Since the jet follows the curve $y = x^2 + 2$,we have $y - 2 = x^2$.
Substituting this into the distance formula,let $z = (AP)^2 = (x - 3)^2 + (x^2)^2 = (x - 3)^2 + x^4$.
To find the minimum distance,we differentiate $z$ with respect to $x$: $\frac{dz}{dx} = 2(x - 3) + 4x^3$.
Setting $\frac{dz}{dx} = 0$,we get $4x^3 + 2x - 6 = 0$,which simplifies to $2x^3 + x - 3 = 0$.
By inspection,$x = 1$ is a root since $2(1)^3 + 1 - 3 = 0$.
Checking the second derivative: $\frac{d^2z}{dx^2} = 12x^2 + 2$. At $x = 1$,$\frac{d^2z}{dx^2} = 14 > 0$,so $x = 1$ gives a minimum.
For $x = 1$,$y = (1)^2 + 2 = 3$.
The minimum distance is $\sqrt{(1 - 3)^2 + (3 - 2)^2} = \sqrt{(-2)^2 + (1)^2} = \sqrt{4 + 1} = \sqrt{5}$ units.

(C) Given the curve $y = -x^{3} + 3x^{2} + 2x - 27$.
To find the slope of the curve,we differentiate with respect to $x$:
$\frac{dy}{dx} = -3x^{2} + 6x + 2$.
Let the slope be $m = -3x^{2} + 6x + 2$.
To find the maximum slope,we differentiate $m$ with respect to $x$:
$\frac{dm}{dx} = -6x + 6$.
Setting $\frac{dm}{dx} = 0$ for critical points:
$-6x + 6 = 0 \implies x = 1$.
Checking the second derivative:
$\frac{d^{2}m}{dx^{2}} = -6 < 0$.
Since the second derivative is negative,the slope is maximum at $x = 1$.
Substituting $x = 1$ into the expression for $m$:
$m_{\text{max}} = -3(1)^{2} + 6(1) + 2 = -3 + 6 + 2 = 5$.

(D) Let the length of the rectangle be $a$ and the breadth be $b$. The diagonal of the rectangle is the diameter of the circle,so $d = 2r = 2(2) = 4$.
In the right-angled triangle formed by the diagonal,$a^2 + b^2 = d^2 = 4^2 = 16$.
The area of the rectangle is $A = a \cdot b$.
Since $b = \sqrt{16 - a^2}$,we have $A = a \sqrt{16 - a^2}$.
To maximize $A$,we maximize $A^2 = f(a) = a^2(16 - a^2) = 16a^2 - a^4$.
Differentiating with respect to $a$: $f'(a) = 32a - 4a^3$.
Setting $f'(a) = 0$,we get $4a(8 - a^2) = 0$. Since $a > 0$,$a^2 = 8$,so $a = 2\sqrt{2}$.
Then $b = \sqrt{16 - 8} = \sqrt{8} = 2\sqrt{2}$.
The maximum area is $A = (2\sqrt{2})(2\sqrt{2}) = 8 \text{ sq units}$.

(B) Let $y = x e^{-x}$.
On differentiating with respect to $x$,we get:
$\frac{dy}{dx} = x \cdot (-e^{-x}) + e^{-x} \cdot (1) = e^{-x}(1 - x)$.
For maximum or minimum values,we set $\frac{dy}{dx} = 0$:
$e^{-x}(1 - x) = 0$.
Since $e^{-x} \neq 0$ for any real $x$,we have $1 - x = 0$,which implies $x = 1$.
Now,we find the second derivative to check for maxima:
$\frac{d^2y}{dx^2} = e^{-x}(-1) + (1 - x)(-e^{-x}) = -e^{-x} - e^{-x} + x e^{-x} = e^{-x}(x - 2)$.
At $x = 1$:
$\frac{d^2y}{dx^2} = e^{-1}(1 - 2) = -\frac{1}{e} < 0$.
Since the second derivative is negative at $x = 1$,the function has a local maximum at $x = 1$.
The maximum value is $y(1) = 1 \cdot e^{-1} = \frac{1}{e}$.

(C) Let the rectangle have sides $x$ and $y$ inscribed in a circle of radius $r = 2 \text{ unit}$.
The diagonal of the rectangle is equal to the diameter of the circle,so $d = 2r = 2 \times 2 = 4 \text{ unit}$.
By the Pythagorean theorem,$x^2 + y^2 = d^2 = 4^2 = 16$.
The area of the rectangle is $A = xy$.
To maximize $A$,we maximize $A^2 = x^2 y^2$.
Since $y^2 = 16 - x^2$,we have $A^2 = x^2(16 - x^2) = 16x^2 - x^4$.
Let $f(x) = 16x^2 - x^4$. To find the maximum,set $f'(x) = 32x - 4x^3 = 0$.
$4x(8 - x^2) = 0$,which gives $x^2 = 8$ (since $x > 0$).
Then $y^2 = 16 - 8 = 8$,so $x = y = \sqrt{8} = 2\sqrt{2}$.
The rectangle is a square with side length $2\sqrt{2}$.
The maximum area is $A = x \times y = \sqrt{8} \times \sqrt{8} = 8 \text{ sq unit}$.

(C) Let $y = \frac{\log x}{x}$.
On differentiating with respect to $x$,we get:
$\frac{dy}{dx} = \frac{x \cdot \frac{1}{x} - \log x \cdot 1}{x^2} = \frac{1 - \log x}{x^2}$.
For maxima,set $\frac{dy}{dx} = 0$:
$\frac{1 - \log x}{x^2} = 0 \implies 1 - \log x = 0 \implies \log x = 1 \implies x = e$.
Since $e \approx 2.718$,$x = e$ lies in the interval $(2, \infty)$.
Now,check the second derivative:
$\frac{d^2y}{dx^2} = \frac{x^2(-\frac{1}{x}) - (1 - \log x)(2x)}{x^4} = \frac{-x - 2x + 2x \log x}{x^4} = \frac{2x \log x - 3x}{x^4}$.
At $x = e$,$\frac{d^2y}{dx^2} = \frac{2e(1) - 3e}{e^4} = \frac{-e}{e^4} = -\frac{1}{e^3} < 0$.
Since the second derivative is negative,the function has a maximum at $x = e$.
The maximum value is $y = \frac{\log e}{e} = \frac{1}{e}$.

(C) Let $f(x) = \frac{\log _{e} x}{x}$.
To find the maximum value,we find the derivative $f'(x)$ using the quotient rule:
$f'(x) = \frac{x \cdot \frac{d}{dx}(\log _{e} x) - \log _{e} x \cdot \frac{d}{dx}(x)}{x^{2}} = \frac{x \cdot \frac{1}{x} - \log _{e} x \cdot 1}{x^{2}} = \frac{1 - \log _{e} x}{x^{2}}$.
For critical points,set $f'(x) = 0$:
$1 - \log _{e} x = 0 \Rightarrow \log _{e} x = 1 \Rightarrow x = e$.
Now,we check the second derivative $f''(x)$ to confirm the maximum:
$f''(x) = \frac{x^{2}(-\frac{1}{x}) - (1 - \log _{e} x)(2x)}{x^{4}} = \frac{-x - 2x(1 - \log _{e} x)}{x^{4}}$.
At $x = e$,$f''(e) = \frac{-e - 2e(1 - 1)}{e^{4}} = \frac{-e}{e^{4}} = -\frac{1}{e^{3}} < 0$.
Since $f''(e) < 0$,the function has a local maximum at $x = e$.
The maximum value is $f(e) = \frac{\log _{e} e}{e} = \frac{1}{e}$.

(C) Given $x+y=20$. We want to maximize $f(x, y) = x^3 y$.
Using the $AM$-$GM$ inequality for the terms $\frac{x}{3}, \frac{x}{3}, \frac{x}{3}, y$:
$\frac{\frac{x}{3} + \frac{x}{3} + \frac{x}{3} + y}{4} \geq \sqrt[4]{\left(\frac{x}{3}\right)^3 y}$
$\frac{x+y}{4} \geq \sqrt[4]{\frac{x^3 y}{27}}$
$\frac{20}{4} \geq \sqrt[4]{\frac{x^3 y}{27}} \Rightarrow 5 \geq \sqrt[4]{\frac{x^3 y}{27}}$
$5^4 \geq \frac{x^3 y}{27} \Rightarrow x^3 y \leq 27 \times 625 = 16875$.
So,$k = 16875$.
The maximum occurs when $\frac{x}{3} = y$.
Since $x+y=20$,we have $3y+y=20$ $\Rightarrow 4y=20$ $\Rightarrow y=5=\beta$ and $x=15=\alpha$.
Now,$\frac{k}{\alpha^2 \beta^2} = \frac{27 \times 625}{15^2 \times 5^2} = \frac{27 \times 625}{225 \times 25} = \frac{16875}{5625} = 3$.
Since $\frac{\alpha}{\beta} = \frac{15}{5} = 3$,the correct option is $\frac{\alpha}{\beta}$.

(B) Given the equation $a^2 x^4 + b^2 y^4 = c^6$.
Applying the Arithmetic Mean-Geometric Mean $(AM \geq GM)$ inequality for the two terms $a^2 x^4$ and $b^2 y^4$:
$\frac{a^2 x^4 + b^2 y^4}{2} \geq \sqrt{(a^2 x^4)(b^2 y^4)}$
Substituting the given value $a^2 x^4 + b^2 y^4 = c^6$:
$\frac{c^6}{2} \geq \sqrt{a^2 b^2 x^4 y^4}$
$\frac{c^6}{2} \geq ab x^2 y^2$
$x^2 y^2 \leq \frac{c^6}{2ab}$
Taking the square root on both sides:
$xy \leq \frac{c^3}{\sqrt{2ab}}$
Thus,the maximum value of $xy$ is $\frac{c^3}{\sqrt{2ab}}$.

(C) The function is $f(x) = \sin (x)$.
In the given interval $[-\pi / 2, \pi / 2]$,the sine function is strictly increasing.
The value at the lower bound is $f(-\pi / 2) = \sin(-\pi / 2) = -1$.
The value at the upper bound is $f(\pi / 2) = \sin(\pi / 2) = 1$.
Therefore,the maximum value of the function in the interval $[-\pi / 2, \pi / 2]$ is $1$.

(B) Let the lengths of the adjacent sides of the rectangle be $x \ cm$ and $y \ cm$.
The perimeter of the rectangle is given by $p = 2(x + y)$,which implies $y = \frac{p}{2} - x$.
The area of the rectangle is $A = x y$.
Substituting $y$,we get $A = x(\frac{p}{2} - x) = \frac{px}{2} - x^2$.
To find the maximum area,we differentiate $A$ with respect to $x$ and set it to zero: $\frac{dA}{dx} = \frac{p}{2} - 2x = 0$.
This gives $x = \frac{p}{4} \ cm$.
Consequently,$y = \frac{p}{2} - \frac{p}{4} = \frac{p}{4} \ cm$.
Thus,the maximum area is $A = \frac{p}{4} \times \frac{p}{4} = \frac{p^2}{16} \ cm^2$.
Therefore,option $B$ is correct.

(A) Given $f(x) = \sin x - x$.
To find the range $f(A)$,we check the derivative $f'(x) = \cos x - 1$.
Since $\cos x < 1$ for all $x \in [\frac{\pi}{4}, \frac{\pi}{3}]$,$f'(x) < 0$,which means $f(x)$ is a strictly decreasing function.
Therefore,the minimum value is at the upper bound $x = \frac{\pi}{3}$ and the maximum value is at the lower bound $x = \frac{\pi}{4}$.
$f(\frac{\pi}{3}) = \sin(\frac{\pi}{3}) - \frac{\pi}{3} = \frac{\sqrt{3}}{2} - \frac{\pi}{3}$.
$f(\frac{\pi}{4}) = \sin(\frac{\pi}{4}) - \frac{\pi}{4} = \frac{1}{\sqrt{2}} - \frac{\pi}{4}$.
Thus,$f(A) = [\frac{\sqrt{3}}{2} - \frac{\pi}{3}, \frac{1}{\sqrt{2}} - \frac{\pi}{4}]$.

(B) Given $A=\{x : 9x \geq x^2+20\}$.
Solving the inequality $x^2-9x+20 \leq 0$,we get $(x-4)(x-5) \leq 0$,which implies $x \in [4, 5]$.
Thus,$A=[4, 5]$.
Given $f(x)=2x^3-15x^2+36x-48$.
Finding the derivative,$f'(x)=6x^2-30x+36=6(x^2-5x+6)=6(x-3)(x-2)$.
For critical points,set $f'(x)=0$,which gives $x=2$ and $x=3$.
Since both critical points $x=2$ and $x=3$ lie outside the interval $A=[4, 5]$,the function $f(x)$ is monotonic in this interval.
Checking the sign of $f'(x)$ for $x \in [4, 5]$: $f'(4)=6(4-3)(4-2)=12 > 0$.
Since $f'(x) > 0$ for all $x \in [4, 5]$,$f(x)$ is an increasing function on $[4, 5]$.
Therefore,the maximum value occurs at the right endpoint $x=5$.
$f(5)=2(5)^3-15(5)^2+36(5)-48 = 2(125)-15(25)+180-48 = 250-375+180-48 = 7$.

(C) Let the intercepts of the line on the $x$-axis and $y$-axis be $a$ and $b$ respectively. The equation of the line is $\frac{x}{a} + \frac{y}{b} = 1$.
Given that the sum of the intercepts is $a + b = 12$,so $b = 12 - a$.
The area of the triangle formed by the line with the coordinate axes is $A = \frac{1}{2}ab$.
Substituting $b = 12 - a$,we get $A = \frac{1}{2}a(12 - a) = 6a - \frac{1}{2}a^2$.
To find the maximum area,we differentiate $A$ with respect to $a$ and set it to zero:
$\frac{dA}{da} = 6 - a = 0 \Rightarrow a = 6$.
Since $a = 6$,then $b = 12 - 6 = 6$.
The equation of the line is $\frac{x}{6} + \frac{y}{6} = 1$,which simplifies to $x + y = 6$.

(B) Given the curve $y=2x^3-15x^2+36x+c$.
Stationary points occur where $\frac{dy}{dx} = 0$.
$\frac{dy}{dx} = 6x^2 - 30x + 36 = 6(x^2 - 5x + 6) = 6(x-2)(x-3)$.
Setting $\frac{dy}{dx} = 0$,we get $x=2$ and $x=3$.
Since $(2, a)$ is a point on the curve,$a = 2(2)^3 - 15(2)^2 + 36(2) + c = 16 - 60 + 72 + c = 28 + c$.
Since $(b, 19)$ is a point on the curve,$b$ must be the other $x$-coordinate,so $b=3$.
Substituting $x=3$ and $y=19$ into the curve equation: $19 = 2(3)^3 - 15(3)^2 + 36(3) + c$.
$19 = 54 - 135 + 108 + c \Rightarrow 19 = 27 + c \Rightarrow c = -8$.
Now,$a = 28 + (-8) = 20$.
Therefore,$a+b+c = 20 + 3 - 8 = 15$.

(C) Given the curve $y = \tan^{-1}(\sin \sqrt{x})$.
To find the points where the tangent is parallel to the $x$-axis,we set the derivative $\frac{dy}{dx} = 0$.
Differentiating with respect to $x$:
$\frac{dy}{dx} = \frac{1}{1 + (\sin \sqrt{x})^2} \cdot \cos \sqrt{x} \cdot \frac{1}{2\sqrt{x}}$.
Setting $\frac{dy}{dx} = 0$ implies $\cos \sqrt{x} = 0$ (since $x > 0$ for the derivative to be defined).
Thus,$\sqrt{x} = (2n+1)\frac{\pi}{2}$ for $n = 0, 1, 2, \dots$.
Given $0 \leq x \leq 8\pi^2$,we have $0 \leq \sqrt{x} \leq 2\sqrt{2}\pi \approx 8.88$.
Possible values for $\sqrt{x}$ are $\frac{\pi}{2}, \frac{3\pi}{2}, \frac{5\pi}{2}$.
At these points,$\sin \sqrt{x} = \sin((2n+1)\frac{\pi}{2}) = \pm 1$.
Therefore,$y = \tan^{-1}(\pm 1) = \pm \frac{\pi}{4}$.

(D) Let $s = x^2 + y^2$.
Given $x + y = 32$,we can write $y = 32 - x$.
Substituting this into the expression for $s$:
$s = x^2 + (32 - x)^2$.
To find the minimum,we differentiate $s$ with respect to $x$:
$\frac{ds}{dx} = 2x + 2(32 - x)(-1) = 2x - 64 + 2x = 4x - 64$.
Setting $\frac{ds}{dx} = 0$ for critical points:
$4x - 64 = 0 \implies x = 16$.
Then $y = 32 - 16 = 16$.
Checking the second derivative:
$\frac{d^2s}{dx^2} = 4 > 0$,which confirms that $s$ is minimum at $x = 16$.
The minimum value is $s = 16^2 + 16^2 = 256 + 256 = 512$.

(B) Given: $f(x) = e^x \sin x$.
The slope of the tangent is given by $m(x) = f'(x)$.
$f'(x) = e^x \sin x + e^x \cos x = e^x (\sin x + \cos x)$.
Let $g(x) = f'(x) = e^x (\sin x + \cos x)$.
To find the maximum slope,we find $g'(x) = 0$.
$g'(x) = e^x (\sin x + \cos x) + e^x (\cos x - \sin x) = 2 e^x \cos x$.
Setting $g'(x) = 0$,we get $2 e^x \cos x = 0$.
Since $e^x \neq 0$ for all $x$,we have $\cos x = 0$.
In the interval $[0, 2 \pi]$,$x = \frac{\pi}{2}$ or $x = \frac{3 \pi}{2}$.
Now,we check the second derivative $g''(x) = 2 e^x \cos x - 2 e^x \sin x = 2 e^x (\cos x - \sin x)$.
At $x = \frac{\pi}{2}$,$g''(\frac{\pi}{2}) = 2 e^{\frac{\pi}{2}} (0 - 1) = -2 e^{\frac{\pi}{2}} < 0$,which indicates a local maximum.
At $x = \frac{3 \pi}{2}$,$g''(\frac{3 \pi}{2}) = 2 e^{\frac{3 \pi}{2}} (0 - (-1)) = 2 e^{\frac{3 \pi}{2}} > 0$,which indicates a local minimum.
Thus,the slope is maximum at $x = \frac{\pi}{2}$.

(B) Let the radius of the cone be $R = 10 \ cm$ and its height be $H$. Let the radius of the inscribed cylinder be $r$ and its height be $h$.
By similar triangles,$\frac{H-h}{r} = \frac{H}{R}$,which implies $h = H(1 - \frac{r}{R})$.
The curved surface area $S$ of the cylinder is $S = 2\pi rh = 2\pi r H(1 - \frac{r}{R}) = 2\pi H(r - \frac{r^2}{R})$.
To maximize $S$,we differentiate with respect to $r$ and set to zero: $\frac{dS}{dr} = 2\pi H(1 - \frac{2r}{R}) = 0$.
This gives $1 - \frac{2r}{R} = 0$,so $r = \frac{R}{2}$.
Given $R = 10 \ cm$,we have $r = \frac{10}{2} = 5 \ cm$.

(A) Given curve is $y = 8x^2 - x^4 - 4$.
To find the stationary points,we differentiate the function with respect to $x$:
$\frac{dy}{dx} = 16x - 4x^3$.
Stationary points occur where $\frac{dy}{dx} = 0$:
$16x - 4x^3 = 0$
$4x(4 - x^2) = 0$
This gives $x = 0$ or $x^2 = 4$,which implies $x = 0, 2, -2$.
Now,we find the corresponding $y$-values:
For $x = 0$: $y = 8(0)^2 - (0)^4 - 4 = -4$.
For $x = 2$: $y = 8(2)^2 - (2)^4 - 4 = 8(4) - 16 - 4 = 32 - 20 = 12$.
For $x = -2$: $y = 8(-2)^2 - (-2)^4 - 4 = 8(4) - 16 - 4 = 32 - 20 = 12$.
Thus,the stationary points are $(0, -4), (2, 12), (-2, 12)$.

(C) Given the function $f(x) = x^3 + 3x^2 - 2$.
To find the stationary points,we calculate the first derivative of the function with respect to $x$:
$f'(x) = \frac{d}{dx}(x^3 + 3x^2 - 2) = 3x^2 + 6x$.
Stationary points occur where the first derivative is equal to zero:
$f'(x) = 0 \Rightarrow 3x^2 + 6x = 0$.
Factoring the expression,we get:
$3x(x + 2) = 0$.
This gives the values $x = 0$ and $x = -2$.
Therefore,the values of $x$ at the stationary points are $0$ and $-2$.

(C) Given the displacement function $S(t) = t^3 - 16t^2 + 64t - 16$.
To find the maximum displacement,we first find the velocity $v(t)$ by differentiating $S$ with respect to $t$:
$v(t) = \frac{dS}{dt} = 3t^2 - 32t + 64$.
Setting $v(t) = 0$ to find the critical points:
$3t^2 - 32t + 64 = 0$.
Using the quadratic formula $t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:
$t = \frac{32 \pm \sqrt{(-32)^2 - 4(3)(64)}}{2(3)} = \frac{32 \pm \sqrt{1024 - 768}}{6} = \frac{32 \pm \sqrt{256}}{6} = \frac{32 \pm 16}{6}$.
This gives two values: $t_1 = \frac{48}{6} = 8$ and $t_2 = \frac{16}{6} = \frac{8}{3}$.
Now,we find the second derivative to check for maxima:
$a(t) = \frac{d^2S}{dt^2} = 6t - 32$.
For $t = 8$: $a(8) = 6(8) - 32 = 48 - 32 = 16 > 0$ (Local minimum).
For $t = \frac{8}{3}$: $a(\frac{8}{3}) = 6(\frac{8}{3}) - 32 = 16 - 32 = -16 < 0$ (Local maximum).
Thus,the displacement is maximum at $t = \frac{8}{3}$.

(C) Let $u = \sin x$. Since $x \in (0, \frac{\pi}{2})$,$u \in (0, 1)$.
Define $g(u) = \frac{4}{u} + \frac{1}{1-u}$.
To find the extreme value,differentiate $g(u)$ with respect to $u$:
$g'(u) = -\frac{4}{u^2} + \frac{1}{(1-u)^2}$.
Set $g'(u) = 0$ to find critical points:
$\frac{1}{(1-u)^2} = \frac{4}{u^2} \implies u^2 = 4(1-u)^2 \implies u^2 = 4(1 - 2u + u^2)$.
$u^2 = 4 - 8u + 4u^2 \implies 3u^2 - 8u + 4 = 0$.
Solving the quadratic equation: $(3u - 2)(u - 2) = 0$.
Since $u \in (0, 1)$,we have $u = \frac{2}{3}$.
Thus,$\sin k = \frac{2}{3}$.
Then $\cos^2 k = 1 - \sin^2 k = 1 - \frac{4}{9} = \frac{5}{9}$,so $\cos k = \frac{\sqrt{5}}{3}$.
The value $m = f(k) = g(\frac{2}{3}) = \frac{4}{2/3} + \frac{1}{1-2/3} = 6 + 3 = 9$.
We need to find $\cos k$ in terms of $m=9$.
$\cos k = \frac{\sqrt{5}}{3} = \frac{\sqrt{5}}{\sqrt{9}} = \frac{\sqrt{5}}{\sqrt{m}}$.
Thus,the correct option is $C$.

(D) Let the sides of the right-angled triangle be $x$ and $y$,and the hypotenuse be $h = 5$.
By Pythagoras theorem,$x^2 + y^2 = 5^2 = 25$.
The area of the triangle is $A = \frac{1}{2}xy$.
To maximize $A$,we maximize $A^2 = \frac{1}{4}x^2y^2$.
Let $x^2 = 25 \cos^2 \theta$ and $y^2 = 25 \sin^2 \theta$,where $\theta \in (0, \pi/2)$.
Then $A = \frac{1}{2} (5 \cos \theta)(5 \sin \theta) = \frac{25}{4} \sin(2\theta)$.
The area $A$ is maximum when $\sin(2\theta) = 1$,which means $2\theta = \pi/2$,so $\theta = \pi/4$.
Thus,$x = 5 \cos(\pi/4) = \frac{5}{\sqrt{2}}$ and $y = 5 \sin(\pi/4) = \frac{5}{\sqrt{2}}$.
The triangle is an isosceles right-angled triangle.
The perimeter $P = x + y + h = \frac{5}{\sqrt{2}} + \frac{5}{\sqrt{2}} + 5 = \frac{10}{\sqrt{2}} + 5 = 5\sqrt{2} + 5 = 5(\sqrt{2} + 1)$.

(A) Given $P(x) = x^4 + ax^3 + bx^2 + cx + d$.
Then $P'(x) = 4x^3 + 3ax^2 + 2bx + c$.
It is given that $x = 0$ is the only real root of $P'(x) = 0$.
Since $P'(x)$ is a cubic polynomial,it must have at least one real root.
If $x = 0$ is the only real root,then $P'(x)$ must be of the form $k x^3$ for some constant $k$.
Comparing coefficients,$4x^3 = k x^3 \implies k = 4$,and $3a = 0, 2b = 0, c = 0$.
Thus,$a = 0, b = 0, c = 0$.
So,$P(x) = x^4 + d$.
$P'(x) = 4x^3$.
For $x \in [-1, 0)$,$P'(x) < 0$,so $P(x)$ is strictly decreasing.
For $x \in (0, 1]$,$P'(x) > 0$,so $P(x)$ is strictly increasing.
Thus,$P(x)$ has a global minimum at $x = 0$.
In the interval $[-1, 1]$,the minimum value is $P(0) = d$.
The maximum value occurs at the endpoints $x = -1$ or $x = 1$.
Since $P(-1) = (-1)^4 + d = 1 + d$ and $P(1) = (1)^4 + d = 1 + d$,we have $P(-1) = P(1)$.
However,the problem states $P(-1) < P(1)$,which contradicts $P(x) = x^4 + d$.
Re-evaluating: If $P'(x) = 4x(x^2 + k)$ where $k > 0$,then $x=0$ is the only real root.
Then $P(x) = x^4 + \frac{k}{2}x^2 + d$.
$P(-1) = 1 + \frac{k}{2} + d$ and $P(1) = 1 + \frac{k}{2} + d$.
This also leads to $P(-1) = P(1)$.
Given the condition $P(-1) < P(1)$,the function must be such that the derivative has only one real root $x=0$ and the function is increasing on average.
Since $P'(x) = 4x^3 + 3ax^2 + 2bx + c = 4x^3$ (as $x=0$ is the only root),the function $P(x) = x^4 + d$ is symmetric.
Given the constraint $P(-1) < P(1)$,the only logical conclusion is that $P(-1)$ is not the minimum and $P(1)$ is the maximum.

(A) To find the maximum value of the function $f(x) = x e^{-x}$, we first find its derivative with respect to $x$.
Using the product rule, $f'(x) = (1)e^{-x} + x(-e^{-x}) = e^{-x}(1 - x)$.
For critical points, set $f'(x) = 0$.
$e^{-x}(1 - x) = 0$.
Since $e^{-x} \neq 0$ for any $x$, we have $1 - x = 0$, which gives $x = 1$.
To confirm this is a maximum, we use the second derivative test.
$f''(x) = -e^{-x}(1 - x) + e^{-x}(-1) = e^{-x}(x - 1 - 1) = e^{-x}(x - 2)$.
At $x = 1$, $f''(1) = e^{-1}(1 - 2) = -e^{-1} < 0$.
Since the second derivative is negative at $x = 1$, the function attains a local maximum at $x = 1$.
Thus, $k = 1$.