Maxima and Minima Questions in English

(B) Let $z = xy$.
Given $x + y = 8$,we have $y = 8 - x$.
Substituting $y$ in $z$,we get $z = x(8 - x) = 8x - x^2$.
To find the maximum value,we differentiate $z$ with respect to $x$:
$\frac{dz}{dx} = 8 - 2x$.
Setting $\frac{dz}{dx} = 0$,we get $8 - 2x = 0$,which implies $x = 4$.
Now,find the second derivative: $\frac{d^2z}{dx^2} = -2$.
Since $\frac{d^2z}{dx^2} < 0$,the function $z$ has a maximum at $x = 4$.
Substituting $x = 4$ into the equation for $y$,we get $y = 8 - 4 = 4$.
The maximum value is $z = xy = 4 \times 4 = 16$.

(B) Given $f(x) = 1 + 2x^2 + 2^2 x^4 + \dots + 2^{10} x^{20}$.
This is a geometric progression with $a = 1$,common ratio $r = 2x^2$,and $n = 11$ terms.
$f(x) = \frac{1((2x^2)^{11} - 1)}{2x^2 - 1} = \frac{(2x^2)^{11} - 1}{2x^2 - 1}$.
Alternatively,differentiating $f(x)$ directly:
$f'(x) = 0 + 2(2x) + 2^2(4x^3) + \dots + 2^{10}(20x^{19})$.
$f'(x) = 4x + 16x^3 + \dots + 2^{10}(20x^{19}) = 4x(1 + 2(2x^2) + 3(2x^2)^2 + \dots + 10(2x^2)^9)$.
Setting $f'(x) = 0$,we get $x = 0$ as the only real critical point because the term in the bracket is a sum of positive terms for all $x \neq 0$ (specifically,it is the derivative of a geometric series which is strictly increasing for $x^2 > 0$).
For $x < 0$,$f'(x) < 0$ and for $x > 0$,$f'(x) > 0$.
By the first derivative test,$f(x)$ has a local minimum at $x = 0$.
Since $x=0$ is the only critical point,$f(x)$ has exactly one minimum.

(C) Let $f(x) = x + \sin(2x)$.
First,find the derivative: $f'(x) = 1 + 2\cos(2x)$.
Set $f'(x) = 0$ to find critical points: $1 + 2\cos(2x) = 0 \Rightarrow \cos(2x) = -\frac{1}{2}$.
In the interval $[0, 2\pi]$,$2x$ can be $\frac{2\pi}{3}, \frac{4\pi}{3}, \frac{8\pi}{3}, \frac{10\pi}{3}$.
Thus,$x = \frac{\pi}{3}, \frac{2\pi}{3}, \frac{4\pi}{3}, \frac{5\pi}{3}$.
Find the second derivative: $f''(x) = -4\sin(2x)$.
Check the sign of $f''(x)$ at critical points:
For $x = \frac{\pi}{3}$,$f''(\frac{\pi}{3}) = -4\sin(\frac{2\pi}{3}) = -4(\frac{\sqrt{3}}{2}) = -2\sqrt{3} < 0$ (Local Maximum).
For $x = \frac{2\pi}{3}$,$f''(\frac{2\pi}{3}) = -4\sin(\frac{4\pi}{3}) = -4(-\frac{\sqrt{3}}{2}) = 2\sqrt{3} > 0$ (Local Minimum).
Calculating the value at $x = \frac{\pi}{3}$: $f(\frac{\pi}{3}) = \frac{\pi}{3} + \sin(\frac{2\pi}{3}) = \frac{\pi}{3} + \frac{\sqrt{3}}{2}$.

(A) Let the two parts be $x$ and $(10 - x)$.
Define the function $f(x) = 2x + (10 - x)^2$.
To find the minimum,differentiate with respect to $x$:
$f'(x) = 2 + 2(10 - x)(-1) = 2 - 20 + 2x = 2x - 18$.
Set $f'(x) = 0$ to find the critical point:
$2x - 18 = 0 \implies x = 9$.
Find the second derivative to check for minima:
$f''(x) = 2$.
Since $f''(9) = 2 > 0$,the function has a local minimum at $x = 9$.
The two parts are $x = 9$ and $10 - x = 1$.

(C) Let the function be $f(x) = x + \frac{1}{x}$.
To find the critical points,we 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 $x = \pm 1$.
Now,we find the second derivative $f''(x) = \frac{2}{x^3}$.
For $x = 1$,$f''(1) = 2 > 0$,so the function has a local minimum at $x = 1$.
The minimum value is $f(1) = 1 + \frac{1}{1} = 2$.
For $x = -1$,$f''(-1) = -2 < 0$,so the function has a local maximum at $x = -1$,where $f(-1) = -2$.

(A) Let the radius of the sphere be $R$. Let the height of the cone be $h$ and the radius of the base of the cone be $x$.
From the geometry of the sphere,the distance from the center of the sphere to the base of the cone is $|h - R|$.
Using the Pythagorean theorem in the triangle formed by the sphere's radius,the cone's base radius,and the distance from the center,we have $x^2 + (h - R)^2 = R^2$.
$x^2 = R^2 - (h^2 - 2hR + R^2) = 2hR - h^2$.
The volume of the cone is $V = \frac{1}{3}\pi x^2 h = \frac{1}{3}\pi (2hR - h^2)h = \frac{1}{3}\pi (2Rh^2 - h^3)$.
To maximize volume,differentiate with respect to $h$: $\frac{dV}{dh} = \frac{1}{3}\pi (4Rh - 3h^2)$.
Setting $\frac{dV}{dh} = 0$,we get $h(4R - 3h) = 0$. Since $h \neq 0$,$h = \frac{4}{3}R$.
The diameter of the sphere is $D = 2R$.
The ratio of the height of the cone to the diameter of the sphere is $\frac{h}{D} = \frac{\frac{4}{3}R}{2R} = \frac{4}{6} = \frac{2}{3}$.

(D) Given $y = f(x) = a \log x + bx^2 + x$.
First,find the derivative: $\frac{dy}{dx} = \frac{a}{x} + 2bx + 1$.
Since the function has extreme values at $x = -1$ and $x = 2$,the derivative must be zero at these points.
For $x = -1$: $\frac{a}{-1} + 2b(-1) + 1 = 0 \implies -a - 2b + 1 = 0 \implies a + 2b = 1$.
For $x = 2$: $\frac{a}{2} + 2b(2) + 1 = 0 \implies \frac{a}{2} + 4b + 1 = 0 \implies a + 8b = -2$.
Subtracting the first equation from the second: $(a + 8b) - (a + 2b) = -2 - 1 \implies 6b = -3 \implies b = -\frac{1}{2}$.
Substituting $b = -\frac{1}{2}$ into $a + 2b = 1$: $a + 2(-\frac{1}{2}) = 1 \implies a - 1 = 1 \implies a = 2$.
Thus,$a = 2$ and $b = -\frac{1}{2}$.

(A) Given $f(x) = 1 + 2 \sin x + 3 \cos^2 x = 1 + 2 \sin x + 3(1 - \sin^2 x) = 4 + 2 \sin x - 3 \sin^2 x$.
Let $u = \sin x$. Since $0 < x < 2\pi / 3$,the range of $u = \sin x$ is $(0, 1]$.
Then $g(u) = 4 + 2u - 3u^2$.
To find critical points,$g'(u) = 2 - 6u = 0 \implies u = 1/3$.
Since $g''(u) = -6 < 0$,$u = 1/3$ is a point of local maximum.
Thus,$f(x)$ has a maximum at $\sin x = 1/3$,i.e.,$x = \sin^{-1}(1/3)$.
At the boundaries of the interval $(0, 2\pi / 3)$:
As $x \to 0^+$,$f(x) \to 1 + 0 + 3(1) = 4$.
At $x = 2\pi / 3$,$f(2\pi / 3) = 1 + 2(\sqrt{3}/2) + 3(1/4) = 1 + \sqrt{3} + 0.75 = 1.75 + \sqrt{3} \approx 3.48$.
Since $g(1/3) = 4 + 2/3 - 3(1/9) = 4 + 2/3 - 1/3 = 4.33$,the maximum is at $x = \sin^{-1}(1/3)$.
Checking the options,the question asks for the behavior. Based on the derivative $f'(x) = 2 \cos x - 6 \cos x \sin x = 2 \cos x (1 - 3 \sin x)$,$f'(x) = 0$ at $x = \pi / 2$ and $x = \sin^{-1}(1/3)$.
At $x = \pi / 2$,$f(\pi / 2) = 1 + 2(1) + 3(0) = 3$. Since $3 < 4$ and $3 < 4.33$,$x = \pi / 2$ is a local minimum.

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

(A) Given $f(x) = \frac{1}{4x^2 + 2x + 1}$.
To find the maximum value,we first find the derivative $f'(x)$:
$f'(x) = -\frac{1}{(4x^2 + 2x + 1)^2} \cdot (8x + 2) = -\frac{2(4x + 1)}{(4x^2 + 2x + 1)^2}$.
Setting $f'(x) = 0$ gives $4x + 1 = 0$,so $x = -1/4$.
Since the denominator $4x^2 + 2x + 1$ is always positive (as its discriminant $D = 2^2 - 4(4)(1) = 4 - 16 = -12 < 0$),the function $f(x)$ is maximized when the denominator $4x^2 + 2x + 1$ is minimized.
The minimum value of the quadratic $g(x) = 4x^2 + 2x + 1$ occurs at $x = -b/(2a) = -2/(2 \cdot 4) = -1/4$.
The minimum value of the denominator is $g(-1/4) = 4(1/16) + 2(-1/4) + 1 = 1/4 - 1/2 + 1 = 3/4$.
Therefore,the maximum value of $f(x)$ is $1 / (3/4) = 4/3$.

(A) Let the two numbers be $x$ and $y$.
Given that $x + y = 24$,so $y = 24 - x$ $(i)$.
Let $P$ be the product of the two numbers,then $P = xy$.
Substituting $y$ from $(i)$,we get $P = x(24 - x) = 24x - x^2$.
To find the maximum product,we differentiate $P$ with respect to $x$:
$\frac{dP}{dx} = 24 - 2x$.
For critical points,set $\frac{dP}{dx} = 0$,which gives $24 - 2x = 0$,so $x = 12$.
Now,find the second derivative: $\frac{d^2P}{dx^2} = -2$.
Since $\frac{d^2P}{dx^2} < 0$,the function $P$ has a local maximum at $x = 12$.
When $x = 12$,$y = 24 - 12 = 12$.
Thus,the two numbers are $12$ and $12$.

(B) Let $f(x) = 4e^{2x} + 9e^{-2x}$.
To find the minimum value,we use the Arithmetic Mean-Geometric Mean $(AM \ge GM)$ inequality.
For any two positive numbers $a$ and $b$,$\frac{a+b}{2} \ge \sqrt{ab}$.
Here,$a = 4e^{2x}$ and $b = 9e^{-2x}$.
$\frac{4e^{2x} + 9e^{-2x}}{2} \ge \sqrt{4e^{2x} \times 9e^{-2x}}$
$\frac{4e^{2x} + 9e^{-2x}}{2} \ge \sqrt{36 \times e^{2x-2x}}$
$\frac{4e^{2x} + 9e^{-2x}}{2} \ge \sqrt{36 \times 1}$
$\frac{4e^{2x} + 9e^{-2x}}{2} \ge 6$
$4e^{2x} + 9e^{-2x} \ge 12$.
Thus,the minimum value is $12$.

(D) Given $f(x) = x^3 + 1$.
Step $1$: Find the derivative $f'(x) = 3x^2$.
Step $2$: For a local extremum,we set $f'(x) = 0$,which gives $3x^2 = 0$,so $x = 0$.
Step $3$: Check the sign of $f'(x)$ around $x = 0$. For $x < 0$,$f'(x) = 3x^2 > 0$. For $x > 0$,$f'(x) = 3x^2 > 0$.
Since the derivative $f'(x)$ does not change sign at $x = 0$,the function $f(x)$ does not have a local extremum at $x = 0$. Thus,Statement-$1$ is false.
Step $4$: The function $f(x) = x^3 + 1$ is a polynomial function,which is continuous and differentiable on $(-\infty, \infty)$. Also,$f'(0) = 3(0)^2 = 0$. Thus,Statement-$2$ is true.

(C) Given the function $y = xe^x$.
First,find the first derivative:
$\frac{dy}{dx} = x \cdot e^x + e^x \cdot 1 = e^x(x + 1)$.
To find the critical points,set $\frac{dy}{dx} = 0$:
$e^x(x + 1) = 0$.
Since $e^x > 0$ for all $x$,we have $x + 1 = 0$,which implies $x = -1$.
Now,find the second derivative:
$\frac{d^2y}{dx^2} = \frac{d}{dx}[e^x(x + 1)] = e^x(1) + (x + 1)e^x = e^x(x + 2)$.
Evaluate the second derivative at $x = -1$:
$\left. \frac{d^2y}{dx^2} \right|_{x=-1} = e^{-1}(-1 + 2) = e^{-1}(1) = \frac{1}{e}$.
Since $\frac{1}{e} > 0$,the function has a local minimum at $x = -1$.

(A) Let the length of the rectangle be $x \ cm$ and the width be $y \ cm$.
Given perimeter $P = 2(x + y) = 176 \ cm$.
Therefore,$x + y = 88$,which implies $y = 88 - x$.
The area $A$ of the rectangle is given by $A = x \times y = x(88 - x) = 88x - x^2$.
To find the maximum area,we differentiate $A$ with respect to $x$ and set it to zero:
$\frac{dA}{dx} = 88 - 2x = 0$.
Solving for $x$,we get $2x = 88$,so $x = 44 \ cm$.
Since $x = 44$,then $y = 88 - 44 = 44 \ cm$.
Thus,the rectangle is a square with side $44 \ cm$.
The maximum area is $A = 44 \times 44 = 1936 \ sq. \ cm$.

(C) Given $f(x) = x^2 e^{-2x}$.
First,find the derivative $f'(x)$ using the product rule:
$f'(x) = x^2 \frac{d}{dx}(e^{-2x}) + e^{-2x} \frac{d}{dx}(x^2)$
$f'(x) = x^2(-2e^{-2x}) + e^{-2x}(2x)$
$f'(x) = 2xe^{-2x}(1 - x) = \frac{2x(1 - x)}{e^{2x}}$.
To find the critical points,set $f'(x) = 0$:
Since $x > 0$ and $e^{2x} > 0$,we have $1 - x = 0$,which gives $x = 1$.
Now,check the second derivative $f''(x)$ to confirm the maximum:
$f''(x) = \frac{d}{dx}[2(x - x^2)e^{-2x}] = 2[(1 - 2x)e^{-2x} + (x - x^2)(-2e^{-2x})]$
$f''(1) = 2[(1 - 2)e^{-2} + (1 - 1)(-2e^{-2})] = 2[-e^{-2}] = -2/e^2 < 0$.
Since $f''(1) < 0$,$f(x)$ has a local maximum at $x = 1$.
The maximum value is $f(1) = (1)^2 e^{-2(1)} = 1/e^2$.

(B) Given function is $f(x) = 2x^3 - 21x^2 + 36x + 7$.
First,find the first derivative $f'(x)$:
$f'(x) = \frac{d}{dx}(2x^3 - 21x^2 + 36x + 7) = 6x^2 - 42x + 36$.
Set $f'(x) = 0$ to find critical points:
$6(x^2 - 7x + 6) = 0$
$6(x - 1)(x - 6) = 0$
So,$x = 1$ and $x = 6$.
Now,find the second derivative $f''(x)$:
$f''(x) = \frac{d}{dx}(6x^2 - 42x + 36) = 12x - 42$.
Check the sign of $f''(x)$ at critical points:
For $x = 1$: $f''(1) = 12(1) - 42 = -30$. Since $f''(1) < 0$,the function has a local maximum at $x = 1$.
For $x = 6$: $f''(6) = 12(6) - 42 = 72 - 42 = 30$. Since $f''(6) > 0$,the function has a local minimum at $x = 6$.
Therefore,the function has a local maximum at $x = 1$.

(C) Given $f(x) = \sin x (1 + \cos x) = \sin x + \sin x \cos x = \sin x + \frac{1}{2} \sin 2x$.
Find the first derivative: $f'(x) = \cos x + \cos 2x$.
Set $f'(x) = 0$ for critical points: $\cos x + (2 \cos^2 x - 1) = 0$.
Rearranging gives $2 \cos^2 x + \cos x - 1 = 0$.
Factoring the quadratic: $(2 \cos x - 1)(\cos x + 1) = 0$.
This gives $\cos x = 1/2$ or $\cos x = -1$.
For $x \in [0, \pi]$,$\cos x = 1/2 \implies x = \pi / 3$ and $\cos x = -1 \implies x = \pi$.
Find the second derivative: $f''(x) = -\sin x - 2 \sin 2x$.
At $x = \pi / 3$: $f''(\pi / 3) = -\sin(\pi / 3) - 2 \sin(2\pi / 3) = -\frac{\sqrt{3}}{2} - 2(\frac{\sqrt{3}}{2}) = -\frac{3\sqrt{3}}{2} < 0$.
Since $f''(\pi / 3) < 0$,the function has a local maximum at $x = \pi / 3$.

(A) Given $xy = 1$,we have $y = \frac{1}{x}$.
Let $f(x) = x + y = x + \frac{1}{x}$.
To find the minimum value,we find the derivative $f'(x) = 1 - \frac{1}{x^2}$.
Setting $f'(x) = 0$,we get $1 - \frac{1}{x^2} = 0$,which implies $x^2 = 1$. Since $x > 0$,we have $x = 1$.
Now,we find the second derivative $f''(x) = \frac{2}{x^3}$.
At $x = 1$,$f''(1) = \frac{2}{1^3} = 2 > 0$.
Since the second derivative is positive,$x = 1$ is a point of local minima.
The minimum value is $f(1) = 1 + \frac{1}{1} = 2$.

(C) Given $h(x) = f(x) + f(-x)$.
To find the extremum,we differentiate $h(x)$ with respect to $x$:
$h'(x) = \frac{d}{dx}(f(x) + f(-x)) = f'(x) - f'(-x)$.
For $h(x)$ to attain an extremum,we must have $h'(x) = 0$.
Therefore,$f'(x) - f'(-x) = 0$ at the point of extremum.
Thus,the value of $f'(x) - f'(-x)$ is $0$.

(D) Given the function $f(x) = ax + \frac{b}{x}$.
To find the critical points,we calculate the first derivative $f'(x)$:
$f'(x) = \frac{d}{dx}(ax + bx^{-1}) = a - \frac{b}{x^2}$.
Setting $f'(x) = 0$ for extrema:
$a - \frac{b}{x^2} = 0 \Rightarrow a = \frac{b}{x^2} \Rightarrow x^2 = \frac{b}{a} \Rightarrow x = \sqrt{\frac{b}{a}}$ (since $x > 0$).
Now,we check the second derivative $f''(x)$ to confirm the nature of the extremum:
$f''(x) = \frac{d}{dx}(a - bx^{-2}) = 0 - b(-2)x^{-3} = \frac{2b}{x^3}$.
Since $b > 0$ and $x > 0$,$f''(x) = \frac{2b}{x^3} > 0$.
Since the second derivative is positive at $x = \sqrt{\frac{b}{a}}$,the function $f(x)$ attains its minimum value at $x = \sqrt{\frac{b}{a}}$.

(C) Given $f(x) = \sin x + \cos 2x$.
First,find the derivative $f'(x) = \cos x - 2 \sin 2x$.
Using $\sin 2x = 2 \sin x \cos x$,we get $f'(x) = \cos x - 4 \sin x \cos x = \cos x (1 - 4 \sin x)$.
For critical points,set $f'(x) = 0$,which gives $\cos x = 0$ or $\sin x = \frac{1}{4}$.
Now,find the second derivative $f''(x) = -\sin x - 4 \cos 2x$.
For $x = (2n+1)\frac{\pi}{2}$,$\sin x = (-1)^n$ and $\cos 2x = \cos((2n+1)\pi) = -1$.
Thus,$f''((2n+1)\frac{\pi}{2}) = -(-1)^n - 4(-1) = 4 - (-1)^n$.
If $n$ is even,$f'' = 4 - 1 = 3 > 0$ (Local Minimum).
If $n$ is odd,$f'' = 4 - (-1) = 5 > 0$ (Local Minimum).
Thus,the function is minimum at $x = (2n+1)\frac{\pi}{2}$.

(C) Given $f(x) = \frac{x}{x^2 + x + 4}$.
To find the critical points,we calculate the derivative $f'(x)$:
$f'(x) = \frac{(x^2 + x + 4)(1) - x(2x + 1)}{(x^2 + x + 4)^2} = \frac{x^2 + x + 4 - 2x^2 - x}{(x^2 + x + 4)^2} = \frac{4 - x^2}{(x^2 + x + 4)^2}$.
Setting $f'(x) = 0$ gives $4 - x^2 = 0$,so $x = 2$ or $x = -2$.
Neither $x = 2$ nor $x = -2$ lies in the interval $[-1, 1]$.
Therefore,the function is monotonic on $[-1, 1]$.
We evaluate the function at the endpoints:
$f(-1) = \frac{-1}{4 - 1 + 1} = \frac{-1}{4} = -0.25$.
$f(1) = \frac{1}{4 + 1 + 1} = \frac{1}{6} \approx 0.1667$.
Comparing the values,the maximum value is $\frac{1}{6}$.

(C) Given $\frac{dy}{dx} = (x - 1)^3 (x - 2)^4$.
To find critical points,set $\frac{dy}{dx} = 0$,which gives $x = 1$ and $x = 2$.
We use the first derivative test to check the nature of these points.
For $x = 1$:
- If $x = 1 - h$ (where $h > 0$ is very small),$\frac{dy}{dx} = (1 - h - 1)^3 (1 - h - 2)^4 = (-h)^3 (-1 - h)^4 = (-h^3)(+) = -ve$.
- If $x = 1 + h$,$\frac{dy}{dx} = (1 + h - 1)^3 (1 + h - 2)^4 = (h)^3 (h - 1)^4 = (+)(+) = +ve$.
Since $\frac{dy}{dx}$ changes sign from negative to positive at $x = 1$,$y$ has a local minimum at $x = 1$.
For $x = 2$:
- If $x = 2 - h$,$\frac{dy}{dx} = (2 - h - 1)^3 (2 - h - 2)^4 = (1 - h)^3 (-h)^4 = (+)(+) = +ve$.
- If $x = 2 + h$,$\frac{dy}{dx} = (2 + h - 1)^3 (2 + h - 2)^4 = (1 + h)^3 (h)^4 = (+)(+) = +ve$.
Since $\frac{dy}{dx}$ does not change sign at $x = 2$,$x = 2$ is a point of inflection.

(A) Given,$x = t^4 - kt^3$.
Velocity $v = \frac{dx}{dt} = 4t^3 - 3kt^2$.
For the velocity to be maximum,its derivative with respect to time must be zero,i.e.,$\frac{dv}{dt} = 0$.
$\frac{dv}{dt} = \frac{d}{dt}(4t^3 - 3kt^2) = 12t^2 - 6kt$.
At $t = 2$,$\frac{dv}{dt} = 0$.
$12(2)^2 - 6k(2) = 0$.
$12(4) - 12k = 0$.
$48 - 12k = 0$.
$12k = 48$.
$k = 4$.

(B) Let $f(x) = e^{(2x^2 - 2x - 1)\sin^2 x}$.
To find the minimum value of $f(x)$,we need to find the minimum value of the exponent $u(x) = (2x^2 - 2x - 1)\sin^2 x$.
Consider the quadratic expression $g(x) = 2x^2 - 2x - 1$. The vertex of this parabola is at $x = -(-2)/(2 \times 2) = 0.5$.
At $x = 0.5$,$g(0.5) = 2(0.25) - 2(0.5) - 1 = 0.5 - 1 - 1 = -1.5$.
Since $\sin^2 x$ is always non-negative and $g(x)$ can be negative,the product $u(x)$ will be minimized when $g(x)$ is at its minimum and $\sin^2 x$ is at its maximum value of $1$ (which occurs at $x = \pi/2 \approx 1.57$).
However,checking the behavior near $x=1$,if we set $2x^2 - 2x - 1 = -1$,we get $2x^2 - 2x = 0$,implying $x=0$ or $x=1$.
For $x=1$,$u(1) = (2 - 2 - 1)\sin^2(1) = -\sin^2(1) \approx -0.708$.
For $x=0.5$,$u(0.5) = -1.5 \times \sin^2(0.5) \approx -1.5 \times 0.23 = -0.345$.
Comparing values,the minimum value of the exponent $u(x)$ is $-1$ when $2x^2 - 2x - 1 = -1$ and $\sin^2 x = 1$ is not simultaneously possible.
Re-evaluating: The function $e^{u(x)}$ reaches its minimum when $u(x)$ is minimum. The minimum value of $2x^2 - 2x - 1$ is $-1.5$. The minimum value of the exponent is $-1$ at $x=1$ (if $\sin^2 x$ is considered). Given the options,the minimum value is $1/e$.

(D) Given $y = a \log x + bx^2 + x$.
Taking the derivative with respect to $x$:
$\frac{dy}{dx} = \frac{a}{x} + 2bx + 1$.
Since the function has extreme values at $x = 1$ and $x = 2$,the derivative must be zero at these points.
For $x = 1$: $\frac{a}{1} + 2b(1) + 1 = 0 \implies a + 2b = -1$.
For $x = 2$: $\frac{a}{2} + 2b(2) + 1 = 0 \implies \frac{a}{2} + 4b = -1 \implies a + 8b = -2$.
Subtracting the first equation from the second:
$(a + 8b) - (a + 2b) = -2 - (-1) \implies 6b = -1 \implies b = -\frac{1}{6}$.
Substituting $b = -\frac{1}{6}$ into $a + 2b = -1$:
$a + 2(-\frac{1}{6}) = -1 \implies a - \frac{1}{3} = -1 \implies a = -1 + \frac{1}{3} = -\frac{2}{3}$.
Thus,$(a, b) = \left( -\frac{2}{3}, -\frac{1}{6} \right)$.

(B) Given $f(x) = \int\limits_0^x {{e^{\frac{{ - {t^2}}}{2}}}} \left( {1 - {t^2}} \right)\,dt$.
Applying the Leibniz rule for differentiation,we get:
$f'(x) = e^{\frac{{ - {x^2}}}{2}} (1 - x^2)$.
For local extrema,set $f'(x) = 0$:
$e^{\frac{{ - {x^2}}}{2}} (1 - x^2) = 0$.
Since $e^{\frac{{ - {x^2}}}{2}} > 0$ for all $x$,we have $1 - x^2 = 0$,which implies $x^2 = 1$,so $x = \pm 1$.
Now,find the second derivative $f''(x)$:
$f''(x) = e^{\frac{{ - {x^2}}}{2}} (-2x) + (1 - x^2) e^{\frac{{ - {x^2}}}{2}} (-x) = e^{\frac{{ - {x^2}}}{2}} (-2x - x + x^3) = e^{\frac{{ - {x^2}}}{2}} (x^3 - 3x)$.
Check the sign of $f''(x)$ at critical points:
At $x = 1$: $f''(1) = e^{-1/2} (1 - 3) = -2e^{-1/2} < 0$ (Local Maximum).
At $x = -1$: $f''(-1) = e^{-1/2} (-1 + 3) = 2e^{-1/2} > 0$ (Local Minimum).
Therefore,$f(x)$ is minimum at $x = -1$.

(B) Given $P(x) = a_0 + a_1x^2 + a_2x^4 + \dots + a_nx^{2n}$.
For maxima and minima,we find the derivative $P'(x) = 0$.
$P'(x) = 2a_1x + 4a_2x^3 + \dots + 2na_nx^{2n-1} = 2x(a_1 + 2a_2x^2 + \dots + na_nx^{2n-2}) = 0$.
This gives $x = 0$ as a critical point.
Now,we find the second derivative $P''(x) = 2a_1 + 12a_2x^2 + \dots + 2n(2n-1)a_nx^{2n-2}$.
At $x = 0$,$P''(0) = 2a_1$.
Since $0 < a_1$,$P''(0) = 2a_1 > 0$.
Since the second derivative is positive at $x = 0$,$P(x)$ has a local minimum at $x = 0$.
Since $P(x)$ is an even function and the coefficients are strictly increasing,$x = 0$ is the only point where $P'(x) = 0$,thus $P(x)$ has exactly one minimum point.

(A) Given $f'(x) = (x - a)^{2m} (x - b)^{2n + 1}$.
Critical points are found by setting $f'(x) = 0$,which gives $x = a$ and $x = b$.
Consider the point $x = a$. For a small positive value $h$,we check the sign of $f'(x)$ around $x = a$:
For $x = a - h$,$f'(a - h) = (a - h - a)^{2m} (a - h - b)^{2n + 1} = (-h)^{2m} (a - b - h)^{2n + 1}$. Since $2m$ is even,$(-h)^{2m} = h^{2m} > 0$. Since $a > b$,$(a - b - h) > 0$ for small $h$,so $f'(a - h) > 0$.
For $x = a + h$,$f'(a + h) = (a + h - a)^{2m} (a + h - b)^{2n + 1} = (h)^{2m} (a - b + h)^{2n + 1}$. Since $a > b$,$(a - b + h) > 0$,so $f'(a + h) > 0$.
Since the sign of $f'(x)$ does not change as $x$ passes through $a$,$x = a$ is a point of inflection and gives neither a maximum nor a minimum.

(C) Given $xy = c^2$,we have $y = \frac{c^2}{x}$.
Let $f(x) = ax + by = ax + b\left(\frac{c^2}{x}\right) = ax + \frac{bc^2}{x}$.
To find the minimum value,we differentiate $f(x)$ with respect to $x$:
$f'(x) = a - \frac{bc^2}{x^2}$.
Setting $f'(x) = 0$ for critical points:
$a = \frac{bc^2}{x^2} \implies x^2 = \frac{bc^2}{a} \implies x = c\sqrt{\frac{b}{a}}$ (since $x, a, b, c > 0$ for a minimum in this context).
Now,find the second derivative:
$f''(x) = \frac{2bc^2}{x^3}$.
Since $x > 0$,$f''(x) > 0$,confirming a minimum at $x = c\sqrt{\frac{b}{a}}$.
Substituting $x$ back into $f(x)$:
$f\left(c\sqrt{\frac{b}{a}}\right) = a\left(c\sqrt{\frac{b}{a}}\right) + \frac{bc^2}{c\sqrt{\frac{b}{a}}} = c\sqrt{ab} + c\sqrt{\frac{b^2}{b/a}} = c\sqrt{ab} + c\sqrt{ab} = 2c\sqrt{ab}$.

(D) Let $g(x) = 3x^4 + 8x^3 - 18x^2 + 60$. The function $f(x) = \frac{40}{g(x)}$ is minimum when $g(x)$ is maximum,and $f(x)$ is maximum when $g(x)$ is minimum.
To find the extrema of $g(x)$,we find its derivative: $g'(x) = 12x^3 + 24x^2 - 36x$.
Setting $g'(x) = 0$,we get $12x(x^2 + 2x - 3) = 0$,which factors to $12x(x+3)(x-1) = 0$.
The critical points are $x = 0, x = 1, x = -3$.
We evaluate $g(x)$ at these points:
$g(0) = 60$.
$g(1) = 3(1)^4 + 8(1)^3 - 18(1)^2 + 60 = 3 + 8 - 18 + 60 = 53$.
$g(-3) = 3(-3)^4 + 8(-3)^3 - 18(-3)^2 + 60 = 3(81) + 8(-27) - 18(9) + 60 = 243 - 216 - 162 + 60 = -75$.
Since $g(x) \to \infty$ as $x \to \pm \infty$,the global minimum of $g(x)$ is $-75$ at $x = -3$. However,$f(x)$ is defined as $40/g(x)$.
Checking the values of $f(x)$ at critical points:
$f(0) = 40/60 = 2/3$.
$f(1) = 40/53$.
$f(-3) = 40/(-75) = -8/15$.
As $x \to \pm \infty$,$f(x) \to 0$.
The minimum value of the function is $-8/15$. Since this is not among the options,the correct choice is $D$.

(C) Let $f(x) = x - x^3$.
To find the critical points,we find the derivative $f'(x) = 1 - 3x^2$.
Setting $f'(x) = 0$,we get $1 - 3x^2 = 0$,which implies $x^2 = 1/3$,so $x = \pm 1/\sqrt{3}$.
Since the interval is $0 \leq x \leq 2$,we only consider $x = 1/\sqrt{3}$.
Now,we check the second derivative $f''(x) = -6x$.
At $x = 1/\sqrt{3}$,$f''(1/\sqrt{3}) = -6/\sqrt{3} < 0$.
Since the second derivative is negative,the function has a local maximum at $x = 1/\sqrt{3}$.
We also check the endpoints of the interval $[0, 2]$:
$f(0) = 0(1 - 0) = 0$.
$f(2) = 2(1 - 4) = 2(-3) = -6$.
$f(1/\sqrt{3}) = (1/\sqrt{3})(1 - 1/3) = (1/\sqrt{3})(2/3) = 2/(3\sqrt{3}) \approx 0.385$.
Comparing the values $0$,$-6$,and $2/(3\sqrt{3})$,the maximum value occurs at $x = 1/\sqrt{3}$.

(D) Given $f(x) = \sin 2x - x$.
Step $1$: Find the derivative $f'(x)$.
$f'(x) = 2\cos 2x - 1$.
Step $2$: Set $f'(x) = 0$ to find critical points.
$2\cos 2x - 1 = 0 \implies \cos 2x = \frac{1}{2}$.
In the interval $[0, \pi]$,$2x$ ranges from $[0, 2\pi]$.
Thus,$2x = \frac{\pi}{3}$ or $2x = \frac{5\pi}{3}$.
So,$x = \frac{\pi}{6}$ or $x = \frac{5\pi}{6}$.
Step $3$: Use the second derivative test.
$f''(x) = -4\sin 2x$.
At $x = \frac{\pi}{6}$,$f''(\frac{\pi}{6}) = -4\sin(\frac{\pi}{3}) = -4(\frac{\sqrt{3}}{2}) = -2\sqrt{3} < 0$. So,$x = \frac{\pi}{6}$ is a point of local maximum.
At $x = \frac{5\pi}{6}$,$f''(\frac{5\pi}{6}) = -4\sin(\frac{5\pi}{3}) = -4(-\frac{\sqrt{3}}{2}) = 2\sqrt{3} > 0$. So,$x = \frac{5\pi}{6}$ is a point of local minimum.
Step $4$: Calculate the values.
Local maximum value $f(\frac{\pi}{6}) = \sin(\frac{\pi}{3}) - \frac{\pi}{6} = \frac{\sqrt{3}}{2} - \frac{\pi}{6} = \frac{3\sqrt{3} - \pi}{6}$.
Local minimum value $f(\frac{5\pi}{6}) = \sin(\frac{5\pi}{3}) - \frac{5\pi}{6} = -\frac{\sqrt{3}}{2} - \frac{5\pi}{6} = \frac{-3\sqrt{3} - 5\pi}{6}$.
Since these values do not match the options provided,the correct answer is $D$.

(D) For the first stone,the equation of motion is $s_1 = 19.6t - 4.9t^2$. The velocity is $v_1 = \frac{ds_1}{dt} = 19.6 - 9.8t$. At maximum height,$v_1 = 0$,so $19.6 - 9.8t = 0$,which gives $t = 2 \text{ s}$.
The maximum height $h$ reached by the first stone is $s_1(2) = 19.6(2) - 4.9(2)^2 = 39.2 - 19.6 = 19.6 \text{ m}$.
Now,consider the second stone with equation $s_2 = 9.8t - 4.9t^2$. The velocity is $v_2 = \frac{ds_2}{dt} = 9.8 - 9.8t$. The second stone reaches its maximum height at $t = 1 \text{ s}$,where $s_2(1) = 9.8(1) - 4.9(1)^2 = 4.9 \text{ m}$.
After $t = 1 \text{ s}$,the second stone starts moving downwards and hits the ground when $s_2 = 0$,i.e.,$9.8t - 4.9t^2 = 0 \implies 4.9t(2 - t) = 0$,so $t = 2 \text{ s}$.
At $t = 2 \text{ s}$,the first stone is at its maximum height $h$,and the second stone is at $s_2(2) = 9.8(2) - 4.9(2)^2 = 19.6 - 19.6 = 0 \text{ m}$.

(A) Given the function $f(x) = \begin{cases} |x| & 0 < |x| \leqslant 2 \\ 1 & x = 0 \end{cases}$.
For $x$ close to $0$ (but $x \neq 0$),$|x| > 0$. Since $f(x) = |x|$ for $x \neq 0$,we have $f(x) > 0$ for all $x$ in the neighborhood of $0$ except at $x=0$.
At $x = 0$,$f(0) = 1$.
For any $x$ in the interval $(-\delta, \delta)$ where $\delta$ is small (e.g.,$\delta = 0.5$),we have $f(x) = |x|$.
Since $|x| < 1$ for all $x \in (-1, 1)$,it follows that $f(x) < f(0)$ for all $x$ in the neighborhood of $0$ (excluding $x=0$).
Specifically,for $x \in (-0.5, 0.5)$ and $x \neq 0$,$f(x) = |x| < 0.5 < 1 = f(0)$.
Since $f(x) < f(0)$ for all $x$ in the neighborhood of $0$,$x = 0$ is a point of local maximum.

(D) Let $PQ = a$ and $PR = b$,and let $\theta$ be the angle between them.
The area of $\Delta PQR$ is given by $A = \frac{1}{2} ab \sin \theta$.
Let $f(\theta) = \frac{1}{2} ab \sin \theta$.
To find the maximum area,we differentiate $f(\theta)$ with respect to $\theta$:
$f'(\theta) = \frac{1}{2} ab \cos \theta$.
For maximum area,set $f'(\theta) = 0$:
$\frac{1}{2} ab \cos \theta = 0 \implies \cos \theta = 0$.
Since $0 < \theta < \pi$,we have $\theta = \frac{\pi}{2}$.
Using the second derivative test:
$f''(\theta) = -\frac{1}{2} ab \sin \theta$.
At $\theta = \frac{\pi}{2}$,$f''(\frac{\pi}{2}) = -\frac{1}{2} ab < 0$.
Since the second derivative is negative,the area is maximum at $\theta = \frac{\pi}{2}$.

(B) Given the constraint $x + y = 8$,we can express $y$ as $y = 8 - x$.
Let the function to be maximized be $f(x) = xy$.
Substituting $y$,we get $f(x) = x(8 - x) = 8x - x^2$.
To find the maximum value,we find the derivative $f'(x) = 8 - 2x$.
Setting $f'(x) = 0$ for critical points,we get $8 - 2x = 0$,which implies $x = 4$.
Since $x = 4$,then $y = 8 - 4 = 4$.
The maximum value of $xy$ is $4 \times 4 = 16$.

(B) Let the two equal sides of the triangle be $x$ and the angle between them be $2\theta$.
From the geometry of the triangle,the height is $h = x \sin \theta$ and the base is $b = 2x \cos \theta$.
The area of the triangle is given by $A = \frac{1}{2} \times \text{base} \times \text{height}$.
$A = \frac{1}{2} \times (2x \cos \theta) \times (x \sin \theta) = x^2 \sin \theta \cos \theta$.
Using the identity $\sin 2\theta = 2 \sin \theta \cos \theta$,we get $A = \frac{1}{2} x^2 \sin 2\theta$.
To maximize the area $A$,we need to maximize $\sin 2\theta$.
The maximum value of $\sin 2\theta$ is $1$,which occurs when $2\theta = 90^\circ$ or $\theta = 45^\circ$.
Therefore,the maximum area is $A_{\max} = \frac{1}{2} x^2 (1) = \frac{1}{2} x^2$.

(D) For Statement-$I$: We examine the behavior of $f(x)$ near $x = 0$.
For $x < 0$,$f(x) = -\frac{x}{2}$. As $x \to 0^-$,$f(x) \to 0$.
For $x \geq 0$,$f(x) = 7x + 8$. As $x \to 0^+$,$f(x) \to 8$.
At $x = 0$,$f(0) = 8$.
Since $f(x)$ is negative for $x < 0$ and $f(0) = 8$,for any small $h > 0$,$f(-h) = \frac{h}{2} > 0$,but $f(0) = 8$. However,we need $f(0) < f(-h)$ for a local minimum,which is $8 < \frac{h}{2}$. This is false for small $h$. Thus,$x=0$ is not a local minimum.
Statement-$I$ is false.
For Statement-$II$: This is the standard definition of a local minimum at $x = a$. Thus,Statement-$II$ is true.

(D) Let $f(x) = x^3 - 12x^2 + 45x$.
First,find the derivative $f'(x) = 3x^2 - 24x + 45 = 3(x^2 - 8x + 15) = 3(x - 3)(x - 5)$.
To find the critical points,set $f'(x) = 0$,which gives $x = 3$ and $x = 5$.
Now,evaluate the function $f(x)$ at the critical points and the endpoints of the interval $[0, 7]$:
$f(0) = (0)^3 - 12(0)^2 + 45(0) = 0$
$f(3) = (3)^3 - 12(3)^2 + 45(3) = 27 - 108 + 135 = 54$
$f(5) = (5)^3 - 12(5)^2 + 45(5) = 125 - 300 + 225 = 50$
$f(7) = (7)^3 - 12(7)^2 + 45(7) = 343 - 588 + 315 = 70$
Comparing these values,the maximum value of $f(x)$ on the interval $[0, 7]$ is $70$.

(C) Let $y = 64 \sec x + 27 \csc x$.
To find the minimum value,differentiate $y$ with respect to $x$:
$\frac{dy}{dx} = 64 \sec x \tan x - 27 \csc x \cot x$.
Set $\frac{dy}{dx} = 0$ for critical points:
$64 \sec x \tan x = 27 \csc x \cot x$.
Substituting trigonometric identities:
$64 \left(\frac{1}{\cos x}\right) \left(\frac{\sin x}{\cos x}\right) = 27 \left(\frac{1}{\sin x}\right) \left(\frac{\cos x}{\sin x}\right)$.
$\frac{\sin^3 x}{\cos^3 x} = \frac{27}{64} \Rightarrow \tan^3 x = \left(\frac{3}{4}\right)^3$.
Thus,$\tan x = \frac{3}{4}$.
For a right-angled triangle with opposite side $3$ and adjacent side $4$,the hypotenuse is $\sqrt{3^2 + 4^2} = 5$.
Therefore,$\sec x = \frac{5}{4}$ and $\csc x = \frac{5}{3}$.
Substituting these values into the original expression:
$y = 64 \left(\frac{5}{4}\right) + 27 \left(\frac{5}{3}\right) = 16(5) + 9(5) = 80 + 45 = 125$.

(B) Let $l$ be the slant height and $\alpha$ be the semi-vertical angle of the right circular cone. Let $h$ be the height and $r$ be the radius of the base.
Then $h = l \cos \alpha$ and $r = l \sin \alpha$.
The volume $V$ of the cone is given by $V = \frac{1}{3} \pi r^2 h$.
Substituting the values of $r$ and $h$,we get $V = \frac{1}{3} \pi (l \sin \alpha)^2 (l \cos \alpha) = \frac{1}{3} \pi l^3 \sin^2 \alpha \cos \alpha$.
To find the maximum volume,differentiate $V$ with respect to $\alpha$:
$\frac{dV}{d\alpha} = \frac{1}{3} \pi l^3 [\sin^2 \alpha (-\sin \alpha) + \cos \alpha (2 \sin \alpha \cos \alpha)]$
$\frac{dV}{d\alpha} = \frac{1}{3} \pi l^3 [-\sin^3 \alpha + 2 \sin \alpha \cos^2 \alpha]$.
Setting $\frac{dV}{d\alpha} = 0$ for critical points:
$-\sin^3 \alpha + 2 \sin \alpha (1 - \sin^2 \alpha) = 0$
$\sin \alpha (2 - 3 \sin^2 \alpha) = 0$.
Since $\alpha \neq 0$,we have $2 - 3 \sin^2 \alpha = 0$,which implies $\sin^2 \alpha = \frac{2}{3}$.
Using $\cos^2 \alpha = 1 - \sin^2 \alpha = 1 - \frac{2}{3} = \frac{1}{3}$,we find $\tan^2 \alpha = \frac{\sin^2 \alpha}{\cos^2 \alpha} = \frac{2/3}{1/3} = 2$.
Thus,$\tan \alpha = \sqrt{2}$,or $\alpha = \tan^{-1}(\sqrt{2})$.
Checking the second derivative $\frac{d^2V}{d\alpha^2}$ confirms that the volume is maximum at this value.

(C) Given function is $f(x) = \frac{\sin(x + a)}{\sin(x + b)}$.
To find the critical points,we differentiate $f(x)$ with respect to $x$ using the quotient rule:
$f'(x) = \frac{\sin(x + b) \cos(x + a) - \sin(x + a) \cos(x + b)}{\sin^2(x + b)}$.
Using the trigonometric identity $\sin(A - B) = \sin A \cos B - \cos A \sin B$,we get:
$f'(x) = \frac{\sin((x + a) - (x + b))}{\sin^2(x + b)} = \frac{\sin(a - b)}{\sin^2(x + b)}$.
Since $a \neq b$,$\sin(a - b) \neq 0$. Therefore,$f'(x) \neq 0$ for any $x$ in the domain.
Since the derivative is never zero,the function $f(x)$ does not have any local maxima or local minima at $x = 0$ or anywhere else in its domain.

(B) 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 $P'(0) = 0$,we have $c = 0$.
Thus,$P'(x) = 4x^3 + 3ax^2 + 2bx = x(4x^2 + 3ax + 2b)$.
Since $x=0$ is the only real root,the quadratic $4x^2 + 3ax + 2b = 0$ must have no real roots or a repeated root at $x=0$.
If $4x^2 + 3ax + 2b = 0$ has no real roots,then the discriminant $D = (3a)^2 - 4(4)(2b) = 9a^2 - 32b < 0$.
Since $P'(x)$ changes sign at $x=0$ (as it is a cubic with only one real root),$x=0$ is a point of local extremum.
For $x < 0$,$P'(x) < 0$ and for $x > 0$,$P'(x) > 0$,so $x=0$ is a local minimum.
Since $P(x)$ is a polynomial of degree $4$ with a positive leading coefficient,it is continuous on $[-1, 1]$.
The absolute minimum must occur at $x=0$ (since it is the only critical point).
Given $P(-1) < P(1)$,the maximum value on $[-1, 1]$ must occur at the endpoint $x=1$.
Therefore,$P(-1)$ is the minimum value and $P(1)$ is the maximum value of $P(x)$ on $[-1, 1]$.

(B) Since the curve passes through the origin $(0, 0)$,we have $0 = 2(0)^3 + a(0)^2 + b(0) + c$,which implies $c = 0$.
The derivative of the curve is $\frac{dy}{dx} = 6x^2 + 2ax + b$.
Given that the tangents at $x = -1$ and $x = 2$ are parallel to the $X$-axis,their slopes must be zero.
At $x = -1$: $6(-1)^2 + 2a(-1) + b = 0 \implies 6 - 2a + b = 0 \implies 2a - b = 6 \quad (1)$.
At $x = 2$: $6(2)^2 + 2a(2) + b = 0 \implies 24 + 4a + b = 0 \implies 4a + b = -24 \quad (2)$.
Adding equations $(1)$ and $(2)$:
$(2a - b) + (4a + b) = 6 - 24$
$6a = -18 \implies a = -3$.
Substituting $a = -3$ into equation $(1)$:
$2(-3) - b = 6 \implies -6 - b = 6 \implies b = -12$.
Thus,the values are $a = -3, b = -12, c = 0$.

(B) Let $f(x) = x^{25}(1 - x)^{75}$.
To find the critical points,we differentiate $f(x)$ with respect to $x$:
$f'(x) = 25x^{24}(1 - x)^{75} + x^{25} \cdot 75(1 - x)^{74}(-1)$
$f'(x) = 25x^{24}(1 - x)^{74} [(1 - x) - 3x]$
$f'(x) = 25x^{24}(1 - x)^{74} (1 - 4x)$.
Setting $f'(x) = 0$,we get critical points at $x = 0$,$x = 1$,and $x = 1/4$.
Evaluating the function at the critical points and endpoints:
$f(0) = 0^{25}(1 - 0)^{75} = 0$
$f(1) = 1^{25}(1 - 1)^{75} = 0$
$f(1/4) = (1/4)^{25}(1 - 1/4)^{75} = (1/4)^{25}(3/4)^{75} > 0$.
Since $f(x) \geq 0$ for all $x \in [0, 1]$ and the value at $x = 1/4$ is positive,the function attains its maximum at $x = 1/4$.

(A) Let $y = \sin^p x \cos^q x$.
Taking the natural logarithm on both sides,we get $z = \ln(y) = p \ln(\sin x) + q \ln(\cos x)$.
Differentiating with respect to $x$,we have $\frac{dz}{dx} = p \cot x - q \tan x$.
For critical points,set $\frac{dz}{dx} = 0$,which gives $p \cot x = q \tan x$,or $\tan^2 x = p/q$. Thus,$x = \tan^{-1} \sqrt{p/q}$.
To check for a maximum,we find the second derivative: $\frac{d^2z}{dx^2} = -p \csc^2 x - q \sec^2 x$.
Since $p, q > 0$,$\frac{d^2z}{dx^2} < 0$ for all $x$ in the domain,confirming that $x = \tan^{-1} \sqrt{p/q}$ is a point of local maximum.

(C) Given $f(x) = 2\sin x + \cos 2x$ for $0 \leq x \leq 2\pi$.
First,find the derivative: $f'(x) = 2\cos x - 2\sin 2x$.
Set $f'(x) = 0$ to find critical points:
$2\cos x - 4\sin x \cos x = 0$
$2\cos x(1 - 2\sin x) = 0$
This gives $\cos x = 0$ or $\sin x = 1/2$.
In the interval $[0, 2\pi]$,$\cos x = 0$ at $x = \pi/2, 3\pi/2$ and $\sin x = 1/2$ at $x = \pi/6, 5\pi/6$.
Now,find the second derivative: $f''(x) = -2\sin x - 4\cos 2x$.
Check the values:
At $x = \pi/6$: $f''(\pi/6) = -2(1/2) - 4\cos(\pi/3) = -1 - 4(1/2) = -3 < 0$ (Local Maximum).
At $x = \pi/2$: $f''(\pi/2) = -2(1) - 4\cos(\pi) = -2 + 4 = 2 > 0$ (Local Minimum).
At $x = 5\pi/6$: $f''(5\pi/6) = -2(1/2) - 4\cos(5\pi/3) = -1 - 4(1/2) = -3 < 0$ (Local Maximum).
At $x = 3\pi/2$: $f''(3\pi/2) = -2(-1) - 4\cos(3\pi) = 2 + 4 = 6 > 0$ (Local Minimum).
Comparing values at critical points: $f(\pi/6) = 2(1/2) + 1/2 = 1.5$,$f(5\pi/6) = 1.5$,$f(\pi/2) = 2(1) - 1 = 1$,$f(3\pi/2) = 2(-1) - 1 = -3$.
The maximum value is $1.5$ at $x = \pi/6$ and $x = 5\pi/6$.

(C) Given $f(x) = \ln|x| + bx^2 + ax$. The derivative is $f'(x) = \frac{1}{x} + 2bx + a$.
Since $f(x)$ has extreme values at $x = -1$ and $x = 2$,we must have $f'(-1) = 0$ and $f'(2) = 0$.
For $x = -1$: $-1 - 2b + a = 0 \implies a - 2b = 1$.
For $x = 2$: $\frac{1}{2} + 4b + a = 0 \implies a + 4b = -\frac{1}{2}$.
Subtracting the equations: $(a + 4b) - (a - 2b) = -\frac{1}{2} - 1 \implies 6b = -\frac{3}{2} \implies b = -\frac{1}{4}$.
Substituting $b = -\frac{1}{4}$ into $a - 2b = 1$: $a - 2(-\frac{1}{4}) = 1 \implies a + \frac{1}{2} = 1 \implies a = \frac{1}{2}$.
Thus,Statement-$2$ is true.
Now,$f''(x) = -\frac{1}{x^2} + 2b = -\frac{1}{x^2} - \frac{1}{2}$.
At $x = -1$,$f''(-1) = -1 - \frac{1}{2} = -\frac{3}{2} < 0$,so $f$ has a local maximum at $x = -1$.
At $x = 2$,$f''(2) = -\frac{1}{4} - \frac{1}{2} = -\frac{3}{4} < 0$,so $f$ has a local maximum at $x = 2$.
Thus,Statement-$1$ is true and Statement-$2$ is a correct explanation for Statement-$1$.