Increasing and Decreasing function Questions in English

We have,$f(x) = x^{2} + 2x - 5$.
First,find the derivative of the function with respect to $x$:
$f'(x) = \frac{d}{dx}(x^{2} + 2x - 5) = 2x + 2$.
To find the critical points,set $f'(x) = 0$:
$2x + 2 = 0 \Rightarrow 2x = -2 \Rightarrow x = -1$.
The point $x = -1$ divides the real line into two disjoint intervals: $(-\infty, -1)$ and $(-1, \infty)$.
Case $1$: In the interval $(-\infty, -1)$,choose a test point,say $x = -2$.
$f'(-2) = 2(-2) + 2 = -4 + 2 = -2 < 0$.
Since $f'(x) < 0$ for all $x \in (-\infty, -1)$,the function $f$ is strictly decreasing in $(-\infty, -1)$.
Case $2$: In the interval $(-1, \infty)$,choose a test point,say $x = 0$.
$f'(0) = 2(0) + 2 = 2 > 0$.
Since $f'(x) > 0$ for all $x \in (-1, \infty)$,the function $f$ is strictly increasing in $(-1, \infty)$.
Conclusion: The function is strictly decreasing on $(-\infty, -1)$ and strictly increasing on $(-1, \infty)$.

(A) Given function is $f(x) = 10 - 6x - 2x^2$.
First,find the derivative $f'(x)$:
$f'(x) = \frac{d}{dx}(10 - 6x - 2x^2) = -6 - 4x$.
To find the critical points,set $f'(x) = 0$:
$-6 - 4x = 0 \Rightarrow 4x = -6 \Rightarrow x = -\frac{3}{2}$.
The point $x = -\frac{3}{2}$ divides the real line into two intervals: $(-\infty, -\frac{3}{2})$ and $(-\frac{3}{2}, \infty)$.
Case $1$: For $x \in (-\infty, -\frac{3}{2})$,let $x = -2$.
$f'(-2) = -6 - 4(-2) = -6 + 8 = 2 > 0$.
Thus,$f(x)$ is strictly increasing on $(-\infty, -\frac{3}{2})$.
Case $2$: For $x \in (-\frac{3}{2}, \infty)$,let $x = 0$.
$f'(0) = -6 - 4(0) = -6 < 0$.
Thus,$f(x)$ is strictly decreasing on $(-\frac{3}{2}, \infty)$.

(N/A) Given function: $f(x) = -2x^{3} - 9x^{2} - 12x + 1$
First,find the derivative $f'(x)$:
$f'(x) = \frac{d}{dx}(-2x^{3} - 9x^{2} - 12x + 1) = -6x^{2} - 18x - 12$
Factorize $f'(x)$:
$f'(x) = -6(x^{2} + 3x + 2) = -6(x + 1)(x + 2)$
To find the critical points,set $f'(x) = 0$:
$-6(x + 1)(x + 2) = 0 \Rightarrow x = -1, x = -2$
The points $x = -2$ and $x = -1$ divide the real line into three intervals: $(-\infty, -2)$,$(-2, -1)$,and $(-1, \infty)$.
$1$. In the interval $(-\infty, -2)$,let $x = -3$:
$f'(-3) = -6(-3 + 1)(-3 + 2) = -6(-2)(-1) = -12 < 0$.
Thus,$f(x)$ is strictly decreasing in $(-\infty, -2)$.
$2$. In the interval $(-2, -1)$,let $x = -1.5$:
$f'(-1.5) = -6(-1.5 + 1)(-1.5 + 2) = -6(-0.5)(0.5) = 1.5 > 0$.
Thus,$f(x)$ is strictly increasing in $(-2, -1)$.
$3$. In the interval $(-1, \infty)$,let $x = 0$:
$f'(0) = -6(0 + 1)(0 + 2) = -12 < 0$.
Thus,$f(x)$ is strictly decreasing in $(-1, \infty)$.
Conclusion:
$f(x)$ is strictly increasing in $(-2, -1)$ and strictly decreasing in $(-\infty, -2) \cup (-1, \infty)$.

(N/A) Given function: $f(x) = 6 - 9x - x^{2}$.
Step $1$: Find the derivative $f'(x)$.
$f'(x) = \frac{d}{dx}(6 - 9x - x^{2}) = -9 - 2x$.
Step $2$: Find the critical point by setting $f'(x) = 0$.
$-9 - 2x = 0 \implies 2x = -9 \implies x = -\frac{9}{2}$.
Step $3$: The point $x = -\frac{9}{2}$ divides the real line into two intervals: $(-\infty, -\frac{9}{2})$ and $(-\frac{9}{2}, \infty)$.
Step $4$: Test the intervals.
For $x \in (-\infty, -\frac{9}{2})$,choose $x = -5$. Then $f'(-5) = -9 - 2(-5) = -9 + 10 = 1 > 0$. Thus,$f(x)$ is strictly increasing on $(-\infty, -\frac{9}{2})$.
For $x \in (-\frac{9}{2}, \infty)$,choose $x = 0$. Then $f'(0) = -9 - 2(0) = -9 < 0$. Thus,$f(x)$ is strictly decreasing on $(-\frac{9}{2}, \infty)$.

(A) Given function is $f(x) = (x+1)^{3}(x-3)^{3}$.
First,we find the derivative $f'(x)$ using the product rule:
$f'(x) = 3(x+1)^{2}(x-3)^{3} + 3(x-3)^{2}(x+1)^{3}$
$f'(x) = 3(x+1)^{2}(x-3)^{2} [(x-3) + (x+1)]$
$f'(x) = 3(x+1)^{2}(x-3)^{2} (2x-2)$
$f'(x) = 6(x+1)^{2}(x-3)^{2}(x-1)$
To find the critical points,set $f'(x) = 0$:
$6(x+1)^{2}(x-3)^{2}(x-1) = 0$
This gives $x = -1, 1, 3$.
These points divide the real line into four intervals: $(-\infty, -1)$,$(-1, 1)$,$(1, 3)$,and $(3, \infty)$.
$1$. For $x \in (-\infty, -1)$,$f'(x) < 0$,so $f$ is strictly decreasing.
$2$. For $x \in (-1, 1)$,$f'(x) < 0$,so $f$ is strictly decreasing.
$3$. For $x \in (1, 3)$,$f'(x) > 0$,so $f$ is strictly increasing.
$4$. For $x \in (3, \infty)$,$f'(x) > 0$,so $f$ is strictly increasing.
Thus,$f$ is strictly decreasing in $(-\infty, -1) \cup (-1, 1)$ and strictly increasing in $(1, 3) \cup (3, \infty)$.

(A) We have,$y=\log (1+x)-\frac{2 x}{2+x}$.
Differentiating with respect to $x$,we get:
$\frac{d y}{d x} = \frac{1}{1+x} - \frac{(2+x)(2) - 2x(1)}{(2+x)^2}$
$= \frac{1}{1+x} - \frac{4 + 2x - 2x}{(2+x)^2}$
$= \frac{1}{1+x} - \frac{4}{(2+x)^2}$
$= \frac{(2+x)^2 - 4(1+x)}{(1+x)(2+x)^2}$
$= \frac{4 + 4x + x^2 - 4 - 4x}{(1+x)(2+x)^2}$
$= \frac{x^2}{(1+x)(2+x)^2}$.
Since $x > -1$,we have $(1+x) > 0$ and $(2+x)^2 > 0$ for all $x$ in the domain.
Also,$x^2 \ge 0$ for all real $x$.
Thus,$\frac{d y}{d x} = \frac{x^2}{(1+x)(2+x)^2} \ge 0$ for all $x > -1$.
Since the derivative is non-negative and only zero at the isolated point $x=0$,the function $y$ is an increasing function throughout its domain $(-1, \infty)$.

(A) We have,$y = [x(x-2)]^2 = (x^2 - 2x)^2$.
To find the intervals where the function is increasing,we calculate the derivative:
$\frac{dy}{dx} = 2(x^2 - 2x)(2x - 2) = 4x(x-2)(x-1)$.
Setting $\frac{dy}{dx} = 0$,we get the critical points: $x = 0, 1, 2$.
These points divide the real line into intervals $(-\infty, 0), (0, 1), (1, 2), (2, \infty)$.
Testing the sign of $\frac{dy}{dx}$ in each interval:
$1$. For $x \in (-\infty, 0)$,$\frac{dy}{dx} < 0$ (decreasing).
$2$. For $x \in (0, 1)$,$\frac{dy}{dx} > 0$ (increasing).
$3$. For $x \in (1, 2)$,$\frac{dy}{dx} < 0$ (decreasing).
$4$. For $x \in (2, \infty)$,$\frac{dy}{dx} > 0$ (increasing).
Thus,the function is increasing for $x \in (0, 1) \cup (2, \infty)$.

(A) Given function: $y = \frac{4 \sin \theta}{2 + \cos \theta} - \theta$
Find the derivative with respect to $\theta$ using the quotient rule:
$\frac{dy}{d\theta} = \frac{(2 + \cos \theta)(4 \cos \theta) - (4 \sin \theta)(-\sin \theta)}{(2 + \cos \theta)^2} - 1$
Simplify the expression:
$\frac{dy}{d\theta} = \frac{8 \cos \theta + 4 \cos^2 \theta + 4 \sin^2 \theta}{(2 + \cos \theta)^2} - 1$
Since $\cos^2 \theta + \sin^2 \theta = 1$:
$\frac{dy}{d\theta} = \frac{8 \cos \theta + 4(1)}{(2 + \cos \theta)^2} - 1 = \frac{8 \cos \theta + 4 - (4 + 4 \cos \theta + \cos^2 \theta)}{(2 + \cos \theta)^2}$
$\frac{dy}{d\theta} = \frac{4 \cos \theta - \cos^2 \theta}{(2 + \cos \theta)^2} = \frac{\cos \theta (4 - \cos \theta)}{(2 + \cos \theta)^2}$
In the interval $\theta \in \left[0, \frac{\pi}{2}\right]$,we know that $\cos \theta \ge 0$ and $4 - \cos \theta > 0$ (since $\cos \theta \le 1$).
Also,$(2 + \cos \theta)^2 > 0$ for all $\theta$.
Therefore,$\frac{dy}{d\theta} \ge 0$ for all $\theta \in \left[0, \frac{\pi}{2}\right]$.
Since the derivative is non-negative and the function is continuous on the closed interval,$y$ is an increasing function in $\left[0, \frac{\pi}{2}\right]$.

(N/A) The given function is $f(x) = \log x$.
To determine the interval of increase,we find the derivative of the function with respect to $x$:
$f'(x) = \frac{d}{dx}(\log x) = \frac{1}{x}$.
For the function to be strictly increasing,we require $f'(x) > 0$.
In the given interval $(0, \infty)$,$x$ is always positive $(x > 0)$.
Therefore,$\frac{1}{x} > 0$ for all $x \in (0, \infty)$.
Since $f'(x) > 0$ for all $x$ in the interval $(0, \infty)$,the function $f(x) = \log x$ is strictly increasing on $(0, \infty)$.

The given function is $f(x)=x^{2}-x+1$.
$\therefore f^{\prime}(x)=2x-1$.
Now,$f^{\prime}(x)=0 \Rightarrow 2x-1=0 \Rightarrow x=\frac{1}{2}$.
The point $x=\frac{1}{2}$ divides the interval $(-1,1)$ into two disjoint intervals,i.e.,$(-1, \frac{1}{2})$ and $(\frac{1}{2}, 1)$.
In the interval $(-1, \frac{1}{2})$,choose a test point $x=0$. Then $f^{\prime}(0)=2(0)-1=-1 < 0$.
Therefore,$f$ is strictly decreasing in the interval $(-1, \frac{1}{2})$.
In the interval $(\frac{1}{2}, 1)$,choose a test point $x=\frac{3}{4}$. Then $f^{\prime}(\frac{3}{4})=2(\frac{3}{4})-1=\frac{3}{2}-1=\frac{1}{2} > 0$.
Therefore,$f$ is strictly increasing in the interval $(\frac{1}{2}, 1)$.
Since the function is strictly decreasing on $(-1, \frac{1}{2})$ and strictly increasing on $(\frac{1}{2}, 1)$,it is neither strictly increasing nor strictly decreasing on the entire interval $(-1, 1)$.

(A, B) Let $f_{1}(x) = \cos x$.
Then $f_{1}^{\prime}(x) = -\sin x$.
In the interval $\left(0, \frac{\pi}{2}\right)$,$\sin x > 0$,so $f_{1}^{\prime}(x) = -\sin x < 0$.
Thus,$f_{1}(x) = \cos x$ is strictly decreasing on $\left(0, \frac{\pi}{2}\right)$.
Let $f_{2}(x) = \cos 2x$.
Then $f_{2}^{\prime}(x) = -2 \sin 2x$.
For $0 < x < \frac{\pi}{2}$,we have $0 < 2x < \pi$,so $\sin 2x > 0$.
Thus,$f_{2}^{\prime}(x) = -2 \sin 2x < 0$ on $\left(0, \frac{\pi}{2}\right)$.
Therefore,$f_{2}(x) = \cos 2x$ is strictly decreasing on $\left(0, \frac{\pi}{2}\right)$.
Let $f_{3}(x) = \cos 3x$.
Then $f_{3}^{\prime}(x) = -3 \sin 3x$.
For $x \in \left(0, \frac{\pi}{2}\right)$,$3x \in (0, \frac{3\pi}{2})$.
In $\left(0, \frac{\pi}{3}\right)$,$\sin 3x > 0$,so $f_{3}^{\prime}(x) < 0$.
In $\left(\frac{\pi}{3}, \frac{\pi}{2}\right)$,$\sin 3x < 0$,so $f_{3}^{\prime}(x) > 0$.
Thus,$f_{3}(x)$ is not strictly decreasing on $\left(0, \frac{\pi}{2}\right)$.
Let $f_{4}(x) = \tan x$.
Then $f_{4}^{\prime}(x) = \sec^{2} x > 0$ for all $x \in \left(0, \frac{\pi}{2}\right)$.
Thus,$f_{4}(x)$ is strictly increasing on $\left(0, \frac{\pi}{2}\right)$.
Conclusion: Both $\cos x$ and $\cos 2x$ are strictly decreasing on $\left(0, \frac{\pi}{2}\right)$.

(B) Given the function $f(x) = x^{100} + \sin x - 1$.
To determine the intervals where the function is decreasing,we find its derivative:
$f'(x) = 100x^{99} + \cos x$.
$1$. In the interval $(0, 1)$: Since $x \in (0, 1)$,$100x^{99} > 0$ and $\cos x > 0$. Thus,$f'(x) > 0$,so the function is strictly increasing.
$2$. In the interval $\left(\frac{\pi}{2}, \pi\right)$: Here,$\cos x < 0$. However,$100x^{99}$ is very large for $x > \frac{\pi}{2} \approx 1.57$. Specifically,$100x^{99} > 1$ and $|\cos x| \le 1$,so $100x^{99} + \cos x > 0$. Thus,$f'(x) > 0$,and the function is strictly increasing.
$3$. In the interval $\left(0, \frac{\pi}{2}\right)$: Here,$100x^{99} > 0$ and $\cos x > 0$. Thus,$f'(x) > 0$,and the function is strictly increasing.
Since $f'(x) > 0$ in all the given intervals,the function is not decreasing in any of them.
Therefore,the correct answer is $B$.

(A) Given the function $f(x) = x^{2} + ax + 1$.
To find where the function is increasing,we calculate the derivative:
$f'(x) = 2x + a$.
The function $f(x)$ is increasing on $[1, 2]$ if $f'(x) \geq 0$ for all $x \in [1, 2]$.
Since $f'(x) = 2x + a$ is a linear function with a positive slope,its minimum value on the interval $[1, 2]$ occurs at the smallest value of $x$,which is $x = 1$.
Therefore,we require $f'(1) \geq 0$:
$2(1) + a \geq 0$
$2 + a \geq 0$
$a \geq -2$.
Thus,the function $f(x)$ is increasing on $[1, 2]$ for all $a \geq -2$.

(N/A) We have,$f(x) = x + \frac{1}{x}$.
Taking the derivative with respect to $x$,we get $f'(x) = 1 - \frac{1}{x^2} = \frac{x^2 - 1}{x^2}$.
For the function to be strictly increasing,we require $f'(x) > 0$.
Since $x^2 > 0$ for all $x \neq 0$,the sign of $f'(x)$ depends on the numerator $x^2 - 1$.
$f'(x) > 0 \iff x^2 - 1 > 0 \iff x^2 > 1 \iff |x| > 1$.
This implies $x > 1$ or $x < -1$.
Thus,$f'(x) > 0$ for $x \in (-\infty, -1) \cup (1, \infty)$.
Given that $I$ is an interval such that $I \cap [-1, 1] = \phi$,it follows that $I \subset (-\infty, -1)$ or $I \subset (1, \infty)$.
In both cases,$f'(x) > 0$ for all $x \in I$.
Therefore,the function $f$ is strictly increasing on $I$.

Given function is $f(x) = \log(\sin x)$.
First,we find the derivative of $f(x)$ with respect to $x$:
$f'(x) = \frac{d}{dx}(\log(\sin x)) = \frac{1}{\sin x} \cdot \cos x = \cot x$.
Case $1$: In the interval $\left(0, \frac{\pi}{2}\right)$,the value of $\cot x$ is positive $(\cot x > 0)$.
Since $f'(x) > 0$ for all $x \in \left(0, \frac{\pi}{2}\right)$,the function $f(x)$ is strictly increasing on $\left(0, \frac{\pi}{2}\right)$.
Case $2$: In the interval $\left(\frac{\pi}{2}, \pi\right)$,the value of $\cot x$ is negative $(\cot x < 0)$.
Since $f'(x) < 0$ for all $x \in \left(\frac{\pi}{2}, \pi\right)$,the function $f(x)$ is strictly decreasing on $\left(\frac{\pi}{2}, \pi\right)$.

Given $f(x) = \log |\cos x|$.
First,we find the derivative $f'(x)$:
$f'(x) = \frac{1}{\cos x} \cdot (-\sin x) = -\tan x$.
Case $1$: For $x \in \left(0, \frac{\pi}{2}\right)$:
In the first quadrant,$\tan x > 0$.
Therefore,$f'(x) = -\tan x < 0$.
Since $f'(x) < 0$ for all $x \in \left(0, \frac{\pi}{2}\right)$,the function $f(x)$ is strictly decreasing on $\left(0, \frac{\pi}{2}\right)$.
Case $2$: For $x \in \left(\frac{3\pi}{2}, 2\pi\right)$:
In the fourth quadrant,$\tan x < 0$.
Therefore,$f'(x) = -\tan x > 0$.
Since $f'(x) > 0$ for all $x \in \left(\frac{3\pi}{2}, 2\pi\right)$,the function $f(x)$ is strictly increasing on $\left(\frac{3\pi}{2}, 2\pi\right)$.

(A) Given function is $f(x) = x^{3} - 3x^{2} + 3x - 100$.
To check if the function is increasing,we find its derivative $f'(x)$ with respect to $x$:
$f'(x) = \frac{d}{dx}(x^{3} - 3x^{2} + 3x - 100)$
$f'(x) = 3x^{2} - 6x + 3$
Factor out $3$ from the expression:
$f'(x) = 3(x^{2} - 2x + 1)$
Recognize the perfect square trinomial:
$f'(x) = 3(x - 1)^{2}$
Since $(x - 1)^{2} \geq 0$ for all $x \in R$,it follows that $f'(x) = 3(x - 1)^{2} \geq 0$ for all $x \in R$.
Since the derivative $f'(x)$ is non-negative for all $x \in R$,the function $f(x)$ is an increasing function in $R$.

(D) We have,$y = x^{2} e^{-x}$.
To find the intervals where the function is increasing,we calculate the derivative:
$\frac{dy}{dx} = 2x e^{-x} - x^{2} e^{-x} = x e^{-x}(2 - x)$.
Setting $\frac{dy}{dx} = 0$ gives critical points:
$x e^{-x}(2 - x) = 0 \Rightarrow x = 0$ or $x = 2$.
These points divide the real line into three intervals: $(-\infty, 0)$,$(0, 2)$,and $(2, \infty)$.
We test the sign of $f'(x) = x e^{-x}(2 - x)$ in each interval:
$1$. For $x \in (-\infty, 0)$,let $x = -1$: $f'(-1) = (-1)e^{1}(2 - (-1)) = -3e < 0$. Thus,$f$ is decreasing.
$2$. For $x \in (0, 2)$,let $x = 1$: $f'(1) = (1)e^{-1}(2 - 1) = e^{-1} > 0$. Thus,$f$ is increasing.
$3$. For $x \in (2, \infty)$,let $x = 3$: $f'(3) = (3)e^{-3}(2 - 3) = -3e^{-3} < 0$. Thus,$f$ is decreasing.
Therefore,the function is increasing in the interval $(0, 2)$.
The correct answer is $D$.

(A) Given $f(x) = \frac{3}{10}x^4 - \frac{4}{5}x^3 - 3x^2 + \frac{36}{5}x + 11$.
Differentiating with respect to $x$:
$f'(x) = \frac{3}{10}(4x^3) - \frac{4}{5}(3x^2) - 3(2x) + \frac{36}{5}$
$f'(x) = \frac{6}{5}x^3 - \frac{12}{5}x^2 - 6x + \frac{36}{5}$
$f'(x) = \frac{6}{5}(x^3 - 2x^2 - 5x + 6)$
Factoring the cubic polynomial:
$f'(x) = \frac{6}{5}(x - 1)(x + 2)(x - 3)$
Setting $f'(x) = 0$,we get critical points $x = 1, x = -2, x = 3$. These points divide the real line into intervals $(-\infty, -2), (-2, 1), (1, 3), (3, \infty)$.
Testing the sign of $f'(x)$ in each interval:
$1.$ For $(-\infty, -2)$,choose $x = -3$: $f'(-3) = \frac{6}{5}(-4)(-1)(-6) < 0$. Thus,$f$ is decreasing.
$2.$ For $(-2, 1)$,choose $x = 0$: $f'(0) = \frac{6}{5}(-1)(2)(-3) > 0$. Thus,$f$ is increasing.
$3.$ For $(1, 3)$,choose $x = 2$: $f'(2) = \frac{6}{5}(1)(4)(-1) < 0$. Thus,$f$ is decreasing.
$4.$ For $(3, \infty)$,choose $x = 4$: $f'(4) = \frac{6}{5}(3)(6)(1) > 0$. Thus,$f$ is increasing.
Conclusion:
$(a)$ The function is increasing in $(-2, 1) \cup (3, \infty)$.
$(b)$ The function is decreasing in $(-\infty, -2) \cup (1, 3)$.

(N/A) Given $f(x) = \tan^{-1}(\sin x + \cos x)$.
Taking the derivative with respect to $x$:
$f'(x) = \frac{1}{1 + (\sin x + \cos x)^2} \cdot (\cos x - \sin x)$
$f'(x) = \frac{\cos x - \sin x}{1 + (\sin^2 x + \cos^2 x + 2\sin x \cos x)}$
Since $\sin^2 x + \cos^2 x = 1$ and $2\sin x \cos x = \sin 2x$,we have:
$f'(x) = \frac{\cos x - \sin x}{1 + (1 + \sin 2x)} = \frac{\cos x - \sin x}{2 + \sin 2x}$
For $x \in \left(0, \frac{\pi}{4}\right)$,we know that $\cos x > \sin x$,so $\cos x - \sin x > 0$.
Also,$2 + \sin 2x > 0$ for all $x$.
Since both the numerator and denominator are positive in the interval $\left(0, \frac{\pi}{4}\right)$,$f'(x) > 0$.
Therefore,the function $f(x)$ is strictly increasing in $\left(0, \frac{\pi}{4}\right)$.

(A) Given $f(x)=\frac{4 \sin x-2 x-x \cos x}{2+\cos x}$.
Using the quotient rule,$f'(x) = \frac{(2+\cos x)(4 \cos x - 2 - \cos x + x \sin x) - (4 \sin x - 2 x - x \cos x)(-\sin x)}{(2+\cos x)^2}$.
Simplifying the numerator: $(2+\cos x)(3 \cos x - 2 + x \sin x) + \sin x(4 \sin x - 2 x - x \cos x)$.
$= 6 \cos x - 4 + 2 x \sin x + 3 \cos^2 x - 2 \cos x + x \sin x \cos x + 4 \sin^2 x - 2 x \sin x - x \sin x \cos x$.
$= 4 \cos x - 4 + 3 \cos^2 x + 4 \sin^2 x = 4 \cos x - 4 + 3 \cos^2 x + 4(1 - \cos^2 x) = 4 \cos x - \cos^2 x$.
Thus,$f'(x) = \frac{\cos x(4 - \cos x)}{(2+\cos x)^2}$.
Since $(2+\cos x)^2 > 0$ and $(4 - \cos x) > 0$ for all $x$,the sign of $f'(x)$ depends on $\cos x$.
Setting $f'(x) = 0$ gives $\cos x = 0$,so $x = \frac{\pi}{2}, \frac{3\pi}{2}$ in $(0, 2\pi)$.
$(i)$ $f'(x) > 0$ when $\cos x > 0$,which occurs in $(0, \frac{\pi}{2}) \cup (\frac{3\pi}{2}, 2\pi)$.
$(ii)$ $f'(x) < 0$ when $\cos x < 0$,which occurs in $(\frac{\pi}{2}, \frac{3\pi}{2})$.

(N/A) $f(x) = x^{3} + x^{-3}$
$\therefore f'(x) = 3x^{2} - 3x^{-4} = 3x^{2} - \frac{3}{x^{4}} = \frac{3(x^{6} - 1)}{x^{4}}$
To find the critical points,set $f'(x) = 0$:
$3(x^{6} - 1) = 0 \Rightarrow x^{6} = 1 \Rightarrow x = \pm 1$
Since $x \neq 0$,the points $x = -1, 0, 1$ divide the real line into intervals $(-\infty, -1), (-1, 0), (0, 1), (1, \infty)$.
For $x \in (-\infty, -1)$,$f'(x) > 0$,so $f$ is increasing.
For $x \in (-1, 0)$,$f'(x) < 0$,so $f$ is decreasing.
For $x \in (0, 1)$,$f'(x) < 0$,so $f$ is decreasing.
For $x \in (1, \infty)$,$f'(x) > 0$,so $f$ is increasing.
Thus,$f$ is increasing in $(-\infty, -1) \cup (1, \infty)$ and decreasing in $(-1, 0) \cup (0, 1)$.

To prove that $f$ is an increasing function on $(a, b)$,we use the Mean Value Theorem $(MVT)$.
Let $x_1$ and $x_2$ be any two points in $(a, b)$ such that $x_1 < x_2$.
Since $f$ is defined on $[a, b]$ and $f^{\prime}(x) > 0$ for all $x \in (a, b)$,$f$ is continuous on $[x_1, x_2]$ and differentiable on $(x_1, x_2)$.
By the Mean Value Theorem,there exists a point $c \in (x_1, x_2)$ such that $f^{\prime}(c) = \frac{f(x_2) - f(x_1)}{x_2 - x_1}$.
Since $f^{\prime}(x) > 0$ for all $x \in (a, b)$,we have $f^{\prime}(c) > 0$.
Since $x_1 < x_2$,we have $x_2 - x_1 > 0$.
Therefore,$\frac{f(x_2) - f(x_1)}{x_2 - x_1} > 0$,which implies $f(x_2) - f(x_1) > 0$,or $f(x_2) > f(x_1)$.
Since $x_1 < x_2$ implies $f(x_1) < f(x_2)$ for any $x_1, x_2 \in (a, b)$,the function $f$ is strictly increasing on $(a, b)$.

(A) For $x \neq 0$,$f'(x) = \frac{d}{dx} \left( \frac{\ln(1+x)}{x} \right) = \frac{x \cdot \frac{1}{1+x} - \ln(1+x)}{x^2} = \frac{x - (1+x)\ln(1+x)}{x^2(1+x)}$.
Let $h(x) = x - (1+x)\ln(1+x)$.
Then $h'(x) = 1 - [\ln(1+x) + (1+x) \cdot \frac{1}{1+x}] = 1 - \ln(1+x) - 1 = -\ln(1+x)$.
For $x \in (-1, 0)$,$1+x \in (0, 1)$,so $\ln(1+x) < 0$,which implies $h'(x) > 0$.
For $x \in (0, \infty)$,$1+x > 1$,so $\ln(1+x) > 0$,which implies $h'(x) < 0$.
Since $h(0) = 0 - (1)\ln(1) = 0$,$h(x)$ increases on $(-1, 0)$ and decreases on $(0, \infty)$.
Thus,$h(x) < h(0) = 0$ for all $x \in (-1, \infty) \setminus \{0\}$.
Since $x^2(1+x) > 0$ for all $x \in (-1, \infty) \setminus \{0\}$,$f'(x) = \frac{h(x)}{x^2(1+x)} < 0$ for all $x \in (-1, \infty) \setminus \{0\}$.
Therefore,the function $f$ is strictly decreasing on $(-1, \infty)$.

(B) Given function: $f(x) = (3x - 7)x^{2/3} = 3x^{5/3} - 7x^{2/3}$.
To find the intervals where the function is increasing,we find the derivative $f'(x)$:
$f'(x) = 3 \cdot \frac{5}{3}x^{2/3} - 7 \cdot \frac{2}{3}x^{-1/3}$
$f'(x) = 5x^{2/3} - \frac{14}{3x^{1/3}}$
Simplify the expression for $f'(x)$:
$f'(x) = \frac{5x^{2/3} \cdot 3x^{1/3} - 14}{3x^{1/3}} = \frac{15x - 14}{3x^{1/3}}$
For the function to be increasing,we require $f'(x) > 0$:
$\frac{15x - 14}{3x^{1/3}} > 0$
We analyze the sign of $f'(x)$ using the critical points $x = 0$ and $x = \frac{14}{15}$:
- For $x < 0$,$15x - 14 < 0$ and $3x^{1/3} < 0$,so $f'(x) > 0$.
- For $0 < x < \frac{14}{15}$,$15x - 14 < 0$ and $3x^{1/3} > 0$,so $f'(x) < 0$.
- For $x > \frac{14}{15}$,$15x - 14 > 0$ and $3x^{1/3} > 0$,so $f'(x) > 0$.
Thus,$f(x)$ is increasing for $x \in (-\infty, 0) \cup \left(\frac{14}{15}, \infty\right)$.

(A) Given $f(x) = 3 \log_{e} \left| \frac{x-1}{x+1} \right| - \frac{2}{x-1}$.
We can rewrite this as $f(x) = 3 \log_{e} |x-1| - 3 \log_{e} |x+1| - \frac{2}{x-1}$.
Now,differentiate $f(x)$ with respect to $x$:
$f'(x) = 3 \left( \frac{1}{x-1} \right) - 3 \left( \frac{1}{x+1} \right) + \frac{2}{(x-1)^2}$.
Simplify the expression:
$f'(x) = 3 \left( \frac{(x+1) - (x-1)}{(x-1)(x+1)} \right) + \frac{2}{(x-1)^2} = 3 \left( \frac{2}{x^2-1} \right) + \frac{2}{(x-1)^2}$.
$f'(x) = \frac{6}{(x-1)(x+1)} + \frac{2}{(x-1)^2} = \frac{6(x-1) + 2(x+1)}{(x-1)^2(x+1)} = \frac{6x - 6 + 2x + 2}{(x-1)^2(x+1)} = \frac{8x - 4}{(x-1)^2(x+1)} = \frac{4(2x-1)}{(x-1)^2(x+1)}$.
For $f(x)$ to be increasing,$f'(x) \geq 0$.
Since $(x-1)^2 > 0$ for all $x \neq 1$,the sign of $f'(x)$ depends on $\frac{2x-1}{x+1}$.
Using the sign scheme for $\frac{2x-1}{x+1} \geq 0$:
The critical points are $x = \frac{1}{2}$ and $x = -1$.
Testing intervals: $(-\infty, -1)$,$(-1, \frac{1}{2}]$,and $[\frac{1}{2}, \infty)$.
$f'(x) > 0$ for $x \in (-\infty, -1) \cup [\frac{1}{2}, 1) \cup (1, \infty)$.

(A) Given $f(x) = \frac{4x^3 - 3x^2}{6} - 2 \sin x + (2x - 1) \cos x$.
First,we find the derivative $f'(x)$:
$f'(x) = \frac{12x^2 - 6x}{6} - 2 \cos x + [2 \cos x + (2x - 1)(-\sin x)]$
$f'(x) = (2x^2 - x) - 2 \cos x + 2 \cos x - (2x - 1) \sin x$
$f'(x) = x(2x - 1) - (2x - 1) \sin x$
$f'(x) = (2x - 1)(x - \sin x)$.
We know that for $x > 0$,$x > \sin x$,so $(x - \sin x) > 0$.
For $x < 0$,$x < \sin x$,so $(x - \sin x) < 0$.
Now,analyze the sign of $f'(x) = (2x - 1)(x - \sin x)$:
$1$. If $x \in [\frac{1}{2}, \infty)$,then $(2x - 1) \geq 0$ and $(x - \sin x) > 0$,so $f'(x) \geq 0$. Thus,$f(x)$ is increasing.
$2$. If $x \in [0, \frac{1}{2}]$,then $(2x - 1) \leq 0$ and $(x - \sin x) \geq 0$,so $f'(x) \leq 0$. Thus,$f(x)$ is decreasing.
$3$. If $x \in (-\infty, 0]$,then $(2x - 1) < 0$ and $(x - \sin x) \leq 0$,so $f'(x) \geq 0$. Thus,$f(x)$ is increasing.
Comparing with the options,$f(x)$ increases in $[\frac{1}{2}, \infty)$.

(D) To find where $f(x)$ is increasing,we calculate $f'(x)$ for each interval:
$1$. For $x < -5$,$f(x) = -55x$,so $f'(x) = -55$. Since $f'(x) < 0$,the function is decreasing on $(-\infty, -5)$.
$2$. For $-5 < x < 4$,$f(x) = 2x^3 - 3x^2 - 120x$,so $f'(x) = 6x^2 - 6x - 120 = 6(x^2 - x - 20) = 6(x - 5)(x + 4)$.
For $f'(x) > 0$,we need $(x - 5)(x + 4) > 0$,which occurs when $x < -4$ or $x > 5$. Within the interval $(-5, 4)$,this is satisfied for $x \in (-5, -4)$.
$3$. For $x > 4$,$f(x) = 2x^3 - 3x^2 - 36x - 336$,so $f'(x) = 6x^2 - 6x - 36 = 6(x^2 - x - 6) = 6(x - 3)(x + 2)$.
For $x > 4$,both $(x - 3)$ and $(x + 2)$ are positive,so $f'(x) > 0$ for all $x > 4$.
Combining these,$f(x)$ is increasing on $(-5, -4) \cup (4, \infty)$.

(A) Given the function $f(x) = x^{2} + ax + 1$.
The derivative is $f'(x) = 2x + a$.
For $f(x)$ to be increasing on $[1, 2]$,we must have $f'(x) \geq 0$ for all $x \in [1, 2]$.
$2x + a \geq 0 \implies a \geq -2x$ for all $x \in [1, 2]$.
The minimum value of $-2x$ on $[1, 2]$ occurs at $x = 2$,which is $-2(2) = -4$.
Thus,$R = -4$.
For $f(x)$ to be decreasing on $[1, 2]$,we must have $f'(x) \leq 0$ for all $x \in [1, 2]$.
$2x + a \leq 0 \implies a \leq -2x$ for all $x \in [1, 2]$.
The maximum value of $a$ is determined by the minimum value of $-2x$ on the interval,but since $a$ must be less than or equal to $-2x$ for all $x$,$a$ must be less than or equal to the minimum value of $-2x$ on $[1, 2]$.
The minimum value of $-2x$ on $[1, 2]$ is $-4$. Wait,let us re-evaluate: $a \leq -2x$ for all $x \in [1, 2]$. The condition $a \leq -2x$ must hold for the entire interval,so $a \leq \min(-2x) = -4$. Thus,the greatest value $S = -4$.
Let us re-check the increasing condition: $a \geq -2x$ for all $x \in [1, 2]$. This means $a \geq \max(-2x) = -2(1) = -2$. So $R = -2$.
Then $|R - S| = |-2 - (-4)| = |-2 + 4| = 2$.

(A) For $x > 0$,$f'(x) = -4x^2 + 4x + 3$.
Setting $f'(x) > 0$,we have $-4x^2 + 4x + 3 > 0$,which implies $4x^2 - 4x - 3 < 0$.
Factoring gives $(2x - 3)(2x + 1) < 0$,so $x \in \left(-\frac{1}{2}, \frac{3}{2}\right)$.
Since $x > 0$,$f(x)$ is increasing in $\left(0, \frac{3}{2}\right)$.
For $x \leq 0$,$f'(x) = 3e^x + 3xe^x = 3e^x(1 + x)$.
Setting $f'(x) > 0$,since $3e^x > 0$ for all $x$,we need $1 + x > 0$,which means $x > -1$.
Thus,for $x \leq 0$,$f(x)$ is increasing in $(-1, 0]$.
Combining both intervals,$f(x)$ is increasing in $(-1, 0] \cup (0, \frac{3}{2}) = \left(-1, \frac{3}{2}\right)$.

(C) Given $f(x)=3 \sin ^{4} x+10 \sin ^{3} x+6 \sin ^{2} x-3$.
Differentiating with respect to $x$:
$f'(x) = 12 \sin^3 x \cos x + 30 \sin^2 x \cos x + 12 \sin x \cos x$
$f'(x) = 6 \sin x \cos x (2 \sin^2 x + 5 \sin x + 2)$
Factoring the quadratic expression:
$f'(x) = 6 \sin x \cos x (2 \sin x + 1)(\sin x + 2)$
For $x \in \left[-\frac{\pi}{6}, \frac{\pi}{2}\right]$,we have:
$1. \cos x > 0$ for all $x \in \left[-\frac{\pi}{6}, \frac{\pi}{2}\right]$.
$2. (\sin x + 2) > 0$ for all $x \in \left[-\frac{\pi}{6}, \frac{\pi}{2}\right]$.
$3. (2 \sin x + 1) \ge 0$ for all $x \in \left[-\frac{\pi}{6}, \frac{\pi}{2}\right]$.
Thus,the sign of $f'(x)$ depends on $\sin x$:
- If $x \in \left(-\frac{\pi}{6}, 0\right)$,then $\sin x < 0$,so $f'(x) < 0$. Thus,$f$ is decreasing in $\left(-\frac{\pi}{6}, 0\right)$.
- If $x \in \left(0, \frac{\pi}{2}\right)$,then $\sin x > 0$,so $f'(x) > 0$. Thus,$f$ is increasing in $\left(0, \frac{\pi}{2}\right)$.

(D) Given $f_{\lambda}(x) = 4\lambda x^{3} - 36\lambda x^{2} + 36x + 48$.
For $f_{\lambda}(x)$ to be increasing for all $x \in \mathbb{R}$,we must have $f_{\lambda}^{\prime}(x) \geq 0$ for all $x \in \mathbb{R}$.
$f_{\lambda}^{\prime}(x) = 12\lambda x^{2} - 72\lambda x + 36$.
Setting $f_{\lambda}^{\prime}(x) \geq 0$,we get $12(\lambda x^{2} - 6\lambda x + 3) \geq 0$,which implies $\lambda x^{2} - 6\lambda x + 3 \geq 0$.
For this quadratic to be non-negative for all $x$,we must have $\lambda > 0$ and the discriminant $D \leq 0$.
$D = (-6\lambda)^{2} - 4(\lambda)(3) = 36\lambda^{2} - 12\lambda \leq 0$.
$12\lambda(3\lambda - 1) \leq 0$,which gives $\lambda \in [0, 1/3]$.
Since $\lambda > 0$,the largest value is $\lambda^{*} = 1/3$.
Now,$f_{\lambda^{*}}(x) = 4(1/3)x^{3} - 36(1/3)x^{2} + 36x + 48 = \frac{4}{3}x^{3} - 12x^{2} + 36x + 48$.
$f_{\lambda^{*}}(1) = \frac{4}{3} - 12 + 36 + 48 = \frac{4}{3} + 72 = \frac{220}{3}$.
$f_{\lambda^{*}}(-1) = \frac{4}{3}(-1) - 12(1) + 36(-1) + 48 = -\frac{4}{3} - 12 - 36 + 48 = -\frac{4}{3}$.
$f_{\lambda^{*}}(1) + f_{\lambda^{*}}(-1) = \frac{220}{3} - \frac{4}{3} = \frac{216}{3} = 72$.

(B) Let $f(x) = x^{7} + 5x^{3} + 3x + 1$.
Find the derivative of the function:
$f'(x) = 7x^{6} + 15x^{2} + 3$.
Since $x^{6} \ge 0$ and $x^{2} \ge 0$ for all real $x$,it follows that $7x^{6} \ge 0$ and $15x^{2} \ge 0$.
Therefore,$f'(x) = 7x^{6} + 15x^{2} + 3 \ge 3 > 0$ for all $x \in \mathbb{R}$.
Since $f'(x) > 0$ for all $x$,the function $f(x)$ is a strictly increasing function.
As $x \to -\infty$,$f(x) \to -\infty$,and as $x \to \infty$,$f(x) \to \infty$.
Since $f(x)$ is a continuous and strictly increasing function that ranges from $-\infty$ to $\infty$,by the Intermediate Value Theorem,it must cross the $x$-axis exactly once.
Thus,the number of real solutions is $1$.

(D) Given the function $f_{a}(x) = \tan^{-1}(2x) - 3ax + 7$.
For the function to be non-decreasing,its derivative must be non-negative: $f_{a}'(x) \geq 0$.
$f_{a}'(x) = \frac{2}{1+4x^2} - 3a \geq 0$.
This implies $3a \leq \frac{2}{1+4x^2}$,or $a \leq \frac{2}{3(1+4x^2)}$.
To satisfy this for all $x \in \left(-\frac{\pi}{6}, \frac{\pi}{6}\right)$,$a$ must be less than or equal to the minimum value of $\frac{2}{3(1+4x^2)}$ on the interval.
The minimum value occurs at the boundaries $x = \pm \frac{\pi}{6}$.
At $x^2 = \frac{\pi^2}{36}$,we have $a \leq \frac{2}{3(1 + 4(\frac{\pi^2}{36}))} = \frac{2}{3(1 + \frac{\pi^2}{9})} = \frac{2}{3(\frac{9+\pi^2}{9})} = \frac{6}{9+\pi^2}$.
Thus,$\bar{a} = \frac{6}{9+\pi^2}$.
Now,calculate $f_{\bar{a}}\left(\frac{\pi}{8}\right) = \tan^{-1}\left(2 \cdot \frac{\pi}{8}\right) - 3 \cdot \left(\frac{6}{9+\pi^2}\right) \cdot \frac{\pi}{8} + 7$.
$f_{\bar{a}}\left(\frac{\pi}{8}\right) = \tan^{-1}\left(\frac{\pi}{4}\right) - \frac{18\pi}{8(9+\pi^2)} + 7 = 7 + \tan^{-1}\left(\frac{\pi}{4}\right) - \frac{9\pi}{4(9+\pi^2)}$.

(A) Given function is $f(x) = x e^{x-x^2}$.
To find the intervals of increase and decrease,we calculate the derivative $f'(x)$:
$f'(x) = 1 \cdot e^{x-x^2} + x \cdot e^{x-x^2} \cdot (1-2x)$
$f'(x) = e^{x-x^2} [1 + x - 2x^2]$
$f'(x) = e^{x-x^2} [-(2x^2 - x - 1)]$
$f'(x) = -e^{x-x^2} (2x+1)(x-1)$.
For the function to be increasing,$f'(x) > 0$:
$-e^{x-x^2} (2x+1)(x-1) > 0$
$(2x+1)(x-1) < 0$.
The roots are $x = -\frac{1}{2}$ and $x = 1$. The inequality holds for $x \in \left(-\frac{1}{2}, 1\right)$.
Thus,the function is increasing in $\left(-\frac{1}{2}, 1\right)$.

(A) Let $f(x) = x^3-3ax^2+(27a^2+9)x+2016$.
To determine the number of real roots,we find the derivative $f'(x)$:
$f'(x) = 3x^2 - 6ax + (27a^2 + 9)$.
We can rewrite the derivative by completing the square:
$f'(x) = 3(x^2 - 2ax + 9a^2 + 3)$
$f'(x) = 3((x-a)^2 + 8a^2 + 3)$.
Since $(x-a)^2 \geq 0$,$8a^2 \geq 0$,and $3 > 0$,it follows that $f'(x) > 0$ for all real $x$ and all real $a$.
Because the derivative $f'(x)$ is always positive,the function $f(x)$ is strictly increasing for all real $a$.
$A$ strictly increasing cubic polynomial crosses the $x$-axis exactly once.
Therefore,the equation has exactly one real root for any real $a$.

(C) Let $f(x) = x^{2n+1} - (2n+1)x + a$.
To find the number of zeroes in the interval $[-1, 1]$,we examine the derivative of $f(x)$.
$f'(x) = (2n+1)x^{2n} - (2n+1) = (2n+1)(x^{2n} - 1)$.
For $x \in (-1, 1)$,we have $|x| < 1$,which implies $x^{2n} < 1$.
Thus,$x^{2n} - 1 < 0$,so $f'(x) < 0$ for all $x \in (-1, 1)$.
Since $f'(x) \leq 0$ for $x \in [-1, 1]$,the function $f(x)$ is strictly decreasing on the interval $[-1, 1]$.
$A$ strictly monotonic function can intersect the $X$-axis at most once.
Therefore,the equation $f(x) = 0$ has at most one root in the interval $[-1, 1]$ for any value of $a$.

(B) Let $f(x) = (x-41)^{49} + (x-49)^{41} + (x-2009)^{2009}$.
To determine the number of real roots,we examine the derivative of the function $f(x)$:
$f'(x) = 49(x-41)^{48} + 41(x-49)^{40} + 2009(x-2009)^{2008}$.
Since the exponents $48$,$40$,and $2008$ are all even,each term $(x-a)^{2n}$ is non-negative for all $x \in \mathbb{R}$.
Specifically,$(x-41)^{48} \ge 0$,$(x-49)^{40} \ge 0$,and $(x-2009)^{2008} \ge 0$.
Since the coefficients $49$,$41$,and $2009$ are all positive,$f'(x) \ge 0$ for all $x \in \mathbb{R}$.
Furthermore,$f'(x)$ is never zero for all $x$ simultaneously,as the terms vanish at different points $(x=41, 49, 2009)$.
Thus,$f'(x) > 0$ for all $x \in \mathbb{R}$,which implies that $f(x)$ is a strictly increasing function.
$A$ strictly increasing continuous function can cross the $x$-axis at most once.
As $x \to \infty$,$f(x) \to \infty$,and as $x \to -\infty$,$f(x) \to -\infty$.
By the Intermediate Value Theorem,there exists exactly one real root.
To check if the root is positive,we evaluate $f(0)$:
$f(0) = (-41)^{49} + (-49)^{41} + (-2009)^{2009} < 0$.
Since $f(0) < 0$ and $\lim_{x \to \infty} f(x) = \infty$,the unique real root must lie in the interval $(0, \infty)$.
Therefore,there is exactly one positive real root.

(C) Given $f(x) = \frac{\cos x}{\cos 2x}$.
Using the quotient rule,$f'(x) = \frac{-\sin x \cos 2x - \cos x (-2 \sin 2x)}{(\cos 2x)^2} = \frac{-\sin x \cos 2x + 2 \sin 2x \cos x}{(\cos 2x)^2}$.
Using the identity $\sin(A-B) = \sin A \cos B - \cos A \sin B$,we have $f'(x) = \frac{\sin(2x-x) + \sin 2x \cos x}{(\cos 2x)^2} = \frac{\sin x + 2 \sin x \cos^2 x}{(\cos 2x)^2}$.
$f'(x) = \frac{\sin x (1 + 2 \cos^2 x)}{(\cos 2x)^2}$.
Since $(1 + 2 \cos^2 x) > 0$ and $(\cos 2x)^2 > 0$ for all $x \in \left(-\frac{\pi}{4}, \frac{\pi}{4}\right)$,the sign of $f'(x)$ depends only on $\sin x$.
For $x \in \left(-\frac{\pi}{4}, 0\right)$,$\sin x < 0$,so $f'(x) < 0$,meaning $f(x)$ is decreasing.
For $x \in \left(0, \frac{\pi}{4}\right)$,$\sin x > 0$,so $f'(x) > 0$,meaning $f(x)$ is increasing.
Thus,the correct option is $C$.

(C) Given $g(x)=3 f\left(\frac{x}{3}\right)+f(3-x)$ and $f^{\prime \prime}(x) > 0$ for all $x \in(0,3)$.
Since $f^{\prime \prime}(x) > 0$,$f^{\prime}(x)$ is an increasing function.
Now,differentiate $g(x)$ with respect to $x$:
$g^{\prime}(x) = 3 \times \frac{1}{3} f^{\prime}\left(\frac{x}{3}\right) - f^{\prime}(3-x) = f^{\prime}\left(\frac{x}{3}\right) - f^{\prime}(3-x)$.
For $g$ to be decreasing in $(0, \alpha)$,we must have $g^{\prime}(x) < 0$ for $x \in (0, \alpha)$.
$f^{\prime}\left(\frac{x}{3}\right) - f^{\prime}(3-x) < 0 \Rightarrow f^{\prime}\left(\frac{x}{3}\right) < f^{\prime}(3-x)$.
Since $f^{\prime}(x)$ is an increasing function,this implies $\frac{x}{3} < 3-x$.
Solving for $x$: $x + \frac{x}{3} < 3 \Rightarrow \frac{4x}{3} < 3 \Rightarrow x < \frac{9}{4}$.
Thus,$\alpha = \frac{9}{4}$.
Finally,$8 \alpha = 8 \times \frac{9}{4} = 18$.

(B) Given $f(x) = \frac{x}{x^2-6x-16}$.
Using the quotient rule,$f'(x) = \frac{(x^2-6x-16)(1) - x(2x-6)}{(x^2-6x-16)^2}$.
Simplifying the numerator: $x^2 - 6x - 16 - 2x^2 + 6x = -x^2 - 16 = -(x^2 + 16)$.
Thus,$f'(x) = \frac{-(x^2+16)}{(x^2-6x-16)^2}$.
Since $x^2+16 > 0$ for all $x \in \mathbb{R}$ and $(x^2-6x-16)^2 > 0$ for all $x \neq -2, 8$,it follows that $f'(x) < 0$ for all $x$ in the domain.
Therefore,the function $f(x)$ is strictly decreasing in its entire domain $(-\infty, -2) \cup (-2, 8) \cup (8, \infty)$.

(B) Given the equation: $5 f(x)+4 f\left(\frac{1}{x}\right)=x^2-2$ ....$(1)$
Replace $x$ with $\frac{1}{x}$ in equation $(1)$:
$5 f\left(\frac{1}{x}\right)+4 f(x)=\frac{1}{x^2}-2$ ....$(2)$
Multiply $(1)$ by $5$ and $(2)$ by $4$:
$25 f(x)+20 f\left(\frac{1}{x}\right)=5x^2-10$
$16 f(x)+20 f\left(\frac{1}{x}\right)=\frac{4}{x^2}-8$
Subtracting the two equations:
$9 f(x) = 5x^2 - 10 - \frac{4}{x^2} + 8 = 5x^2 - 2 - \frac{4}{x^2}$
Given $y = 9x^2 f(x)$,substitute $9 f(x)$:
$y = x^2 \left(5x^2 - 2 - \frac{4}{x^2}\right) = 5x^4 - 2x^2 - 4$
To find where $y$ is strictly increasing,calculate the derivative:
$\frac{dy}{dx} = 20x^3 - 4x = 4x(5x^2 - 1)$
For strictly increasing,$\frac{dy}{dx} > 0$:
$4x(\sqrt{5}x - 1)(\sqrt{5}x + 1) > 0$
The critical points are $x = -\frac{1}{\sqrt{5}}, 0, \frac{1}{\sqrt{5}}$.
Testing intervals,we find $\frac{dy}{dx} > 0$ for $x \in \left(-\frac{1}{\sqrt{5}}, 0\right) \cup \left(\frac{1}{\sqrt{5}}, \infty\right)$.

(D) Given $f(x) = \sin x + 3x - \frac{2}{\pi}(x^2 + x)$.
Step $1$: Analyze statement $(I)$.
$f'(x) = \cos x + 3 - \frac{2}{\pi}(2x + 1)$.
For $x \in (0, \pi/2)$,$\cos x \in (0, 1)$ and $2x+1 \in (1, \pi+1)$.
$f'(x) = \cos x + 3 - \frac{4x}{\pi} - \frac{2}{\pi}$.
At $x=0$,$f'(0) = 1 + 3 - 2/\pi = 4 - 2/\pi > 0$.
At $x=\pi/2$,$f'(\pi/2) = 0 + 3 - \frac{2}{\pi}(\pi + 1) = 3 - 2 - 2/\pi = 1 - 2/\pi > 0$.
Since $f''(x) = -\sin x - 4/\pi < 0$,$f'(x)$ is decreasing. The minimum value of $f'(x)$ on $[0, \pi/2]$ is $f'(\pi/2) = 1 - 2/\pi > 0$. Thus,$f'(x) > 0$ for all $x \in (0, \pi/2)$,so $f$ is increasing.
Step $2$: Analyze statement $(II)$.
$f''(x) = -\sin x - 4/\pi$.
Since $\sin x > 0$ for $x \in (0, \pi/2)$,$f''(x) = -(\sin x + 4/\pi) < 0$.
Since $f''(x) < 0$,$f'(x)$ is decreasing in $(0, \pi/2)$.
Conclusion: Both $(I)$ and $(II)$ are true.

(D) Given the function $f(x) = x^x$ for $x > 0$.
Taking the natural logarithm on both sides,we get $\ln(f(x)) = x \ln(x)$.
Differentiating both sides with respect to $x$ using the product rule:
$\frac{1}{f(x)} f'(x) = 1 \cdot \ln(x) + x \cdot \frac{1}{x} = \ln(x) + 1$.
Thus,$f'(x) = x^x(1 + \ln(x))$.
For the function to be strictly increasing,we require $f'(x) > 0$.
Since $x^x > 0$ for all $x > 0$,the condition $f'(x) > 0$ implies $1 + \ln(x) > 0$.
$\ln(x) > -1$.
$x > e^{-1}$,which means $x > \frac{1}{e}$.
Therefore,the interval in which the function is strictly increasing is $\left(\frac{1}{e}, \infty\right)$.
Note: The option $\left[\frac{1}{e}, \infty\right)$ is the standard representation for the interval where the function is non-decreasing.

(B) Given $f(x) = \cos x - x + 1$.
Find the derivative: $f'(x) = -\sin x - 1$.
Since $-1 \le \sin x \le 1$,we have $f'(x) = -(\sin x + 1) \le 0$ for all $x \in R$.
Specifically,$f'(x) < 0$ for all $x$ except where $\sin x = -1$.
Thus,$f(x)$ is a strictly decreasing function on $R$.
For $(S1)$: Evaluate at endpoints of $[0, \pi]$: $f(0) = \cos(0) - 0 + 1 = 2$ and $f(\pi) = \cos(\pi) - \pi + 1 = -1 - \pi + 1 = -\pi$.
Since $f(0) = 2 > 0$ and $f(\pi) = -\pi < 0$,by the Intermediate Value Theorem,there exists exactly one root in $(0, \pi)$ because $f$ is strictly decreasing. So,$(S1)$ is correct.
For $(S2)$: Since $f'(x) \le 0$ for all $x$,$f(x)$ is decreasing on the entire interval $[0, \pi]$. Therefore,the statement that it is increasing in $[\frac{\pi}{2}, \pi]$ is incorrect. So,$(S2)$ is incorrect.
Conclusion: Only $(S1)$ is correct.

(B) For $f(x)$ to be strictly increasing on $(2, 3)$,we require $f'(x) \geq 0$ for $x \in (2, 3)$.
$f'(x) = \frac{2}{x-2} - 2x + a \geq 0$.
Since $f''(x) = -\frac{2}{(x-2)^2} - 2 < 0$,$f'(x)$ is a strictly decreasing function.
Thus,$f'(x) \geq 0$ on $(2, 3)$ implies $f'(3) \geq 0$.
$f'(3) = \frac{2}{3-2} - 2(3) + a = 2 - 6 + a = a - 4 \geq 0$,so $a \geq 4$. The smallest value is $a = 4$.
Now,$g(x) = (x-1)^3(x+2-4)^2 = (x-1)^3(x-2)^2$.
$g'(x) = 3(x-1)^2(x-2)^2 + (x-1)^3 \cdot 2(x-2) = (x-1)^2(x-2)[3(x-2) + 2(x-1)]$.
$g'(x) = (x-1)^2(x-2)(3x - 6 + 2x - 2) = (x-1)^2(x-2)(5x - 8)$.
For $g(x)$ to be strictly decreasing,$g'(x) < 0$.
Since $(x-1)^2 > 0$ for $x \neq 1$,we need $(x-2)(5x-8) < 0$.
The roots are $x = 8/5$ and $x = 2$. The interval is $(8/5, 2)$.
Thus,$b = 8/5$ and $c = 2$.
$100(a + b - c) = 100(4 + 8/5 - 2) = 100(2 + 1.6) = 100(3.6) = 360$.

(D) Given $f(x) = \frac{x}{3} + \frac{3}{x} + 3$,$x \neq 0$.
To find the intervals of increase and decrease,we calculate the derivative:
$f'(x) = \frac{1}{3} - \frac{3}{x^2} = \frac{x^2 - 9}{3x^2} = \frac{(x-3)(x+3)}{3x^2}$.
Setting $f'(x) = 0$,we get $x = 3$ and $x = -3$.
The critical points are $x = -3, 0, 3$.
Testing the intervals:
For $x \in (-\infty, -3)$,$f'(x) > 0$,so $f(x)$ is strictly increasing.
For $x \in (-3, 0)$,$f'(x) < 0$,so $f(x)$ is strictly decreasing.
For $x \in (0, 3)$,$f'(x) < 0$,so $f(x)$ is strictly decreasing.
For $x \in (3, \infty)$,$f'(x) > 0$,so $f(x)$ is strictly increasing.
Comparing with the given intervals:
Increasing in $(-\infty, -3) \cup (3, \infty) \Rightarrow \alpha_1 = -3, \alpha_2 = 3$.
Decreasing in $(-3, 0) \cup (0, 3) \Rightarrow \alpha_3 = -3, \alpha_4 = 0, \alpha_5 = 3$.
Thus,$\sum_{i=1}^5 \alpha_i^2 = (-3)^2 + (3)^2 + (-3)^2 + (0)^2 + (3)^2 = 9 + 9 + 9 + 0 + 9 = 36$.

(A) Given $2f(x) + 3f\left(\frac{1}{x}\right) = x^2 + 1$ (Equation $1$).
Replace $x$ with $\frac{1}{x}$: $2f\left(\frac{1}{x}\right) + 3f(x) = \frac{1}{x^2} + 1$ (Equation $2$).
Multiply Equation $1$ by $2$ and Equation $2$ by $3$:
$4f(x) + 6f\left(\frac{1}{x}\right) = 2x^2 + 2$
$9f(x) + 6f\left(\frac{1}{x}\right) = \frac{3}{x^2} + 3$
Subtracting the first from the second: $5f(x) = \frac{3}{x^2} + 3 - 2x^2 - 2 = \frac{3}{x^2} - 2x^2 + 1$.
Then $y = 5x^2 f(x) = x^2 \left(\frac{3}{x^2} - 2x^2 + 1\right) = 3 - 2x^4 + x^2$.
To find where $y$ is strictly increasing,calculate $\frac{dy}{dx} = -8x^3 + 2x = 2x(1 - 4x^2) = 2x(1 - 2x)(1 + 2x)$.
Set $\frac{dy}{dx} > 0$: $2x(1 - 2x)(1 + 2x) > 0$.
Critical points are $x = 0, \frac{1}{2}, -\frac{1}{2}$.
Testing intervals: $(-\infty, -1/2) \implies (-)(-)(-) < 0$; $(-1/2, 0) \implies (-)(-)(+) > 0$; $(0, 1/2) \implies (+)(+)(+) > 0$; $(1/2, \infty) \implies (+)(-)(+) < 0$.
Thus,$y$ is strictly increasing in $(-\infty, -1/2) \cup (0, 1/2)$.
Comparing with options,the interval $(0, 1/2)$ is correct.

(C) To determine the intervals of increase and decrease,we find the derivative $f'(x)$.
Given $f(x) = (x + 2) e^{-x}$.
Using the product rule,$f'(x) = (1) e^{-x} + (x + 2) (-e^{-x})$.
$f'(x) = e^{-x} (1 - x - 2) = e^{-x} (-x - 1) = -(x + 1) e^{-x}$.
For the function to be increasing,$f'(x) > 0$,which implies $-(x + 1) e^{-x} > 0$. Since $e^{-x} > 0$ for all $x$,we must have $-(x + 1) > 0$,or $x + 1 < 0$,which means $x < -1$.
Thus,the function is increasing in $(-\infty, -1)$.
For the function to be decreasing,$f'(x) < 0$,which implies $-(x + 1) e^{-x} < 0$. This means $x + 1 > 0$,or $x > -1$.
Thus,the function is decreasing in $(-1, \infty)$.
Therefore,the correct option is $C$.

(D) For $f(x)$ to be strictly increasing,we must have $f'(x) > 0$ for all $x$.
Given $f(x) = \frac{k \sin x + 2 \cos x}{\sin x + \cos x}$.
Using the quotient rule,$f'(x) = \frac{(k \cos x - 2 \sin x)(\sin x + \cos x) - (k \sin x + 2 \cos x)(\cos x - \sin x)}{(\sin x + \cos x)^2}$.
Simplifying the numerator:
$N = (k \sin x \cos x + k \cos^2 x - 2 \sin^2 x - 2 \sin x \cos x) - (k \sin x \cos x - k \sin^2 x + 2 \cos^2 x - 2 \sin x \cos x)$.
$N = k \cos^2 x - 2 \sin^2 x + k \sin^2 x - 2 \cos^2 x$.
$N = (k - 2) \cos^2 x + (k - 2) \sin^2 x = (k - 2)(\cos^2 x + \sin^2 x) = k - 2$.
Thus,$f'(x) = \frac{k - 2}{(\sin x + \cos x)^2}$.
For $f(x)$ to be strictly increasing,$f'(x) > 0$.
Since $(\sin x + \cos x)^2 > 0$ for all $x$ where the function is defined,we require $k - 2 > 0$,which implies $k > 2$.