Increasing and Decreasing function Questions in English

(D) First,we find $f'(x)$ for both intervals:
For $x \le 0$,$f(x) = x e^{3x}$. Using the product rule,$f'(x) = e^{3x} + 3x e^{3x} = e^{3x}(1 + 3x)$.
For $x > 0$,$f(x) = 2x^3 + x$. Then $f'(x) = 6x^2 + 1$.
Thus,$f'(x) = \begin{cases} e^{3x}(1 + 3x), & x \le 0 \\ 6x^2 + 1, & x > 0 \end{cases}$.
For $f'(x)$ to be an increasing function,its derivative $f''(x)$ must be greater than $0$.
For $x < 0$,$f''(x) = \frac{d}{dx}[e^{3x}(1 + 3x)] = 3e^{3x}(1 + 3x) + 3e^{3x} = 3e^{3x}(1 + 3x + 1) = 3e^{3x}(2 + 3x)$.
Setting $f''(x) > 0$,we get $3e^{3x}(2 + 3x) > 0$. Since $3e^{3x} > 0$ for all $x$,we need $2 + 3x > 0$,which implies $x > -\frac{2}{3}$.
For $x > 0$,$f''(x) = \frac{d}{dx}[6x^2 + 1] = 12x$. Setting $f''(x) > 0$,we get $12x > 0$,which implies $x > 0$.
Combining these intervals for $x < 0$ and $x > 0$,we get $x \in (-\frac{2}{3}, 0) \cup (0, \infty)$.
Checking continuity of $f'(x)$ at $x=0$: $\lim_{x \to 0^-} f'(x) = e^0(1+0) = 1$ and $\lim_{x \to 0^+} f'(x) = 6(0)^2 + 1 = 1$. Since $f'(0) = 1$,$f'(x)$ is continuous at $x=0$.
Thus,$f'(x)$ is increasing for $x \in (-\frac{2}{3}, \infty)$.

(A) Given $f(x) = e^x - e^{-x} + \cos x$.
First,we find the derivative $f'(x)$:
$f'(x) = \frac{d}{dx}(e^x - e^{-x} + \cos x) = e^x + e^{-x} - \sin x$.
We know that $e^x + e^{-x} = 2 \cosh x$.
Since $e^x + e^{-x} \ge 2$ for all $x \in \mathbb{R}$ (by $AM$-$GM$ inequality) and $\sin x \le 1$ for all $x \in \mathbb{R}$,we have:
$f'(x) = (e^x + e^{-x}) - \sin x \ge 2 - 1 = 1$.
Since $f'(x) > 0$ for all $x \in \mathbb{R}$,the function $f(x)$ is always increasing.

(C) Given $f(x) = \int e^x (x - 1)(x - 2) dx$.
To find the interval where $f(x)$ decreases,we need to find $f'(x)$ and set $f'(x) < 0$.
By the Fundamental Theorem of Calculus,$f'(x) = \frac{d}{dx} \int e^x (x - 1)(x - 2) dx = e^x (x - 1)(x - 2)$.
For $f(x)$ to be a decreasing function,we require $f'(x) < 0$.
So,$e^x (x - 1)(x - 2) < 0$.
Since $e^x > 0$ for all real $x$,the inequality simplifies to $(x - 1)(x - 2) < 0$.
The roots of the quadratic $(x - 1)(x - 2) = 0$ are $x = 1$ and $x = 2$.
The expression $(x - 1)(x - 2)$ is negative between its roots.
Therefore,$f(x)$ decreases in the interval $(1, 2)$.

(C) Given the function $f(x) = 2x^2 - \ln |x|$.
For the function to be monotonically decreasing,we must have $f'(x) < 0$.
First,find the derivative: $f'(x) = 4x - \frac{1}{x}$.
Set the derivative to be less than zero: $4x - \frac{1}{x} < 0$.
This simplifies to $\frac{4x^2 - 1}{x} < 0$,which is $\frac{(2x - 1)(2x + 1)}{x} < 0$.
To solve this inequality,we find the critical points where the numerator or denominator is zero: $x = -1/2, 0, 1/2$.
Testing the intervals:
For $x \in (-\infty, -1/2)$,$f'(x) < 0$.
For $x \in (-1/2, 0)$,$f'(x) > 0$.
For $x \in (0, 1/2)$,$f'(x) < 0$.
For $x \in (1/2, \infty)$,$f'(x) > 0$.
Thus,the function is monotonically decreasing in the interval $(-\infty, -1/2) \cup (0, 1/2)$.

(D) Given $f(x) = x^3 - x^2 + 100x + 1001$.
To determine the nature of the function,we find its derivative:
$f'(x) = 3x^2 - 2x + 100$.
The discriminant of this quadratic is $D = (-2)^2 - 4(3)(100) = 4 - 1200 = -1196$.
Since $D < 0$ and the coefficient of $x^2$ is positive,$f'(x) > 0$ for all $x \in \mathbb{R}$.
Thus,$f(x)$ is a strictly increasing function.
For a strictly increasing function,if $a > b$,then $f(a) > f(b)$.
Since $\frac{1}{1999} > \frac{1}{2000}$,it follows that $f\left(\frac{1}{1999}\right) > f\left(\frac{1}{2000}\right)$.
Therefore,option $D$ is correct.

(D) Given $f(x) = \frac{|x - 1|}{x^2}$.
Case $1$: If $x < 1$ and $x \neq 0$,then $f(x) = \frac{-(x - 1)}{x^2} = \frac{1 - x}{x^2} = x^{-2} - x^{-1}$.
$f'(x) = -2x^{-3} + x^{-2} = \frac{-2 + x}{x^3} = \frac{x - 2}{x^3}$.
For $f(x)$ to be decreasing,$f'(x) < 0$,so $\frac{x - 2}{x^3} < 0$.
This holds for $x \in (0, 1)$ and $x \in (2, \infty)$ is not applicable here as we assumed $x < 1$.
Thus,for $x \in (0, 1)$,$f'(x) < 0$.
Case $2$: If $x > 1$,then $f(x) = \frac{x - 1}{x^2} = x^{-1} - x^{-2}$.
$f'(x) = -x^{-2} + 2x^{-3} = \frac{-x + 2}{x^3} = \frac{-(x - 2)}{x^3}$.
For $f(x)$ to be decreasing,$f'(x) < 0$,so $\frac{-(x - 2)}{x^3} < 0$,which means $\frac{x - 2}{x^3} > 0$.
This holds for $x \in (2, \infty)$.
Combining both cases,the function is decreasing in $(0, 1) \cup (2, \infty)$.

(C) Let $f(x) = |x^2 - |x| - 2|$.
Since $f(x)$ is an even function,we analyze it for $x \ge 0$,where $f(x) = |x^2 - x - 2| = |(x-2)(x+1)|$.
For $x \ge 0$,$x+1 > 0$,so $f(x) = |x-2|(x+1)$.
If $0 \le x < 2$,$f(x) = (2-x)(x+1) = -x^2 + x + 2$.
The derivative is $f'(x) = -2x + 1$.
Setting $f'(x) = 0$ gives $x = 1/2$.
The function increases on $(0, 1/2)$ and decreases on $(1/2, 2)$.
Thus,the function is non-monotonic in any interval containing $1/2$,such as $(0, 2)$.
Looking at the options,$x \in (0, 2)$ is the correct interval where the function is non-monotonic.

(C) Analysis of $S_1$: Let $f(x) = e^x$. Then $f'(x) = e^x > 0$. The function $g(x) = \frac{f(x)}{f'(x)} = \frac{e^x}{e^x} = 1$. Since $g(x) = 1$ is a constant function,it is not strictly increasing. Thus,$S_1$ is false.
Analysis of $S_2$: For $x \in (0, \frac{\pi}{2})$,$\frac{d}{dx}(\sin x) = \cos x > 0$ and $\frac{d}{dx}(\tan x) = \sec^2 x > 0$. Since the derivatives are positive,both $\sin x$ and $\tan x$ are strictly increasing in $(0, \frac{\pi}{2})$. Thus,$S_2$ is true.
Therefore,$S_1$ is wrong and $S_2$ is right.

(D) The function is defined as $f(x) = \begin{cases} \frac{-(x-1)}{x^2} & x < 1 \\ \frac{x-1}{x^2} & x \geq 1 \end{cases}$.
Calculating the derivative $f'(x)$:
For $x < 1$,$f(x) = \frac{-x+1}{x^2} = -x^{-1} + x^{-2}$,so $f'(x) = x^{-2} - 2x^{-3} = \frac{x-2}{x^3}$.
For $x > 1$,$f(x) = \frac{x-1}{x^2} = x^{-1} - x^{-2}$,so $f'(x) = -x^{-2} + 2x^{-3} = \frac{-(x-2)}{x^3}$.
For $f(x)$ to be monotonically decreasing,we require $f'(x) < 0$.
Case $1$: $x < 1$. We need $\frac{x-2}{x^3} < 0$. Since $x^3$ has the same sign as $x$,this implies $x(x-2) < 0$,which gives $x \in (0, 2)$. Combining with $x < 1$,we get $x \in (0, 1)$.
Case $2$: $x > 1$. We need $\frac{-(x-2)}{x^3} < 0$,which implies $\frac{x-2}{x^3} > 0$. This holds when $x \in (-\infty, 0) \cup (2, \infty)$. Combining with $x > 1$,we get $x \in (2, \infty)$.
Therefore,the function is monotonically decreasing in $(0, 1) \cup (2, \infty)$.

(B) For a function of the form $f(x) = \frac{a \sin x + b \cos x}{c \sin x + d \cos x}$ to be monotonically increasing,its derivative $f'(x)$ must be greater than $0$.
The derivative of $f(x) = \frac{u}{v}$ is $f'(x) = \frac{u'v - uv'}{v^2}$.
Here,$u = \lambda \sin x + 3 \cos x$ and $v = 2 \sin x + 6 \cos x$.
$u' = \lambda \cos x - 3 \sin x$ and $v' = 2 \cos x - 6 \sin x$.
$u'v - uv' = (\lambda \cos x - 3 \sin x)(2 \sin x + 6 \cos x) - (\lambda \sin x + 3 \cos x)(2 \cos x - 6 \sin x)$.
Expanding this,we get $(2\lambda \sin x \cos x + 6\lambda \cos^2 x - 6 \sin^2 x - 18 \sin x \cos x) - (2\lambda \sin x \cos x - 6\lambda \sin^2 x + 6 \cos^2 x - 18 \sin x \cos x)$.
Simplifying,the terms $2\lambda \sin x \cos x$ and $-18 \sin x \cos x$ cancel out.
We are left with $6\lambda \cos^2 x - 6 \sin^2 x + 6\lambda \sin^2 x - 6 \cos^2 x = 6(\lambda - 1)(\cos^2 x + \sin^2 x) = 6(\lambda - 1)$.
For $f'(x) > 0$,we must have $6(\lambda - 1) > 0$,which implies $\lambda - 1 > 0$,or $\lambda > 1$.

(A) The function is defined as $f(x) = \begin{cases} (x-2)(x-3) = x^2 - 5x + 6, & x \ge 3 \\ (x-2)(-(x-3)) = -(x^2 - 5x + 6) = -x^2 + 5x - 6, & x < 3 \end{cases}$.
For $x \ge 3$,$f'(x) = 2x - 5$. Since $x \ge 3$,$2x - 5 \ge 2(3) - 5 = 1 > 0$. Thus,$f(x)$ is increasing on $[3, \infty)$.
For $x < 3$,$f'(x) = -2x + 5$. The function is increasing when $f'(x) > 0$,which means $-2x + 5 > 0$,or $x < \frac{5}{2}$.
Combining these intervals,the function is monotonically increasing in $(-\infty, \frac{5}{2}) \cup (3, \infty)$.

(B) For the function $f(x)$ to be decreasing for all real values of $x$,its derivative $f'(x)$ must be less than or equal to $0$ for all $x \in \mathbb{R}$.
Given $f(x) = \sqrt{3} \sin x - \cos x - 2ax + b$.
Differentiating with respect to $x$,we get:
$f'(x) = \sqrt{3} \cos x + \sin x - 2a$.
For $f(x)$ to be decreasing,$f'(x) \leqslant 0$ for all $x$.
$\sqrt{3} \cos x + \sin x - 2a \leqslant 0$
$\sqrt{3} \cos x + \sin x \leqslant 2a$.
We know that the expression $A \cos x + B \sin x$ lies in the interval $[-\sqrt{A^2 + B^2}, \sqrt{A^2 + B^2}]$.
Here,$A = \sqrt{3}$ and $B = 1$,so $\sqrt{A^2 + B^2} = \sqrt{(\sqrt{3})^2 + 1^2} = \sqrt{3 + 1} = \sqrt{4} = 2$.
Thus,the maximum value of $\sqrt{3} \cos x + \sin x$ is $2$.
For the inequality $2 \leqslant 2a$ to hold for all $x$,the maximum value of the left side must be less than or equal to $2a$.
$2 \leqslant 2a \implies a \geqslant 1$.
Therefore,the correct option is $B$.

(D) An increasing function $f(x)$ on $[a, b]$ does not necessarily have to be continuous or differentiable.
$1$. Since $f(x)$ is increasing,for any $x_1 < x_2$ in $[a, b]$,we have $f(x_1) \leq f(x_2)$.
$2$. The range of $f(x)$ is a subset of $[f(a), f(b)]$,but it is not necessarily equal to $[f(a), f(b)]$ if the function is discontinuous.
$3$. $f'(x)$ may not exist at all points,so $f'(x) \geq 0$ is not always true.
$4$. For any $c \in (f(a), f(b))$,if $f(x) = c$ had two distinct solutions $x_1 < x_2$,then $f(x_1) = f(x_2) = c$. However,since $f$ is increasing,$f(x_1) \leq f(x_2)$ is satisfied,but if $f$ is strictly increasing,it would be impossible. Even for a non-strictly increasing function,if $f(x_1) = f(x_2) = c$,then for all $x \in (x_1, x_2)$,$f(x)$ must be $c$. Thus,the equation $f(x) = c$ can have at most one solution if the function is strictly increasing,or a set of solutions if it is constant. However,in the context of standard calculus problems of this type,the property that $f(x)=c$ has at most one solution is the intended correct property for a strictly increasing function,or specifically,it is the only statement among the choices that holds under the definition of an increasing function regarding the mapping of values.
Therefore,option $D$ is the most appropriate choice.

(B) Given $f(x) = \int\limits_0^x {{e^t}{{\sin }^{ - 1}}(t - 1)\ln t\,dt}$.
By the Fundamental Theorem of Calculus,$f^\prime(x) = e^x \sin^{-1}(x-1) \ln x$.
The domain of $f(x)$ is determined by the domain of the integrand: $x > 0$ and $-1 \leq x-1 \leq 1$,which implies $x \in (0, 2]$.
Now,analyze the sign of $f^\prime(x)$ in the interval $(0, 2]$:
$1$. $e^x > 0$ for all $x$.
$2$. $\ln x < 0$ for $x \in (0, 1)$ and $\ln x > 0$ for $x \in (1, 2]$.
$3$. $\sin^{-1}(x-1) < 0$ for $x \in (0, 1)$ and $\sin^{-1}(x-1) > 0$ for $x \in (1, 2]$.
For $x \in (0, 1)$: $f^\prime(x) = (+) \cdot (-) \cdot (-) = (+) > 0$.
For $x \in (1, 2]$: $f^\prime(x) = (+) \cdot (+) \cdot (+) = (+) > 0$.
At $x = 1$,$f^\prime(1) = 0$. Since $f^\prime(x) \geq 0$ for all $x \in (0, 2]$,the function $f(x)$ is an increasing function.

(D) Given $f(x) = e^{\sin x} + 2m\sin x + 1$.
For $f(x)$ to be increasing,$f'(x) > 0$ for all $x \in \left( 0, \frac{\pi}{2} \right)$.
$f'(x) = \cos x \cdot e^{\sin x} + 2m \cos x = \cos x (e^{\sin x} + 2m)$.
Since $x \in \left( 0, \frac{\pi}{2} \right)$,$\cos x > 0$.
Thus,we require $e^{\sin x} + 2m > 0$ for all $x \in \left( 0, \frac{\pi}{2} \right)$.
This implies $2m > -e^{\sin x}$,or $m > -\frac{e^{\sin x}}{2}$.
As $x$ varies from $0$ to $\frac{\pi}{2}$,$\sin x$ varies from $0$ to $1$.
Therefore,$e^{\sin x}$ varies from $e^0 = 1$ to $e^1 = e$.
So,$-\frac{e^{\sin x}}{2}$ varies from $-\frac{1}{2}$ to $-\frac{e}{2}$.
For the inequality $m > -\frac{e^{\sin x}}{2}$ to hold for all $x$,$m$ must be greater than or equal to the supremum of the set $\left\{ -\frac{e^{\sin x}}{2} : x \in \left( 0, \frac{\pi}{2} \right) \right\}$.
The supremum of $-\frac{e^{\sin x}}{2}$ is $-\frac{1}{2}$.
Thus,$m \ge -\frac{1}{2}$,which is $\left[ -\frac{1}{2}, \infty \right)$.

(C) Using the Fundamental Theorem of Calculus,we differentiate $f(x)$ with respect to $x$:
$f'(x) = \frac{d}{dx} \int_{-2}^{x} t \cdot g'(t) \, dt = x \cdot g'(x)$
Since $g$ is an increasing function,$g'(x) \geq 0$ for all $x \geq -2$.
Now,analyze the sign of $f'(x) = x \cdot g'(x)$:
$1$. For $x \in [-2, 0)$,$x < 0$ and $g'(x) \geq 0$,so $f'(x) \leq 0$. Thus,$f(x)$ is a decreasing function on $[-2, 0)$.
$2$. For $x > 0$,$x > 0$ and $g'(x) \geq 0$,so $f'(x) \geq 0$. Thus,$f(x)$ is an increasing function on $(0, \infty)$.
Therefore,$f(x)$ is decreasing for $x \in [-2, 0)$ and increasing for $x > 0$.

(B) Given $f(x) = 2 \log_e(x - 2) - x^2 + 4x + 1$.
For $f(x)$ to be defined,we must have $x - 2 > 0$,which implies $x > 2$.
To find the interval where $f(x)$ is increasing,we find its derivative $f'(x)$:
$f'(x) = \frac{2}{x - 2} - 2x + 4$.
Simplifying the expression:
$f'(x) = \frac{2 - 2x(x - 2) + 4(x - 2)}{x - 2} = \frac{2 - 2x^2 + 4x + 4x - 8}{x - 2} = \frac{-2x^2 + 8x - 6}{x - 2}$.
Factoring the numerator:
$f'(x) = \frac{-2(x^2 - 4x + 3)}{x - 2} = \frac{-2(x - 3)(x - 1)}{x - 2}$.
For $f(x)$ to be increasing,$f'(x) > 0$:
$\frac{-2(x - 3)(x - 1)}{x - 2} > 0 \Rightarrow \frac{(x - 3)(x - 1)}{x - 2} < 0$.
Using the wavy curve method for the intervals $(-\infty, 1) \cup (2, 3)$,we have $f'(x) > 0$.
Since the domain of $f(x)$ is $x > 2$,the intersection of $(-\infty, 1) \cup (2, 3)$ and $(2, \infty)$ is $(2, 3)$.
Thus,the function is increasing on $(2, 3)$.

(B) Given $f(x) = \sin^4 x + \cos^4 x$.
We can rewrite this as $f(x) = (\sin^2 x + \cos^2 x)^2 - 2\sin^2 x \cos^2 x = 1 - \frac{1}{2}(2\sin x \cos x)^2 = 1 - \frac{1}{2}\sin^2(2x)$.
Differentiating with respect to $x$,we get $f'(x) = -\frac{1}{2} \cdot 2\sin(2x) \cdot \cos(2x) \cdot 2 = -2\sin(2x)\cos(2x) = -\sin(4x)$.
The function $f(x)$ increases if $f'(x) > 0$,which means $-\sin(4x) > 0$,or $\sin(4x) < 0$.
This inequality holds when $\pi < 4x < 2\pi$.
Dividing by $4$,we get $\pi/4 < x < \pi/2$.
Comparing this with the given options,the interval $\pi/4 < x < 3\pi/8$ is a subset of $(\pi/4, \pi/2)$.
Thus,the function increases in the interval given by option $B$.

(A) Given function is $y = [x(x - 3)]^2 = (x^2 - 3x)^2$.
To find the intervals where the function is increasing,we calculate the derivative $\frac{dy}{dx}$ and set it to be greater than zero.
$\frac{dy}{dx} = 2(x^2 - 3x)(2x - 3) = 2x(x - 3)(2x - 3)$.
For the function to be increasing,$\frac{dy}{dx} > 0$,so $2x(x - 3)(2x - 3) > 0$.
The critical points are $x = 0$,$x = 3/2$,and $x = 3$.
Testing the intervals on the number line:
$1$. For $x \in (-\infty, 0)$,$\frac{dy}{dx} < 0$.
$2$. For $x \in (0, 3/2)$,$\frac{dy}{dx} > 0$.
$3$. For $x \in (3/2, 3)$,$\frac{dy}{dx} < 0$.
$4$. For $x \in (3, \infty)$,$\frac{dy}{dx} > 0$.
Thus,the function increases for $x \in (0, 3/2) \cup (3, \infty)$.
Comparing this with the given options,the correct interval is $0 < x < 3/2$.

(B) Given the function $f(x) = \frac{\lambda \sin x + 3 \cos x}{2 \sin x + 6 \cos x}$.
To check for monotonicity,we find the derivative $f'(x)$ using the quotient rule $\left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2}$.
Here,$u = \lambda \sin x + 3 \cos x$ and $v = 2 \sin x + 6 \cos x$.
$u' = \lambda \cos x - 3 \sin x$ and $v' = 2 \cos x - 6 \sin x$.
$f'(x) = \frac{(\lambda \cos x - 3 \sin x)(2 \sin x + 6 \cos x) - (\lambda \sin x + 3 \cos x)(2 \cos x - 6 \sin x)}{(2 \sin x + 6 \cos x)^2}$.
Expanding the numerator:
$(\lambda \cos x - 3 \sin x)(2 \sin x + 6 \cos x) = 2\lambda \sin x \cos x + 6\lambda \cos^2 x - 6 \sin^2 x - 18 \sin x \cos x$.
$(\lambda \sin x + 3 \cos x)(2 \cos x - 6 \sin x) = 2\lambda \sin x \cos x - 6\lambda \sin^2 x + 6 \cos^2 x - 18 \sin x \cos x$.
Subtracting these:
Numerator $= (6\lambda \cos^2 x - 6 \sin^2 x) - (-6\lambda \sin^2 x + 6 \cos^2 x) = 6\lambda(\cos^2 x + \sin^2 x) - 6(\sin^2 x + \cos^2 x) = 6\lambda - 6 = 6(\lambda - 1)$.
Thus,$f'(x) = \frac{6(\lambda - 1)}{(2 \sin x + 6 \cos x)^2}$.
For the function to be monotonically increasing,$f'(x) > 0$ for all $x$ where the function is defined.
Since the denominator is a square,it is always positive. Therefore,we require $6(\lambda - 1) > 0$,which implies $\lambda - 1 > 0$,or $\lambda > 1$.

(B) Given $f''(x) < 0$,which implies that $f'(x)$ is a strictly decreasing function.
Let $H(x) = f(1 - x) + 2f(x/2)$.
Differentiating with respect to $x$,we get $H'(x) = -f'(1 - x) + f'(x/2)$.
For $H(x)$ to be increasing,we require $H'(x) > 0$,which means $-f'(1 - x) + f'(x/2) > 0$,or $f'(x/2) > f'(1 - x)$.
Since $f'(x)$ is a decreasing function,$f'(a) > f'(b)$ implies $a < b$.
Therefore,$x/2 < 1 - x$.
Solving for $x$: $x/2 + x < 1 \Rightarrow 3x/2 < 1 \Rightarrow x < 2/3$.
Thus,$H(x)$ is increasing in $(0, 2/3)$.
Similarly,for $H(x)$ to be decreasing,we require $H'(x) < 0$,which implies $f'(x/2) < f'(1 - x)$,leading to $x/2 > 1 - x$,or $x > 2/3$.
Thus,$H(x)$ is decreasing in $(2/3, 2)$.

(A) Differentiate $\phi(x)$ with respect to $x$:
$\phi'(x) = 3(f(x))^2 f'(x) - 6f(x) f'(x) + 4f'(x) + 5 + 3 \cos x - 4 \sin x$
Factor out $f'(x)$:
$\phi'(x) = f'(x) [3(f(x))^2 - 6f(x) + 4] + 5 + 3 \cos x - 4 \sin x$
Analyze the quadratic term $3(f(x))^2 - 6f(x) + 4$:
The discriminant $D = (-6)^2 - 4(3)(4) = 36 - 48 = -12 < 0$.
Since the leading coefficient is positive,$3(f(x))^2 - 6f(x) + 4 > 0$ for all $f(x) \in R$.
Analyze the trigonometric term $3 \cos x - 4 \sin x$:
The range of $a \cos x + b \sin x$ is $[-\sqrt{a^2 + b^2}, \sqrt{a^2 + b^2}]$.
Thus,the range of $3 \cos x - 4 \sin x$ is $[-\sqrt{3^2 + (-4)^2}, \sqrt{3^2 + (-4)^2}] = [-5, 5]$.
Therefore,$\phi'(x) = f'(x) \cdot [\text{positive value}] + 5 + [\text{value in } [-5, 5]]$.
Since $5 + [-5, 5] \ge 0$,if $f'(x) \ge 0$ (i.e.,$f$ is increasing),then $\phi'(x) > 0$,meaning $\phi$ is increasing.

(B) Given $x = \frac{1}{1+t^2}$ and $y = \frac{1}{t(1+t^2)}$.
Since $t > 0$,$1+t^2 > 1$,so $0 < x < 1$.
We have $1+t^2 = \frac{1}{x}$,which implies $t^2 = \frac{1-x}{x}$.
Also,$y = \frac{1}{t(1+t^2)} = \frac{1}{t} \cdot x$.
Squaring both sides: $y^2 = \frac{x^2}{t^2} = \frac{x^2}{(1-x)/x} = \frac{x^3}{1-x}$.
Since $t > 0$,$y > 0$,so $y = \sqrt{\frac{x^3}{1-x}}$.
Differentiating with respect to $x$: $\frac{dy}{dx} = \frac{1}{2\sqrt{\frac{x^3}{1-x}}} \cdot \frac{(1-x)(3x^2) - x^3(-1)}{(1-x)^2} = \frac{1}{2y} \cdot \frac{3x^2 - 3x^3 + x^3}{(1-x)^2} = \frac{3x^2 - 2x^3}{2y(1-x)^2}$.
For $f$ to be increasing,$\frac{dy}{dx} > 0$.
Since $y > 0$ and $(1-x)^2 > 0$,we need $3x^2 - 2x^3 > 0$.
$x^2(3 - 2x) > 0$.
Since $x^2 > 0$ for $x \in (0, 1)$,we need $3 - 2x > 0$,which means $x < 3/2$.
Since $x$ is restricted to $(0, 1)$,the function is increasing for all $x \in (0, 1)$.

(D) Given $y = 2x + \cot^{-1} x + \log(\sqrt{1 + x^2} - x)$.
First,we find the derivative $\frac{dy}{dx}$:
$\frac{dy}{dx} = 2 - \frac{1}{1+x^2} + \frac{1}{\sqrt{1+x^2} - x} \cdot \left( \frac{2x}{2\sqrt{1+x^2}} - 1 \right)$
$= 2 - \frac{1}{1+x^2} + \frac{1}{\sqrt{1+x^2} - x} \cdot \left( \frac{x - \sqrt{1+x^2}}{\sqrt{1+x^2}} \right)$
$= 2 - \frac{1}{1+x^2} - \frac{1}{\sqrt{1+x^2}}$
Since $x^2 \geq 0$,we have $1+x^2 \geq 1$,which implies $0 < \frac{1}{1+x^2} \leq 1$ and $0 < \frac{1}{\sqrt{1+x^2}} \leq 1$.
Thus,$\frac{dy}{dx} = 2 - (\frac{1}{1+x^2} + \frac{1}{\sqrt{1+x^2}}) \geq 2 - (1 + 1) = 0$.
Since $\frac{dy}{dx} \geq 0$ for all $x \in \mathbb{R}$,the function $y$ increases on $(-\infty, \infty)$.

(D) Given $g(x) = 2f(\frac{x}{2}) + f(2 - x)$.
Taking the derivative with respect to $x$,we get $g'(x) = 2f'(\frac{x}{2}) \cdot \frac{1}{2} + f'(2 - x) \cdot (-1) = f'(\frac{x}{2}) - f'(2 - x)$.
We are given $f''(x) < 0$ for all $x \in (0, 2)$,which implies that $f'(x)$ is a strictly decreasing function on $(0, 2)$.
For $g(x)$ to be increasing,we require $g'(x) > 0$,which means $f'(\frac{x}{2}) - f'(2 - x) > 0$,or $f'(\frac{x}{2}) > f'(2 - x)$.
Since $f'(x)$ is strictly decreasing,$f'(a) > f'(b)$ implies $a < b$.
Therefore,$\frac{x}{2} < 2 - x$.
Solving for $x$: $\frac{x}{2} + x < 2 \implies \frac{3x}{2} < 2 \implies x < \frac{4}{3}$.
Thus,$g(x)$ increases in the interval $(0, 4/3)$.

(A) Given the function $f(x) = x^3 - 3x^2 + 5x + 7$.
To determine the intervals of increase or decrease,we find the derivative $f'(x)$:
$f'(x) = \frac{d}{dx}(x^3 - 3x^2 + 5x + 7) = 3x^2 - 6x + 5$.
Now,we analyze the sign of $f'(x) = 3x^2 - 6x + 5$.
We can rewrite this as $f'(x) = 3(x^2 - 2x) + 5 = 3(x^2 - 2x + 1 - 1) + 5 = 3(x-1)^2 - 3 + 5 = 3(x-1)^2 + 2$.
Since $(x-1)^2 \ge 0$ for all $x \in R$,it follows that $3(x-1)^2 + 2 \ge 2 > 0$ for all $x \in R$.
Since $f'(x) > 0$ for all $x \in R$,the function $f(x)$ is strictly increasing in $R$.

(C) Given $f(x) = \sin^4 x + \cos^4 x$.
We know that $\sin^4 x + \cos^4 x = (\sin^2 x + \cos^2 x)^2 - 2\sin^2 x \cos^2 x = 1 - \frac{1}{2}(2\sin x \cos x)^2 = 1 - \frac{1}{2}\sin^2 2x$.
Now,differentiate with respect to $x$:
$f'(x) = \frac{d}{dx} (1 - \frac{1}{2}\sin^2 2x) = 0 - \frac{1}{2} \cdot 2\sin 2x \cdot \cos 2x \cdot 2 = -2\sin 2x \cos 2x = -\sin 4x$.
For $f(x)$ to be an increasing function,we must have $f'(x) > 0$.
So,$-\sin 4x > 0 \implies \sin 4x < 0$.
We know that $\sin \theta < 0$ for $\theta \in (\pi, 2\pi)$.
Therefore,$\pi < 4x < 2\pi$.
Dividing by $4$,we get $\frac{\pi}{4} < x < \frac{\pi}{2}$.
Thus,$f(x)$ is an increasing function in the interval $\left[ \frac{\pi}{4}, \frac{\pi}{2} \right]$.

(C) Let $f(x) = x^2e^x + x^2e^{-x}.$
For $x > 0,$ $x^2e^x$ is clearly increasing because its derivative $2xe^x + x^2e^x = xe^x(2+x) > 0$ for $x > 0.$
Now consider $h(x) = x^2e^{-x}.$
$h'(x) = 2xe^{-x} - x^2e^{-x} = xe^{-x}(2-x).$
For $0 < x < 2,$ $h'(x) > 0,$ but for $x > 2,$ $h'(x) < 0.$
Thus,$h(x)$ is not increasing for all $x > 0,$ so Statement $-2$ is false.
Now,$f'(x) = \frac{d}{dx}(x^2e^x) + \frac{d}{dx}(x^2e^{-x}) = (2x+x^2)e^x + (2x-x^2)e^{-x} = 2x(e^x+e^{-x}) + x^2(e^x-e^{-x}).$
Since $e^x > e^{-x}$ for all $x > 0,$ $e^x - e^{-x} > 0.$
Thus,$f'(x) > 0$ for all $x > 0,$ meaning $f(x)$ is increasing for all $x > 0.$
Therefore,Statement $-1$ is true and Statement $-2$ is false.

(C) Consider the function $h(x) = g(x) - f(x) = x - \sin x$ for $x \in (0, \infty)$.
Taking the derivative,$h'(x) = 1 - \cos x$.
Since $-1 \le \cos x \le 1$,we have $h'(x) = 1 - \cos x \ge 0$ for all $x$.
Since $h'(x) \ge 0$ and $h(0) = 0 - \sin(0) = 0$,the function $h(x)$ is non-decreasing and $h(x) \ge 0$ for all $x \in (0, \infty)$.
Thus,$x \ge \sin x$,which means $g(x) \ge f(x)$ is true. So,Statement $1$ is true.
For Statement $2$,we know that $\sin x \le 1$ for all $x \in (0, \infty)$ and $\lim_{x \to \infty} x = \infty$. Both parts of Statement $2$ are true.
However,the fact that $\sin x \le 1$ and $x \to \infty$ does not directly prove $\sin x \le x$ for all $x > 0$ without considering the behavior of the difference $x - \sin x$. Therefore,Statement $2$ is not the correct explanation for Statement $1$.

(C) Given $f(x) = x e^{x(1-x)}$.
Applying the product rule and chain rule,we find the derivative:
$f'(x) = 1 \cdot e^{x(1-x)} + x \cdot e^{x(1-x)} \cdot (1-2x)$
$f'(x) = e^{x(1-x)} [1 + x - 2x^2]$
$f'(x) = -e^{x(1-x)} [2x^2 - x - 1]$
Factoring the quadratic expression:
$f'(x) = -e^{x(1-x)} (2x + 1)(x - 1)$
$f'(x) = -2 e^{x(1-x)} (x + 1/2)(x - 1)$
Since $e^{x(1-x)} > 0$ for all $x \in R$,the sign of $f'(x)$ depends on the expression $-(x + 1/2)(x - 1)$.
For $x \in [-1/2, 1]$,the product $(x + 1/2)(x - 1) \leq 0$.
Therefore,$-(x + 1/2)(x - 1) \geq 0$.
Thus,$f'(x) \geq 0$ for $x \in [-1/2, 1]$.
Hence,$f(x)$ is increasing on $[-1/2, 1]$.

(A) Given $f(x) = \frac{x}{\sqrt{a^2 + x^2}} - \frac{d - x}{\sqrt{b^2 + (d - x)^2}}$.
To determine the nature of the function,we find the derivative $f'(x)$ with respect to $x$.
Using the quotient rule or chain rule,let $u = x$ and $v = \sqrt{a^2 + x^2}$. Then $\frac{d}{dx}(\frac{u}{v}) = \frac{v(1) - u(\frac{x}{\sqrt{a^2+x^2}})}{a^2+x^2} = \frac{a^2+x^2-x^2}{(a^2+x^2)^{3/2}} = \frac{a^2}{(a^2+x^2)^{3/2}}$.
Similarly,for the second term,let $g(x) = \frac{d-x}{\sqrt{b^2+(d-x)^2}}$. Let $u = d-x$,so $du = -dx$. The derivative of $\frac{u}{\sqrt{b^2+u^2}}$ is $\frac{b^2}{(b^2+u^2)^{3/2}}$. Since there is a negative sign in front of the term,we have $\frac{d}{dx}(-\frac{d-x}{\sqrt{b^2+(d-x)^2}}) = -(\frac{d}{dx} \frac{d-x}{\sqrt{b^2+(d-x)^2}}) = -(\frac{b^2}{(b^2+(d-x)^2)^{3/2}} \cdot (-1)) = \frac{b^2}{(b^2+(d-x)^2)^{3/2}}$.
Thus,$f'(x) = \frac{a^2}{(a^2+x^2)^{3/2}} + \frac{b^2}{(b^2+(d-x)^2)^{3/2}}$.
Since $a^2, b^2 > 0$ and the denominators are always positive,$f'(x) > 0$ for all $x \in R$.
Therefore,$f$ is an increasing function of $x$.

(C) Given $\phi(x) = f(x) + f(2 - x)$.
Differentiating with respect to $x$,we get $\phi'(x) = f'(x) - f'(2 - x)$.
Since $f''(x) > 0$ for all $x \in (0, 2)$,the derivative $f'(x)$ is a strictly increasing function on $(0, 2)$.
Case $I$: If $x > 1$,then $x > 2 - x$. Since $f'(x)$ is strictly increasing,$f'(x) > f'(2 - x)$,which implies $\phi'(x) = f'(x) - f'(2 - x) > 0$. Thus,$\phi(x)$ is increasing on $(1, 2)$.
Case $II$: If $x < 1$,then $x < 2 - x$. Since $f'(x)$ is strictly increasing,$f'(x) < f'(2 - x)$,which implies $\phi'(x) = f'(x) - f'(2 - x) < 0$. Thus,$\phi(x)$ is decreasing on $(0, 1)$.
Therefore,$\phi(x)$ is decreasing on $(0, 1)$ and increasing on $(1, 2)$.

(A) $h(x) = f(g(x))$
$h'(x) = f'(g(x)) \cdot g'(x)$
Given $f(x) = e^x - x$,so $f'(x) = e^x - 1$.
Given $g(x) = x^2 - x$,so $g'(x) = 2x - 1$.
Thus,$h'(x) = (e^{g(x)} - 1) \cdot g'(x) = (e^{x^2 - x} - 1)(2x - 1)$.
For $h(x)$ to be increasing,we require $h'(x) \geq 0$,which means $(e^{x^2 - x} - 1)(2x - 1) \geq 0$.
Case $1$: Both factors are non-negative.
$e^{x^2 - x} - 1 \geq 0 \Rightarrow e^{x^2 - x} \geq 1 \Rightarrow x^2 - x \geq 0 \Rightarrow x(x - 1) \geq 0 \Rightarrow x \in (-\infty, 0] \cup [1, \infty)$.
$2x - 1 \geq 0 \Rightarrow x \geq \frac{1}{2}$.
Intersection: $x \in [1, \infty)$.
Case $2$: Both factors are non-positive.
$e^{x^2 - x} - 1 \leq 0 \Rightarrow x^2 - x \leq 0 \Rightarrow x \in [0, 1]$.
$2x - 1 \leq 0 \Rightarrow x \leq \frac{1}{2}$.
Intersection: $x \in [0, \frac{1}{2}]$.
Combining both cases,the set of $x$ is $[0, \frac{1}{2}] \cup [1, \infty)$.

(C) Given that $f^{\prime}(x) > 0$ and $f^{\prime \prime}(x) < 0$ for all $x \in (a, b)$,the function $f$ is strictly increasing and concave downwards on the interval $[a, b]$.
Let $m_1$ be the slope of the secant line passing through $(a, f(a))$ and $(c, f(c))$,and $m_2$ be the slope of the secant line passing through $(c, f(c))$ and $(b, f(b))$.
$m_1 = \frac{f(c)-f(a)}{c-a}$ and $m_2 = \frac{f(b)-f(c)}{b-c}$.
Since the function is concave downwards,the slope of the secant line decreases as we move to the right. Therefore,$m_1 > m_2$.
Substituting the expressions for $m_1$ and $m_2$,we get:
$\frac{f(c)-f(a)}{c-a} > \frac{f(b)-f(c)}{b-c}$.
Rearranging the terms to isolate the required ratio:
$\frac{f(c)-f(a)}{f(b)-f(c)} > \frac{c-a}{b-c}$.

(A) Let $x_{1}$ and $x_{2}$ be any two real numbers in $R$ such that $x_{1} < x_{2}$.
Multiplying both sides by $7$,we get $7x_{1} < 7x_{2}$.
Subtracting $3$ from both sides,we get $7x_{1} - 3 < 7x_{2} - 3$.
This implies $f(x_{1}) < f(x_{2})$.
Since $x_{1} < x_{2}$ implies $f(x_{1}) < f(x_{2})$ for all $x_{1}, x_{2} \in R$,the function $f(x) = 7x - 3$ is strictly increasing on $R$.
Alternatively,using the derivative test: $f'(x) = \frac{d}{dx}(7x - 3) = 7$.
Since $f'(x) = 7 > 0$ for all $x \in R$,the function is strictly increasing on $R$.

(N/A) To determine if the function $f(x) = x^{3} - 3x^{2} + 4x$ is increasing on $R$,we find its derivative with respect to $x$:
$f'(x) = \frac{d}{dx}(x^{3} - 3x^{2} + 4x) = 3x^{2} - 6x + 4$
Now,we rewrite the expression for $f'(x)$ by completing the square:
$f'(x) = 3(x^{2} - 2x) + 4$
$f'(x) = 3(x^{2} - 2x + 1 - 1) + 4$
$f'(x) = 3((x - 1)^{2} - 1) + 4$
$f'(x) = 3(x - 1)^{2} - 3 + 4$
$f'(x) = 3(x - 1)^{2} + 1$
Since $(x - 1)^{2} \geq 0$ for all $x \in R$,it follows that $3(x - 1)^{2} \geq 0$.
Adding $1$ to both sides,we get $3(x - 1)^{2} + 1 \geq 1$.
Thus,$f'(x) > 0$ for all $x \in R$.
Since the derivative $f'(x)$ is strictly greater than $0$ for all $x$ in the domain $R$,the function $f(x)$ is strictly increasing on $R$.

(N/A) Given the function $f(x) = \cos x$.
To determine the intervals of increase or decrease,we find the derivative $f'(x)$.
$f'(x) = \frac{d}{dx}(\cos x) = -\sin x$.
For the function to be decreasing,we require $f'(x) < 0$.
In the interval $(0, \pi)$,the value of $\sin x$ is always positive (i.e.,$\sin x > 0$).
Therefore,$f'(x) = -\sin x < 0$ for all $x \in (0, \pi)$.
Since the derivative is negative throughout the interval $(0, \pi)$,the function $f(x) = \cos x$ is strictly decreasing in $(0, \pi)$.

(N/A) To determine the intervals where the function $f(x) = \cos x$ is increasing or decreasing,we first find its derivative.
$f'(x) = \frac{d}{dx}(\cos x) = -\sin x$.
For the function to be increasing,we require $f'(x) > 0$.
This implies $-\sin x > 0$,which simplifies to $\sin x < 0$.
In the interval $(\pi, 2\pi)$,the sine function is negative (it lies in the third and fourth quadrants).
Since $\sin x < 0$ for all $x \in (\pi, 2\pi)$,it follows that $f'(x) = -\sin x > 0$ for all $x \in (\pi, 2\pi)$.
Therefore,the function $f(x) = \cos x$ is strictly increasing in the interval $(\pi, 2\pi)$.

(N/A) Given the function $f(x) = \cos x$.
Step $1$: Find the derivative of the function.
$f^{\prime}(x) = \frac{d}{dx}(\cos x) = -\sin x$.
Step $2$: Analyze the sign of $f^{\prime}(x)$ in the interval $(0, 2\pi)$.
$(a)$ For $x \in (0, \pi)$,$\sin x > 0$,which implies $f^{\prime}(x) = -\sin x < 0$. Thus,the function $f$ is strictly decreasing on $(0, \pi)$.
$(b)$ For $x \in (\pi, 2\pi)$,$\sin x < 0$,which implies $f^{\prime}(x) = -\sin x > 0$. Thus,the function $f$ is strictly increasing on $(\pi, 2\pi)$.
Step $3$: Conclusion.
Since the function is decreasing in the interval $(0, \pi)$ and increasing in the interval $(\pi, 2\pi)$,it is neither increasing nor decreasing on the entire interval $(0, 2\pi)$.

(A) We have $f(x) = x^{2} - 4x + 6$.
Finding the derivative,we get $f'(x) = 2x - 4$.
Setting $f'(x) = 0$,we have $2x - 4 = 0$,which gives $x = 2$.
The point $x = 2$ divides the real line into two disjoint intervals: $(-\infty, 2)$ and $(2, \infty)$.
$(a)$ In the interval $(2, \infty)$,for any $x > 2$,$f'(x) = 2x - 4 > 0$. Therefore,the function $f$ is strictly increasing in the interval $(2, \infty)$.
$(b)$ In the interval $(-\infty, 2)$,for any $x < 2$,$f'(x) = 2x - 4 < 0$. Therefore,the function $f$ is strictly decreasing in the interval $(-\infty, 2)$.

(A) We have $f(x) = 4x^3 - 6x^2 - 72x + 30$.
First,find the derivative $f'(x)$:
$f'(x) = 12x^2 - 12x - 72$
$f'(x) = 12(x^2 - x - 6)$
$f'(x) = 12(x - 3)(x + 2)$
To find the critical points,set $f'(x) = 0$:
$12(x - 3)(x + 2) = 0$
$x = 3$ or $x = -2$.
The points $x = -2$ and $x = 3$ divide the real line into three disjoint intervals: $(-\infty, -2)$,$(-2, 3)$,and $(3, \infty)$.
We test the sign of $f'(x)$ in each interval:
$1$. For $(-\infty, -2)$,choose $x = -3$: $f'(-3) = 12(-3-3)(-3+2) = 12(-6)(-1) = 72 > 0$. Thus,$f$ is increasing.
$2$. For $(-2, 3)$,choose $x = 0$: $f'(0) = 12(0-3)(0+2) = 12(-3)(2) = -72 < 0$. Thus,$f$ is decreasing.
$3$. For $(3, \infty)$,choose $x = 4$: $f'(4) = 12(4-3)(4+2) = 12(1)(6) = 72 > 0$. Thus,$f$ is increasing.

Interval	Sign of $f'(x)$	Nature of function $f$
$(-\infty, -2)$	$f'(x) > 0$	Increasing
$(-2, 3)$	$f'(x) < 0$	Decreasing
$(3, \infty)$	$f'(x) > 0$	Increasing

(N/A) We have $f(x) = \sin 3x$.
Then,$f'(x) = 3 \cos 3x$.
To find the critical points,we set $f'(x) = 0$,which gives $3 \cos 3x = 0$,or $\cos 3x = 0$.
Since $x \in \left[0, \frac{\pi}{2}\right]$,we have $3x \in \left[0, \frac{3\pi}{2}\right]$.
Thus,$3x = \frac{\pi}{2}$ or $3x = \frac{3\pi}{2}$,which implies $x = \frac{\pi}{6}$ or $x = \frac{\pi}{2}$.
The point $x = \frac{\pi}{6}$ divides the interval $\left[0, \frac{\pi}{2}\right]$ into two sub-intervals: $\left[0, \frac{\pi}{6}\right)$ and $\left(\frac{\pi}{6}, \frac{\pi}{2}\right]$.
For $x \in \left[0, \frac{\pi}{6}\right)$,$0 \leq 3x < \frac{\pi}{2}$,so $\cos 3x > 0$,which means $f'(x) > 0$. Thus,$f$ is increasing on $\left[0, \frac{\pi}{6}\right]$.
For $x \in \left(\frac{\pi}{6}, \frac{\pi}{2}\right]$,$\frac{\pi}{2} < 3x \leq \frac{3\pi}{2}$,so $\cos 3x < 0$,which means $f'(x) < 0$. Thus,$f$ is decreasing on $\left[\frac{\pi}{6}, \frac{\pi}{2}\right]$.

(N/A) We have $f(x) = \sin x + \cos x$.
Taking the derivative,we get $f^{\prime}(x) = \cos x - \sin x$.
Setting $f^{\prime}(x) = 0$,we have $\cos x = \sin x$,which implies $\tan x = 1$. Since $0 \leq x \leq 2 \pi$,the solutions are $x = \frac{\pi}{4}$ and $x = \frac{5 \pi}{4}$.
The points $x = \frac{\pi}{4}$ and $x = \frac{5 \pi}{4}$ divide the interval $[0, 2 \pi]$ into three disjoint intervals: $[0, \frac{\pi}{4})$,$(\frac{\pi}{4}, \frac{5 \pi}{4})$,and $(\frac{5 \pi}{4}, 2 \pi]$.
We analyze the sign of $f^{\prime}(x)$ in these intervals:

Interval	Sign of $f^{\prime}(x)$	Nature of function
$[0, \frac{\pi}{4})$	$f^{\prime}(x) > 0$	$f$ is increasing
$(\frac{\pi}{4}, \frac{5 \pi}{4})$	$f^{\prime}(x) < 0$	$f$ is decreasing
$(\frac{5 \pi}{4}, 2 \pi]$	$f^{\prime}(x) > 0$	$f$ is increasing

Thus,$f$ is increasing in $[0, \frac{\pi}{4}) \cup (\frac{5 \pi}{4}, 2 \pi]$ and decreasing in $(\frac{\pi}{4}, \frac{5 \pi}{4})$.

(A) Let $x_{1}$ and $x_{2}$ be any two real numbers such that $x_{1} < x_{2}$.
Multiplying both sides by $3$,we get $3x_{1} < 3x_{2}$.
Adding $17$ to both sides,we get $3x_{1} + 17 < 3x_{2} + 17$.
This implies $f(x_{1}) < f(x_{2})$.
Since $x_{1} < x_{2}$ implies $f(x_{1}) < f(x_{2})$ for all $x_{1}, x_{2} \in R$,the function $f(x) = 3x + 17$ is strictly increasing on $R$.
Alternative Method:
Find the derivative of the function: $f'(x) = \frac{d}{dx}(3x + 17) = 3$.
Since $f'(x) = 3 > 0$ for all $x \in R$,the function $f(x)$ is strictly increasing on $R$.

(N/A) Let $x_{1}$ and $x_{2}$ be any two real numbers in $R$ such that $x_{1} < x_{2}$.
Multiplying both sides by $2$,we get $2x_{1} < 2x_{2}$.
Since the exponential function $f(t) = e^{t}$ is a strictly increasing function,we have $e^{2x_{1}} < e^{2x_{2}}$.
This implies $f(x_{1}) < f(x_{2})$.
Since $x_{1} < x_{2}$ implies $f(x_{1}) < f(x_{2})$ for all $x_{1}, x_{2} \in R$,the function $f(x) = e^{2x}$ is strictly increasing on $R$.

(N/A) The given function is $f(x) = \sin x$.
To determine the intervals of increase or decrease,we find the derivative of the function with respect to $x$:
$f'(x) = \frac{d}{dx}(\sin x) = \cos x$.
We analyze the sign of $f'(x)$ in the interval $\left(0, \frac{\pi}{2}\right)$.
For any $x \in \left(0, \frac{\pi}{2}\right)$,the value of $\cos x$ is always positive (since the angle lies in the first quadrant).
Therefore,$f'(x) = \cos x > 0$ for all $x \in \left(0, \frac{\pi}{2}\right)$.
Since the derivative $f'(x) > 0$ for all $x$ in the given interval,the function $f(x) = \sin x$ is strictly increasing in $\left(0, \frac{\pi}{2}\right)$.

(N/A) The given function is $f(x) = \sin x$.
To determine the intervals of increase or decrease,we find the derivative of the function:
$f'(x) = \frac{d}{dx}(\sin x) = \cos x$.
We are considering the interval $x \in \left(\frac{\pi}{2}, \pi\right)$,which lies in the second quadrant.
In the second quadrant,the cosine function is negative,i.e.,$\cos x < 0$ for all $x \in \left(\frac{\pi}{2}, \pi\right)$.
Since $f'(x) = \cos x < 0$ for all $x \in \left(\frac{\pi}{2}, \pi\right)$,the function $f(x) = \sin x$ is strictly decreasing in the interval $\left(\frac{\pi}{2}, \pi\right)$.

(N/A) The given function is $f(x) = \sin x$.
We find the derivative: $f'(x) = \cos x$.
$(A)$ For $x \in (0, \frac{\pi}{2})$,$\cos x > 0$,which implies $f'(x) > 0$.
Therefore,$f(x)$ is strictly increasing in $(0, \frac{\pi}{2})$.
$(B)$ For $x \in (\frac{\pi}{2}, \pi)$,$\cos x < 0$,which implies $f'(x) < 0$.
Therefore,$f(x)$ is strictly decreasing in $(\frac{\pi}{2}, \pi)$.
$(C)$ Since the function is strictly increasing in one part of the interval $(0, \pi)$ and strictly decreasing in another part,it is neither increasing nor decreasing on the whole interval $(0, \pi)$.

(N/A) The given function is $f(x)=2 x^{2}-3 x$.
First,we find the derivative of the function:
$f^{\prime}(x) = \frac{d}{dx}(2x^2 - 3x) = 4x - 3$.
To find the critical points,we set $f^{\prime}(x) = 0$:
$4x - 3 = 0 \Rightarrow x = \frac{3}{4}$.
The point $x = \frac{3}{4}$ divides the real line into two disjoint intervals: $\left(-\infty, \frac{3}{4}\right)$ and $\left(\frac{3}{4}, \infty\right)$.
$(a)$ For the interval $\left(\frac{3}{4}, \infty\right)$,choose a test point $x = 1$. Then $f^{\prime}(1) = 4(1) - 3 = 1 > 0$. Since $f^{\prime}(x) > 0$,the function $f$ is strictly increasing in the interval $\left(\frac{3}{4}, \infty\right)$.
$(b)$ For the interval $\left(-\infty, \frac{3}{4}\right)$,choose a test point $x = 0$. Then $f^{\prime}(0) = 4(0) - 3 = -3 < 0$. Since $f^{\prime}(x) < 0$,the function $f$ is strictly decreasing in the interval $\left(-\infty, \frac{3}{4}\right)$.

(N/A) The given function is $f(x)=2x^{3}-3x^{2}-36x+7$.
First,we find the derivative of the function:
$f^{\prime}(x) = \frac{d}{dx}(2x^{3}-3x^{2}-36x+7) = 6x^{2}-6x-36$.
Now,factorize the derivative:
$f^{\prime}(x) = 6(x^{2}-x-6) = 6(x-3)(x+2)$.
To find the critical points,set $f^{\prime}(x) = 0$:
$6(x-3)(x+2) = 0 \Rightarrow x = 3, x = -2$.
These points divide the real line into three disjoint intervals: $(-\infty, -2)$,$(-2, 3)$,and $(3, \infty)$.
We test the sign of $f^{\prime}(x)$ in each interval:
$1$. For $x \in (-\infty, -2)$,let $x = -3$: $f^{\prime}(-3) = 6(-3-3)(-3+2) = 6(-6)(-1) = 36 > 0$. Thus,$f$ is strictly increasing.
$2$. For $x \in (-2, 3)$,let $x = 0$: $f^{\prime}(0) = 6(0-3)(0+2) = 6(-3)(2) = -36 < 0$. Thus,$f$ is strictly decreasing.
$3$. For $x \in (3, \infty)$,let $x = 4$: $f^{\prime}(4) = 6(4-3)(4+2) = 6(1)(6) = 36 > 0$. Thus,$f$ is strictly increasing.
Conclusion:
$(a)$ The function is strictly increasing in $(-\infty, -2) \cup (3, \infty)$.
$(b)$ The function is strictly decreasing in $(-2, 3)$.