Increasing and Decreasing function Questions in English

(B) Given the function $f(x) = x^3 + 6x^2 + px + 2$.
To find the interval where the function is decreasing,we calculate the derivative:
$f'(x) = 3x^2 + 12x + p$.
For $f(x)$ to be a decreasing function on the interval $(-3, -1)$,we must have $f'(x) \leq 0$ for all $x \in (-3, -1)$.
The quadratic expression $3x^2 + 12x + p$ must be less than or equal to $0$ between its roots $-3$ and $-1$.
Thus,the roots of the equation $3x^2 + 12x + p = 0$ are $x = -3$ and $x = -1$.
Using the product of roots formula for a quadratic equation $ax^2 + bx + c = 0$,the product of roots is $\frac{c}{a}$.
Therefore,$(-3) \times (-1) = \frac{p}{3}$.
$3 = \frac{p}{3} \implies p = 9$.

(A) Given $f(x) = 3 \sin x - 4 \sin^3 x$.
Using the trigonometric identity $\sin 3x = 3 \sin x - 4 \sin^3 x$,we have $f(x) = \sin 3x$.
For $f(x)$ to be an increasing function,its derivative $f'(x)$ must be greater than $0$.
$f'(x) = \frac{d}{dx}(\sin 3x) = 3 \cos 3x$.
For $f(x)$ to be increasing,$3 \cos 3x > 0$,which implies $\cos 3x > 0$.
This occurs when $-\frac{\pi}{2} < 3x < \frac{\pi}{2}$.
Dividing by $3$,we get $-\frac{\pi}{6} < x < \frac{\pi}{6}$.
The length of this interval is $\frac{\pi}{6} - (-\frac{\pi}{6}) = \frac{2\pi}{6} = \frac{\pi}{3}$.

(B) Given the function $f(x) = x + \frac{1}{x}$.
To find where the function is strictly increasing,we calculate the derivative $f'(x)$ and set $f'(x) > 0$.
$f'(x) = \frac{d}{dx}(x + x^{-1}) = 1 - x^{-2} = 1 - \frac{1}{x^2}$.
For strictly increasing,$f'(x) > 0$,so $1 - \frac{1}{x^2} > 0$.
$\frac{x^2 - 1}{x^2} > 0$.
Since $x^2 > 0$ for all $x \neq 0$,the inequality holds if $x^2 - 1 > 0$.
$x^2 > 1$,which implies $|x| > 1$.

(B) Given the function $y = 8x^3 - 60x^2 + 144x + 27$.
To find the interval where the function is decreasing,we calculate the derivative $\frac{dy}{dx}$.
$\frac{dy}{dx} = 24x^2 - 120x + 144$.
For the function to be decreasing,$\frac{dy}{dx} < 0$.
$24x^2 - 120x + 144 < 0$.
Dividing by $24$,we get $x^2 - 5x + 6 < 0$.
Factoring the quadratic expression,we get $(x - 3)(x - 2) < 0$.
This inequality holds when $x$ lies between the roots $2$ and $3$.
Therefore,the interval is $(2, 3)$.

(B) Given $f(x) = \frac{1}{x + 1} - \log(1 + x)$ for $x > 0$.
To determine the nature of the function,we find its derivative $f'(x)$:
$f'(x) = \frac{d}{dx} \left( \frac{1}{x + 1} \right) - \frac{d}{dx} (\log(1 + x))$
$f'(x) = -\frac{1}{(x + 1)^2} - \frac{1}{1 + x}$
$f'(x) = -\left( \frac{1 + (x + 1)}{(x + 1)^2} \right) = -\frac{x + 2}{(x + 1)^2}$
Since $x > 0$,we have $x + 2 > 0$ and $(x + 1)^2 > 0$.
Therefore,$f'(x) = -\frac{x + 2}{(x + 1)^2} < 0$ for all $x > 0$.
Since the derivative $f'(x) < 0$ for all $x$ in the domain,the function $f$ is a decreasing function.

(B) Let $f(x) = \frac{a \sin x + b \cos x}{c \sin x + d \cos x}$.
For $f(x)$ to be a strictly increasing function,its derivative $f'(x)$ must be greater than $0$.
$f'(x) = \frac{(a \cos x - b \sin x)(c \sin x + d \cos x) - (a \sin x + b \cos x)(c \cos x - d \sin x)}{(c \sin x + d \cos x)^2}$.
Simplifying the numerator,we get $ad - bc > 0$.
In the given function $f(x) = \frac{1 \sin x + b \cos x}{1 \sin x + 4 \cos x}$,we have $a=1, b=b, c=1, d=4$.
Applying the condition $ad - bc > 0$:
$(1)(4) - (b)(1) > 0$
$4 - b > 0$
$4 > b$ or $b < 4$.

(B) Given the function $f(x) = (x - 1)^2 (x - 2)$.
First,find the derivative $f'(x)$ using the product rule:
$f'(x) = \frac{d}{dx}[(x - 1)^2] \cdot (x - 2) + (x - 1)^2 \cdot \frac{d}{dx}[(x - 2)]$
$f'(x) = 2(x - 1)(x - 2) + (x - 1)^2$
$f'(x) = (x - 1) [2(x - 2) + (x - 1)]$
$f'(x) = (x - 1) (2x - 4 + x - 1)$
$f'(x) = (x - 1) (3x - 5)$
For the function to be monotonically decreasing,we must have $f'(x) < 0$:
$(x - 1)(3x - 5) < 0$
The critical points are $x = 1$ and $x = 5/3$.
Testing the intervals $(-\infty, 1)$,$(1, 5/3)$,and $(5/3, \infty)$:
For $x \in (1, 5/3)$,the product $(x - 1)(3x - 5)$ is negative.
Thus,the function decreases on the interval $x \in (1, 5/3)$.

(B) function $f(x)$ increases monotonically if its derivative $f'(x) \ge 0$ for all $x \in \mathbb{R}$.
Given $f(x) = \lambda x - \sin x$.
Find the derivative: $f'(x) = \lambda - \cos x$.
For $f(x)$ to be monotonically increasing,we require $f'(x) \ge 0$,which means $\lambda - \cos x \ge 0$,or $\lambda \ge \cos x$.
Since the maximum value of $\cos x$ is $1$,for the inequality $\lambda \ge \cos x$ to hold for all $x$,we must have $\lambda \ge 1$.
Thus,the function increases monotonically when $\lambda \ge 1$.

(B) To determine if a function $f(x)$ is increasing,we check its derivative $f'(x)$. If $f'(x) > 0$ for all $x$ in the domain,the function is strictly increasing.
$1$. For $f(x) = e^{x^2}$,$f'(x) = e^{x^2} \cdot 2x$. This is positive for $x > 0$ and negative for $x < 0$. Thus,it is not increasing on the entire real line.
$2$. For $f(x) = e^{x^3}$,$f'(x) = e^{x^3} \cdot 3x^2$. Since $e^{x^3} > 0$ and $3x^2 \ge 0$ for all $x$,$f'(x) \ge 0$. The function is strictly increasing on $\mathbb{R}$.
$3$. For $f(x) = e^0 = 1$,$f'(x) = 0$. This is a constant function,not strictly increasing.
Therefore,$e^{x^3}$ is the correct choice as it is an increasing function.

(A) We want to find the interval where $f(x)$ increases less rapidly than $g(x)$,which means $f'(x) < g'(x)$.
Let $h(x) = g(x) - f(x) = -2x^3 + 9x^2 - 12x + 6$.
We want $h'(x) > 0$ for $f(x)$ to increase less rapidly than $g(x)$.
$h'(x) = -6x^2 + 18x - 12 = -6(x^2 - 3x + 2) = -6(x - 1)(x - 2)$.
For $h'(x) > 0$,we need $(x - 1)(x - 2) < 0$.
This inequality holds when $x \in (1, 2)$.
Thus,in the interval $(1, 2)$,$f(x)$ increases less rapidly than $g(x)$.

(C) For a function $f(x)$ to be increasing on an interval,its derivative $f'(x)$ must be greater than or equal to $0$ on that interval.
Given $f(x) = x^2 + kx + 1$,we find the derivative:
$f'(x) = 2x + k$
For $f(x)$ to be increasing on $[1, 2]$,we require $f'(x) \geq 0$ for all $x \in [1, 2]$.
$2x + k \geq 0$
$k \geq -2x$
Since this must hold for all $x$ in the interval $[1, 2]$,$k$ must be greater than or equal to the maximum value of $-2x$ on this interval.
Let $g(x) = -2x$. On the interval $[1, 2]$,the function $g(x)$ is decreasing.
Therefore,the maximum value of $g(x)$ occurs at the left endpoint $x = 1$.
$g(1) = -2(1) = -2$.
Thus,$k \geq -2$.
The minimum value of $k$ is $-2$.

(D) Let us analyze each function:
$A) f(x) = x + |x| = \begin{cases} 2x, & x \ge 0 \\ 0, & x < 0 \end{cases}$. This is not strictly increasing because it is constant for $x < 0$.
$B) f(x) = x - |x| = \begin{cases} 0, & x \ge 0 \\ 2x, & x < 0 \end{cases}$. This is not strictly increasing because it is constant for $x \ge 0$.
$C) f(x) = [x]$ (Greatest Integer Function). This is a step function and is not strictly increasing as it remains constant between integers.
$D) f(x) = x|x| = \begin{cases} x^2, & x \ge 0 \\ -x^2, & x < 0 \end{cases}$.
For $x_1 < x_2$,if both are positive,$x_1^2 < x_2^2$. If both are negative,$-x_1^2 < -x_2^2$. If $x_1 < 0$ and $x_2 \ge 0$,then $f(x_1) < 0$ and $f(x_2) \ge 0$,so $f(x_1) < f(x_2)$. Thus,$f(x)$ is strictly increasing.

(B) Given $f(x) = \cos |x| - 2ax + b$.
Since $\cos |x| = \cos x$,we have $f(x) = \cos x - 2ax + b$.
For $f(x)$ to be an increasing function,$f'(x) \ge 0$ for all $x \in \mathbb{R}$.
$f'(x) = -\sin x - 2a$.
For $f'(x) \ge 0$,we must have $-\sin x - 2a \ge 0$,which implies $2a \le -\sin x$,or $a \le -\frac{1}{2} \sin x$.
Since this must hold for all $x \in \mathbb{R}$,$a$ must be less than or equal to the minimum value of $-\frac{1}{2} \sin x$.
The minimum value of $\sin x$ is $-1$,so the maximum value of $-\frac{1}{2} \sin x$ is $\frac{1}{2}$. However,for the function to be non-decreasing for all $x$,we consider the condition $f'(x) \ge 0$.
Since $-\sin x$ ranges from $-1$ to $1$,for $-\sin x - 2a \ge 0$ to hold for all $x$,we need $2a \le \min(-\sin x) = -1$.
Thus,$a \le -1/2$.

(A) Given the function $f(x) = x^{100} + \sin x - 1$.
To determine the nature of the function,we find its derivative $f'(x)$.
$f'(x) = \frac{d}{dx}(x^{100} + \sin x - 1) = 100x^{99} + \cos x$.
For $x \in (0, 1)$,we observe that $x^{99} > 0$,so $100x^{99} > 0$.
Also,for $x \in (0, 1)$,$\cos x > 0$ because $x$ is in the first quadrant.
Since both terms $100x^{99}$ and $\cos x$ are positive for all $x \in (0, 1)$,their sum $f'(x) = 100x^{99} + \cos x > 0$.
Since $f'(x) > 0$ for all $x \in (0, 1)$,the function $f(x)$ is monotonically increasing on the interval $(0, 1)$.

(B) Given the function $f(x) = 3x + \frac{2}{x}$.
To determine the nature of the function,we find its derivative $f'(x)$.
$f'(x) = \frac{d}{dx}(3x + 2x^{-1}) = 3 - 2x^{-2} = 3 - \frac{2}{x^2}$.
Simplifying the expression,we get $f'(x) = \frac{3x^2 - 2}{x^2}$.
For the interval $x \in (1, 3)$,we observe that $x^2 > 1$,which implies $3x^2 > 3$.
Therefore,$3x^2 - 2 > 3 - 2 = 1$,which is always positive.
Since $f'(x) > 0$ for all $x \in (1, 3)$,the function $f(x)$ is strictly increasing on the interval $(1, 3)$.

(B) Consider the function $h(x) = g(x) - f(x) = x - \sin x$.
To check if $f(x) \leq g(x)$ for $x > 0$,we analyze $h(x)$.
The derivative is $h'(x) = 1 - \cos x$.
Since $\cos x \leq 1$ for all $x$,$h'(x) = 1 - \cos x \geq 0$.
This means $h(x)$ is a non-decreasing function for $x \geq 0$.
Since $h(0) = 0 - \sin(0) = 0$,it follows that $h(x) \geq 0$ for all $x \geq 0$.
Thus,$x - \sin x \geq 0$,which implies $x \geq \sin x$ or $g(x) \geq f(x)$ for $x \in (0, \infty)$.
Therefore,Statement-$1$ is true.
Now consider Statement-$2$: For $x \in (0, \infty)$,$f(x) = \sin x \leq 1$ is true.
Also,as $x \rightarrow \infty$,$g(x) = x \rightarrow \infty$ is true.
However,the fact that $f(x) \leq 1$ and $g(x) \rightarrow \infty$ does not directly prove $f(x) \leq g(x)$ without comparing the growth rates or using the derivative test.
Thus,Statement-$2$ is true,but it is not the direct logical explanation for Statement-$1$ because the inequality $x \geq \sin x$ holds for all $x > 0$ regardless of the limit at infinity.

(B) Let $f(x) = \tan x - x$.
To determine the nature of the function,we find its derivative with respect to $x$:
$f'(x) = \frac{d}{dx}(\tan x - x) = \sec^2 x - 1$.
Using the trigonometric identity $1 + \tan^2 x = \sec^2 x$,we have $\sec^2 x - 1 = \tan^2 x$.
Thus,$f'(x) = \tan^2 x$.
Since the square of any real number is non-negative,$f'(x) \geq 0$ for all $x$ in the domain.
Specifically,$f'(x) > 0$ for all $x \neq n\pi$ (where $n \in Z$).
Since the derivative is non-negative and is not identically zero on any interval,the function is strictly increasing.
Therefore,the function is always increasing.

(C) Given the function $f(x) = \tan^{-1}(\sin x + \cos x)$.
Differentiating with respect to $x$,we get:
$f'(x) = \frac{1}{1 + (\sin x + \cos x)^2} \cdot (\cos x - \sin x)$.
For the function to be increasing,we require $f'(x) > 0$.
Since the denominator $1 + (\sin x + \cos x)^2$ is always positive,the sign of $f'(x)$ depends on the numerator $(\cos x - \sin x)$.
We need $\cos x - \sin x > 0$,which implies $\cos x > \sin x$.
Dividing by $\cos x$ (assuming $\cos x > 0$ in the first quadrant),we get $1 > \tan x$,or $\tan x < 1$.
This holds true for $x \in (0, \frac{\pi}{4})$.
Thus,the function is strictly increasing in the interval $(0, \frac{\pi}{4})$.

(B) function $f(x)$ is decreasing when its derivative $f'(x) < 0$.
Given $f(x) = \int e^x (x - 1)(x - 2) dx$.
By the Fundamental Theorem of Calculus,the derivative is $f'(x) = e^x (x - 1)(x - 2)$.
Since $e^x$ is always positive for all real $x$,the sign of $f'(x)$ depends on the product $(x - 1)(x - 2)$.
We set $f'(x) < 0$,which implies $(x - 1)(x - 2) < 0$.
This inequality holds when $x$ lies between the roots $1$ and $2$.
Therefore,$f(x)$ is a decreasing function in the interval $(1, 2)$.

(B) Given function is $f(x) = x^{3/2}(3x - 10) = 3x^{5/2} - 10x^{3/2}$.
To find the interval where the function is increasing,we find the derivative $f'(x)$:
$f'(x) = \frac{d}{dx}(3x^{5/2} - 10x^{3/2}) = 3 \cdot \frac{5}{2} x^{3/2} - 10 \cdot \frac{3}{2} x^{1/2} = \frac{15}{2} x^{3/2} - 15 x^{1/2}$.
Setting $f'(x) = 0$ for critical points:
$\frac{15}{2} x^{1/2} (x - 2) = 0$.
Since $x \geq 0$,the critical point is $x = 2$.
For $x > 2$,$f'(x) > 0$,so the function is increasing on $[2, \infty)$.
For $0 < x < 2$,$f'(x) < 0$,so the function is decreasing on $(0, 2)$.

(C) Let the function be $f(x) = e^{ax}$.
To find the intervals where the function is monotonically decreasing,we calculate the first derivative:
$f'(x) = \frac{d}{dx}(e^{ax}) = a \cdot e^{ax}$.
$A$ function is monotonically decreasing if $f'(x) < 0$.
Since the exponential function $e^{ax}$ is always positive for all real $x$ $(e^{ax} > 0)$,the sign of $f'(x)$ depends entirely on the constant $a$.
Therefore,$f'(x) < 0$ if and only if $a < 0$.
Thus,the function $f(x) = e^{ax}$ is monotonically decreasing when $a < 0$.

(D) To find the intervals where the function $f(x) = 2x^3 - 9x^2 + 12x + 29$ is decreasing,we find its derivative $f'(x)$.
$f'(x) = \frac{d}{dx}(2x^3 - 9x^2 + 12x + 29) = 6x^2 - 18x + 12$.
For the function to be decreasing,we must have $f'(x) < 0$.
$6x^2 - 18x + 12 < 0$.
Dividing by $6$,we get $x^2 - 3x + 2 < 0$.
Factoring the quadratic expression,we get $(x - 1)(x - 2) < 0$.
The inequality $(x - 1)(x - 2) < 0$ holds when $x$ lies between the roots $1$ and $2$.
Therefore,the function is decreasing for $1 < x < 2$.

(C) Given the function $f(x) = 1 - e^{-\frac{x^2}{2}}$.
To determine the intervals of increase and decrease,we find the derivative $f'(x)$:
$f'(x) = \frac{d}{dx}(1 - e^{-\frac{x^2}{2}}) = 0 - e^{-\frac{x^2}{2}} \cdot \frac{d}{dx}(-\frac{x^2}{2}) = -e^{-\frac{x^2}{2}} \cdot (-x) = x e^{-\frac{x^2}{2}}$.
Since $e^{-\frac{x^2}{2}} > 0$ for all $x \in R$,the sign of $f'(x)$ depends only on the sign of $x$.
Case $1$: If $x > 0$,then $f'(x) > 0$,which means the function $f(x)$ is increasing.
Case $2$: If $x < 0$,then $f'(x) < 0$,which means the function $f(x)$ is decreasing.
Therefore,the function is decreasing for $x < 0$ and increasing for $x > 0$.

(A) Given $f(x) = xe^{x(1 - x)}$.
Taking the derivative with respect to $x$ using the product rule and chain rule:
$f'(x) = 1 \cdot e^{x(1 - x)} + x \cdot e^{x(1 - x)} \cdot (1 - 2x)$
$f'(x) = e^{x(1 - x)} [1 + x(1 - 2x)]$
$f'(x) = e^{x(1 - x)} [1 + x - 2x^2]$
$f'(x) = e^{x(1 - x)} (-2x^2 + x + 1)$
$f'(x) = e^{x(1 - x)} (-2x^2 + 2x - x + 1)$
$f'(x) = e^{x(1 - x)} [-2x(x - 1) - 1(x - 1)]$
$f'(x) = e^{x(1 - x)} (1 - 2x)(1 + x)$
For $f(x)$ to be increasing,$f'(x) \geq 0$.
Since $e^{x(1 - x)} > 0$ for all $x \in R$,we need $(1 - 2x)(1 + x) \geq 0$.
Solving the inequality $(2x - 1)(x + 1) \leq 0$,we get $x \in [-1, 1/2]$.
However,checking the interval $[-1/2, 1]$: for $x \in [-1/2, 1/2]$,$f'(x) \geq 0$,and for $x \in [1/2, 1]$,$f'(x) \leq 0$.
Re-evaluating the derivative: $f'(x) = e^{x(1-x)}(1+x)(1-2x)$.
At $x = 0$,$f'(0) = 1 > 0$. At $x = 1$,$f'(1) = 0$. At $x = -1/2$,$f'(-1/2) = e^{-3/4}(1/2)(2) = e^{-3/4} > 0$.
Thus,$f(x)$ is increasing on $[-1/2, 1/2]$ and decreasing on $[1/2, 1]$.
Given the options,the function is increasing on $[-1/2, 1/2]$,which is a subset of $[-1/2, 1]$. The most appropriate choice is $A$.

(C) Given the function $f(x) = \cos |x| - 2ax + b$.
Since $\cos |x| = \cos x$,we have $f(x) = \cos x - 2ax + b$.
For the function to be increasing on the entire real line,we must have $f'(x) \geq 0$ for all $x \in \mathbb{R}$.
Calculating the derivative: $f'(x) = -\sin x - 2a$.
Setting the condition $f'(x) \geq 0$ gives $-\sin x - 2a \geq 0$,which simplifies to $\sin x + 2a \leq 0$.
For this inequality to hold for all $x \in \mathbb{R}$,the maximum value of $\sin x + 2a$ must be less than or equal to $0$.
The maximum value of $\sin x$ is $1$,so we require $1 + 2a \leq 0$.
Solving for $a$,we get $2a \leq -1$,which implies $a \leq -1/2$.

(C) Given the function $f(x) = [x(x - 3)]^2 = (x^2 - 3x)^2$.
To find where the function is increasing,we calculate the derivative $f'(x)$ and set $f'(x) > 0$.
$f'(x) = 2(x^2 - 3x) \cdot (2x - 3) = 2x(x - 3)(2x - 3)$.
Setting $f'(x) > 0$,we get $2x(x - 3)(2x - 3) > 0$,which simplifies to $x(x - 3)(2x - 3) > 0$.
The critical points are $x = 0$,$x = 3/2$,and $x = 3$.
Using the wavy curve method (sign scheme) on the number line:
For $x \in (0, 3/2)$,$f'(x) > 0$.
For $x \in (3/2, 3)$,$f'(x) < 0$.
For $x \in (3, \infty)$,$f'(x) > 0$.
Thus,the function is increasing for $x \in (0, 3/2) \cup (3, \infty)$.

(B) Let $f(x) = \frac{\log(\pi + x)}{\log(e + x)}$.
Using the quotient rule,$f'(x) = \frac{\log(e + x) \cdot \frac{1}{\pi + x} - \log(\pi + x) \cdot \frac{1}{e + x}}{[\log(e + x)]^2}$.
$f'(x) = \frac{(e + x)\log(e + x) - (\pi + x)\log(\pi + x)}{(\pi + x)(e + x)[\log(e + x)]^2}$.
Since $e < \pi$,for $x > 0$,we have $e + x < \pi + x$ and $\log(e + x) < \log(\pi + x)$.
Thus,$(e + x)\log(e + x) < (\pi + x)\log(\pi + x)$,which implies $f'(x) < 0$ for all $x \in (0, \infty)$.
Therefore,$f(x)$ is a decreasing function on $(0, \infty)$.
Hence,option $B$ is the correct answer.

(A) Given $f(x) = \int_{x^2}^{x^2+1} e^{-t^2} dt$.
Applying the Leibniz rule for differentiation under the integral sign:
$f'(x) = e^{-(x^2+1)^2} \cdot \frac{d}{dx}(x^2+1) - e^{-(x^2)^2} \cdot \frac{d}{dx}(x^2)$
$f'(x) = e^{-(x^2+1)^2} \cdot (2x) - e^{-x^4} \cdot (2x)$
$f'(x) = 2x \left( e^{-(x^4+2x^2+1)} - e^{-x^4} \right)$
$f'(x) = 2x \cdot e^{-x^4} \left( e^{-(2x^2+1)} - 1 \right)$
Since $e^{-x^4} > 0$ for all $x$,and $e^{-(2x^2+1)} < 1$ for all $x$ (because $2x^2+1 > 0$),the term $(e^{-(2x^2+1)} - 1)$ is always negative.
For $f(x)$ to be an increasing function,we require $f'(x) \geq 0$.
$2x \cdot (\text{negative value}) \geq 0 \implies x \leq 0$.
Thus,$f(x)$ is increasing on the interval $(-\infty, 0]$.

(B) Given $f(x) = \frac{\log(\pi + x)}{\log(e + x)}$.
Taking the derivative using the quotient rule:
$f'(x) = \frac{\frac{1}{\pi + x} \cdot \log(e + x) - \frac{1}{e + x} \cdot \log(\pi + x)}{(\log(e + x))^2}$.
$f'(x) = \frac{(e + x)\log(e + x) - (\pi + x)\log(\pi + x)}{(\pi + x)(e + x)(\log(e + x))^2}$.
Let $g(t) = t \log t$. Then $g'(t) = 1 + \log t$. For $t > 0$,$g(t)$ is increasing for $t > 1/e$.
Since $x > 0$,$e + x > e > 1/e$ and $\pi + x > \pi > 1/e$.
Since $\pi > e$,we have $\pi + x > e + x$.
Because $g(t)$ is an increasing function for $t > 1/e$,$g(\pi + x) > g(e + x)$.
Therefore,$(\pi + x)\log(\pi + x) > (e + x)\log(e + x)$.
This implies $(e + x)\log(e + x) - (\pi + x)\log(\pi + x) < 0$.
Thus,$f'(x) < 0$ for all $x \in (0, \infty)$.
Hence,$f(x)$ is a decreasing function on $(0, \infty)$.

(B) Given function is $f(x) = x^3 - 6x^2 + 9x + 3$.
To find the interval where the function is decreasing,we first find the derivative $f'(x)$.
$f'(x) = \frac{d}{dx}(x^3 - 6x^2 + 9x + 3) = 3x^2 - 12x + 9$.
Factorizing the derivative:
$f'(x) = 3(x^2 - 4x + 3) = 3(x - 3)(x - 1)$.
$A$ function is decreasing when $f'(x) < 0$.
So,$3(x - 3)(x - 1) < 0$.
This inequality holds when $x$ lies between the roots $1$ and $3$.
Therefore,$x \in (1, 3)$.

(A) To determine the nature of the function,we find its derivative $f'(x)$.
$f'(x) = \frac{d}{dx}(2x^3 + 6x^2 + 7x - 19) = 6x^2 + 12x + 7$.
We can rewrite this as $f'(x) = 6(x^2 + 2x) + 7$.
Completing the square,$f'(x) = 6(x^2 + 2x + 1 - 1) + 7 = 6(x + 1)^2 - 6 + 7 = 6(x + 1)^2 + 1$.
Since $(x + 1)^2 \geq 0$ for all real $x$,it follows that $6(x + 1)^2 + 1 \geq 1$.
Thus,$f'(x) > 0$ for all $x \in \mathbb{R}$.
Since the derivative is strictly positive for all $x$,the function $f(x)$ is strictly increasing.

(A) To determine the nature of the function,we find its derivative:
$f'(x) = -\frac{1}{2} + \cos x$
For the interval $x \in \left( -\frac{\pi}{3}, \frac{\pi}{3} \right)$,we know that $\cos x$ ranges from $\frac{1}{2}$ to $1$.
Specifically,$\frac{1}{2} < \cos x \le 1$.
Subtracting $\frac{1}{2}$ from all parts,we get:
$\frac{1}{2} - \frac{1}{2} < \cos x - \frac{1}{2} \le 1 - \frac{1}{2}$
$0 < f'(x) \le \frac{1}{2}$
Since $f'(x) > 0$ for all $x$ in the given interval,the function $f(x)$ is strictly increasing.

(B) To find the interval where the function $f(x) = \tan^{-1}(\sin x + \cos x)$ is increasing,we need to find the derivative $f'(x)$ and set it to be greater than $0$.
Given $f(x) = \tan^{-1}(\sin x + \cos x)$.
Differentiating with respect to $x$ using the chain rule:
$f'(x) = \frac{1}{1 + (\sin x + \cos x)^2} \times (\cos x - \sin x)$.
For the function to be increasing,$f'(x) > 0$.
Since $1 + (\sin x + \cos x)^2$ is always positive,we require:
$\cos x - \sin x > 0$
$\cos x > \sin x$
Dividing by $\cos x$ (assuming $\cos x > 0$ in the relevant domain),we get:
$1 > \tan x$
$\tan x < 1$.
Since $\tan x = 1$ at $x = \pi/4$,the condition $\tan x < 1$ holds for $x < \pi/4$.
Considering the standard domain of the trigonometric functions involved,the interval is $(-\pi/2, \pi/4)$.

(A) To find the intervals where the function $f(x) = x^2 e^{-x}$ is strictly increasing,we calculate its derivative $f'(x)$.
Using the product rule:
$f'(x) = \frac{d}{dx}(x^2) \cdot e^{-x} + x^2 \cdot \frac{d}{dx}(e^{-x})$
$f'(x) = 2x e^{-x} - x^2 e^{-x}$
$f'(x) = x e^{-x} (2 - x)$
For the function to be strictly increasing,we must have $f'(x) > 0$.
Since $e^{-x} > 0$ for all $x \in \mathbb{R}$,the inequality $x e^{-x} (2 - x) > 0$ simplifies to:
$x(2 - x) > 0$
Multiplying by $-1$ reverses the inequality sign:
$x(x - 2) < 0$
The roots of the quadratic expression are $x = 0$ and $x = 2$. The expression is negative between the roots.
Therefore,the function is strictly increasing for $0 < x < 2$.

(D) Given $f(x) = \sin x - \cos x$.
To find the intervals where $f(x)$ is strictly decreasing,we find the derivative $f'(x)$.
$f'(x) = \cos x + \sin x$.
For $f(x)$ to be strictly decreasing,we must have $f'(x) < 0$.
$\cos x + \sin x < 0$.
$\sqrt{2} \left( \frac{1}{\sqrt{2}} \cos x + \frac{1}{\sqrt{2}} \sin x \right) < 0$.
$\sqrt{2} \cos \left( x - \frac{\pi}{4} \right) < 0$.
Since $\sqrt{2} > 0$,we need $\cos \left( x - \frac{\pi}{4} \right) < 0$.
We know that $\cos \theta < 0$ when $\frac{\pi}{2} < \theta < \frac{3\pi}{2}$.
Substituting $\theta = x - \frac{\pi}{4}$,we get $\frac{\pi}{2} < x - \frac{\pi}{4} < \frac{3\pi}{2}$.
Adding $\frac{\pi}{4}$ to all parts: $\frac{\pi}{2} + \frac{\pi}{4} < x < \frac{3\pi}{2} + \frac{\pi}{4}$.
$\frac{3\pi}{4} < x < \frac{7\pi}{4}$.
Comparing this with the given options,none of the intervals provided are subsets of $\left( \frac{3\pi}{4}, \frac{7\pi}{4} \right)$ except for the possibility of none of these being correct.

(B) Given the function $y = ax^3 + 3x^2 + (2a + 1)x + 1000$.
For $y$ to be a strictly increasing function for all $x \in \mathbb{R}$,its derivative $y'$ must be greater than $0$ for all $x$.
$y' = 3ax^2 + 6x + (2a + 1) > 0$.
For a quadratic expression $Ax^2 + Bx + C > 0$ to hold for all $x$,we must have $A > 0$ and the discriminant $D < 0$.
Here,$A = 3a$,$B = 6$,and $C = 2a + 1$.
Condition $1$: $3a > 0 \implies a > 0$.
Condition $2$: $D = B^2 - 4AC < 0$.
$6^2 - 4(3a)(2a + 1) < 0$.
$36 - 12a(2a + 1) < 0$.
Divide by $12$: $3 - a(2a + 1) < 0$.
$3 - 2a^2 - a < 0$.
$2a^2 + a - 3 > 0$.
Factoring the quadratic: $(2a + 3)(a - 1) > 0$.
This inequality holds when $a > 1$ or $a < -3/2$.
Combining with Condition $1$ $(a > 0)$,we get $a > 1$.

(A) Given $f(x) = x^3 + bx^2 + cx + d$.
Find the derivative: $f'(x) = 3x^2 + 2bx + c$.
For a quadratic expression $ax^2 + bx + c$ to be always positive,the discriminant $D$ must be less than $0$ and $a > 0$.
Here,$D = (2b)^2 - 4(3)(c) = 4b^2 - 12c = 4(b^2 - 3c)$.
Since $0 < b^2 < c$,it follows that $b^2 - 3c < 0$ (because $b^2 < c < 3c$).
Thus,$D < 0$ and the coefficient of $x^2$ is $3 > 0$.
Therefore,$f'(x) > 0$ for all $x \in R$.
Since $f'(x) > 0$ for all $x \in R$,$f(x)$ is a strictly increasing function on $R$.

(D) function $f(x)$ is strictly increasing if $f'(x) > 0$ for all $x \in \mathbb{R}$.
Given $f(x) = \lambda x + \cos x$.
Differentiating with respect to $x$,we get $f'(x) = \lambda - \sin x$.
For $f(x)$ to be strictly increasing,we require $f'(x) > 0$,which means $\lambda - \sin x > 0$,or $\lambda > \sin x$ for all $x$.
Since the maximum value of $\sin x$ is $1$,for $\lambda > \sin x$ to hold for all $x$,we must have $\lambda > 1$.

(A) Given the function $f(x) = x^3 - 3x^2 - 9x + 22$.
To find the interval where the function is strictly decreasing,we calculate the derivative $f'(x)$.
$f'(x) = \frac{d}{dx}(x^3 - 3x^2 - 9x + 22) = 3x^2 - 6x - 9$.
For the function to be strictly decreasing,$f'(x) < 0$.
$3x^2 - 6x - 9 < 0$.
Dividing by $3$,we get $x^2 - 2x - 3 < 0$.
Factoring the quadratic expression: $(x - 3)(x + 1) < 0$.
The roots of the equation $(x - 3)(x + 1) = 0$ are $x = 3$ and $x = -1$.
Testing the intervals $(-\infty, -1)$,$(-1, 3)$,and $(3, \infty)$,we find that the expression is negative in the interval $(-1, 3)$.
Therefore,the function is strictly decreasing for $-1 < x < 3$.

(D) Given $f(x) = \tan^{-1}(\sin x + \cos x)$.
To find where the function is increasing,we calculate the derivative $f'(x)$.
$f'(x) = \frac{1}{1 + (\sin x + \cos x)^2} \cdot (\cos x - \sin x)$.
$f'(x) = \frac{\cos x - \sin x}{1 + (\sin^2 x + \cos^2 x + 2\sin x \cos x)} = \frac{\cos x - \sin x}{2 + \sin 2x}$.
For the function to be increasing,we require $f'(x) > 0$.
Since $2 + \sin 2x > 0$ for all $x$,we must have $\cos x - \sin x > 0$.
$\cos x > \sin x$.
Dividing by $\cos x$ (assuming $\cos x > 0$),we get $1 > \tan x$,which implies $x < \frac{\pi}{4}$.
Considering the domain of the function,the interval where $f'(x) > 0$ is $(-\frac{\pi}{2}, \frac{\pi}{4})$.

(C) Given the function $f(x) = x^5 - 20x^3 + 240x$.
To determine the monotonicity,we find the derivative $f'(x)$:
$f'(x) = \frac{d}{dx}(x^5 - 20x^3 + 240x) = 5x^4 - 60x^2 + 240$.
We can factor out $5$:
$f'(x) = 5(x^4 - 12x^2 + 48)$.
Now,we complete the square for the expression inside the parenthesis:
$x^4 - 12x^2 + 48 = (x^2 - 6)^2 - 36 + 48 = (x^2 - 6)^2 + 12$.
Thus,$f'(x) = 5[(x^2 - 6)^2 + 12]$.
Since $(x^2 - 6)^2 \ge 0$ for all $x \in \mathbb{R}$,it follows that $(x^2 - 6)^2 + 12 \ge 12$.
Therefore,$f'(x) \ge 5 \times 12 = 60 > 0$ for all $x \in \mathbb{R}$.
Since $f'(x) > 0$ for all $x$,the function $f(x)$ is monotonically increasing everywhere.

(B) Given $f(x) = \int {e^x}(x - 1)(x - 2)dx$.
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 the function $f$ to be decreasing,we must have $f'(x) < 0$.
So,${e^x}(x - 1)(x - 2) < 0$.
Since ${e^x} > 0$ for all real $x$,the inequality simplifies to $(x - 1)(x - 2) < 0$.
Using the sign scheme for the quadratic expression $(x - 1)(x - 2)$,the product is negative between the roots $x = 1$ and $x = 2$.
Thus,$f$ decreases in the interval $(1, 2)$.

(C) To determine the monotonicity of $f(x)$,we find its derivative $f'(x)$.
Given $f(x) = \int {\left( {2 - \frac{1}{{1 + {x^2}}} - \frac{1}{{\sqrt {1 + {x^2}} }}} \right)} \,dx$,by the Fundamental Theorem of Calculus,$f'(x) = 2 - \frac{1}{{1 + {x^2}}} - \frac{1}{{\sqrt {1 + {x^2}} }}$.
We analyze the sign of $f'(x)$ for all $x \in \mathbb{R}$.
$f'(x) = \frac{2(1 + {x^2}) - 1 - \sqrt{1 + {x^2}}}{1 + {x^2}} = \frac{2{x^2} + 1 - \sqrt{1 + {x^2}}}{1 + {x^2}}$.
Let $u = \sqrt{1 + {x^2}}$. Since ${x^2} \ge 0$,we have $u \ge 1$. Then ${x^2} = {u^2} - 1$.
Substituting this into the numerator: $2({u^2} - 1) + 1 - u = 2{u^2} - u - 1 = (2u + 1)(u - 1)$.
Since $u \ge 1$,$(u - 1) \ge 0$ and $(2u + 1) > 0$,so the numerator is $\ge 0$.
Specifically,$f'(x) > 0$ for all $x \neq 0$,and $f'(0) = 0$.
Since $f'(x) \ge 0$ for all $x \in \mathbb{R}$ and is not identically zero on any interval,$f(x)$ is strictly increasing on $(-\infty, \infty)$.

(D) Given $f(x) = \int\limits_1^x {\left( {t\ln(t) - \frac{{\ln(t)}}{t}} \right)dt}$.
Using the Fundamental Theorem of Calculus,$f'(x) = x\ln(x) - \frac{{\ln(x)}}{x}$.
Factoring the expression,$f'(x) = \ln(x) \left( x - \frac{1}{x} \right) = \ln(x) \left( \frac{x^2 - 1}{x} \right) = \frac{{\ln(x)(x - 1)(x + 1)}}{x}$.
For $x > 1$,we know that $\ln(x) > 0$,$(x - 1) > 0$,$(x + 1) > 0$,and $x > 0$.
Therefore,$f'(x) > 0$ for all $x > 1$.
Since the derivative $f'(x)$ is strictly positive for all $x$ in the domain $(1, \infty)$,the function $f(x)$ is strictly increasing.
Thus,$f(x)$ is monotonic.

(A) The domain of $f(x)$ is $x > 0$ because the logarithmic function $\ln t$ is defined only for $t > 0$.
Given $f(x) = 1 + x + \int_{1}^{x} (\ln^2 t + 2 \ln t) \, dt$.
Differentiating both sides with respect to $x$ using the Leibniz integral rule:
$f'(x) = \frac{d}{dx}(1) + \frac{d}{dx}(x) + \frac{d}{dx} \int_{1}^{x} (\ln^2 t + 2 \ln t) \, dt$
$f'(x) = 0 + 1 + (\ln^2 x + 2 \ln x)$
$f'(x) = \ln^2 x + 2 \ln x + 1$
$f'(x) = (\ln x + 1)^2$
Since $(\ln x + 1)^2 \ge 0$ for all $x > 0$,and $f'(x) = 0$ only at $x = e^{-1}$,the function $f(x)$ is strictly increasing on its entire domain $(0, \infty)$.

(A) Let $f(x) = \frac{\ln(\pi + x)}{\ln(e + x)}$.
To determine the monotonicity,we find the derivative $f'(x)$ using the quotient rule:
$f'(x) = \frac{\frac{1}{\pi + x} \cdot \ln(e + x) - \frac{1}{e + x} \cdot \ln(\pi + x)}{(\ln(e + x))^2}$.
The sign of $f'(x)$ depends on the numerator $g(x) = \frac{\ln(e + x)}{\pi + x} - \frac{\ln(\pi + x)}{e + x}$.
Consider the function $h(t) = \frac{\ln(t)}{t}$. Its derivative is $h'(t) = \frac{1 - \ln(t)}{t^2}$.
$h'(t) > 0$ for $t < e$ and $h'(t) < 0$ for $t > e$.
Since $\pi > e$,the function $h(t)$ is decreasing for $t \ge e$.
Comparing $h(e+x)$ and $h(\pi+x)$,since $\pi+x > e+x > e$,we have $h(\pi+x) < h(e+x)$.
Thus,$\frac{\ln(\pi + x)}{\pi + x} < \frac{\ln(e + x)}{e + x}$,which implies $f'(x) > 0$.
Therefore,the function is increasing on $[0, \infty)$.

(D) Given $g(x) = 2f(x/2) + f(1 - x)$.
Taking the derivative with respect to $x$,we get $g'(x) = 2f'(x/2) \cdot (1/2) + f'(1 - x) \cdot (-1) = f'(x/2) - f'(1 - x)$.
Given $f''(x) < 0$,which implies that $f'(x)$ is a strictly decreasing function.
For $g(x)$ to be increasing,we require $g'(x) > 0$.
$f'(x/2) - f'(1 - x) > 0 \implies f'(x/2) > f'(1 - x)$.
Since $f'(x)$ is strictly decreasing,$f'(a) > f'(b) \implies a < b$.
Therefore,$x/2 < 1 - x \implies 3x/2 < 1 \implies x < 2/3$.
Thus,$g(x)$ increases in $[0, 2/3)$.
For $g(x)$ to be decreasing,we require $g'(x) < 0$.
$f'(x/2) - f'(1 - x) < 0 \implies f'(x/2) < f'(1 - x)$.
Since $f'(x)$ is strictly decreasing,$f'(a) < f'(b) \implies a > b$.
Therefore,$x/2 > 1 - x \implies 3x/2 > 1 \implies x > 2/3$.
Thus,$g(x)$ decreases in $(2/3, 1]$.
Both $(B)$ and $(C)$ are correct.

(D) To determine where the function $f(x) = x^{100} + \sin x - 1$ is strictly increasing,we find its derivative:
$f'(x) = 100x^{99} + \cos x$
For the interval $(0, 1)$:
Since $x \in (0, 1)$,$x^{99} > 0$ and $\cos x > 0$. Thus,$f'(x) = 100x^{99} + \cos x > 0$.
For the interval $(0, \pi / 2)$:
Since $x \in (0, \pi / 2)$,$x^{99} > 0$ and $\cos x > 0$. Thus,$f'(x) = 100x^{99} + \cos x > 0$.
For the interval $(\pi / 2, \pi )$:
Here,$x > \pi / 2 \approx 1.57$,so $x^{99}$ is very large. $100x^{99} > 100(1.57)^{99}$,while $\cos x$ lies between $-1$ and $0$. Therefore,$100x^{99} + \cos x > 0$.
Since the derivative is positive in all the given intervals,the function is strictly increasing in all of them.
Therefore,the correct option is $D$.

(D) Given $h(x) = f(x) - \{f(x)\}^2 + \{f(x)\}^3$.
Differentiating with respect to $x$,we get:
$h'(x) = f'(x) - 2f(x)f'(x) + 3\{f(x)\}^2f'(x)$
Factoring out $f'(x)$:
$h'(x) = f'(x) [1 - 2f(x) + 3\{f(x)\}^2]$
Let $y = f(x)$. Then the expression becomes:
$h'(x) = f'(x) (3y^2 - 2y + 1)$
Consider the quadratic expression $Q(y) = 3y^2 - 2y + 1$. The discriminant $D = (-2)^2 - 4(3)(1) = 4 - 12 = -8$.
Since $D < 0$ and the coefficient of $y^2$ is $3 > 0$,the quadratic $3y^2 - 2y + 1$ is always positive for all real $y$.
Thus,$h'(x) = f'(x) \times (\text{a positive quantity})$.
This implies that the sign of $h'(x)$ is the same as the sign of $f'(x)$.
Therefore,$h(x)$ is increasing whenever $f(x)$ is increasing $(f'(x) > 0 \implies h'(x) > 0)$ and $h(x)$ is decreasing whenever $f(x)$ is decreasing $(f'(x) < 0 \implies h'(x) < 0)$.
Hence,both $(A)$ and $(C)$ are correct.

(A) The function is defined as $f(x) = \begin{cases} 2\ln(x) - x^2 & \text{if } x > 0 \\ 2\ln(-x) + x^2 & \text{if } x < 0 \end{cases}$.
To find the intervals where the function is increasing,we calculate the derivative $f'(x)$:
For $x > 0$,$f'(x) = \frac{2}{x} - 2x = \frac{2(1 - x^2)}{x}$.
For $x < 0$,$f'(x) = \frac{2}{-x}(-1) + 2x = \frac{2}{x} + 2x = \frac{2(1 + x^2)}{x}$.
For the function to be increasing,we require $f'(x) > 0$.
Case $1$: $x > 0$. We need $\frac{2(1 - x^2)}{x} > 0$. Since $x > 0$,this implies $1 - x^2 > 0$,which means $x^2 < 1$,so $0 < x < 1$.
Case $2$: $x < 0$. We need $\frac{2(1 + x^2)}{x} > 0$. Since $x < 0$ and $(1 + x^2) > 0$,the expression $\frac{2(1 + x^2)}{x}$ is always negative for $x < 0$.
Thus,the function is increasing only on the interval $(0, 1)$.