Increasing and Decreasing function Questions in English

(C) Let $y = x^{1/x}$.
Taking the natural logarithm on both sides,we get $\ln y = \frac{1}{x} \ln x$.
Differentiating with respect to $x$ using the product rule:
$\frac{1}{y} \frac{dy}{dx} = \left(-\frac{1}{x^2}\right) \ln x + \frac{1}{x} \left(\frac{1}{x}\right) = \frac{1 - \ln x}{x^2}$.
Thus,$\frac{dy}{dx} = x^{1/x} \left(\frac{1 - \ln x}{x^2}\right)$.
Since $x^{1/x} > 0$ for all $x > 0$ and $x^2 > 0$,the sign of $\frac{dy}{dx}$ depends on $(1 - \ln x)$.
For $x \in (1, e)$,$\ln x < 1$,so $1 - \ln x > 0$,which implies $\frac{dy}{dx} > 0$. Thus,$f(x)$ is increasing in $(1, e)$.
For $x \in (e, \infty)$,$\ln x > 1$,so $1 - \ln x < 0$,which implies $\frac{dy}{dx} < 0$. Thus,$f(x)$ is decreasing in $(e, \infty)$.

(D) Given the function $f(x) = 1 - x^3 - x^5$.
To determine where the function is decreasing,we find its derivative $f'(x)$.
$f'(x) = \frac{d}{dx}(1 - x^3 - x^5) = -3x^2 - 5x^4$.
We can factor out $-x^2$ from the expression: $f'(x) = -x^2(3 + 5x^2)$.
Since $x^2 \ge 0$ for all real $x$ and $(3 + 5x^2) > 0$ for all real $x$,the product $-x^2(3 + 5x^2)$ is always less than or equal to $0$.
Specifically,$f'(x) < 0$ for all $x \neq 0$ and $f'(0) = 0$.
Since the derivative is non-positive for all $x$ and does not vanish on any interval,the function $f(x)$ is strictly decreasing for all real values of $x$.

(A) Let $f(x) = x^x$ 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 increasing,we require $f'(x) > 0$.
Since $x^x > 0$ for all $x > 0$,the condition $f'(x) > 0$ simplifies to $1 + \ln x > 0$.
$\ln x > -1$.
$\ln x > \ln(e^{-1})$.
$x > \frac{1}{e}$.

(C) Let $f(x) = 2{x^3} - 6x + 5$.
To find the intervals where the function is increasing,we calculate the derivative $f'(x)$.
$f'(x) = \frac{d}{dx}(2{x^3} - 6x + 5) = 6{x^2} - 6$.
$A$ function is increasing when $f'(x) > 0$.
So,$6{x^2} - 6 > 0$.
Dividing by $6$,we get ${x^2} - 1 > 0$.
This can be factored as $(x - 1)(x + 1) > 0$.
Using the sign scheme method,the inequality holds when $x > 1$ or $x < -1$.
Thus,the function is increasing for $x \in (-\infty, -1) \cup (1, \infty)$.

(A) Given function is $f(x) = 3\sin x - 4\sin^3 x$.
Using the trigonometric identity $\sin(3x) = 3\sin x - 4\sin^3 x$,we can write $f(x) = \sin(3x)$.
For the function to be increasing,its derivative must be positive: $f'(x) > 0$.
$f'(x) = \frac{d}{dx}(\sin(3x)) = 3\cos(3x)$.
Setting $3\cos(3x) > 0$,we get $\cos(3x) > 0$.
The cosine function is positive in the interval $(-\frac{\pi}{2}, \frac{\pi}{2})$ for its argument.
Thus,$-\frac{\pi}{2} < 3x < \frac{\pi}{2}$.
Dividing by $3$,we get $-\frac{\pi}{6} < x < \frac{\pi}{6}$.
The length of this interval is $\frac{\pi}{6} - (-\frac{\pi}{6}) = \frac{2\pi}{6} = \frac{\pi}{3}$.

(D) Given the function $f(x) = x^3 + bx^2 + cx + d$.
First,we find the derivative of the function:
$f'(x) = 3x^2 + 2bx + c$.
To determine the nature of the function,we examine the discriminant $D$ of the quadratic expression $f'(x)$:
$D = (2b)^2 - 4(3)(c) = 4b^2 - 12c$.
We are given that $0 < b^2 < c$.
Since $b^2 < c$,it follows that $4b^2 < 4c$.
Therefore,$D = 4b^2 - 12c < 4c - 12c = -8c$.
Since $b^2 > 0$,$c$ must also be greater than $0$ (because $c > b^2$).
Thus,$D < -8c < 0$.
Since the discriminant $D < 0$ and the coefficient of $x^2$ (which is $3$) is positive,$f'(x) > 0$ for all $x \in \mathbb{R}$.
Since the derivative $f'(x)$ is strictly positive for all real $x$,the function $f(x)$ is strictly increasing.

(A) Given the function $f(x) = x$ defined on the interval $[-1, 1]$.
To determine the nature of the function,we find its derivative with respect to $x$:
$f'(x) = \frac{d}{dx}(x) = 1$.
Since $f'(x) = 1 > 0$ for all $x \in [-1, 1]$,the function $f(x)$ is strictly increasing on the given interval.
Therefore,the correct option is $A$.

(D) To determine the interval where the function $f(x) = 2x^3 - 3x^2 + 90x + 174$ is increasing,we find its derivative $f'(x)$.
$f'(x) = \frac{d}{dx}(2x^3 - 3x^2 + 90x + 174) = 6x^2 - 6x + 90$.
We can factor out $6$ from the expression: $f'(x) = 6(x^2 - x + 15)$.
For the function to be increasing,we require $f'(x) > 0$,which means $6(x^2 - x + 15) > 0$,or $x^2 - x + 15 > 0$.
We examine the discriminant $D$ of the quadratic $x^2 - x + 15$. Here $a = 1, b = -1, c = 15$.
$D = b^2 - 4ac = (-1)^2 - 4(1)(15) = 1 - 60 = -59$.
Since the discriminant $D < 0$ and the coefficient of $x^2$ is positive $(1 > 0)$,the quadratic $x^2 - x + 15$ is always positive for all real values of $x$.
Therefore,$f'(x) > 0$ for all $x \in (-\infty, \infty)$.
Thus,the function is increasing on the interval $(-\infty, \infty)$.

(C) Given $f(x) = \tan^{-1}(\sin x + \cos x)$.
We can rewrite the expression inside the inverse tangent as:
$f(x) = \tan^{-1}\left(\sqrt{2} \left(\frac{1}{\sqrt{2}}\sin x + \frac{1}{\sqrt{2}}\cos x\right)\right) = \tan^{-1}\left(\sqrt{2} \sin\left(x + \frac{\pi}{4}\right)\right)$.
To find where the function is increasing,we calculate the derivative $f'(x)$:
$f'(x) = \frac{1}{1 + (\sqrt{2} \sin(x + \pi/4))^2} \cdot \sqrt{2} \cos(x + \pi/4)$.
For $f(x)$ to be an increasing function,we require $f'(x) > 0$.
Since the denominator $1 + 2 \sin^2(x + \pi/4)$ is always positive,we only need the numerator to be positive:
$\sqrt{2} \cos(x + \pi/4) > 0$.
This implies $\cos(x + \pi/4) > 0$.
For $x > 0$,the cosine function is positive when the angle is in the first quadrant:
$0 < x + \frac{\pi}{4} < \frac{\pi}{2}$.
Subtracting $\frac{\pi}{4}$ from all sides,we get:
$-\frac{\pi}{4} < x < \frac{\pi}{4}$.
Given the condition $x > 0$,the interval is $(0, \pi/4)$.

(A) Given $f(x) = \frac{e^{2x} - 1}{e^{2x} + 1}$.
To check if the function is increasing or decreasing,we find its derivative $f'(x)$ using the quotient rule $\left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2}$.
Let $u = e^{2x} - 1$ and $v = e^{2x} + 1$. Then $u' = 2e^{2x}$ and $v' = 2e^{2x}$.
$f'(x) = \frac{(2e^{2x})(e^{2x} + 1) - (e^{2x} - 1)(2e^{2x})}{(e^{2x} + 1)^2}$
$f'(x) = \frac{2e^{4x} + 2e^{2x} - (2e^{4x} - 2e^{2x})}{(e^{2x} + 1)^2}$
$f'(x) = \frac{4e^{2x}}{(e^{2x} + 1)^2}$.
Since $e^{2x} > 0$ for all $x \in \mathbb{R}$,$f'(x) > 0$ for all $x \in \mathbb{R}$.
Therefore,$f(x)$ is an increasing function.

(B) Given the function $f(x) = \frac{4x^2 + 1}{x} = 4x + \frac{1}{x}$.
To find the intervals where the function is decreasing,we find the derivative $f'(x)$:
$f'(x) = \frac{d}{dx} \left( 4x + \frac{1}{x} \right) = 4 - \frac{1}{x^2}$.
The function is decreasing when $f'(x) < 0$:
$4 - \frac{1}{x^2} < 0$
$4 < \frac{1}{x^2}$
$x^2 < \frac{1}{4}$
$|x| < \frac{1}{2}$.
This implies $-\frac{1}{2} < x < \frac{1}{2}$.
Since the function is defined for $x \neq 0$,the interval where the function is decreasing is $\left( -\frac{1}{2}, 0 \right) \cup \left( 0, \frac{1}{2} \right)$. Among the given options,the interval $\left[ -\frac{1}{2}, \frac{1}{2} \right]$ is the most appropriate representation of the decreasing behavior,noting that the function is continuous on its domain.

(A) To determine if a function $f(x)$ is increasing on an interval,we check if $f'(x) \ge 0$ for all $x$ in that interval.
$(a)$ For $f(x) = 3x^2 - 2x + 1$,$f'(x) = 6x - 2$. Setting $f'(x) \ge 0$ gives $6x \ge 2$,so $x \ge \frac{1}{3}$. The function is increasing on $[\frac{1}{3}, \infty)$,not $(-\infty, \frac{1}{3}]$. Thus,option $(a)$ is incorrectly matched.
$(b)$ For $f(x) = x^3 + 6x^2 + 6$,$f'(x) = 3x^2 + 12x = 3x(x + 4)$. $f'(x) \ge 0$ when $x \in (-\infty, -4] \cup [0, \infty)$. Thus,it is increasing on $(-\infty, -4]$.
$(c)$ For $f(x) = x^3 - 3x^2 + 3x + 3$,$f'(x) = 3x^2 - 6x + 3 = 3(x - 1)^2$. Since $3(x - 1)^2 \ge 0$ for all $x \in \mathbb{R}$,the function is increasing on $(-\infty, \infty)$.
$(d)$ For $f(x) = 2x^3 - 3x^2 - 12x + 6$,$f'(x) = 6x^2 - 6x - 12 = 6(x^2 - x - 2) = 6(x - 2)(x + 1)$. $f'(x) \ge 0$ when $x \in (-\infty, -1] \cup [2, \infty)$. Thus,it is increasing on $[2, \infty)$.

(B) Let $f(x) = \frac{\ln(\pi + x)}{\ln(e + x)}$.
Using the quotient rule,$f'(x) = \frac{\ln(e + x) \cdot \frac{1}{\pi + x} - \ln(\pi + x) \cdot \frac{1}{e + x}}{(\ln(e + x))^2}$.
$f'(x) = \frac{(e + x)\ln(e + x) - (\pi + x)\ln(\pi + x)}{(\ln(e + x))^2 (e + x)(\pi + x)}$.
Consider the function $g(t) = t \ln(t)$ for $t > 0$. Then $g'(t) = \ln(t) + 1$. For $t > 1/e$,$g'(t) > 0$,so $g(t)$ is an increasing function.
Since $\pi > e$,for $x \ge 0$,we have $\pi + x > e + x > e > 1$. Thus,$g(\pi + x) > g(e + x)$,which implies $(\pi + x)\ln(\pi + x) > (e + x)\ln(e + x)$.
Therefore,the numerator $(e + x)\ln(e + x) - (\pi + x)\ln(\pi + x) < 0$ for all $x \ge 0$.
Since the denominator is always positive,$f'(x) < 0$ for all $x \in [0, \infty)$.
Hence,$f(x)$ is decreasing on $[0, \infty)$.

(B) Given $f(x) = \sin^4 x + \cos^4 x$.
We can rewrite this as $f(x) = (\sin^2 x + \cos^2 x)^2 - 2\sin^2 x \cos^2 x$.
Since $\sin^2 x + \cos^2 x = 1$,we have $f(x) = 1 - 2\sin^2 x \cos^2 x$.
Using the identity $\sin 2x = 2\sin x \cos x$,we get $f(x) = 1 - \frac{1}{2}(2\sin x \cos x)^2 = 1 - \frac{1}{2}\sin^2 2x$.
Using $\sin^2 \theta = \frac{1 - \cos 2\theta}{2}$,we have $f(x) = 1 - \frac{1}{2} \left( \frac{1 - \cos 4x}{2} \right) = 1 - \frac{1}{4} + \frac{1}{4}\cos 4x = \frac{3}{4} + \frac{1}{4}\cos 4x$.
For the function to be increasing,$f'(x) > 0$.
$f'(x) = \frac{d}{dx} \left( \frac{3}{4} + \frac{1}{4}\cos 4x \right) = -\sin 4x$.
Setting $f'(x) > 0$,we get $-\sin 4x > 0$,which implies $\sin 4x < 0$.
The sine function is negative in the interval $(\pi, 2\pi)$.
Thus,$\pi < 4x < 2\pi$.
Dividing by $4$,we get $\frac{\pi}{4} < x < \frac{\pi}{2}$.
Comparing this with the given options,the interval $\frac{\pi}{4} < x < \frac{3\pi}{8}$ is a subset of this range,so the function is increasing in this interval.

(A) Given $h(x) = f(x) - (f(x))^2 + (f(x))^3$.
Differentiating with respect to $x$,we get:
$h'(x) = f'(x) - 2f(x)f'(x) + 3(f(x))^2 f'(x)$
Factoring out $f'(x)$:
$h'(x) = f'(x) [1 - 2f(x) + 3(f(x))^2]$
To analyze the quadratic expression $3(f(x))^2 - 2f(x) + 1$,we complete the square:
$3(f(x))^2 - 2f(x) + 1 = 3 \left( (f(x))^2 - \frac{2}{3}f(x) + \frac{1}{3} \right)$
$= 3 \left( (f(x) - \frac{1}{3})^2 - \frac{1}{9} + \frac{1}{3} \right)$
$= 3 \left( (f(x) - \frac{1}{3})^2 + \frac{2}{9} \right)$
Since $(f(x) - \frac{1}{3})^2 \ge 0$,the expression $(f(x) - \frac{1}{3})^2 + \frac{2}{9}$ is always positive.
Therefore,$h'(x) = 3f'(x) \left( (f(x) - \frac{1}{3})^2 + \frac{2}{9} \right)$.
Since the term in the bracket is always positive,the sign of $h'(x)$ is the same as the sign of $f'(x)$.
Thus,$h$ is increasing whenever $f$ is increasing.

(D) To find where $f(x)$ increases,we calculate the derivative $f'(x)$ using Leibniz's rule:
$f'(x) = \frac{d}{dx} \int_{x^2}^{x^2 + 1} e^{-t^2} dt = e^{-(x^2+1)^2} \cdot \frac{d}{dx}(x^2+1) - e^{-(x^2)^2} \cdot \frac{d}{dx}(x^2)$
$f'(x) = e^{-(x^4 + 2x^2 + 1)} \cdot (2x) - e^{-x^4} \cdot (2x)$
$f'(x) = 2x e^{-x^4} \left( e^{-(2x^2 + 1)} - 1 \right)$
Since $e^{-(2x^2 + 1)} < 1$ for all real $x$,the term $(e^{-(2x^2 + 1)} - 1)$ is always negative.
For $f'(x) > 0$,we need $2x$ to be negative,which implies $x < 0$.
Thus,$f(x)$ increases in $(-\infty, 0)$.

(A) To find the interval where the function $f(x) = x^3 + 5x^2 - 1$ is decreasing,we need to find the interval where its derivative $f'(x) < 0$.
First,find the derivative of $f(x)$:
$f'(x) = \frac{d}{dx}(x^3 + 5x^2 - 1) = 3x^2 + 10x$
Now,set the derivative to be less than zero:
$3x^2 + 10x < 0$
Factor out $x$:
$x(3x + 10) < 0$
The critical points are $x = 0$ and $x = -\frac{10}{3}$.
Using the sign scheme method,we test the intervals:
For $x < -\frac{10}{3}$,$f'(x) > 0$.
For $-\frac{10}{3} < x < 0$,$f'(x) < 0$.
For $x > 0$,$f'(x) > 0$.
Thus,the function is decreasing in the interval $(-\frac{10}{3}, 0)$.

(A) Given the function $f(x) = \sin x$.
To determine the nature of the function,we find its derivative: $f'(x) = \cos x$.
Now,we evaluate the derivative at $x = 2\pi / 3$:
$f'(2\pi / 3) = \cos(2\pi / 3) = -1/2$.
Since $f'(2\pi / 3) < 0$,the function is strictly decreasing in the neighborhood of $x = 2\pi / 3$.
Therefore,the function is monotonically decreasing at this point.

(C) To determine where the function $f(x) = x^9 + 3x^7 + 64$ is strictly increasing,we find its derivative $f'(x)$.
$f'(x) = \frac{d}{dx}(x^9 + 3x^7 + 64) = 9x^8 + 21x^6$.
Since $x^8 \geq 0$ and $x^6 \geq 0$ for all real $x$,it follows that $9x^8 + 21x^6 \geq 0$ for all $x \in \mathbb{R}$.
Specifically,$f'(x) > 0$ for all $x \neq 0$,and $f'(0) = 0$.
Since the derivative is non-negative for all $x$ and does not vanish on any interval,the function $f(x)$ is strictly increasing for all real values of $x$.

(B) Given the function $f(x) = \begin{cases} 0, & x = 0 \\ x - 3, & x > 0 \end{cases}$.
For $x > 0$,the derivative is $f'(x) = \frac{d}{dx}(x - 3) = 1$.
Since $f'(x) = 1 > 0$ for all $x > 0$,the function $f(x)$ is strictly increasing on the interval $(0, \infty)$.
Therefore,$f(x)$ is a strictly increasing function for $x > 0$.

(D) For statement $S$: In the interval $\left( \frac{\pi}{2}, \pi \right)$,the derivative of $\sin x$ is $\cos x$,which is negative. Thus,$\sin x$ is decreasing. The derivative of $\cos x$ is $-\sin x$,which is also negative in this interval. Thus,$\cos x$ is also decreasing. Therefore,$S$ is true.
For statement $R$: Consider the function $f(x) = \frac{1}{x}$ for $x \in (1, 2)$. Here,$f'(x) = -\frac{1}{x^2}$,which is negative,so $f(x)$ is decreasing. However,$f''(x) = \frac{2}{x^3}$,which is positive for $x \in (1, 2)$,meaning $f'(x)$ is an increasing function. Thus,the statement $R$ is false.

(A) Given the function $f(x) = e^{ax} + e^{-ax}$.
To find where the function is increasing,we calculate the derivative $f'(x)$:
$f'(x) = \frac{d}{dx}(e^{ax} + e^{-ax}) = a e^{ax} - a e^{-ax}$.
Factor out $a e^{-ax}$:
$f'(x) = a e^{-ax} (e^{2ax} - 1)$.
For the function to be increasing,we require $f'(x) > 0$.
Since $a > 0$ and $e^{-ax} > 0$ for all real $x$,the sign of $f'(x)$ depends on $(e^{2ax} - 1)$.
We set $e^{2ax} - 1 > 0$,which implies $e^{2ax} > 1$.
Taking the natural logarithm on both sides:
$2ax > \ln(1) = 0$.
Since $a > 0$,we divide by $2a$ to get $x > 0$.
Thus,the function is increasing for $x > 0$.

(B) To find the interval where the function $f(x) = \log x - \frac{2x}{2 + x}$ is increasing,we first find its derivative $f'(x)$.
$f'(x) = \frac{d}{dx}(\log x) - \frac{d}{dx}\left(\frac{2x}{2 + x}\right)$
Using the quotient rule for the second term: $\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2}$
$f'(x) = \frac{1}{x} - \frac{(2)(2 + x) - (2x)(1)}{(2 + x)^2}$
$f'(x) = \frac{1}{x} - \frac{4 + 2x - 2x}{(2 + x)^2}$
$f'(x) = \frac{1}{x} - \frac{4}{(2 + x)^2}$
$f'(x) = \frac{(2 + x)^2 - 4x}{x(2 + x)^2}$
$f'(x) = \frac{4 + x^2 + 4x - 4x}{x(2 + x)^2} = \frac{x^2 + 4}{x(2 + x)^2}$
Since $x^2 + 4 > 0$ and $(2 + x)^2 > 0$ for all $x$ in the domain of $\log x$ (which is $x > 0$),the sign of $f'(x)$ depends only on $x$.
For $f(x)$ to be increasing,we require $f'(x) > 0$.
$\frac{x^2 + 4}{x(2 + x)^2} > 0 \implies x > 0$.
Thus,the function is increasing in the interval $(0, \infty)$.

(A) Given the function $f(x) = 3\sin x - 4\sin^3 x$.
Using the trigonometric identity,we know that $3\sin x - 4\sin^3 x = \sin(3x)$.
So,$f(x) = \sin(3x)$.
For the function to be increasing,its derivative must be positive: $f'(x) > 0$.
$f'(x) = \frac{d}{dx}(\sin(3x)) = 3\cos(3x)$.
We require $3\cos(3x) > 0$,which means $\cos(3x) > 0$.
The cosine function is positive in the interval $(-\pi/2, \pi/2)$.
Therefore,$-\frac{\pi}{2} < 3x < \frac{\pi}{2}$.
Dividing by $3$,we get $-\frac{\pi}{6} < x < \frac{\pi}{6}$.
The length of this interval is $\frac{\pi}{6} - (-\frac{\pi}{6}) = \frac{2\pi}{6} = \frac{\pi}{3}$.

(A) To determine where $f(x) = \sin x$ increases less rapidly than $g(x) = \cos x$,we compare their derivatives.
$f'(x) = \cos x$ and $g'(x) = -\sin x$.
We want to find the interval where $f'(x) < g'(x)$,which means $\cos x < -\sin x$.
This is equivalent to $\sin x + \cos x < 0$.
Dividing by $\sqrt{2}$,we get $\frac{1}{\sqrt{2}} \sin x + \frac{1}{\sqrt{2}} \cos x < 0$,which is $\sin(x + \frac{\pi}{4}) < 0$.
This inequality holds when $\pi < x + \frac{\pi}{4} < 2\pi$ or $-\pi < x + \frac{\pi}{4} < 0$.
Solving for $x$,we get $-\frac{5\pi}{4} < x < -\frac{\pi}{4}$.
Among the given options,the interval $\left( -\frac{\pi}{2}, -\frac{\pi}{4} \right)$ is a subset of this range.
Therefore,in the interval $\left( -\frac{\pi}{2}, -\frac{\pi}{4} \right)$,$f(x)$ increases less rapidly than $g(x)$.

(B) function $f(x)$ is strictly increasing if $f'(x) > 0$.
Given $f(x) = x^3 - 27x + 5$.
Differentiating with respect to $x$,we get $f'(x) = 3x^2 - 27$.
For the function to be strictly increasing,set $f'(x) > 0$:
$3x^2 - 27 > 0$
$3(x^2 - 9) > 0$
$x^2 - 9 > 0$
$x^2 > 9$
Taking the square root on both sides,we get $|x| > 3$.
Thus,the function is strictly increasing when $|x| > 3$.

(D) Given the function $y = x^4$.
Find the derivative with respect to $x$: $\frac{dy}{dx} = 4x^3$.
For the function to be increasing,$\frac{dy}{dx} > 0$,which implies $4x^3 > 0$,so $x > 0$.
Thus,the function is increasing in the interval $(0, \infty)$.
For the function to be decreasing,$\frac{dy}{dx} < 0$,which implies $4x^3 < 0$,so $x < 0$.
Thus,the function is decreasing in the interval $(-\infty, 0)$.

(C) Given $f(x) = \sin x - \cos x$.
To find the interval where the function is strictly increasing,we find the derivative $f'(x)$.
$f'(x) = \cos x + \sin x$.
We can rewrite this as $f'(x) = \sqrt{2} \left( \frac{1}{\sqrt{2}} \cos x + \frac{1}{\sqrt{2}} \sin x \right) = \sqrt{2} \sin(x + \pi /4)$.
The function is strictly increasing when $f'(x) > 0$.
$\sqrt{2} \sin(x + \pi /4) > 0 \implies \sin(x + \pi /4) > 0$.
Since $\sin \theta > 0$ for $\theta \in (0, \pi)$,we have $0 < x + \pi /4 < \pi$.
Subtracting $\pi /4$ from all parts,we get $-\pi /4 < x < 3\pi /4$.
Thus,the function is strictly increasing for $x \in (-\pi /4, 3\pi /4)$.

(D) To find the interval where the function $f(x) = 2x^3 - 15x^2 + 36x + 1$ is strictly decreasing,we find its derivative $f'(x)$.
$f'(x) = \frac{d}{dx}(2x^3 - 15x^2 + 36x + 1) = 6x^2 - 30x + 36$.
For the function to be strictly decreasing,we must have $f'(x) < 0$.
$6x^2 - 30x + 36 < 0$.
Dividing by $6$,we get $x^2 - 5x + 6 < 0$.
Factoring the quadratic expression,we get $(x - 2)(x - 3) < 0$.
The inequality $(x - 2)(x - 3) < 0$ holds when $x$ lies between the roots $2$ and $3$.
Therefore,the function is strictly decreasing in the interval $(2, 3)$.

(A) Given the function $f(x) = \frac{e^{2x} - 1}{e^{2x} + 1}$.
To determine if the function is increasing or decreasing,we find its derivative $f'(x)$ using the quotient rule $\left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2}$.
Let $u = e^{2x} - 1$ and $v = e^{2x} + 1$. Then $u' = 2e^{2x}$ and $v' = 2e^{2x}$.
$f'(x) = \frac{(2e^{2x})(e^{2x} + 1) - (e^{2x} - 1)(2e^{2x})}{(e^{2x} + 1)^2}$
$f'(x) = \frac{2e^{4x} + 2e^{2x} - (2e^{4x} - 2e^{2x})}{(e^{2x} + 1)^2}$
$f'(x) = \frac{4e^{2x}}{(e^{2x} + 1)^2}$.
Since $e^{2x} > 0$ for all $x \in \mathbb{R}$,it follows that $4e^{2x} > 0$ and $(e^{2x} + 1)^2 > 0$.
Therefore,$f'(x) > 0$ for all $x \in \mathbb{R}$.
Since the derivative is strictly positive,the function $f(x)$ is an increasing function.

(A) Given $f(x) = (a + 2)x^3 - 3ax^2 + 9ax - 1$.
For $f(x)$ to be a decreasing function for all $x \in R$,we must have $f'(x) \leq 0$ for all $x \in R$.
$f'(x) = 3(a + 2)x^2 - 6ax + 9a$.
For $f'(x) \leq 0$ for all $x \in R$,the coefficient of $x^2$ must be negative and the discriminant $D$ must be less than or equal to $0$.
$1$) $a + 2 < 0 \implies a < -2$.
$2$) $D = (-6a)^2 - 4(3(a + 2))(9a) \leq 0$.
$36a^2 - 108a(a + 2) \leq 0$.
$36a^2 - 108a^2 - 216a \leq 0$.
$-72a^2 - 216a \leq 0$.
Divide by $-72$ (reversing the inequality sign):
$a^2 + 3a \geq 0$.
$a(a + 3) \geq 0$.
This implies $a \in (-\infty, -3] \cup [0, \infty)$.
Considering both conditions $a < -2$ and $a \in (-\infty, -3] \cup [0, \infty)$,the intersection is $a \in (-\infty, -3]$.

(C) Given $f(x) = Kx^3 + 9x^2 + 9x + 3$.
For $f(x)$ to be an increasing function on $R$,we must have $f'(x) \geq 0$ for all $x \in R$.
First,find the derivative: $f'(x) = 3Kx^2 + 18x + 9$.
We require $3Kx^2 + 18x + 9 \geq 0$ for all $x \in R$.
Dividing by $3$,we get $Kx^2 + 6x + 3 \geq 0$.
For a quadratic $ax^2 + bx + c \geq 0$ to hold for all $x \in R$,we must have $a > 0$ and the discriminant $D = b^2 - 4ac \leq 0$.
Here,$a = K$,$b = 6$,and $c = 3$.
Condition $1$: $K > 0$.
Condition $2$: $D = 6^2 - 4(K)(3) \leq 0$.
$36 - 12K \leq 0$.
$36 \leq 12K$.
$K \geq 3$.
Combining $K > 0$ and $K \geq 3$,we get $K \geq 3$.
Thus,the function is increasing for $K \geq 3$.

(C) Given the function $f(x) = x^2 - x + 1$.
First,find the derivative: $f'(x) = 2x - 1$.
For the function to be monotonic,$f'(x)$ must be either non-negative $(f'(x) \ge 0)$ or non-positive $(f'(x) \le 0)$ throughout the interval.
Setting $f'(x) = 0$,we get $2x - 1 = 0$,which implies $x = 1/2$.
For $x < 1/2$,$f'(x) < 0$ (the function is strictly decreasing).
For $x > 1/2$,$f'(x) > 0$ (the function is strictly increasing).
In any interval that contains $x = 1/2$ in its interior,such as $(0, 1)$,the derivative changes sign. Therefore,the function is not monotonic in any interval containing $1/2$ as an interior point.

(B) Given the function $f(x) = \frac{x}{\ln x}$.
To find the interval where the function is increasing,we calculate the derivative $f'(x)$:
$f'(x) = \frac{(\ln x)(1) - x(\frac{1}{x})}{(\ln x)^2} = \frac{\ln x - 1}{(\ln x)^2}$.
For the function to be increasing,we must have $f'(x) > 0$.
Since $(\ln x)^2 > 0$ for all $x$ in the domain (where $x > 0$ and $x \neq 1$),the condition $f'(x) > 0$ implies:
$\ln x - 1 > 0$
$\ln x > 1$
$x > e$.
Thus,the function is increasing in the interval $(e, \infty)$.

(A) Given function is $f(x) = \frac{e^{2x} - 1}{e^{2x} + 1}$.
To determine if the function is increasing or decreasing,we find its derivative $f'(x)$ using the quotient rule $\left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2}$.
$f'(x) = \frac{(e^{2x} + 1)(2e^{2x}) - (e^{2x} - 1)(2e^{2x})}{(e^{2x} + 1)^2}$
$f'(x) = \frac{2e^{4x} + 2e^{2x} - (2e^{4x} - 2e^{2x})}{(e^{2x} + 1)^2}$
$f'(x) = \frac{4e^{2x}}{(e^{2x} + 1)^2}$
Since $e^{2x} > 0$ for all $x \in \mathbb{R}$,the numerator $4e^{2x}$ is always positive and the denominator $(e^{2x} + 1)^2$ is always positive.
Therefore,$f'(x) > 0$ for all $x \in \mathbb{R}$.
Since the derivative is strictly positive,the function $f(x)$ is a strictly increasing function.

(C) Given the function $f(x) = (x(x - 2))^2 = (x^2 - 2x)^2$.
To find the intervals where the function is increasing,we calculate the derivative $f'(x)$:
$f'(x) = 2(x^2 - 2x)(2x - 2) = 2x(x - 2) \cdot 2(x - 1) = 4x(x - 1)(x - 2)$.
For the function to be increasing,we require $f'(x) > 0$:
$4x(x - 1)(x - 2) > 0$.
We analyze the sign of $f'(x)$ using the intervals determined by the critical points $x = 0, 1, 2$:
$1$. For $x \in (0, 1)$: $f'(x) = 4(+)(-)(-) > 0$.
$2$. For $x \in (1, 2)$: $f'(x) = 4(+)(+)(-) < 0$.
$3$. For $x \in (2, \infty)$: $f'(x) = 4(+)(+)(+) > 0$.
Thus,$f(x)$ is increasing on $(0, 1) \cup (2, \infty)$.

(A) Given the function $f(x) = x - \log x$.
To find the interval where the function is strictly decreasing,we calculate the derivative $f'(x)$.
$f'(x) = \frac{d}{dx}(x - \log x) = 1 - \frac{1}{x}$.
For the function to be strictly decreasing,we must have $f'(x) < 0$.
$1 - \frac{1}{x} < 0$
$\frac{x - 1}{x} < 0$.
Since the domain of $\log x$ is $x > 0$,the denominator $x$ is always positive.
Therefore,for the fraction to be negative,the numerator must be negative: $x - 1 < 0$,which implies $x < 1$.
Combining this with the domain $x > 0$,we get $x \in (0, 1)$.

(D) Given the function $f(x) = x^2 + kx + 1$.
For $f(x)$ to be strictly increasing on $[1, 2]$,its derivative must satisfy $f'(x) > 0$ for all $x \in [1, 2]$.
Calculating the derivative: $f'(x) = 2x + k$.
Setting the condition: $2x + k > 0$ for all $x \in [1, 2]$.
This implies $k > -2x$ for all $x \in [1, 2]$.
To satisfy this for all $x$ in the interval,$k$ must be greater than the maximum value of $-2x$ on $[1, 2]$.
The function $g(x) = -2x$ is a decreasing function,so its maximum value on $[1, 2]$ occurs at $x = 1$.
Thus,$g(1) = -2(1) = -2$.
Therefore,$k > -2$. The infimum (minimum value) of $k$ is $-2$.

(D) Given $f(x) = xe^{x(1-x)}$.
Taking the derivative with respect to $x$:
$f'(x) = e^{x(1-x)} \cdot (1) + x \cdot e^{x(1-x)} \cdot (1-2x)$
$f'(x) = e^{x(1-x)} [1 + x - 2x^2]$
$f'(x) = -e^{x(1-x)} (2x^2 - x - 1)$
$f'(x) = -e^{x(1-x)} (2x+1)(x-1)$
$f'(x) = -2e^{x(1-x)} (x + \frac{1}{2})(x - 1)$.
For $x \in [-\frac{1}{2}, 1]$,the term $(x + \frac{1}{2}) \ge 0$ and $(x - 1) \le 0$.
Thus,the product $(x + \frac{1}{2})(x - 1) \le 0$.
Since $-2e^{x(1-x)}$ is always negative,$f'(x) \le 0$ for all $x \in [-\frac{1}{2}, 1]$.
Therefore,$f(x)$ is a decreasing function on $\left[ -\frac{1}{2}, 1 \right]$.

(D) To find where the function $f(x) = \tan^{-1}(\sin x + \cos x)$ is strictly increasing,we calculate its derivative $f'(x)$.
$f'(x) = \frac{1}{1 + (\sin x + \cos x)^2} \cdot (\cos x - \sin x)$
For the function to be strictly increasing,we must have $f'(x) > 0$.
Since the denominator $1 + (\sin x + \cos x)^2$ is always positive,the condition $f'(x) > 0$ simplifies to:
$\cos x - \sin x > 0$
$\cos x > \sin x$
Dividing by $\cos x$ (assuming $\cos x > 0$),we get $\tan x < 1$.
Alternatively,using the identity $\cos x - \sin x = \sqrt{2} \cos(x + \pi/4)$:
$\sqrt{2} \cos(x + \pi/4) > 0$
This holds when $-\pi/2 < x + \pi/4 < \pi/2$.
Subtracting $\pi/4$ from all sides:
$-\pi/2 - \pi/4 < x < \pi/2 - \pi/4$
$-3\pi/4 < x < \pi/4$
Comparing this with the given options,the interval $0 < x < \pi/4$ is a subset of the interval where the function is strictly increasing.

(D) Given the function $f(x) = 1 - x^3 - x^5$.
To determine where the function is decreasing,we find its derivative $f'(x)$.
$f'(x) = \frac{d}{dx}(1 - x^3 - x^5) = -3x^2 - 5x^4$.
We can factor out $-x^2$:
$f'(x) = -x^2(3 + 5x^2)$.
Since $x^2 \geq 0$ and $(3 + 5x^2) > 0$ for all $x \in \mathbb{R}$,it follows that $f'(x) = -x^2(3 + 5x^2) \leq 0$ for all $x \in \mathbb{R}$.
Since $f'(x) \leq 0$ for all $x \in \mathbb{R}$,the function $f(x)$ is a decreasing function for all $x \in \mathbb{R}$.

(A) Given $f(x) = \sin x - \cos x - ax + b$.
For $f(x)$ to be a decreasing function,its derivative $f'(x)$ must be less than or equal to $0$ for all $x \in R$.
$f'(x) = \cos x + \sin x - a$.
Since $f(x)$ is decreasing,$f'(x) \le 0 \implies \cos x + \sin x - a \le 0$.
This implies $a \ge \sin x + \cos x$ for all $x \in R$.
The maximum value of the expression $(\sin x + \cos x)$ is $\sqrt{1^2 + 1^2} = \sqrt{2}$.
Therefore,for the inequality $a \ge \sin x + \cos x$ to hold for all $x$,$a$ must be greater than or equal to the maximum value of $(\sin x + \cos x)$.
Thus,$a \ge \sqrt{2}$.

(C) Let $y = x^x$. Taking the logarithm on both sides,we get $\log y = x \log x$.
Differentiating with respect to $x$,we have $\frac{1}{y} \frac{dy}{dx} = 1 \cdot \log x + x \cdot \frac{1}{x} = \log x + 1$.
Thus,$\frac{dy}{dx} = x^x(1 + \log x)$.
For the function to be decreasing,we must have $\frac{dy}{dx} < 0$.
Since $x^x > 0$ for $x > 0$,the condition $\frac{dy}{dx} < 0$ implies $1 + \log x < 0$.
This simplifies to $\log x < -1$,which means $\log x < \log(e^{-1})$.
Therefore,$x < 1/e$.
Since the domain of $x^x$ is $x > 0$,the interval in which the function is decreasing is $(0, 1/e)$.

(D) Let us analyze each function:
$1$. For $f(x) = x + |x|$:
If $x \ge 0$,$f(x) = x + x = 2x$,which is increasing.
If $x < 0$,$f(x) = x - x = 0$,which is constant.
Thus,$f(x)$ is monotonically increasing.
$2$. For $f(x) = x - |x|$:
If $x \ge 0$,$f(x) = x - x = 0$,which is constant.
If $x < 0$,$f(x) = x - (-x) = 2x$,which is increasing.
Thus,$f(x)$ is monotonically increasing.
$3$. For $f(x) = x|x|$:
If $x \ge 0$,$f(x) = x^2$,which is increasing for $x \ge 0$.
If $x < 0$,$f(x) = -x^2$,which is increasing for $x < 0$.
Since the derivative $f'(x) = 2|x| \ge 0$ for all $x$,it is monotonically increasing.
Since all three functions are monotonically increasing,the correct option is $D$.

(D) The function is $f(x) = 2x^2 - \log |x|$.
To find the interval of monotonic increase,we find the derivative $f'(x)$.
$f'(x) = \frac{d}{dx}(2x^2 - \log |x|) = 4x - \frac{1}{x}$.
For the function to be monotonically increasing,we require $f'(x) > 0$.
$4x - \frac{1}{x} > 0 \implies \frac{4x^2 - 1}{x} > 0$.
This inequality holds when the numerator and denominator have the same sign.
Case $1$: $x > 0$. Then $4x^2 - 1 > 0 \implies x^2 > 1/4 \implies x > 1/2$.
Case $2$: $x < 0$. Then $4x^2 - 1 < 0 \implies x^2 < 1/4 \implies |x| < 1/2 \implies -1/2 < x < 0$.
Combining these,the function increases in the interval $(-1/2, 0) \cup (1/2, \infty)$.

(D) Given $f(x) = \frac{\lambda \sin x + 6 \cos x}{2 \sin x + 3 \cos x}$.
Using the quotient rule $\left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2}$,we find $f'(x)$:
$f'(x) = \frac{(\lambda \cos x - 6 \sin x)(2 \sin x + 3 \cos x) - (\lambda \sin x + 6 \cos x)(2 \cos x - 3 \sin x)}{(2 \sin x + 3 \cos x)^2}$
Expanding the numerator:
$= \frac{(2\lambda \sin x \cos x + 3\lambda \cos^2 x - 12 \sin^2 x - 18 \sin x \cos x) - (2\lambda \sin x \cos x - 3\lambda \sin^2 x + 12 \cos^2 x - 18 \sin x \cos x)}{(2 \sin x + 3 \cos x)^2}$
$= \frac{3\lambda \cos^2 x - 12 \sin^2 x + 3\lambda \sin^2 x - 12 \cos^2 x}{(2 \sin x + 3 \cos x)^2}$
$= \frac{3\lambda(\cos^2 x + \sin^2 x) - 12(\sin^2 x + \cos^2 x)}{(2 \sin x + 3 \cos x)^2} = \frac{3\lambda - 12}{(2 \sin x + 3 \cos x)^2}$
Since $f(x)$ is strictly increasing,$f'(x) > 0$ for all $x$ in the domain.
Since the denominator $(2 \sin x + 3 \cos x)^2$ is always positive,we must have $3\lambda - 12 > 0$.
$3\lambda > 12 \implies \lambda > 4$.

(A) Given the function $f(x) = \frac{x - 2}{x + 1}$,where $x \neq -1$.
To determine the nature of the function,we find its derivative $f'(x)$ using the quotient rule:
$f'(x) = \frac{(x + 1)(1) - (x - 2)(1)}{(x + 1)^2} = \frac{x + 1 - x + 2}{(x + 1)^2} = \frac{3}{(x + 1)^2}$.
Since $(x + 1)^2 > 0$ for all $x \in R - \{-1\}$,it follows that $f'(x) = \frac{3}{(x + 1)^2} > 0$ for all $x \in R - \{-1\}$.
Since the derivative is strictly positive for all $x$ in the domain,the function $f(x)$ is monotonically increasing.

(A) For a function $f(x)$ to be strictly increasing for all $x \in \mathbb{R}$,its derivative $f'(x)$ must be greater than $0$ for all $x \in \mathbb{R}$.
Given $f(x) = x^3 + 6x^2 + (9 + 2K)x + 1$,we find the derivative:
$f'(x) = 3x^2 + 12x + (9 + 2K)$.
For $f'(x) > 0$ for all $x$,the quadratic expression $ax^2 + bx + c$ must have $a > 0$ and the discriminant $D < 0$.
Here,$a = 3 > 0$,which is satisfied.
The discriminant $D = b^2 - 4ac < 0$:
$D = (12)^2 - 4(3)(9 + 2K) < 0$
$144 - 12(9 + 2K) < 0$
Divide by $12$:
$12 - (9 + 2K) < 0$
$12 - 9 - 2K < 0$
$3 - 2K < 0$
$3 < 2K$
$K > \frac{3}{2}$.

(B) Given the function $f(x) = e^{ax} + e^{-ax}$.
Find the derivative with respect to $x$:
$f'(x) = a e^{ax} - a e^{-ax} = a(e^{ax} - e^{-ax})$.
For the function to be monotonically decreasing,we require $f'(x) < 0$.
Since $a < 0$,the condition $a(e^{ax} - e^{-ax}) < 0$ implies that $(e^{ax} - e^{-ax}) > 0$.
This inequality holds when $e^{ax} > e^{-ax}$.
Taking the natural logarithm on both sides:
$ax > -ax$
$2ax > 0$.
Since $a < 0$,dividing by $2a$ reverses the inequality sign:
$x < 0$.
Therefore,the function is monotonically decreasing for $x < 0$.