Increasing and Decreasing function Questions in English

(B) Given $f(x) = x \cdot e^{x-x^2}$.
To find the intervals of increase or decrease,we compute the derivative $f'(x)$.
Using the product rule and chain rule: $f'(x) = 1 \cdot e^{x-x^2} + x \cdot e^{x-x^2} \cdot (1-2x)$.
$f'(x) = e^{x-x^2} [1 + x(1-2x)] = e^{x-x^2} [1 + x - 2x^2]$.
Factoring the quadratic expression: $1 + x - 2x^2 = -(2x^2 - x - 1) = -(2x+1)(x-1) = (2x+1)(1-x)$.
So,$f'(x) = e^{x-x^2} (2x+1)(1-x)$.
Since $e^{x-x^2} > 0$ for all $x \in R$,the sign of $f'(x)$ depends on $(2x+1)(1-x)$.
$f'(x) > 0$ when $(2x+1)(1-x) > 0$,which implies $-\frac{1}{2} < x < 1$.
Thus,$f(x)$ is increasing in the interval $\left(-\frac{1}{2}, 1\right)$.

(B) Given the function $f(x) = \sin^4 x + \cos^4 x$.
We can rewrite this as $f(x) = (\sin^2 x + \cos^2 x)^2 - 2\sin^2 x \cos^2 x = 1 - \frac{1}{2}(2\sin x \cos x)^2 = 1 - \frac{1}{2}\sin^2(2x)$.
To find where the function increases,we find the derivative $f'(x) = -\frac{1}{2} \cdot 2\sin(2x) \cdot \cos(2x) \cdot 2 = -\sin(4x)$.
The function increases when $f'(x) > 0$,which means $-\sin(4x) > 0$,or $\sin(4x) < 0$.
This occurs when $\pi < 4x < 2\pi$,which simplifies to $\frac{\pi}{4} < x < \frac{\pi}{2}$.

(A) Given function is $f(x) = \frac{x}{2} + \frac{2}{x}$.
To find the intervals where the function is strictly decreasing,we find the derivative $f'(x)$.
$f'(x) = \frac{d}{dx} (\frac{x}{2} + 2x^{-1}) = \frac{1}{2} - 2x^{-2} = \frac{1}{2} - \frac{2}{x^2}$.
For the function to be strictly decreasing,we require $f'(x) < 0$.
$\frac{1}{2} - \frac{2}{x^2} < 0 \implies \frac{x^2 - 4}{2x^2} < 0$.
Since $2x^2 > 0$ for all $x \neq 0$,the inequality holds when $x^2 - 4 < 0$.
$(x - 2)(x + 2) < 0$.
This inequality holds for $x \in (-2, 2)$.
Since $x \neq 0$,the function is strictly decreasing in the interval $(-2, 0) \cup (0, 2)$.

(A) To determine where $f(x)$ is increasing,we find its derivative $f'(x)$.
Given $f(x) = \log(1+x) - \frac{2x}{2+x}$.
The domain of $f(x)$ is $x > -1$.
$f'(x) = \frac{d}{dx} [\log(1+x)] - \frac{d}{dx} [\frac{2x}{2+x}]$.
Using the quotient rule for the second term: $\frac{d}{dx} [\frac{2x}{2+x}] = \frac{(2+x)(2) - 2x(1)}{(2+x)^2} = \frac{4+2x-2x}{(2+x)^2} = \frac{4}{(2+x)^2}$.
So,$f'(x) = \frac{1}{1+x} - \frac{4}{(2+x)^2}$.
$f'(x) = \frac{(2+x)^2 - 4(1+x)}{(1+x)(2+x)^2} = \frac{4+4x+x^2-4-4x}{(1+x)(2+x)^2} = \frac{x^2}{(1+x)(2+x)^2}$.
For $f(x)$ to be increasing,we require $f'(x) > 0$.
Since $x^2 \ge 0$ and $(2+x)^2 > 0$ for all $x$ in the domain,$f'(x) > 0$ whenever $1+x > 0$,which means $x > -1$.
Thus,$f(x)$ is increasing in $(-1, \infty)$.

(D) Given the function $f(x) = [x(x-2)]^2 = (x^2 - 2x)^2$.
To find the intervals where the function is increasing,we find the derivative $f'(x)$.
$f'(x) = 2(x^2 - 2x)(2x - 2) = 4x(x-2)(x-1)$.
For the function to be increasing,we set $f'(x) > 0$.
$4x(x-1)(x-2) > 0$.
Using the wavy curve method (sign scheme) for the critical points $x = 0, 1, 2$:
- For $x \in (0, 1)$,$f'(x) < 0$.
- For $x \in (1, 2)$,$f'(x) > 0$.
- For $x \in (2, \infty)$,$f'(x) > 0$.
Wait,let's re-evaluate the sign scheme:
If $x > 2$,all factors $(x), (x-1), (x-2)$ are positive,so $f'(x) > 0$.
If $1 < x < 2$,$(x)$ is positive,$(x-1)$ is positive,$(x-2)$ is negative,so $f'(x) < 0$.
If $0 < x < 1$,$(x)$ is positive,$(x-1)$ is negative,$(x-2)$ is negative,so $f'(x) > 0$.
If $x < 0$,all factors are negative,so $f'(x) < 0$.
Thus,$f'(x) > 0$ in $(0, 1) \cup (2, \infty)$.
Therefore,the function is increasing in $(0, 1) \cup (2, \infty)$.

(A) Given $f(x) = x^3 + bx^2 + cx + d$.
Taking the derivative with respect to $x$,we get $f'(x) = 3x^2 + 2bx + c$.
For $f(x)$ to be strictly increasing,we need $f'(x) > 0$ for all $x \in \mathbb{R}$.
The discriminant of the quadratic $f'(x)$ is $D = (2b)^2 - 4(3)(c) = 4b^2 - 12c$.
Since $b^2 < c$,we have $4b^2 < 4c$.
Thus,$D = 4b^2 - 12c < 4c - 12c = -8c$.
Since $0 < b^2 < c$,it implies $c > 0$. Therefore,$D < 0$.
Since the leading coefficient of $f'(x)$ is $3 > 0$ and $D < 0$,$f'(x) > 0$ for all $x \in \mathbb{R}$.
Hence,$f(x)$ is a strictly increasing function on $(-\infty, \infty)$.

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

(A) Given the function $f(x) = \frac{\log x}{x}$.
To determine where the function is increasing,we find its derivative $f'(x)$ using the quotient rule:
$f'(x) = \frac{x \cdot \frac{d}{dx}(\log x) - \log x \cdot \frac{d}{dx}(x)}{x^2} = \frac{x \cdot \frac{1}{x} - \log x \cdot 1}{x^2} = \frac{1 - \log x}{x^2}$.
For the function to be increasing,we require $f'(x) > 0$.
Since $x^2 > 0$ for all $x > 0$,the condition $f'(x) > 0$ implies $1 - \log x > 0$.
This simplifies to $1 > \log x$,which means $\log x < 1$.
Taking the exponential of both sides,we get $x < e^1$,or $x < e$.
Given the domain $x > 0$,the function is increasing in the interval $(0, e)$.

(C) Given the function $f(x) = 2x^3 - 6x + 5$.
To find the intervals where the function is increasing,we first find the derivative $f'(x)$.
$f'(x) = \frac{d}{dx}(2x^3 - 6x + 5) = 6x^2 - 6$.
$A$ function is increasing when $f'(x) > 0$.
So,$6x^2 - 6 > 0$.
Dividing by $6$,we get $x^2 - 1 > 0$,which factors as $(x - 1)(x + 1) > 0$.
Using the sign scheme for the inequality $(x - 1)(x + 1) > 0$,the expression is positive when $x > 1$ or $x < -1$.
Thus,the function is increasing for $x \in (-\infty, -1) \cup (1, \infty)$.

(B) Given function: $f(x) = x^3 - 6x^2 + 9x + 3$.
Find the derivative: $f'(x) = 3x^2 - 12x + 9$.
For a function to be monotonically decreasing,the condition is $f'(x) < 0$.
So,$3x^2 - 12x + 9 < 0$.
Divide by $3$: $x^2 - 4x + 3 < 0$.
Factorize the quadratic expression: $(x - 3)(x - 1) < 0$.
Using the sign scheme for the inequality,the expression is negative between the roots $1$ and $3$.
Therefore,$x \in (1, 3)$.

(C) Given $f(x) = \frac{x}{\sqrt{a^2+x^2}} - \frac{d-x}{\sqrt{b^2+(d-x)^2}}$.
To determine if $f(x)$ is increasing or decreasing,we find its derivative $f'(x)$.
Using the quotient rule $\frac{d}{dx}(\frac{u}{v}) = \frac{u'v - uv'}{v^2}$,where $\frac{d}{dx}(\frac{x}{\sqrt{a^2+x^2}}) = \frac{1 \cdot \sqrt{a^2+x^2} - x \cdot \frac{x}{\sqrt{a^2+x^2}}}{a^2+x^2} = \frac{a^2+x^2-x^2}{(a^2+x^2)^{3/2}} = \frac{a^2}{(a^2+x^2)^{3/2}}$.
Similarly,for the second term,let $u = d-x$,then $\frac{d}{dx}(-\frac{u}{\sqrt{b^2+u^2}}) = -\frac{d}{du}(\frac{u}{\sqrt{b^2+u^2}}) \cdot \frac{du}{dx} = -\frac{b^2}{(b^2+u^2)^{3/2}} \cdot (-1) = \frac{b^2}{(b^2+(d-x)^2)^{3/2}}$.
Thus,$f'(x) = \frac{a^2}{(a^2+x^2)^{3/2}} + \frac{b^2}{(b^2+(d-x)^2)^{3/2}}$.
Since $a^2 > 0$ and $b^2 > 0$,the terms $(a^2+x^2)^{3/2}$ and $(b^2+(d-x)^2)^{3/2}$ are always positive.
Therefore,$f'(x) > 0$ for all $x \in R$.
Since $f'(x) > 0$,$f(x)$ is an increasing function of $x$.

(C) Given function is $f(x) = x^3 - 10x^2 + 200x - 10$.
To determine the intervals of increase or decrease,we find the derivative $f'(x)$.
$f'(x) = \frac{d}{dx}(x^3 - 10x^2 + 200x - 10) = 3x^2 - 20x + 200$.
To check the sign of $f'(x)$,we analyze the quadratic expression $3x^2 - 20x + 200$.
The discriminant $D = b^2 - 4ac = (-20)^2 - 4(3)(200) = 400 - 2400 = -2000$.
Since $D < 0$ and the coefficient of $x^2$ (which is $3$) is positive,the quadratic $3x^2 - 20x + 200$ is always positive for all real values of $x$.
Since $f'(x) > 0$ for all $x \in \mathbb{R}$,the function $f(x)$ is strictly increasing throughout the real line.

(A) Given function: $f(x) = 2x^3 - 9x^2 + 12x + 2$.
Find the derivative: $f'(x) = 6x^2 - 18x + 12$.
For the function to be decreasing,we must have $f'(x) < 0$.
$6x^2 - 18x + 12 < 0$.
Divide by $6$: $x^2 - 3x + 2 < 0$.
Factor the quadratic expression: $(x - 1)(x - 2) < 0$.
Using the sign scheme for the inequality,the expression is negative between the roots $x = 1$ and $x = 2$.
Thus,the function is decreasing in the interval $(1, 2)$.

(A) Given $f(x) = x^2 e^{-x}$.
To find the interval where $f(x)$ is strictly increasing,we calculate the derivative $f'(x)$:
$f'(x) = \frac{d}{dx}(x^2) e^{-x} + x^2 \frac{d}{dx}(e^{-x}) = 2x e^{-x} - x^2 e^{-x} = x e^{-x}(2 - x)$.
For $f(x)$ to be strictly increasing,we must have $f'(x) > 0$.
Since $e^{-x} > 0$ for all real $x$,the inequality $x e^{-x}(2 - x) > 0$ simplifies to $x(2 - x) > 0$.
Multiplying by $-1$ reverses the inequality: $x(x - 2) < 0$.
This inequality holds when $x$ lies between the roots $0$ and $2$.
Thus,$x \in (0, 2)$.

(B) Given $f(x) = \frac{a \sin x + b \cos x}{c \sin x + d \cos x}$.
For $f(x)$ to be decreasing,we must have $f'(x) < 0$.
Using the quotient rule,$f'(x) = \frac{(c \sin x + d \cos x)(a \cos x - b \sin x) - (a \sin x + b \cos x)(c \cos x - d \sin x)}{(c \sin x + d \cos x)^2}$.
Expanding the numerator:
$= (ac \sin x \cos x - bc \sin^2 x + ad \cos^2 x - bd \sin x \cos x) - (ac \sin x \cos x - ad \sin^2 x + bc \cos^2 x - bd \sin x \cos x)$.
$= ac \sin x \cos x - bc \sin^2 x + ad \cos^2 x - bd \sin x \cos x - ac \sin x \cos x + ad \sin^2 x - bc \cos^2 x + bd \sin x \cos x$.
$= ad(\sin^2 x + \cos^2 x) - bc(\sin^2 x + \cos^2 x)$.
$= ad - bc$.
Since the denominator $(c \sin x + d \cos x)^2$ is always positive,for $f'(x) < 0$,we require $ad - bc < 0$.

(A) Given $f(x) = x e^{x(1-x)}$.
Applying the product rule,$f'(x) = x \cdot e^{x(1-x)} \cdot \frac{d}{dx}(x-x^2) + e^{x(1-x)} \cdot 1$.
$f'(x) = x e^{x(1-x)}(1-2x) + e^{x(1-x)}$.
$f'(x) = e^{x(1-x)} [x(1-2x) + 1] = e^{x(1-x)} (x - 2x^2 + 1)$.
$f'(x) = e^{x(1-x)} (-2x^2 + x + 1) = -e^{x(1-x)} (2x^2 - x - 1)$.
$f'(x) = -e^{x(1-x)} (2x+1)(x-1) = e^{x(1-x)} (2x+1)(1-x)$.
For $f(x)$ to be increasing,$f'(x) \geq 0$.
Since $e^{x(1-x)} > 0$ for all $x \in R$,we need $(2x+1)(1-x) \geq 0$.
This inequality holds when $x \in \left[-\frac{1}{2}, 1\right]$.
Thus,$f(x)$ is increasing in $\left[-\frac{1}{2}, 1\right]$.

(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 $f'(x) = -\frac{1}{2} \cdot 2\sin 2x \cdot \cos 2x \cdot 2 = -2\sin 2x \cos 2x = -\sin 4x$.
For $f(x)$ to be increasing,$f'(x) > 0$,so $-\sin 4x > 0$,which implies $\sin 4x < 0$.
The sine function is negative in the third and fourth quadrants,so $\pi < 4x < 2\pi$.
Dividing by $4$,we get $\frac{\pi}{4} < x < \frac{\pi}{2}$.

(B) Given $f(x) = x^3 + bx^2 + cx + d$.
Taking the derivative with respect to $x$,we get $f'(x) = 3x^2 + 2bx + c$.
For $f(x)$ to be strictly increasing,we need $f'(x) > 0$ for all $x \in \mathbb{R}$.
The quadratic expression $3x^2 + 2bx + c$ is always positive if its discriminant $D < 0$ and the coefficient of $x^2$ is positive.
Here,$D = (2b)^2 - 4(3)(c) = 4b^2 - 12c = 4(b^2 - 3c)$.
Since $0 < b^2 < c$,it follows that $b^2 - 3c < c - 3c = -2c < 0$ (assuming $c > 0$).
More directly,since $b^2 < c$,then $3b^2 < 3c$,so $b^2 - 3c < -2b^2 < 0$.
Thus,$D < 0$,which implies $f'(x) > 0$ for all $x \in \mathbb{R}$.
Therefore,$f(x)$ is a strictly increasing function.

(C) Given $f(x) = \int \frac{x^2-3x+2}{x^4+1} \, dx$.
By the Fundamental Theorem of Calculus,the derivative is $f'(x) = \frac{x^2-3x+2}{x^4+1}$.
For the function $f(x)$ to be decreasing,we must have $f'(x) < 0$.
Since $x^4+1 > 0$ for all real $x$,the condition $f'(x) < 0$ is equivalent to $x^2-3x+2 < 0$.
Factoring the quadratic expression,we get $(x-1)(x-2) < 0$.
The roots of the quadratic are $x=1$ and $x=2$.
Testing the intervals,the expression $(x-1)(x-2)$ is negative when $x \in (1, 2)$.
Therefore,the function decreases in the interval $(1, 2)$.

(A) Given the function $f(x) = (x+2)e^{-x}$.
To determine the intervals of increase and decrease,we find the derivative $f'(x)$ using the product rule:
$f'(x) = \frac{d}{dx}(x+2) \cdot e^{-x} + (x+2) \cdot \frac{d}{dx}(e^{-x})$
$f'(x) = 1 \cdot e^{-x} + (x+2) \cdot (-e^{-x})$
$f'(x) = e^{-x}(1 - (x+2))$
$f'(x) = e^{-x}(1 - x - 2)$
$f'(x) = -e^{-x}(x+1)$
Now,we analyze the sign of $f'(x)$:
Since $e^{-x} > 0$ for all real $x$,the sign of $f'(x)$ depends on $-(x+1)$.
$1$. If $x < -1$,then $(x+1) < 0$,so $-(x+1) > 0$. Thus,$f'(x) > 0$,and the function is monotonically increasing in $(-\infty, -1)$.
$2$. If $x > -1$,then $(x+1) > 0$,so $-(x+1) < 0$. Thus,$f'(x) < 0$,and the function is monotonically decreasing in $(-1, \infty)$.
Therefore,the function is increasing in $(-\infty, -1)$ and decreasing in $(-1, \infty)$.

(D) Given $f(x) = \frac{\log(\pi + x)}{\log(e + x)}$.
Applying the quotient rule,$f'(x) = \frac{\frac{1}{\pi + x} \log(e + x) - \log(\pi + x) \cdot \frac{1}{e + x}}{\{\log(e + x)\}^2}$.
$f'(x) = \frac{(e + x) \log(e + x) - (\pi + x) \log(\pi + x)}{(\pi + x)(e + x) \{\log(e + x)\}^2}$.
Let $g(x) = (e + x) \log(e + x) - (\pi + x) \log(\pi + x)$.
Then $g'(x) = \log(e + x) + 1 - (\log(\pi + x) + 1) = \log(e + x) - \log(\pi + x)$.
Since $\pi > e$,for $x > 0$,$\pi + x > e + x$,so $\log(\pi + x) > \log(e + x)$.
Thus,$g'(x) < 0$,which means $g(x)$ is a decreasing function.
Since $g(0) = e \log e - \pi \log \pi = e - \pi \log \pi < 0$ (as $e < \pi \log \pi$),and $g(x)$ is decreasing,$g(x) < 0$ for all $x > 0$.
Therefore,$f'(x) < 0$ for all $x \in (0, \infty)$,implying $f(x)$ is decreasing on $(0, \infty)$.

(A) Given function: $f(x)=2 x^3-9 x^2+12 x+29$
To find the intervals where the function is monotonically increasing,we find the derivative $f^{\prime}(x)$:
$f^{\prime}(x) = \frac{d}{dx}(2 x^3-9 x^2+12 x+29) = 6 x^2-18 x+12$
Factor the derivative:
$f^{\prime}(x) = 6(x^2-3 x+2) = 6(x-1)(x-2)$
For the function to be monotonically increasing,we require $f^{\prime}(x) > 0$:
$6(x-1)(x-2) > 0$
Using the sign scheme method on the number line with critical points $x=1$ and $x=2$:
- For $x < 1$,$f^{\prime}(x) > 0$ (positive)
- For $1 < x < 2$,$f^{\prime}(x) < 0$ (negative)
- For $x > 2$,$f^{\prime}(x) > 0$ (positive)
Thus,$f(x)$ is monotonically increasing in the interval $(-\infty, 1) \cup(2, \infty)$.

(B) Given $f(x) = \frac{x}{\log x}$.
To find the interval where $f(x)$ is increasing,we calculate the derivative $f'(x)$:
$f'(x) = \frac{(\log x)(1) - (x)(\frac{1}{x})}{(\log x)^2} = \frac{\log x - 1}{(\log x)^2}$.
For $f(x)$ to be increasing,we must have $f'(x) > 0$.
Since $(\log x)^2 > 0$ for all $x > 0$ (and $x \neq 1$),the sign of $f'(x)$ depends on the numerator $\log x - 1$.
$\log x - 1 > 0 \Rightarrow \log x > 1 \Rightarrow x > e$.
Thus,$f(x)$ is increasing in the interval $(e, \infty)$.
Comparing this with the given options,the function is increasing in $(e, \infty)$.

(C) Given function is $f(x) = \frac{1}{8^x} = 8^{-x}$.
To determine the nature of the function,we find its derivative with respect to $x$:
$f'(x) = \frac{d}{dx}(8^{-x}) = 8^{-x} \cdot \ln(8) \cdot (-1)$.
$f'(x) = -\frac{\ln(8)}{8^x}$.
Since $8^x > 0$ for all $x \in [1, 3]$ and $\ln(8) > 0$,the expression $f'(x) = -\frac{\ln(8)}{8^x}$ is always negative for all $x \in [1, 3]$.
Since $f'(x) < 0$ for all $x \in [1, 3]$,the function $f(x)$ is a decreasing function on the interval $[1, 3]$.

(A) Given function is $f(x) = \cot^{-1} x + x$.
To find the interval where the function is increasing,we calculate the derivative $f'(x)$.
$f'(x) = \frac{d}{dx}(\cot^{-1} x) + \frac{d}{dx}(x) = -\frac{1}{1+x^2} + 1$.
Simplifying the expression: $f'(x) = \frac{-1 + (1+x^2)}{1+x^2} = \frac{x^2}{1+x^2}$.
Since $x^2 \geq 0$ for all $x \in \mathbb{R}$ and $1+x^2 > 0$,it follows that $f'(x) \geq 0$ for all $x \in \mathbb{R}$.
Since $f'(x) \geq 0$ for all $x \in \mathbb{R}$,the function $f(x)$ is strictly increasing on the entire real line $(-\infty, \infty)$.

(C) Given the function $f(x) = \frac{\lambda \sin x + 6 \cos x}{2 \sin x + 3 \cos x}$.
For the function to be increasing,we must have $f'(x) \geq 0$.
Using the quotient rule $\left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2}$,we calculate $f'(x)$:
$f'(x) = \frac{(\lambda \cos x - 6 \sin x)(2 \sin x + 3 \cos x) - (\lambda \sin x + 6 \cos x)(2 \cos x - 3 \sin x)}{(2 \sin x + 3 \cos x)^2}$.
Expanding the numerator:
Numerator $= (2\lambda \sin x \cos x + 3\lambda \cos^2 x - 12 \sin^2 x - 18 \sin x \cos x) - (2\lambda \sin x \cos x - 3\lambda \sin^2 x + 12 \cos^2 x - 18 \sin x \cos x)$.
Simplifying the numerator:
Numerator $= 3\lambda \cos^2 x - 12 \sin^2 x + 3\lambda \sin^2 x - 12 \cos^2 x$.
Numerator $= 3\lambda(\sin^2 x + \cos^2 x) - 12(\sin^2 x + \cos^2 x) = 3\lambda - 12$.
Since the denominator $(2 \sin x + 3 \cos x)^2$ is always positive,$f'(x) \geq 0$ implies $3\lambda - 12 \geq 0$.
Therefore,$3\lambda \geq 12$,which gives $\lambda \geq 4$.

(C) Given the function $f(x) = \log |\sin x|$ for $x \in (0, \pi)$.
Since $x \in (0, \pi)$,$\sin x$ is always positive,so we can write $f(x) = \log(\sin x)$.
To find the intervals where the function is strictly increasing,we calculate the derivative $f'(x)$:
$f'(x) = \frac{1}{\sin x} \cdot \cos x = \cot x$.
For the function to be strictly increasing,we must have $f'(x) > 0$.
Therefore,$\cot x > 0$.
In the interval $(0, \pi)$,$\cot x$ is positive when $x \in \left(0, \frac{\pi}{2}\right)$.
Thus,the function is strictly increasing on $\left(0, \frac{\pi}{2}\right)$.

(D) Given the function $f(x) = e^{-1/x}$.
To find the intervals where the function is strictly increasing,we calculate the derivative $f'(x)$:
$f'(x) = \frac{d}{dx}(e^{-1/x}) = e^{-1/x} \cdot \frac{d}{dx}(-x^{-1}) = e^{-1/x} \cdot (x^{-2}) = \frac{1}{x^2 e^{1/x}}$.
For the function to be strictly increasing,we must have $f'(x) > 0$.
Since $x^2 > 0$ for all $x \neq 0$ and $e^{1/x} > 0$ for all $x \neq 0$,the derivative $f'(x) = \frac{1}{x^2 e^{1/x}}$ is always positive for all $x$ in its domain.
The domain of the function $f(x) = e^{-1/x}$ is all real numbers except $x = 0$.
Therefore,the function is strictly increasing for all $x \in \mathbb{R} \setminus \{0\}$.

(D) Given $f(x) = \log x - \frac{2x}{x+2}$.
The domain of the function is $x > 0$.
Differentiating with respect to $x$:
$f'(x) = \frac{1}{x} - \frac{(x+2)(2) - 2x(1)}{(x+2)^2}$
$f'(x) = \frac{1}{x} - \frac{4}{(x+2)^2}$
$f'(x) = \frac{(x+2)^2 - 4x}{x(x+2)^2} = \frac{x^2 + 4x + 4 - 4x}{x(x+2)^2} = \frac{x^2 + 4}{x(x+2)^2}$
Since $x^2 + 4 > 0$ and $(x+2)^2 > 0$ for all $x$ in the domain,$f'(x) > 0$ for all $x > 0$.
Therefore,the function is strictly increasing for $x \in (0, \infty)$.

(A) Given the function $f(x) = \frac{1}{a^x} = a^{-x}$,where $a > 0$.
To determine the nature of the function,we find its derivative with respect to $x$:
$f'(x) = \frac{d}{dx}(a^{-x}) = -a^{-x} \cdot \ln(a)$.
Since $a^x > 0$ for all $x$ and $a > 0$,the sign of $f'(x)$ depends on the value of $a$:
$1$. If $a > 1$,then $\ln(a) > 0$,which implies $f'(x) < 0$,so the function is strictly decreasing.
$2$. If $0 < a < 1$,then $\ln(a) < 0$,which implies $f'(x) > 0$,so the function is strictly increasing.
$3$. If $a = 1$,then $f(x) = 1$,which is a constant function.
Since the question asks for the nature of the function for every value of $x$ and $a > 0$ (without specifying $a > 1$),the function is not strictly increasing or decreasing for all $a$. However,in standard textbook contexts where $a$ is assumed to be a base such that $a > 1$,it is decreasing. Given the options,if $a > 1$,the function is decreasing.

(C) Given function: $f(x)=3x^{4}+16x^{3}-30x^{2}+10$
Find the derivative: $f'(x)=12x^{3}+48x^{2}-60x$
For the function to be increasing,we require $f'(x) > 0$:
$12x^{3}+48x^{2}-60x > 0$
Factor out $12x$:
$12x(x^{2}+4x-5) > 0$
Factor the quadratic expression:
$12x(x+5)(x-1) > 0$
To find the intervals,we identify the critical points: $x = -5, 0, 1$.
Testing the intervals:
For $x \in (-\infty, -5)$,$f'(x) < 0$.
For $x \in (-5, 0)$,$f'(x) > 0$.
For $x \in (0, 1)$,$f'(x) < 0$.
For $x \in (1, \infty)$,$f'(x) > 0$.
Thus,the function is increasing for $x \in (-5, 0) \cup (1, \infty)$.

(A) Given the function $f(x) = x^3 - 3x$.
First,we find the derivative of the function with respect to $x$:
$f'(x) = \frac{d}{dx}(x^3 - 3x) = 3x^2 - 3$.
Factoring the derivative,we get:
$f'(x) = 3(x^2 - 1) = 3(x - 1)(x + 1)$.
To determine the intervals of increase and decrease,we find the critical points by setting $f'(x) = 0$:
$3(x - 1)(x + 1) = 0 \implies x = 1, x = -1$.
These points divide the real number line into three intervals: $(-\infty, -1)$,$(-1, 1)$,and $(1, \infty)$.
$1$. For $x \in (-\infty, -1)$,choose $x = -2$: $f'(-2) = 3((-2)^2 - 1) = 3(4 - 1) = 9 > 0$. Thus,$f(x)$ is increasing.
$2$. For $x \in (-1, 1)$,choose $x = 0$: $f'(0) = 3(0^2 - 1) = -3 < 0$. Thus,$f(x)$ is decreasing.
$3$. For $x \in (1, \infty)$,choose $x = 2$: $f'(2) = 3(2^2 - 1) = 3(3) = 9 > 0$. Thus,$f(x)$ is increasing.
Therefore,$f(x)$ is increasing in $(-\infty, -1) \cup (1, \infty)$ and decreasing in $(-1, 1)$.

(D) Given $f(x) = \frac{x}{x^2+1}$.
To find the interval where $f(x)$ is increasing,we find its derivative $f'(x)$ using the quotient rule:
$f'(x) = \frac{(x^2+1)(1) - x(2x)}{(x^2+1)^2} = \frac{x^2+1-2x^2}{(x^2+1)^2} = \frac{1-x^2}{(x^2+1)^2}$.
For $f(x)$ to be an increasing function,we must have $f'(x) > 0$.
Since $(x^2+1)^2$ is always positive for all real $x$,the condition $f'(x) > 0$ implies $1-x^2 > 0$.
This simplifies to $x^2 - 1 < 0$,which is $(x-1)(x+1) < 0$.
Solving this inequality,we get $x \in (-1, 1)$.

(A) function $f(x)$ is monotonically increasing if $f'(x) \geq 0$ for all $x \in R$.
Given $f(x) = kx - \sin x$.
Differentiating with respect to $x$,we get $f'(x) = k - \cos x$.
For $f(x)$ to be monotonically increasing,$f'(x) \geq 0$ for all $x \in R$.
$k - \cos x \geq 0 \implies k \geq \cos x$.
Since the maximum value of $\cos x$ is $1$,for $k \geq \cos x$ to hold for all $x$,$k$ must be greater than or equal to the maximum value of $\cos x$.
Therefore,$k \geq 1$.

(A) Given,$f(x) = \log(1+x) - \frac{2x}{2+x}$.
First,we find the derivative $f'(x)$ with respect to $x$:
$f'(x) = \frac{1}{1+x} - \frac{(2+x)(2) - 2x(1)}{(2+x)^2}$
$f'(x) = \frac{1}{1+x} - \frac{4 + 2x - 2x}{(2+x)^2}$
$f'(x) = \frac{1}{1+x} - \frac{4}{(2+x)^2}$
$f'(x) = \frac{(2+x)^2 - 4(1+x)}{(1+x)(2+x)^2}$
$f'(x) = \frac{4 + 4x + x^2 - 4 - 4x}{(1+x)(2+x)^2} = \frac{x^2}{(1+x)(2+x)^2}$.
For the function to be increasing,we require $f'(x) > 0$.
Since $x^2 \geq 0$ and $(2+x)^2 > 0$ for $x > -1$ (domain of $\log(1+x)$),the sign of $f'(x)$ depends on $(1+x)$.
Thus,$f'(x) > 0$ when $1+x > 0$,which means $x > -1$.
However,checking the options provided,$f'(x) > 0$ for all $x > 0$ is clearly satisfied.
Therefore,the function is increasing on $(0, \infty)$.

(A) Given function: $f(x) = x^3 + 6x^2 - 36x + 7$
Find the derivative: $f'(x) = 3x^2 + 12x - 36$
Factor the derivative: $f'(x) = 3(x^2 + 4x - 12) = 3(x + 6)(x - 2)$
For the function to be increasing,we require $f'(x) > 0$:
$3(x + 6)(x - 2) > 0$
$(x + 6)(x - 2) > 0$
Using the sign scheme for the quadratic inequality,the expression is positive when $x < -6$ or $x > 2$.
Thus,the range of values is $x \in (-\infty, -6) \cup (2, \infty)$.

(C) Given equation is $f(x) = x^3+x-1 = 0$.
To determine the number of real roots,we analyze the derivative of the function $f(x)$.
The derivative is $f'(x) = 3x^2 + 1$.
Since $x^2 \geq 0$ for all $x \in \mathbb{R}$,it follows that $3x^2 + 1 \geq 1 > 0$ for all $x \in \mathbb{R}$.
Because $f'(x) > 0$ for all $x$,the function $f(x)$ is strictly increasing on the entire real line.
$A$ strictly increasing function can cross the $x$-axis at most once.
We observe the values of the function at specific points:
$f(0) = 0^3 + 0 - 1 = -1$
$f(1) = 1^3 + 1 - 1 = 1$
Since $f(0) < 0$ and $f(1) > 0$,by the Intermediate Value Theorem,there must exist at least one real root $x \in (0, 1)$ such that $f(x) = 0$.
Since the function is strictly increasing,this root is unique.
Therefore,the equation has exactly one real root.

(B) Given the function $f(x)=2x^3-15x^2-144x-7$.
To find the intervals where $f(x)$ is strictly decreasing,we find the derivative $f'(x)$:
$f'(x) = \frac{d}{dx}(2x^3-15x^2-144x-7) = 6x^2-30x-144$.
For $f(x)$ to be strictly decreasing,$f'(x) < 0$:
$6x^2-30x-144 < 0$.
Dividing by $6$,we get $x^2-5x-24 < 0$.
Factoring the quadratic expression: $(x-8)(x+3) < 0$.
The roots are $x=8$ and $x=-3$.
The inequality holds for $x$ in the interval $(-3, 8)$.
Thus,$f(x)$ is strictly decreasing in $(-3, 8)$.

(C) Given $f(x) = (x + 2) e^{-x}$.
To determine the intervals of increase and decrease,we find the derivative $f'(x)$:
$f'(x) = (x + 2) \frac{d}{dx}(e^{-x}) + e^{-x} \frac{d}{dx}(x + 2)$
$f'(x) = (x + 2)(-e^{-x}) + e^{-x}(1)$
$f'(x) = e^{-x} [-(x + 2) + 1] = e^{-x}(-x - 1) = -e^{-x}(x + 1)$.
Since $e^{-x} > 0$ for all $x \in \mathbb{R}$,the sign of $f'(x)$ depends on $-(x + 1)$.
For $f(x)$ to be increasing,$f'(x) > 0 \Rightarrow -(x + 1) > 0 \Rightarrow x + 1 < 0 \Rightarrow x < -1$.
Thus,$f(x)$ is increasing in $(-\infty, -1)$.
For $f(x)$ to be decreasing,$f'(x) < 0 \Rightarrow -(x + 1) < 0 \Rightarrow x + 1 > 0 \Rightarrow x > -1$.
Thus,$f(x)$ is decreasing in $(-1, \infty)$.
Therefore,the correct option is $C$.

(A) Given $f(x) = e^x - x$ and $g(x) = x^2 - x$.
$h(x) = (fog)(x) = f(g(x)) = e^{x^2-x} - (x^2-x) = e^{x^2-x} - x^2 + x$.
Now,differentiate $h(x)$ with respect to $x$:
$h'(x) = e^{x^2-x}(2x-1) - 2x + 1$.
$h'(x) = (2x-1)(e^{x^2-x} - 1)$.
For $h(x)$ to be increasing,we require $h'(x) \geq 0$.
$(2x-1)(e^{x^2-x} - 1) \geq 0$.
Let $u = x^2-x$. Since $e^u - 1$ has the same sign as $u$,we have $(2x-1)(x^2-x) \geq 0$.
$(2x-1)x(x-1) \geq 0$.
Using the sign scheme for the critical points $0, \frac{1}{2}, 1$:
For $x \in [0, \frac{1}{2}]$,$(2x-1) \leq 0$ and $x(x-1) \leq 0$,so the product is $\geq 0$.
For $x \in [1, \infty)$,$(2x-1) > 0$ and $x(x-1) \geq 0$,so the product is $\geq 0$.
Thus,$h(x)$ is increasing for $x \in [0, \frac{1}{2}] \cup [1, \infty)$.

(C) Let $f(x) = x^3 + x - 1$.
Since $f(0) = -1 < 0$ and $f(1) = 1 > 0$,by the Intermediate Value Theorem,there exists at least one real root $c$ in the interval $(0, 1)$.
Alternatively,consider the derivative $f'(x) = 3x^2 + 1$.
Since $3x^2 + 1 > 0$ for all $x \in \mathbb{R}$,$f(x)$ is a strictly increasing function.
$A$ strictly increasing cubic function crosses the $X$-axis exactly once.
Therefore,the equation has exactly one real root.

(A) Given $f(x) = 3 \sin x - 4 \sin^3 x$.
Using the identity $\sin 3x = 3 \sin x - 4 \sin^3 x$,we have $f(x) = \sin 3x$.
To find the interval where $f(x)$ is increasing,we find the derivative $f'(x) = 3 \cos 3x$.
For $f(x)$ to be increasing,$f'(x) \geq 0$,which implies $3 \cos 3x \geq 0$,or $\cos 3x \geq 0$.
The cosine function is non-negative in the interval $[-\frac{\pi}{2}, \frac{\pi}{2}]$.
Thus,$-\frac{\pi}{2} \leq 3x \leq \frac{\pi}{2}$.
Dividing by $3$,we get $-\frac{\pi}{6} \leq x \leq \frac{\pi}{6}$.
The length of this interval is $\frac{\pi}{6} - (-\frac{\pi}{6}) = \frac{2\pi}{6} = \frac{\pi}{3}$.

(D) To determine the intervals of increase and decrease,we find the derivative of $f(x) = \sin 3x$.
$f'(x) = \frac{d}{dx}(\sin 3x) = 3 \cos 3x$.
For the function to be increasing,$f'(x) > 0$,so $3 \cos 3x > 0$,which implies $\cos 3x > 0$.
Since $x \in [0, \frac{\pi}{2}]$,$3x \in [0, \frac{3\pi}{2}]$.
$\cos 3x > 0$ when $3x \in [0, \frac{\pi}{2})$,which means $x \in [0, \frac{\pi}{6})$.
For the function to be decreasing,$f'(x) < 0$,so $3 \cos 3x < 0$,which implies $\cos 3x < 0$.
$\cos 3x < 0$ when $3x \in (\frac{\pi}{2}, \frac{3\pi}{2}]$,which means $x \in (\frac{\pi}{6}, \frac{\pi}{2}]$.
Thus,the function is increasing on $[0, \frac{\pi}{6})$ and decreasing on $(\frac{\pi}{6}, \frac{\pi}{2}]$.
Therefore,the correct option is $D$.

(C) To determine which function is decreasing in the interval $I = \left(0, \frac{\pi}{8}\right)$,we examine the derivative of each function:
$A) f(x) = \tan 4x \implies f'(x) = 4 \sec^2 4x$. Since $\sec^2 4x > 0$ for all $x \in I$,$f'(x) > 0$,so the function is increasing.
$B) f(x) = \sin x \implies f'(x) = \cos x$. Since $\cos x > 0$ for all $x \in \left(0, \frac{\pi}{8}\right)$,$f'(x) > 0$,so the function is increasing.
$C) f(x) = \cos 4x \implies f'(x) = -4 \sin 4x$. For $x \in \left(0, \frac{\pi}{8}\right)$,$4x \in (0, \frac{\pi}{2})$. In this interval,$\sin 4x > 0$,so $f'(x) = -4 \sin 4x < 0$. Thus,the function is decreasing.
$D) f(x) = -\cos x \implies f'(x) = \sin x$. Since $\sin x > 0$ for all $x \in I$,$f'(x) > 0$,so the function is increasing.
Therefore,the correct option is $C$.

(D) To find the intervals where the function $y = f(x) = x^2 e^{-x}$ is increasing,we find its derivative $f'(x)$.
Using the product rule,$f'(x) = \frac{d}{dx}(x^2) \cdot e^{-x} + x^2 \cdot \frac{d}{dx}(e^{-x})$.
$f'(x) = 2x e^{-x} - x^2 e^{-x} = x(2 - x) e^{-x}$.
For the function to be increasing,we require $f'(x) > 0$.
Since $e^{-x} > 0$ for all real $x$,the sign of $f'(x)$ depends on $x(2 - x)$.
$x(2 - x) > 0$ implies $x(x - 2) < 0$.
This inequality holds when $x$ is between the roots $0$ and $2$.
Thus,the function is increasing on the interval $(0, 2)$.

(D) Given the function $f(x) = \frac{x}{\log_x e}$.
Using the property $\log_x e = \frac{1}{\ln x}$,we can rewrite the function as:
$f(x) = x \cdot \ln x$.
To determine the interval where the function is increasing,we find the derivative $f'(x)$:
$f'(x) = \frac{d}{dx}(x \ln x) = 1 \cdot \ln x + x \cdot \frac{1}{x} = \ln x + 1$.
For the function to be increasing,we require $f'(x) > 0$:
$\ln x + 1 > 0$
$\ln x > -1$
$x > e^{-1}$
$x > \frac{1}{e}$.
Since the domain of the function is $x \in \mathbb{R}^+ - \{1\}$,the function is increasing on the interval $(\frac{1}{e}, 1) \cup (1, \infty)$. However,looking at the provided options,the most appropriate interval representing the increasing behavior is $(\frac{1}{e}, \infty)$ excluding the point $x=1$ where the function is undefined. Thus,the correct option is $D$.

(A) To find where the function $f(x) = x + \sqrt{1 - x}$ decreases,we find its derivative $f'(x)$.
$f'(x) = \frac{d}{dx}(x) + \frac{d}{dx}((1 - x)^{1/2}) = 1 + \frac{1}{2}(1 - x)^{-1/2}(-1) = 1 - \frac{1}{2\sqrt{1 - x}}$.
For the function to decrease,we set $f'(x) < 0$.
$1 - \frac{1}{2\sqrt{1 - x}} < 0 \implies 1 < \frac{1}{2\sqrt{1 - x}}$.
Since $0 < x < 1$,$\sqrt{1 - x}$ is positive,so we can multiply by $2\sqrt{1 - x}$ without changing the inequality sign:
$2\sqrt{1 - x} < 1 \implies \sqrt{1 - x} < \frac{1}{2}$.
Squaring both sides:
$1 - x < \frac{1}{4} \implies 1 - \frac{1}{4} < x \implies x > \frac{3}{4}$.
Given the domain $0 < x < 1$,the function decreases on the interval $\left(\frac{3}{4}, 1\right)$.
Thus,the correct option is $A$.

(A) To determine the interval where the function $y = f(x) = 6 - 9x - x^2$ is strictly increasing,we find its derivative:
$f'(x) = \frac{d}{dx}(6 - 9x - x^2) = -9 - 2x$.
For the function to be strictly increasing,we must have $f'(x) > 0$.
So,$-9 - 2x > 0$.
$-2x > 9$.
Dividing by $-2$ reverses the inequality: $x < -4.5$.
Thus,the function is strictly increasing on the interval $(-\infty, -4.5)$.