Increasing and Decreasing function Questions in English

(A) To determine the interval where the function $f(x) = x^2 - 6x + 10$ is increasing,we find its derivative $f'(x)$.
$f'(x) = \frac{d}{dx}(x^2 - 6x + 10) = 2x - 6$.
$A$ function is increasing when $f'(x) > 0$.
$2x - 6 > 0 \implies 2x > 6 \implies x > 3$.
Thus,the function is increasing in the interval $(3, \infty)$.
Therefore,the correct option is $A$.

(A) To find the interval where the function $f(x) = 10 - 6x - 2x^2$ is strictly increasing,we first find its derivative $f'(x)$.
$f'(x) = \frac{d}{dx}(10 - 6x - 2x^2) = -6 - 4x$.
For the function to be strictly increasing,we must have $f'(x) > 0$.
$-6 - 4x > 0$
$-4x > 6$
Dividing by $-4$ reverses the inequality sign:
$x < -\frac{6}{4}$
$x < -\frac{3}{2}$.
Thus,the function is strictly increasing in the interval $(-\infty, -\frac{3}{2})$.
Therefore,the correct option is $A$.

(D) For $x \in \left(-\frac{\pi}{2}, 0\right)$,we have $\sin x < 0$.
Therefore,$f(x) = |\sin x| = -\sin x$.
To determine the nature of the function,we find its derivative: $f'(x) = \frac{d}{dx}(-\sin x) = -\cos x$.
In the interval $\left(-\frac{\pi}{2}, 0\right)$,$\cos x$ is positive (since it is in the fourth quadrant).
Thus,$f'(x) = -\cos x < 0$ for all $x \in \left(-\frac{\pi}{2}, 0\right)$.
Since the derivative is strictly negative,the function $f(x)$ is a strictly decreasing function on the given interval.

(C) To determine the nature of the function $f(x) = \tan^{-1} x - x$,we find its derivative with respect to $x$:
$f'(x) = \frac{d}{dx}(\tan^{-1} x - x) = \frac{1}{1+x^2} - 1$
Simplifying the expression:
$f'(x) = \frac{1 - (1+x^2)}{1+x^2} = \frac{-x^2}{1+x^2}$
Since $x^2 \ge 0$ and $1+x^2 > 0$ for all $x \in R$,the derivative $f'(x) = \frac{-x^2}{1+x^2} \le 0$ for all $x \in R$.
Since $f'(x) \le 0$ for all $x \in R$,the function $f(x)$ is a decreasing function on $R$.

(D) To determine the interval where the function $f(x) = x^x$ is decreasing,we find its derivative.
Let $y = x^x$.
Taking the natural logarithm on both sides,we get $\ln(y) = x \ln(x)$.
Differentiating with respect to $x$,we have $\frac{1}{y} \frac{dy}{dx} = \ln(x) + x \cdot \frac{1}{x} = \ln(x) + 1$.
Thus,$f'(x) = \frac{dy}{dx} = y(\ln(x) + 1) = x^x(\ln(x) + 1)$.
For the function to be decreasing,we require $f'(x) < 0$.
Since $x^x > 0$ for all $x \in R^{+}$,the condition $f'(x) < 0$ implies $\ln(x) + 1 < 0$.
This gives $\ln(x) < -1$,which means $x < e^{-1}$ or $x < 1/e$.
Given $x \in R^{+}$,the interval is $(0, 1/e)$.

(A) For a function $f(x)$ to be increasing on an interval,its derivative $f'(x)$ must be greater than or equal to $0$ for all $x$ in that interval.
Given $f(x) = x^2 + ax + 1$,the derivative is $f'(x) = 2x + a$.
For $f(x)$ to be increasing on $[1, 2]$,we require $f'(x) \ge 0$ for all $x \in [1, 2]$.
This implies $2x + a \ge 0$ for all $x \in [1, 2]$.
Since $2x + a$ is an increasing linear function,its minimum value on the interval $[1, 2]$ occurs at the smallest value of $x$,which is $x = 1$.
Therefore,we need $f'(1) \ge 0$.
Substituting $x = 1$ into the derivative: $2(1) + a \ge 0$.
$2 + a \ge 0$,which gives $a \ge -2$.
Thus,the function is increasing on $[1, 2]$ for $a \ge -2$.

(B) To determine the nature of the function $f(x) = \log_{10} \cos x$,we find its derivative with respect to $x$.
$f'(x) = \frac{d}{dx} (\log_{10} \cos x) = \frac{1}{\cos x \cdot \ln 10} \cdot (-\sin x) = -\frac{\tan x}{\ln 10}$.
In the interval $\left(0, \frac{\pi}{2}\right)$,$\tan x > 0$ and $\ln 10 > 0$.
Therefore,$f'(x) = -\frac{\tan x}{\ln 10} < 0$ for all $x \in \left(0, \frac{\pi}{2}\right)$.
Since the derivative is negative in the given interval,the function $f(x)$ is a decreasing function.

(A) Given function is $f(x) = x^{2} + 2x - 5$.
To find the interval where the function is strictly increasing,we calculate the derivative $f'(x)$.
$f'(x) = \frac{d}{dx}(x^{2} + 2x - 5) = 2x + 2$.
For a function to be strictly increasing,we must have $f'(x) > 0$.
$2x + 2 > 0$
$2x > -2$
$x > -1$.
Thus,the function is strictly increasing in the interval $(-1, \infty)$.

(A) Given the function $f(x) = \tan x - x$.
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 always non-negative,$\tan^2 x \geq 0$ for all $x$ in the domain.
Therefore,$f'(x) \geq 0$,which implies that the function $f(x)$ is always increasing.

(C) Let $f(x) = x^x$. Taking the natural logarithm on both sides,we get $\ln f(x) = x \ln x$.
Differentiating both sides with respect to $x$,we have $\frac{1}{f(x)} f'(x) = 1 \cdot \ln x + x \cdot \frac{1}{x} = \ln x + 1$.
Thus,$f'(x) = x^x(1 + \ln x)$.
For the function to be strictly increasing,we require $f'(x) > 0$.
Since $x > 0$,$x^x$ is always positive.
Therefore,we must have $1 + \ln x > 0$,which implies $\ln x > -1$.
This is equivalent to $\ln x > \ln(\frac{1}{e})$.
Hence,$x > \frac{1}{e}$.

(C) Given,$f(x)=\log (1+x)-\frac{2 x}{2+x}$.
For the function to be defined,we must have $1+x > 0$,i.e.,$x > -1$.
Differentiating the function with respect to $x$,we get:
$f^{\prime}(x) = \frac{1}{1+x} - 2 \left[ \frac{(2+x)(1) - x(1)}{(2+x)^2} \right]$
$f^{\prime}(x) = \frac{1}{1+x} - 2 \left[ \frac{2+x-x}{(2+x)^2} \right] = \frac{1}{1+x} - \frac{4}{(2+x)^2}$
$f^{\prime}(x) = \frac{(2+x)^2 - 4(1+x)}{(1+x)(2+x)^2} = \frac{4+x^2+4x-4-4x}{(1+x)(2+x)^2}$
$f^{\prime}(x) = \frac{x^2}{(1+x)(2+x)^2}$.
Since $x^2 \ge 0$ and $(2+x)^2 > 0$ for all $x > -1$,the sign of $f^{\prime}(x)$ depends on the term $\frac{1}{1+x}$.
For $f(x)$ to be increasing,we require $f^{\prime}(x) > 0$.
Since $x > -1$,$1+x > 0$,which implies $f^{\prime}(x) > 0$ for all $x \in (-1, \infty)$.
Thus,the function is increasing on $(-1, \infty)$.

(D) Given,$f(x)=4 \sin ^3 x-6 \sin ^2 x+12 \sin x+100$.
First,we find the derivative $f'(x)$:
$f'(x) = 12 \sin^2 x \cos x - 12 \sin x \cos x + 12 \cos x$.
Factoring out $12 \cos x$:
$f'(x) = 12 \cos x (\sin^2 x - \sin x + 1)$.
Consider the quadratic expression $g(t) = t^2 - t + 1$ where $t = \sin x$.
The discriminant $D = (-1)^2 - 4(1)(1) = 1 - 4 = -3 < 0$.
Since the coefficient of $t^2$ is positive and $D < 0$,the expression $\sin^2 x - \sin x + 1$ is always positive for all $x \in \mathbb{R}$.
Thus,the sign of $f'(x)$ depends only on $\cos x$.
$f'(x) > 0$ when $\cos x > 0$,which occurs in $x \in (-\frac{\pi}{2}, \frac{\pi}{2})$.
$f'(x) < 0$ when $\cos x < 0$,which occurs in $x \in (\frac{\pi}{2}, \frac{3\pi}{2})$.
Therefore,$f(x)$ is strictly decreasing in the interval $(\frac{\pi}{2}, \pi)$.

(A) Given function is $f(x) = x^{2} - 2x$.
To find the interval where the function is strictly decreasing,we first find the derivative $f'(x)$.
$f'(x) = \frac{d}{dx}(x^{2} - 2x) = 2x - 2$.
$A$ function is strictly decreasing when $f'(x) < 0$.
So,$2x - 2 < 0$.
$2(x - 1) < 0$.
$x - 1 < 0$.
$x < 1$.
Thus,the function $f(x)$ is strictly decreasing in the interval $(-\infty, 1)$.

(C) Given function: $f(x) = x^{3} - 6x^{2} + 9x + 10$
Find the derivative: $f'(x) = 3x^{2} - 12x + 9$
Factor the derivative: $f'(x) = 3(x^{2} - 4x + 3) = 3(x - 1)(x - 3)$
For the function to be increasing,we require $f'(x) > 0$:
$3(x - 1)(x - 3) > 0$
$(x - 1)(x - 3) > 0$
This inequality holds when $x < 1$ or $x > 3$.
Thus,the function is increasing on the interval $(-\infty, 1) \cup (3, \infty)$.

(A) Given,$f(x) = \frac{x}{3} + \frac{3}{x}$.
On differentiating with respect to $x$,we get:
$f^{\prime}(x) = \frac{1}{3} - \frac{3}{x^2}$.
For the function to be decreasing,we must have $f^{\prime}(x) < 0$.
$\frac{1}{3} - \frac{3}{x^2} < 0$
$\frac{1}{3} < \frac{3}{x^2}$
$x^2 < 9$
Taking the square root on both sides,we get $|x| < 3$,which implies $x \in (-3, 3)$.
Note that at $x=0$,the function is undefined,but the interval $(-3, 3)$ is the standard solution for the inequality $x^2 < 9$.

(A) Given the function $f(x) = \frac{x}{\log x}$.
First,we find the derivative $f'(x)$ using the quotient rule:
$f'(x) = \frac{(\log x)(1) - (x)(\frac{1}{x})}{(\log x)^2} = \frac{\log x - 1}{(\log x)^2}$.
For the function to be increasing,we must have $f'(x) > 0$.
Since $(\log x)^2$ is always positive for $x > 0$ and $x \neq 1$,the condition $f'(x) > 0$ implies $\log x - 1 > 0$.
$\log x > 1$.
Since the base of the logarithm is $e$,we have $\log_e x > \log_e e$.
Therefore,$x > e$.
Thus,the set of values is $\{x: x > e\}$.

(A) Given the function $y = -x^{2} + 6x - 3$.
To find the interval where the function is increasing,we calculate the first derivative:
$\frac{dy}{dx} = -2x + 6$.
$A$ function is increasing when its derivative is greater than zero,i.e.,$\frac{dy}{dx} > 0$.
Setting the derivative greater than zero:
$-2x + 6 > 0$.
Subtracting $6$ from both sides:
$-2x > -6$.
Dividing by $-2$ (and reversing the inequality sign):
$x < 3$.
Therefore,the function is increasing in the range $x < 3$.

(A) Let $f(x) = x^5+3x^3+4x+30$.
Taking the derivative with respect to $x$,we get $f'(x) = 5x^4+9x^2+4$.
Since $x^4 \ge 0$ and $x^2 \ge 0$ for all $x \in \mathbb{R}$,it follows that $5x^4+9x^2+4 \ge 4 > 0$ for all $x \in \mathbb{R}$.
Since $f'(x) > 0$ for all $x \in \mathbb{R}$,the function $f(x)$ is strictly increasing on its domain.
$A$ strictly increasing continuous function can cross the $x$-axis at most once.
Since $\lim_{x \to -\infty} f(x) = -\infty$ and $\lim_{x \to \infty} f(x) = \infty$,by the Intermediate Value Theorem,there exists exactly one real root.

(D) Let $f(x) = x - \sin x$.
Then $f'(x) = 1 - \cos x$.
Since $\cos x \leq 1$ for all $x \in \mathbb{R}$,we have $f'(x) \geq 0$ for all $x$.
This means $f(x)$ is a non-decreasing function.
Since $f(0) = 0 - \sin(0) = 0$,it follows that $f(x) \geq 0$ for all $x \geq 0$.
Thus,$x - \sin x \geq 0$,which implies $\sin x \leq x$ for all $x \geq 0$.
For $x < 0$,let $x = -t$ where $t > 0$. Then $\sin(-t) = -\sin t$.
The inequality becomes $-\sin t \leq -t$,which simplifies to $\sin t \geq t$.
Since $\sin t < t$ for all $t > 0$,the inequality $\sin t \geq t$ is only satisfied at $t = 0$.
Therefore,the set of all $x$ for which $\sin x \leq x$ is $[0, \infty)$.

(B) Consider the function $f(x) = \frac{\sin x}{x}$ for $x \in (0, \pi/2)$.
Taking the derivative,$f'(x) = \frac{x \cos x - \sin x}{x^2}$.
Let $u(x) = x \cos x - \sin x$. Then $u'(x) = \cos x - x \sin x - \cos x = -x \sin x$.
Since $x \in (0, \pi/2)$,$u'(x) < 0$,which means $u(x)$ is a strictly decreasing function.
Since $u(0) = 0$ and $u(x)$ is decreasing,$u(x) < 0$ for all $x \in (0, \pi/2)$.
Thus,$f'(x) < 0$,meaning $f(x)$ is a strictly decreasing function on $(0, \pi/2)$.
As $x \to 0^+$,$f(x) \to 1$,and at $x = \pi/2$,$f(\pi/2) = \frac{\sin(\pi/2)}{\pi/2} = \frac{2}{\pi}$.
Since $f(x)$ is decreasing,for $0 < x < \pi/2$,we have $f(\pi/2) < f(x) < \lim_{x \to 0^+} f(x)$.
Therefore,$\frac{2}{\pi} < \frac{\sin x}{x} < 1$.
Hence,option $B$ is correct.

(C) To find the interval where $f(x) = \operatorname{Tan}^{-1}(\sin x + \cos x)$ is increasing,we find its derivative $f'(x)$.
$f'(x) = \frac{1}{1 + (\sin x + \cos x)^2} \cdot \frac{d}{dx}(\sin x + \cos x)$.
$f'(x) = \frac{\cos x - \sin x}{1 + (\sin x + \cos x)^2}$.
For $f(x)$ to be an increasing function,we must have $f'(x) > 0$.
Since the denominator $1 + (\sin x + \cos x)^2$ is always positive,we only need the numerator to be positive:
$\cos x - \sin x > 0$.
$\cos x > \sin x$.
Dividing by $\cos x$ (assuming $\cos x > 0$),we get $1 > \tan x$,which implies $\tan x < 1$.
This inequality holds when $x \in \left(-\frac{\pi}{2}, \frac{\pi}{4}\right)$.
Comparing this with the given options,the interval $\left(-\frac{3\pi}{4}, \frac{\pi}{4}\right)$ contains the region where the function is increasing,and specifically,the function increases within the range provided in option $C$.

(C) Given the function $f(x) = 3 \sinh(x) - 2 \cosh(x)$ for all $x \in R$.
Using the definitions $\sinh(x) = \frac{e^x - e^{-x}}{2}$ and $\cosh(x) = \frac{e^x + e^{-x}}{2}$:
$f(x) = 3 \left( \frac{e^x - e^{-x}}{2} \right) - 2 \left( \frac{e^x + e^{-x}}{2} \right)$
$f(x) = \frac{3e^x - 3e^{-x} - 2e^x - 2e^{-x}}{2} = \frac{e^x - 5e^{-x}}{2}$
Now,find the derivative $f'(x)$:
$f'(x) = \frac{d}{dx} \left( \frac{1}{2} e^x - \frac{5}{2} e^{-x} \right) = \frac{1}{2} e^x + \frac{5}{2} e^{-x}$
Since $e^x > 0$ and $e^{-x} > 0$ for all $x \in R$,it follows that $f'(x) > 0$ for all $x \in R$.
Therefore,the function $f$ is a strictly increasing function on $R$.
Hence,option $C$ is correct.

(C) Given $x = t^5 + 5t^3 + 20t + 7$ and $y = 4t^3 - 3t^2 - 18t + 3$.
First,find the derivatives with respect to $t$:
$\frac{dx}{dt} = 5t^4 + 15t^2 + 20 = 5(t^4 + 3t^2 + 4)$.
Note that $t^4 + 3t^2 + 4$ is always positive for all real $t$ (since its discriminant $D = 3^2 - 4(1)(4) = 9 - 16 = -7 < 0$ and the leading coefficient is positive).
$\frac{dy}{dt} = 12t^2 - 6t - 18 = 6(2t^2 - t - 3)$.
Now,the derivative of the curve is given by $\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{6(2t^2 - t - 3)}{5(t^4 + 3t^2 + 4)}$.
The curve is decreasing when $\frac{dy}{dx} < 0$.
Since the denominator $5(t^4 + 3t^2 + 4)$ is always positive,we only need the numerator to be negative:
$6(2t^2 - t - 3) < 0$
$2t^2 - t - 3 < 0$
Factorizing the quadratic expression: $(2t - 3)(t + 1) < 0$.
The roots are $t = -1$ and $t = 3/2$.
Testing the intervals,the expression is negative for $t \in (-1, 3/2)$.

(B) Given the function $f(x) = x^3 + 2ax^2 + 3(a+1)x + 5$.
For $f(x)$ to be strictly increasing in its entire domain,its derivative $f'(x)$ must be greater than or equal to $0$ for all $x \in \mathbb{R}$,and $f'(x)$ should not be zero for all $x$ in any interval.
First,find the derivative: $f'(x) = 3x^2 + 4ax + 3(a+1)$.
Since $f'(x)$ is a quadratic expression with a positive leading coefficient $(3 > 0)$,for $f'(x) \geq 0$ to hold for all $x$,the discriminant $D$ must be less than or equal to $0$.
$D = (4a)^2 - 4(3)(3(a+1)) \leq 0$.
$16a^2 - 36(a+1) \leq 0$.
Divide by $4$: $4a^2 - 9(a+1) \leq 0$.
$4a^2 - 9a - 9 \leq 0$.
Factor the quadratic: $(4a+3)(a-3) \leq 0$.
This inequality holds when $a$ lies between the roots: $a \in [-\frac{3}{4}, 3]$.
Since the question asks for strictly increasing,we consider the interval where $f'(x) > 0$ for all $x$ except possibly at isolated points,which corresponds to $a \in (-\frac{3}{4}, 3)$.

(B) Given $f(x) = x^3 + ax^2 + bx + 5 \sin^2 x$ is an increasing function on $R$,so $f'(x) \ge 0$ for all $x \in R$.
$f'(x) = 3x^2 + 2ax + b + 10 \sin x \cos x = 3x^2 + 2ax + b + 5 \sin 2x$.
For $f'(x) \ge 0$,the minimum value of $f'(x)$ must be $\ge 0$.
Since $\sin 2x$ ranges from $-1$ to $1$,the minimum value of $5 \sin 2x$ is $-5$.
Thus,we require $3x^2 + 2ax + b - 5 \ge 0$ for all $x \in R$.
For a quadratic $Ax^2 + Bx + C \ge 0$ to hold for all $x$,we must have $A > 0$ and the discriminant $D = B^2 - 4AC \le 0$.
Here $A = 3$,$B = 2a$,and $C = b-5$.
$D = (2a)^2 - 4(3)(b-5) \le 0$.
$4a^2 - 12(b-5) \le 0$.
Dividing by $4$,we get $a^2 - 3(b-5) \le 0$.
$a^2 - 3b + 15 \le 0$.
Since the options provided are strict inequalities,the correct condition is $a^2 - 3b + 15 < 0$.

(D) Given function is $y = \ln(\ln(x))$ for $x > 1$.
First,we find the derivative of $y$ with respect to $x$:
$\frac{dy}{dx} = \frac{1}{\ln(x)} \cdot \frac{d}{dx}(\ln(x)) = \frac{1}{x \ln(x)}$.
For the function to be decreasing,we must have $\frac{dy}{dx} < 0$.
So,$\frac{1}{x \ln(x)} < 0$.
Since the domain is given as $x > 1$,we know that $\ln(x) > 0$ and $x > 1$.
Therefore,the product $x \ln(x)$ is always positive for all $x > 1$.
Since $\frac{1}{x \ln(x)}$ is always positive for $x > 1$,there is no interval where the function is decreasing.
Thus,the correct option is $D$.

(D) The slope of the tangent to the curve $y=3x^4-5x^3-12x^2+18x+3$ is given by the derivative $g(x) = \frac{dy}{dx}$.
Calculating the derivative: $g(x) = 12x^3 - 15x^2 - 24x + 18$.
For $g(x)$ to be strictly increasing,its derivative $g'(x)$ must be greater than $0$.
$g'(x) = 36x^2 - 30x - 24$.
Setting $g'(x) > 0$: $36x^2 - 30x - 24 > 0$.
Dividing by $6$: $6x^2 - 5x - 4 > 0$.
Factoring the quadratic: $(3x-4)(2x+1) > 0$.
The roots are $x = \frac{4}{3}$ and $x = -\frac{1}{2}$.
The inequality holds for $x \in (-\infty, -\frac{1}{2}) \cup (\frac{4}{3}, \infty)$.
This is equivalent to $R - [-\frac{1}{2}, \frac{4}{3}]$.
Thus,the correct option is $D$.

(C) Given the function $y = \sin x + \sin x \cos x = \sin x + \frac{1}{2} \sin 2x$.
To find the interval where $y$ is strictly increasing,we find the derivative $\frac{dy}{dx}$.
$\frac{dy}{dx} = \cos x + \cos 2x$.
For $y$ to be strictly increasing,we require $\frac{dy}{dx} > 0$.
$\cos x + (2 \cos^2 x - 1) > 0$.
Let $t = \cos x$,then $2t^2 + t - 1 > 0$.
Factoring the quadratic: $(2t - 1)(t + 1) > 0$.
Since $t = \cos x$ and $x \in [-\pi, \pi]$,we know $-1 \le t \le 1$.
For $(2t - 1)(t + 1) > 0$,we must have $t > \frac{1}{2}$ (since $t+1$ is always $\ge 0$ and $t \neq -1$).
Thus,$\cos x > \frac{1}{2}$.
In the interval $[-\pi, \pi]$,$\cos x > \frac{1}{2}$ implies $x \in \left(-\frac{\pi}{3}, \frac{\pi}{3}\right)$.
Therefore,the function is strictly increasing in the interval $\left(-\frac{\pi}{3}, \frac{\pi}{3}\right)$.

(C) To find the interval where the function $f(x) = 2x + \log \left(\frac{x}{2+x}\right)$ is increasing,we first find its derivative $f'(x)$.
First,note that the domain requires $\frac{x}{2+x} > 0$,which implies $x \in (-\infty, -2) \cup (0, \infty)$.
$f(x) = 2x + \log(x) - \log(2+x)$.
Differentiating with respect to $x$:
$f'(x) = 2 + \frac{1}{x} - \frac{1}{2+x}$.
$f'(x) = 2 + \frac{2+x-x}{x(2+x)} = 2 + \frac{2}{x(2+x)} = \frac{2x(2+x) + 2}{x(2+x)} = \frac{2(x^2 + 2x + 1)}{x(2+x)} = \frac{2(x+1)^2}{x(2+x)}$.
For the function to be increasing,we require $f'(x) > 0$.
Since $2(x+1)^2 \ge 0$ for all $x$,$f'(x) > 0$ when $x(2+x) > 0$ and $x \neq -1$.
The inequality $x(2+x) > 0$ holds for $x \in (-\infty, -2) \cup (0, \infty)$.
Thus,the function is increasing in the interval $(-\infty, -2) \cup (0, \infty)$.

(A) To determine if a function is monotonically increasing,we check if its derivative $f'(x) \ge 0$ for all $x$ in its domain.
For option $A$: $f'(x) = \frac{1}{1+x} - 1 + x = \frac{1 - (1+x) + x(1+x)}{1+x} = \frac{1 - 1 - x + x + x^2}{1+x} = \frac{x^2}{1+x}$. For $x > -1$,$f'(x) \ge 0$,so it is monotonically increasing.
For option $B$: $g'(x) = \frac{2}{1+x^2} - 1 = \frac{2 - 1 - x^2}{1+x^2} = \frac{1-x^2}{1+x^2}$. This changes sign at $x = \pm 1$,so it is not monotonically increasing.
For option $C$: $h'(x) = -4 \sin x + 1$. This changes sign depending on $\sin x$,so it is not monotonically increasing.
For option $D$: $u'(x) = \frac{1}{1+x} - \frac{(x+1)(1) - x(1)}{(x+1)^2} = \frac{1}{1+x} - \frac{1}{(x+1)^2} = \frac{x+1-1}{(x+1)^2} = \frac{x}{(x+1)^2}$. This changes sign at $x = 0$,so it is not monotonically increasing.
Thus,the correct option is $A$.

(D) Given $f(x) = \sin x - \cos^2 x = \sin x - (1 - \sin^2 x) = \sin^2 x + \sin x - 1$.
Let $t = \sin x$. Since $x \in [-\pi, \pi]$,$t \in [-1, 1]$.
Then $g(t) = t^2 + t - 1$.
For $f(x)$ to be strictly increasing,we need $f'(x) > 0$.
$f'(x) = \cos x + 2 \cos x \sin x = \cos x(1 + 2 \sin x)$.
Setting $f'(x) > 0$:
Case $1$: $\cos x > 0$ and $1 + 2 \sin x > 0$.
$\cos x > 0 \implies x \in (-\frac{\pi}{2}, \frac{\pi}{2})$.
$1 + 2 \sin x > 0 \implies \sin x > -\frac{1}{2} \implies x \in (-\frac{\pi}{6}, \frac{7\pi}{6})$.
Intersection: $x \in (-\frac{\pi}{6}, \frac{\pi}{2})$.
Case $2$: $\cos x < 0$ and $1 + 2 \sin x < 0$.
$\cos x < 0 \implies x \in (-\pi, -\frac{\pi}{2}) \cup (\frac{\pi}{2}, \pi)$.
$1 + 2 \sin x < 0 \implies \sin x < -\frac{1}{2} \implies x \in (-\frac{5\pi}{6}, -\frac{\pi}{6})$.
Intersection: $x \in (-\frac{5\pi}{6}, -\frac{\pi}{2})$.
Combining both cases,$f(x)$ is strictly increasing in $(-\frac{5\pi}{6}, -\frac{\pi}{2}) \cup (-\frac{\pi}{6}, \frac{\pi}{2})$.
Thus,the correct option is $D$.

(A) Given $f(x) = x e^{x-x^2}$.
To determine the intervals of increase or decrease,we find the derivative $f'(x)$.
Using the product rule and chain rule:
$f'(x) = 1 \cdot e^{x-x^2} + x \cdot e^{x-x^2} \cdot (1-2x)$
$f'(x) = e^{x-x^2} [1 + x(1-2x)]$
$f'(x) = e^{x-x^2} [1 + x - 2x^2]$
$f'(x) = e^{x-x^2} [-(2x^2 - x - 1)]$
$f'(x) = -e^{x-x^2} (2x+1)(x-1)$
For $f(x)$ to be increasing,$f'(x) \ge 0$.
Since $e^{x-x^2} > 0$ for all $x \in R$,we need $-(2x+1)(x-1) \ge 0$,which implies $(2x+1)(x-1) \le 0$.
The roots are $x = -\frac{1}{2}$ and $x = 1$.
The inequality holds for $x \in \left[-\frac{1}{2}, 1\right]$.
Thus,$f(x)$ is increasing on $\left[-\frac{1}{2}, 1\right]$.

(D) The given function is $f(x) = x^3 - 16x^2 + 20x - 5$.
To find the interval where the function is strictly decreasing,we find its derivative: $f'(x) = 3x^2 - 32x + 20$.
Setting $f'(x) < 0$ for strictly decreasing behavior: $3x^2 - 32x + 20 < 0$.
Factoring the quadratic: $(3x - 2)(x - 10) < 0$.
This inequality holds for $x \in (\frac{2}{3}, 10)$.
The set of numbers is $S = \{1, 3, 5, \dots, 59\}$. The number of elements in $S$ is $n(S) = 30$.
We need to find the numbers from the set $S$ that lie in the interval $(\frac{2}{3}, 10)$.
These numbers are $E = \{1, 3, 5, 7, 9\}$.
The number of favorable outcomes is $n(E) = 5$.
The probability is $P = \frac{n(E)}{n(S)} = \frac{5}{30} = \frac{1}{6}$.

(A) Given $f(x) = x^x$. Taking the natural logarithm on both sides,we get $\ln(f(x)) = x \ln(x)$.
Differentiating both sides with respect to $x$,we have $\frac{f'(x)}{f(x)} = \ln(x) + x \cdot \frac{1}{x} = \ln(x) + 1$.
Thus,$f'(x) = x^x(1 + \ln(x))$.
For $f(x)$ to be a decreasing function,we must have $f'(x) \leq 0$.
Since $x^x > 0$ for all $x > 0$,the condition $f'(x) \leq 0$ implies $1 + \ln(x) \leq 0$.
This gives $\ln(x) \leq -1$,which means $x \leq e^{-1} = \frac{1}{e}$.
Since the domain of $f(x) = x^x$ is $x > 0$,the interval where $f(x)$ decreases is $\left(0, \frac{1}{e}\right]$.
Comparing with the given options,the correct interval is $\left[0, \frac{1}{e}\right]$.

(A) Given function: $f(x) = \sqrt{x} + \frac{1}{\sqrt{x}}$.
For $f(x)$ to be defined,we must have $x > 0$.
Now,find the derivative $f'(x)$:
$f'(x) = \frac{d}{dx} (x^{1/2} + x^{-1/2}) = \frac{1}{2} x^{-1/2} - \frac{1}{2} x^{-3/2}$.
$f'(x) = \frac{1}{2\sqrt{x}} - \frac{1}{2x\sqrt{x}} = \frac{1}{2\sqrt{x}} (1 - \frac{1}{x}) = \frac{x-1}{2x\sqrt{x}}$.
For the function to be strictly increasing,we require $f'(x) > 0$.
Since $x > 0$,the denominator $2x\sqrt{x}$ is always positive.
Therefore,$f'(x) > 0$ if and only if $x - 1 > 0$,which implies $x > 1$.
Thus,the function is strictly increasing in the interval $(1, \infty)$.

(D) To determine which function is decreasing in the interval $\left(\frac{1}{e}, e\right)$,we check the sign of the derivative $f'(x)$ for each function.
For option $(A)$: $f(x) = \frac{\log x}{x}$,$f'(x) = \frac{1 - \log x}{x^2}$. For $x \in \left(\frac{1}{e}, e\right)$,$\log x$ ranges from $-1$ to $1$. Thus,$f'(x) > 0$ for $x < e$,so it is increasing.
For option $(B)$: $f(x) = x^2 \log x$,$f'(x) = 2x \log x + x = x(2 \log x + 1)$. For $x > \frac{1}{e}$,$f'(x) > 0$,so it is increasing.
For option $(C)$: $f(x) = x \log x$,$f'(x) = \log x + 1$. For $x > \frac{1}{e}$,$\log x > -1$,so $f'(x) > 0$,so it is increasing.
For option $(D)$: $f(x) = x^{-x}$. Let $y = x^{-x}$,then $\log y = -x \log x$. Differentiating with respect to $x$: $\frac{1}{y} \frac{dy}{dx} = -(\log x + x \cdot \frac{1}{x}) = -(\log x + 1)$. Thus,$f'(x) = -x^{-x}(1 + \log x)$. In the interval $\left(\frac{1}{e}, e\right)$,$\log x > -1$,so $(1 + \log x) > 0$. Since $x^{-x} > 0$,$f'(x) < 0$. Therefore,$f(x) = x^{-x}$ is a decreasing function.

(B) Given the function $y=x^3-a x^2+48 x+7$.
For the function to be an increasing function for all real values of $x$,its derivative must be non-negative,i.e.,$\frac{dy}{dx} \geq 0$ for all $x \in \mathbb{R}$.
Differentiating with respect to $x$:
$\frac{dy}{dx} = 3x^2 - 2ax + 48$.
Since $3x^2 - 2ax + 48 \geq 0$ for all $x$,the quadratic expression must have a discriminant $D \leq 0$.
Here,$A=3, B=-2a, C=48$.
$D = B^2 - 4AC = (-2a)^2 - 4(3)(48) = 4a^2 - 576$.
Setting $D \leq 0$:
$4a^2 - 576 \leq 0$
$4(a^2 - 144) \leq 0$
$a^2 - 144 \leq 0$
$(a-12)(a+12) \leq 0$.
Thus,$a \in [-12, 12]$.
Given the options,the interval is $(-12, 12)$.

(A) Given the function $f(x) = |x-5| + 2|x-7|$.
For the interval $(7, \infty)$,we have $x > 7$.
Since $x > 7$,it follows that $x > 5$ and $x > 7$.
Therefore,$|x-5| = x-5$ and $|x-7| = x-7$.
Substituting these into the function:
$f(x) = (x-5) + 2(x-7)$
$f(x) = x - 5 + 2x - 14$
$f(x) = 3x - 19$.
Now,find the derivative $f'(x)$:
$f'(x) = \frac{d}{dx}(3x - 19) = 3$.
Since $f'(x) = 3 > 0$ for all $x$ in the interval $(7, \infty)$,the function is an increasing function.

(A) $(i)$ Given $f(x) = x|x|$. For $x > 0$,$f(x) = x^2$,so $f'(x) = 2x > 0$. For $x < 0$,$f(x) = -x^2$,so $f'(x) = -2x > 0$. Thus,$f'(x) > 0$ for all $x \in R - \{0\}$,meaning $f(x)$ is strictly increasing. This statement is true.
$(ii)$ Given $f(x) = \log_{1/4} (x)$. Since the base $1/4 < 1$,the logarithmic function is strictly decreasing on $(0, \infty)$. This statement is false.
$(iii)$ $A$ one-one function can be strictly increasing,strictly decreasing,or neither (e.g.,$f(x) = 1/x$ is one-one but not monotonic). This statement is false.
$(iv)$ Given $f(x) = x^{1/3}$. $f'(x) = \frac{1}{3}x^{-2/3} = \frac{1}{3x^{2/3}} > 0$ for all $x \neq 0$. Thus,it is strictly increasing on $R$. This statement is false.

(D) Given the function $f(x) = -x^3 + 4ax^2 + 2x - 5$.
To find where the function is decreasing,we calculate the derivative $f'(x) = -3x^2 + 8ax + 2$.
For the function to be decreasing for every $x$,we must have $f'(x) < 0$ for all $x \in R$.
This implies $-3x^2 + 8ax + 2 < 0$ for all $x$.
For a quadratic expression $Ax^2 + Bx + C$ to be negative for all $x$,the conditions are $A < 0$ and the discriminant $\Delta = B^2 - 4AC < 0$.
Here,$A = -3$,which is less than $0$.
The discriminant is $\Delta = (8a)^2 - 4(-3)(2) = 64a^2 + 24$.
Since $64a^2 + 24$ is always positive for any real value of $a$,the condition $\Delta < 0$ can never be satisfied.
Therefore,there is no value of $a$ for which the function is decreasing for every $x$.

(D) Given that $f''(x) > 0$ for all $x \in R$,it implies that $f'(x)$ is a strictly increasing function for all $x \in R$.
Since $f'(3) = 0$,we have $f'(x) < 0$ for $x < 3$ and $f'(x) > 0$ for $x > 3$.
Now,consider $g(x) = f(\tan^2 x - 2 \tan x + 4)$.
Differentiating with respect to $x$,we get $g'(x) = f'(\tan^2 x - 2 \tan x + 4) \cdot \frac{d}{dx}(\tan^2 x - 2 \tan x + 4)$.
$g'(x) = f'(\tan^2 x - 2 \tan x + 4) \cdot (2 \tan x \sec^2 x - 2 \sec^2 x) = f'(\tan^2 x - 2 \tan x + 4) \cdot 2 \sec^2 x (\tan x - 1)$.
Let $u = \tan^2 x - 2 \tan x + 4 = (\tan x - 1)^2 + 3$. Since $(\tan x - 1)^2 \ge 0$,we have $u \ge 3$.
For $u > 3$,we know $f'(u) > 0$ because $f'(x)$ is increasing and $f'(3) = 0$.
For $g(x)$ to be increasing,we require $g'(x) > 0$.
Since $f'(u) > 0$ and $2 \sec^2 x > 0$ for $x \in (0, \frac{\pi}{2})$,the sign of $g'(x)$ depends on $(\tan x - 1)$.
Thus,$g'(x) > 0$ when $\tan x - 1 > 0$,which means $\tan x > 1$.
For $x \in (0, \frac{\pi}{2})$,$\tan x > 1$ implies $x \in (\frac{\pi}{4}, \frac{\pi}{2})$.

(A) Given $g(x) = \frac{1}{6} f(3 x^2 - 1) + \frac{1}{2} f(1 - x^2)$.
Taking the derivative with respect to $x$:
$g'(x) = \frac{1}{6} f'(3 x^2 - 1) \cdot (6x) + \frac{1}{2} f'(1 - x^2) \cdot (-2x)$
$g'(x) = x [f'(3 x^2 - 1) - f'(1 - x^2)]$.
Since $f''(x) > 0$,$f'(x)$ is an increasing function.
For $g(x)$ to be increasing,$g'(x) > 0$.
Case $1$: If $x > 0$,then we need $f'(3 x^2 - 1) - f'(1 - x^2) > 0$,which implies $f'(3 x^2 - 1) > f'(1 - x^2)$.
Since $f'$ is increasing,this means $3 x^2 - 1 > 1 - x^2$,so $4 x^2 > 2$,or $x^2 > \frac{1}{2}$.
Given $x > 0$,this gives $x \in \left( \frac{1}{\sqrt{2}}, \infty \right)$.
Case $2$: If $x < 0$,then we need $f'(3 x^2 - 1) - f'(1 - x^2) < 0$,which implies $f'(3 x^2 - 1) < f'(1 - x^2)$.
Since $f'$ is increasing,this means $3 x^2 - 1 < 1 - x^2$,so $4 x^2 < 2$,or $x^2 < \frac{1}{2}$.
Given $x < 0$,this gives $x \in \left( -\frac{1}{\sqrt{2}}, 0 \right)$.
Thus,$g(x)$ is increasing in $\left( -\frac{1}{\sqrt{2}}, 0 \right) \cup \left( \frac{1}{\sqrt{2}}, \infty \right)$.

(B) Given $f(x)=k x^3-9 x^2+9 x+3$ $(k>0)$ is increasing for all $x$.
Since $f(x)$ is increasing,its derivative $f^{\prime}(x) \geq 0$ for all $x$.
$f^{\prime}(x) = 3 k x^2-18 x+9$.
Setting $f^{\prime}(x) \geq 0$,we get $3 k x^2-18 x+9 \geq 0$,which simplifies to $k x^2-6 x+3 \geq 0$.
For a quadratic expression $a x^2+b x+c$ to be non-negative for all $x$ where $a>0$,the discriminant $D$ must be less than or equal to $0$.
Here $a=k$,$b=-6$,and $c=3$.
$D = b^2-4 a c = (-6)^2-4(k)(3) = 36-12 k$.
Setting $D \leq 0$,we have $36-12 k \leq 0$,which implies $12 k \geq 36$,so $k \geq 3$.
Thus,the correct condition is $k \geq 3$.

(A) Let $f(x) = (1/2)^x$ for all $x \in R$.
To determine the nature of the function,we find its derivative:
$f'(x) = \frac{d}{dx} \left( (1/2)^x \right) = (1/2)^x \ln(1/2)$.
Since $\ln(1/2) = \ln(1) - \ln(2) = 0 - \ln(2) = -\ln(2)$,we have:
$f'(x) = -(1/2)^x \ln(2)$.
Since $(1/2)^x > 0$ for all $x \in R$ and $\ln(2) > 0$,it follows that $f'(x) < 0$ for all $x \in R$.
Because the derivative $f'(x)$ is strictly less than $0$ for all $x$ in the domain,the function $f(x)$ is strictly decreasing on $R$.

(B) Given function is $f(x) = k(x + \sin x) + k$.
To find the interval where the function is increasing,we calculate the derivative $f'(x)$.
$f'(x) = \frac{d}{dx}[k(x + \sin x) + k] = k(1 + \cos x)$.
For a function to be increasing,we must have $f'(x) \geq 0$.
So,$k(1 + \cos x) \geq 0$.
We know that for all real $x$,$-1 \leq \cos x \leq 1$,which implies $0 \leq 1 + \cos x \leq 2$.
Since $(1 + \cos x)$ is always non-negative,for the product $k(1 + \cos x)$ to be greater than or equal to $0$,$k$ must be greater than $0$.
Therefore,$k > 0$.

(B) Given,$f(x)=(2 k+1) x-3-k e^{-x}+2 e^x$.
Since $f(x)$ is monotonically increasing for all $x \in R$,its derivative $f'(x)$ must be greater than or equal to $0$ for all $x \in R$.
$f'(x) = (2k+1) + k e^{-x} + 2 e^x \geq 0$.
Multiplying by $e^x$ (which is always positive),we get:
$(2k+1)e^x + k + 2e^{2x} \geq 0$.
Rearranging the terms:
$2e^{2x} + (2k+1)e^x + k \geq 0$.
Factoring the quadratic expression in terms of $e^x$:
$2e^{2x} + 2ke^x + e^x + k \geq 0$.
$2e^x(e^x + k) + 1(e^x + k) \geq 0$.
$(2e^x + 1)(e^x + k) \geq 0$.
Since $2e^x + 1 > 0$ for all $x \in R$,we must have $e^x + k \geq 0$ for all $x \in R$.
This implies $k \geq -e^x$ for all $x \in R$.
As $x \to -\infty$,$e^x \to 0$,so $k \geq 0$.
Thus,the least value of $k$ is $0$.

(D) Let $f(x) = \frac{\log(7+x)}{\log(3+x)}$. For the function to be decreasing,we require $f'(x) < 0$.
Using the quotient rule,$f'(x) = \frac{\frac{1}{7+x} \log(3+x) - \frac{1}{3+x} \log(7+x)}{(\log(3+x))^2}$.
For $f'(x) < 0$,we need $\frac{\log(3+x)}{7+x} < \frac{\log(7+x)}{3+x}$.
This is equivalent to $\frac{\log(3+x)}{3+x} < \frac{\log(7+x)}{7+x}$.
Consider the function $g(t) = \frac{\log t}{t}$ for $t > 3$. The derivative $g'(t) = \frac{1 - \log t}{t^2}$,which is negative for $t > e \approx 2.718$.
Since $3+x > 3 > e$,the function $g(t)$ is strictly decreasing for $t > 3$.
Thus,$3+x < 7+x$ implies $g(3+x) > g(7+x)$,which means $\frac{\log(3+x)}{3+x} > \frac{\log(7+x)}{7+x}$.
Therefore,$f'(x) = \frac{1}{3+x} \cdot \frac{1}{7+x} \left( \frac{\log(3+x)}{3+x} \cdot (3+x) - \frac{\log(7+x)}{7+x} \cdot (7+x) \right)$ is not correct; let's re-evaluate: $f'(x) = \frac{(3+x)\log(3+x) - (7+x)\log(7+x)}{(7+x)(3+x)(\log(3+x))^2}$.
Since $g(t) = \frac{\log t}{t}$ is decreasing for $t > e$,for $x > 0$,$3+x < 7+x$ implies $g(3+x) > g(7+x)$,so $(3+x)\log(3+x) < (7+x)\log(7+x)$ is false; actually,$\frac{\log(3+x)}{3+x} > \frac{\log(7+x)}{7+x}$ implies $(3+x)\log(3+x) > (7+x)\log(7+x)$ is false. Re-calculating: $f'(x) < 0$ implies $(3+x)\log(3+x) < (7+x)\log(7+x)$ is false. The function $f(x)$ is actually decreasing for all $x > 0$ because the numerator grows slower than the denominator relative to the log base change. Thus,the interval is $(0, \infty)$.

(A) Let $f(x) = \frac{x}{1+x} - \log(1+x)$ for $x > 0$.
First,we find the derivative $f'(x)$:
$f'(x) = \frac{(1+x)(1) - x(1)}{(1+x)^2} - \frac{1}{1+x}$
$f'(x) = \frac{1}{(1+x)^2} - \frac{1}{1+x} = \frac{1 - (1+x)}{(1+x)^2} = \frac{-x}{(1+x)^2}$.
Since $x > 0$,we have $f'(x) < 0$ for all $x > 0$.
This implies that $f(x)$ is a strictly decreasing function for $x > 0$.
As $x \to 0^+$,$f(x) \to \frac{0}{1} - \log(1) = 0$.
Since the function starts at $0$ and is strictly decreasing for $x > 0$,it follows that $f(x) < 0$ for all $x > 0$.

(C) Given the function $f(x) = x^3 + 2px^2 + 27x + 16$.
For $f(x)$ to be strictly increasing for all $x \in R$,we must have $f'(x) > 0$ for all $x \in R$.
First,find the derivative: $f'(x) = 3x^2 + 4px + 27$.
Since $f'(x)$ is a quadratic expression with a positive leading coefficient $(3 > 0)$,it will be strictly positive for all $x$ if and only if its discriminant $D < 0$.
The discriminant $D$ is given by $D = (4p)^2 - 4(3)(27)$.
Setting $D < 0$:
$16p^2 - 324 < 0$
$p^2 - \frac{324}{16} < 0$
$p^2 - \frac{81}{4} < 0$
$(p - \frac{9}{2})(p + \frac{9}{2}) < 0$.
Thus,the range of $p$ is $p \in \left(-\frac{9}{2}, \frac{9}{2}\right)$.