Increasing and Decreasing function Questions in English

(B) To find the intervals where $f(x)$ is increasing,we find the derivative $f'(x)$ and set $f'(x) > 0$.
Given $f(x) = 3x^4 - 8x^3 - 6x^2 + 24x - 12$.
$f'(x) = 12x^3 - 24x^2 - 12x + 24$.
Factor out $12$: $f'(x) = 12(x^3 - 2x^2 - x + 2)$.
Factor by grouping: $f'(x) = 12[x^2(x - 2) - 1(x - 2)] = 12(x^2 - 1)(x - 2) = 12(x - 1)(x + 1)(x - 2)$.
For $f(x)$ to be increasing,$f'(x) > 0$,so $(x - 1)(x + 1)(x - 2) > 0$.
Using the wavy curve method (sign scheme) for the critical points $x = -1, 1, 2$:
For $x \in (-1, 1)$,$f'(x) > 0$.
For $x \in (2, \infty)$,$f'(x) > 0$.
Thus,the function is increasing on $(-1, 1) \cup (2, \infty)$.

(D) Given function is $f(x) = 2x^2 - \log x$ for $x > 0$.
To find the interval where the function decreases,we find the derivative $f'(x)$.
$f'(x) = \frac{d}{dx}(2x^2 - \log x) = 4x - \frac{1}{x}$.
The function $f(x)$ is decreasing when $f'(x) < 0$.
$4x - \frac{1}{x} < 0$
$\Rightarrow \frac{4x^2 - 1}{x} < 0$.
Since $x > 0$,we can multiply by $x$ without changing the inequality sign:
$4x^2 - 1 < 0$
$\Rightarrow 4x^2 < 1$
$\Rightarrow x^2 < \frac{1}{4}$
$\Rightarrow |x| < \frac{1}{2}$.
This gives $x \in (-\frac{1}{2}, \frac{1}{2})$.
Given the condition $x > 0$,the interval is $x \in (0, \frac{1}{2})$.
Thus,the function decreases in the interval $(0, \frac{1}{2})$.

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

(B) Given $f(x) = \frac{x}{3} + \frac{3}{x}$.
First,we find the derivative $f'(x)$:
$f'(x) = \frac{1}{3} - \frac{3}{x^2} = \frac{x^2 - 9}{3x^2}$.
For the function to be decreasing,we need $f'(x) < 0$.
Since $3x^2 > 0$ for all $x \neq 0$,the sign of $f'(x)$ depends on the numerator $x^2 - 9$.
We have $x^2 - 9 < 0$ when $x^2 < 9$,which implies $|x| < 3$,or $x \in (-3, 3)$.
Since $f'(x) < 0$ for all $x \in (-3, 3) \setminus \{0\}$,the function $f(x)$ is decreasing on the interval $(-3, 3)$.

(A) Given function: $f(x) = 2x^3 - 9x^2 + 12x - 3$ $(i)$
Differentiating with respect to $x$:
$f^{\prime}(x) = 6x^2 - 18x + 12 = 6(x^2 - 3x + 2) = 6(x - 1)(x - 2)$ (ii)
For $f(x)$ to be increasing,$f^{\prime}(x) > 0$.

$6(x - 1)(x - 2) > 0$

This holds when $x < 1$ or $x > 2$.

Thus,$f(x)$ is increasing outside the interval $(1, 2)$. Hence,statement $A$ is true.
For statement $R$,we check the condition for decreasing function: $f^{\prime}(x) < 0$.

$6(x - 1)(x - 2) < 0$

This holds when $1 < x < 2$.

Thus,$f^{\prime}(x) < 0$ for $x \in (1, 2)$. Hence,statement $R$ is true.
Since statement $A$ describes the increasing behavior and statement $R$ describes the decreasing behavior,$R$ is not the reason for $A$.
Therefore,both $A$ and $R$ are true,and $R$ is not the correct reason for $A$.

(A) Let $f(x) = \log(1+x) - \frac{2x}{2+x}$.
For the function to be defined,we must have $1+x > 0$,which implies $x > -1$.
Now,find the derivative $f'(x)$:
$f'(x) = \frac{1}{1+x} - \frac{(2+x)(2) - 2x(1)}{(2+x)^2}$
$f'(x) = \frac{1}{1+x} - \frac{4+2x-2x}{(2+x)^2} = \frac{1}{1+x} - \frac{4}{(2+x)^2}$
For the function to be increasing,$f'(x) > 0$:
$\frac{(2+x)^2 - 4(1+x)}{(1+x)(2+x)^2} > 0$
$\frac{4+x^2+4x - 4 - 4x}{(1+x)(2+x)^2} > 0$
$\frac{x^2}{(1+x)(2+x)^2} > 0$
Since $x^2 \ge 0$ and $(2+x)^2 > 0$ for $x \neq -2$,the inequality holds when $1+x > 0$ and $x \neq 0$.
Thus,$x > -1$ and $x \neq 0$.

(D) Given the curve $y = x^3 - ax^2 + x + 1$.
The slope of the tangent is given by $\frac{dy}{dx} = 3x^2 - 2ax + 1$.
Since the tangent makes an acute angle with the positive direction of the $X$-axis,the slope must be positive for all $x \in R$.
Thus,$3x^2 - 2ax + 1 > 0$ for all $x \in R$.
For a quadratic expression $Ax^2 + Bx + C > 0$ to be true for all $x \in R$,we must have $A > 0$ and the discriminant $D < 0$.
Here,$A = 3 > 0$,which is satisfied.
Now,we require $D = B^2 - 4AC < 0$.
Substituting the values,$(-2a)^2 - 4(3)(1) < 0$.
$4a^2 - 12 < 0$.
$a^2 - 3 < 0$.
$(a - \sqrt{3})(a + \sqrt{3}) < 0$.
Therefore,$a \in (-\sqrt{3}, \sqrt{3})$.

(A) Given the function $f(x)=\sqrt{3} \sin x-\cos x-2 a x+b$.
To find the condition for $f(x)$ to be decreasing,we first find its derivative $f^{\prime}(x)$.
$f^{\prime}(x)=\sqrt{3} \cos x+\sin x-2 a$.
We can rewrite this as $f^{\prime}(x)=2 \left( \frac{\sqrt{3}}{2} \cos x+\frac{1}{2} \sin x \right)-2 a$.
Using the identity $\sin(A+B) = \sin A \cos B + \cos A \sin B$,we get $f^{\prime}(x)=2 \sin \left(x+\frac{\pi}{3}\right)-2 a$.
For $f(x)$ to decrease for all $x$,we must have $f^{\prime}(x) \leq 0$ for all $x$.
$2 \sin \left(x+\frac{\pi}{3}\right)-2 a \leq 0$.
$2 a \geq 2 \sin \left(x+\frac{\pi}{3}\right)$.
$a \geq \sin \left(x+\frac{\pi}{3}\right)$.
Since the maximum value of $\sin \left(x+\frac{\pi}{3}\right)$ is $1$,for the inequality to hold for all $x$,we must have $a \geq 1$.

(D) Let $f(x) = y = x^4+5x^3+9x^2+6x+2$.
The slope of the tangent to the curve is given by $m(x) = \frac{dy}{dx} = 4x^3+15x^2+18x+6$.
For the slope $m(x)$ to be increasing,its derivative must be greater than or equal to zero,i.e.,$\frac{dm}{dx} = \frac{d^2y}{dx^2} \ge 0$.
Calculating the second derivative: $\frac{d^2y}{dx^2} = 12x^2+30x+18$.
Factoring the quadratic expression: $12x^2+30x+18 = 6(2x^2+5x+3) = 6(2x+3)(x+1)$.
We need $6(2x+3)(x+1) \ge 0$.
The critical points are $x = -\frac{3}{2}$ and $x = -1$.
Testing the intervals:
$1$) For $x \in (-\infty, -\frac{3}{2}]$,$\frac{d^2y}{dx^2} \ge 0$.
$2$) For $x \in [-\frac{3}{2}, -1]$,$\frac{d^2y}{dx^2} \le 0$.
$3$) For $x \in [-1, \infty)$,$\frac{d^2y}{dx^2} \ge 0$.
Thus,the slope increases in the interval $(-\infty, -\frac{3}{2}] \cup [-1, \infty)$,which can be written as $R - (-\frac{3}{2}, -1)$.

(B) To determine the nature of the function $f(x) = x + \log \left( \frac{x-1}{x+1} \right)$,we first find its derivative $f'(x)$.
First,note that for the function to be well-defined,we require $\frac{x-1}{x+1} > 0$,which implies $x \in (-\infty, -1) \cup (1, \infty)$.
Now,differentiate $f(x)$ with respect to $x$:
$f'(x) = 1 + \frac{d}{dx} [\log(x-1) - \log(x+1)]$
$f'(x) = 1 + \left( \frac{1}{x-1} - \frac{1}{x+1} \right)$
$f'(x) = 1 + \frac{(x+1) - (x-1)}{(x-1)(x+1)}$
$f'(x) = 1 + \frac{2}{x^2 - 1}$
$f'(x) = \frac{x^2 - 1 + 2}{x^2 - 1} = \frac{x^2 + 1}{x^2 - 1}$
Since $x^2 + 1 > 0$ for all real $x$,the sign of $f'(x)$ depends on the denominator $x^2 - 1$.
For $x \in (1, \infty)$,$x^2 > 1$,so $x^2 - 1 > 0$,which means $f'(x) > 0$. Thus,$f$ is increasing in $(1, \infty)$.
For $x \in (-\infty, -1)$,$x^2 > 1$,so $x^2 - 1 > 0$,which means $f'(x) > 0$. Thus,$f$ is increasing in $(-\infty, -1)$.
Therefore,$f$ is a monotonically increasing function in its domain.

(B) Given $f(x) = (x^2+x+1)^{(x^2-3x-4)}$.
Taking the natural logarithm on both sides,we get $\ln(f(x)) = (x^2-3x-4) \ln(x^2+x+1)$.
Differentiating with respect to $x$:
$\frac{f'(x)}{f(x)} = (2x-3) \ln(x^2+x+1) + (x^2-3x-4) \cdot \frac{2x+1}{x^2+x+1}$.
For $x \in [4, \infty)$,$x^2+x+1 > 0$ and $x^2-3x-4 = (x-4)(x+1) \ge 0$.
Since $x \ge 4$,$(2x-3) > 0$ and $\ln(x^2+x+1) > 0$,and $(x^2-3x-4) \ge 0$ and $(2x+1) > 0$.
Thus,$f'(x) = f(x) [ (2x-3) \ln(x^2+x+1) + \frac{(x^2-3x-4)(2x+1)}{x^2+x+1} ] > 0$ for all $x > 4$.
Therefore,$f(x)$ is a monotonically increasing function.

(A) function $f(x)$ is monotonic if it is either non-increasing or non-decreasing for all $x \in \mathbb{R}$.
This implies that its derivative $f'(x)$ must not change sign.
Given $f(x) = 4x^3 + ax^2 + 3x - 2$,we find the derivative:
$f'(x) = 12x^2 + 2ax + 3$.
For $f(x)$ to be monotonic,$f'(x) \geq 0$ or $f'(x) \leq 0$ for all $x$.
Since the coefficient of $x^2$ is $12 > 0$,the parabola $f'(x)$ opens upwards,so we must have $f'(x) \geq 0$ for all $x$.
This condition holds if the discriminant $D \leq 0$.
The discriminant $D = (2a)^2 - 4(12)(3) = 4a^2 - 144$.
Setting $D \leq 0$ gives $4a^2 - 144 \leq 0$,which simplifies to $a^2 \leq 36$.
Thus,$-6 \leq a \leq 6$,or $a \in [-6, 6]$.
Comparing this with the given options,the interval $(-6, 6)$ is the most appropriate subset provided.

(B) To determine the intervals of increase and decrease,we find the derivative $f'(x)$.
$f'(x) = \frac{d}{dx} \left( \frac{x^2}{2} - \log(x^2 + x + 1) \right) = x - \frac{2x + 1}{x^2 + x + 1}$.
$f'(x) = \frac{x(x^2 + x + 1) - (2x + 1)}{x^2 + x + 1} = \frac{x^3 + x^2 + x - 2x - 1}{x^2 + x + 1} = \frac{x^3 + x^2 - x - 1}{x^2 + x + 1}$.
Factoring the numerator: $x^2(x + 1) - 1(x + 1) = (x^2 - 1)(x + 1) = (x - 1)(x + 1)^2$.
So,$f'(x) = \frac{(x - 1)(x + 1)^2}{x^2 + x + 1}$.
Since $x^2 + x + 1 > 0$ and $(x + 1)^2 \ge 0$ for all real $x$,the sign of $f'(x)$ depends on $(x - 1)$.
$f'(x) > 0$ when $x > 1$,so the function is strictly increasing in $(1, \infty)$.
Thus,option $B$ is correct.

(A) The domain of the function $f(x) = \log \left(\frac{1+x}{1-x}\right) - 2x - \frac{x^3}{1-x^2}$ is determined by $\frac{1+x}{1-x} > 0$,which implies $x \in (-1, 1)$.
First,we simplify the expression for $f(x)$:
$f(x) = \log(1+x) - \log(1-x) - 2x - \frac{x^3}{1-x^2}$.
Now,we find the derivative $f'(x)$:
$f'(x) = \frac{1}{1+x} + \frac{1}{1-x} - 2 - \frac{d}{dx} \left( \frac{x^3}{1-x^2} \right)$.
$f'(x) = \frac{(1-x) + (1+x)}{1-x^2} - 2 - \frac{3x^2(1-x^2) - x^3(-2x)}{(1-x^2)^2}$.
$f'(x) = \frac{2}{1-x^2} - 2 - \frac{3x^2 - 3x^4 + 2x^4}{(1-x^2)^2}$.
$f'(x) = \frac{2(1-x^2) - 2(1-x^2)^2 - (3x^2 - x^4)}{(1-x^2)^2}$.
$f'(x) = \frac{2 - 2x^2 - 2(1 - 2x^2 + x^4) - 3x^2 + x^4}{(1-x^2)^2}$.
$f'(x) = \frac{2 - 2x^2 - 2 + 4x^2 - 2x^4 - 3x^2 + x^4}{(1-x^2)^2} = \frac{-x^4 - x^2}{(1-x^2)^2} = -\frac{x^2(x^2+1)}{(1-x^2)^2}$.
Since $x^2(x^2+1) > 0$ for all $x \in (-1, 1) \setminus \{0\}$ and $(1-x^2)^2 > 0$,$f'(x) < 0$ for all $x \in (-1, 1)$.
Thus,the function is decreasing on the entire interval $(-1, 1)$.
Here,$a = -1$ and $b = 1$,so $\frac{a}{b} = \frac{-1}{1} = -1$.

(A) Given the function $f(x) = kx^3 - 3x^2 - 12x + 8$.
For $f(x)$ to be strictly decreasing for all $x \in R$,we must have $f'(x) < 0$ for all $x \in R$.
Calculating the derivative: $f'(x) = 3kx^2 - 6x - 12$.
For a quadratic expression $ax^2 + bx + c$ to be strictly negative for all $x$,we require $a < 0$ and the discriminant $D < 0$.
Here,$a = 3k$,$b = -6$,and $c = -12$.
Condition $1$: $3k < 0 \Rightarrow k < 0$.
Condition $2$: $D = b^2 - 4ac < 0$.
$(-6)^2 - 4(3k)(-12) < 0$.
$36 + 144k < 0$.
$144k < -36$.
$k < -\frac{36}{144} \Rightarrow k < -\frac{1}{4}$.
Since $k < -\frac{1}{4}$ already satisfies $k < 0$,the final condition is $k < -\frac{1}{4}$.

(A) Given: $f(x) = \tan x - x$.
Differentiating with respect to $x$,we get $f'(x) = \sec^2 x - 1$.
We know that $\sec^2 x \geq 1$ for all $x$ in the domain $R^*$.
Therefore,$f'(x) = \sec^2 x - 1 \geq 0$.
Since the derivative $f'(x) \geq 0$ for all $x \in R^*$,the function $f(x)$ is an increasing function.
Hence,option $A$ is correct.

(C) Given,$f(x) = \frac{1}{2}[|\sin x| + \sin x]$ for $0 < x \leq 2\pi$.
Case $I$: When $0 < x \leq \pi$,$\sin x \geq 0$,so $|\sin x| = \sin x$.
Thus,$f(x) = \frac{1}{2}[\sin x + \sin x] = \sin x$.
The derivative is $f'(x) = \cos x$.
For $0 < x < \frac{\pi}{2}$,$f'(x) = \cos x > 0$,so $f(x)$ is increasing.
For $\frac{\pi}{2} < x < \pi$,$f'(x) = \cos x < 0$,so $f(x)$ is decreasing.
Case $II$: When $\pi < x \leq 2\pi$,$\sin x \leq 0$,so $|\sin x| = -\sin x$.
Thus,$f(x) = \frac{1}{2}[-\sin x + \sin x] = 0$.
Therefore,$f(x)$ is increasing in $\left(0, \frac{\pi}{2}\right)$ and decreasing in $\left(\frac{\pi}{2}, \pi\right)$.

(A) For the function $f(x) = 2x^3 - 9x^2 + 12x - 3$,the derivative is $f'(x) = 6x^2 - 18x + 12 = 6(x^2 - 3x + 2) = 6(x-1)(x-2)$.
For $f(x)$ to be increasing,we require $f'(x) > 0$,which implies $6(x-1)(x-2) > 0$. This inequality holds when $x < 1$ or $x > 2$. Thus,$f(x)$ is increasing outside the interval $(1,2)$. Hence,statement $A$ is true.
For statement $R$,we check the condition $f'(x) < 0$. This inequality holds when $6(x-1)(x-2) < 0$,which occurs for $x \in (1,2)$. Thus,statement $R$ is true.
Since statement $A$ describes the increasing behavior and statement $R$ describes the decreasing behavior,$R$ is not the reason for $A$. Therefore,both $A$ and $R$ are true,but $R$ is not the correct explanation for $A$.

(A) Given function: $f(x) = x - \log \left(\frac{1+x}{x}\right), x > 0$.
Differentiating with respect to $x$:
$f^{\prime}(x) = \frac{d}{dx}(x) - \frac{d}{dx} \log \left(\frac{1+x}{x}\right) = 1 - \frac{x}{1+x} \cdot \frac{d}{dx} \left( \frac{1}{x} + 1 \right) = 1 - \frac{x}{1+x} \cdot \left( -\frac{1}{x^2} \right) = 1 + \frac{1}{x(1+x)}$.
Since $x > 0$,$x(1+x) > 0$,therefore $f^{\prime}(x) = 1 + \frac{1}{x(1+x)} > 1 > 0$ for all $x > 0$.
Since $f^{\prime}(x) > 0$ for all $x$ in the domain,the function $f(x)$ is strictly increasing and has no local maximum or minimum.
The Reason $(R)$ states that if a function is strictly increasing,then $f^{\prime}(x) \neq 0$. This is a standard property of strictly increasing functions.
Thus,both $(A)$ and $(R)$ are true,and $(R)$ correctly explains $(A)$.

(A) Given $f(x) = \int_x^{x+1} e^{-t^2} dt$.
Using Leibniz's rule for differentiation under the integral sign,we have:
$f'(x) = e^{-(x+1)^2} \cdot \frac{d}{dx}(x+1) - e^{-x^2} \cdot \frac{d}{dx}(x)$
$f'(x) = e^{-(x+1)^2} - e^{-x^2}$
For $f(x)$ to be a decreasing function,we must have $f'(x) < 0$.
$e^{-(x+1)^2} - e^{-x^2} < 0$
$e^{-(x+1)^2} < e^{-x^2}$
Since the exponential function $e^u$ is strictly increasing,we have:
$-(x+1)^2 < -x^2$
Multiplying by $-1$ reverses the inequality:
$(x+1)^2 > x^2$
$x^2 + 2x + 1 > x^2$
$2x + 1 > 0$
$2x > -1$
$x > -\frac{1}{2}$
Thus,the interval in which $f(x)$ is decreasing is $\left(-\frac{1}{2}, \infty\right)$.

(C) Given $f(x) = \cos x - 1 + \frac{x^2}{2!}$.
Taking the derivative with respect to $x$,we get $f'(x) = -\sin x + x$.
For $x > 0$,we know that $x > \sin x$,which implies $x - \sin x > 0$. Thus,$f'(x) > 0$ for $x > 0$.
For $x < 0$,we know that $x < \sin x$,which implies $x - \sin x < 0$. Thus,$f'(x) < 0$ for $x < 0$.
Since the derivative $f'(x)$ changes sign at $x = 0$,the function $f(x)$ is neither increasing nor decreasing on the entire domain $R$.

(C) Given $f(x) = x^{13} + x^{11} + x^{9} + x^{7} + x^{5} + x^{3} + x + 19$.
Taking the derivative with respect to $x$:
$f'(x) = 13x^{12} + 11x^{10} + 9x^{8} + 7x^{6} + 5x^{4} + 3x^{2} + 1$.
Since all exponents of $x$ in $f'(x)$ are even and the coefficients are positive,$f'(x) \geq 1$ for all real $x$.
Thus,$f'(x) > 0$ for all $x \in \mathbb{R}$,which means $f(x)$ is a strictly increasing function.
$A$ strictly increasing function can intersect the $x$-axis at most once.
Therefore,$f(x) = 0$ has not more than one real root.

(B) Given $\phi(x) = f(x) + f(2a - x)$.
Taking the derivative with respect to $x$,we get $\phi^{\prime}(x) = f^{\prime}(x) - f^{\prime}(2a - x)$.
We are given $f^{\prime \prime}(x) > 0$ for all $x \in [0, a]$,which implies that $f^{\prime}(x)$ is a strictly increasing function.
For $x \in [0, a]$,we have $x < 2a - x$.
Since $f^{\prime}(x)$ is strictly increasing,$x < 2a - x$ implies $f^{\prime}(x) < f^{\prime}(2a - x)$.
Therefore,$\phi^{\prime}(x) = f^{\prime}(x) - f^{\prime}(2a - x) < 0$ for all $x \in [0, a]$.
Since $\phi^{\prime}(x) < 0$ on $[0, a]$,$\phi(x)$ is a decreasing function on $[0, a]$.

(A) Given $\phi(x) = f(x) + f(1-x)$.
To determine the monotonicity,we find the derivative $\phi^{\prime}(x) = f^{\prime}(x) - f^{\prime}(1-x)$.
Since $f^{\prime \prime}(x) < 0$,the function $f^{\prime}(x)$ is a strictly decreasing function.
If $x < \frac{1}{2}$,then $x < 1-x$,which implies $f^{\prime}(x) > f^{\prime}(1-x)$ because $f^{\prime}$ is decreasing.
Thus,$\phi^{\prime}(x) = f^{\prime}(x) - f^{\prime}(1-x) > 0$ for $x \in \left[0, \frac{1}{2}\right]$,so $\phi(x)$ is increasing in $\left[0, \frac{1}{2}\right]$.
If $x > \frac{1}{2}$,then $x > 1-x$,which implies $f^{\prime}(x) < f^{\prime}(1-x)$ because $f^{\prime}$ is decreasing.
Thus,$\phi^{\prime}(x) = f^{\prime}(x) - f^{\prime}(1-x) < 0$ for $x \in \left[\frac{1}{2}, 1\right]$,so $\phi(x)$ is decreasing in $\left[\frac{1}{2}, 1\right]$.

(B) Let $f(x) = \cos x + x \sin x - 1$.
Find the derivative of $f(x)$ with respect to $x$:
$f'(x) = -\sin x + (1 \cdot \sin x + x \cos x) = x \cos x$.
For $x \in \left(0, \frac{\pi}{2}\right)$,both $x > 0$ and $\cos x > 0$.
Therefore,$f'(x) = x \cos x > 0$ for all $x \in \left(0, \frac{\pi}{2}\right)$.
Since $f'(x) > 0$,the function $f(x)$ is strictly increasing on the interval $\left(0, \frac{\pi}{2}\right)$.
As $f(x)$ is increasing,for any $x > 0$,we have $f(x) > f(0)$.
Calculate $f(0)$:
$f(0) = \cos(0) + 0 \cdot \sin(0) - 1 = 1 + 0 - 1 = 0$.
Thus,$f(x) > 0$ for all $x \in \left(0, \frac{\pi}{2}\right)$.
$\cos x + x \sin x - 1 > 0$
$\cos x + x \sin x > 1$.

(A) Given $f(x) = \cos \left(\frac{\pi}{x}\right)$.
Taking the derivative with respect to $x$:
$f'(x) = -\sin \left(\frac{\pi}{x}\right) \cdot \left(-\frac{\pi}{x^2}\right) = \frac{\pi}{x^2} \sin \left(\frac{\pi}{x}\right)$.
For $f(x)$ to be increasing,$f'(x) > 0$.
Since $\frac{\pi}{x^2} > 0$ for all $x \neq 0$,we need $\sin \left(\frac{\pi}{x}\right) > 0$.
This occurs when $2k\pi < \frac{\pi}{x} < (2k+1)\pi$ for some integer $k$.
Dividing by $\pi$,we get $2k < \frac{1}{x} < 2k+1$.
Taking the reciprocal,the inequality reverses: $\frac{1}{2k+1} < x < \frac{1}{2k}$.
Thus,$f(x)$ increases in the interval $\left(\frac{1}{2k+1}, \frac{1}{2k}\right)$.
For $f(x)$ to be decreasing,$f'(x) < 0$,which implies $\sin \left(\frac{\pi}{x}\right) < 0$.
This occurs when $(2k+1)\pi < \frac{\pi}{x} < (2k+2)\pi$.
Dividing by $\pi$,we get $2k+1 < \frac{1}{x} < 2k+2$.
Taking the reciprocal,we get $\frac{1}{2k+2} < x < \frac{1}{2k+1}$.
Thus,$f(x)$ decreases in the interval $\left(\frac{1}{2k+2}, \frac{1}{2k+1}\right)$.

(B) Given $f(x) = \sin x - \cos x - Kx + 5$.
Differentiating with respect to $x$,we get $f'(x) = \cos x + \sin x - K$.
For the function to be decreasing for all $x > 0$,we must have $f'(x) \leq 0$ for all $x$.
This implies $\cos x + \sin x - K \leq 0$,or $K \geq \cos x + \sin x$.
We know that the maximum value of $\cos x + \sin x$ is $\sqrt{1^2 + 1^2} = \sqrt{2}$.
Therefore,for $K \geq \cos x + \sin x$ to hold for all $x$,$K$ must be greater than or equal to the maximum value of the expression $\cos x + \sin x$.
Thus,$K \geq \sqrt{2}$.

(A) Given function is $f(x)=e^{x}(x-2)^{2}$.
To find the intervals of increase and decrease,we find the derivative $f^{\prime}(x)$:
$f^{\prime}(x) = \frac{d}{dx}[e^{x}(x-2)^{2}]$
Using the product rule:
$f^{\prime}(x) = e^{x}(x-2)^{2} + e^{x} \cdot 2(x-2)$
$f^{\prime}(x) = e^{x}(x-2)[(x-2) + 2]$
$f^{\prime}(x) = x(x-2)e^{x}$
Now,we determine the sign of $f^{\prime}(x)$:
Since $e^{x} > 0$ for all $x \in \mathbb{R}$,the sign of $f^{\prime}(x)$ depends on $x(x-2)$.
- For $x \in (-\infty, 0)$,$x < 0$ and $(x-2) < 0$,so $f^{\prime}(x) > 0$ (increasing).
- For $x \in (0, 2)$,$x > 0$ and $(x-2) < 0$,so $f^{\prime}(x) < 0$ (decreasing).
- For $x \in (2, \infty)$,$x > 0$ and $(x-2) > 0$,so $f^{\prime}(x) > 0$ (increasing).
Thus,$f$ is increasing in $(-\infty, 0)$ and $(2, \infty)$ and decreasing in $(0, 2)$.

(D) Given the function $F(x) = \int_{0}^{x} \frac{\cos t}{1+t^{2}} dt$ for $0 \leq x \leq 2\pi$.
To determine the intervals of increase and decrease,we find the derivative $F'(x)$ using the Leibniz rule:
$F'(x) = \frac{\cos x}{1+x^{2}}$.
Since $1+x^{2} > 0$ for all $x$,the sign of $F'(x)$ depends solely on $\cos x$.
$F(x)$ is increasing when $F'(x) > 0$,which occurs when $\cos x > 0$. In the interval $[0, 2\pi]$,$\cos x > 0$ for $x \in (0, \frac{\pi}{2}) \cup (\frac{3\pi}{2}, 2\pi)$.
$F(x)$ is decreasing when $F'(x) < 0$,which occurs when $\cos x < 0$. In the interval $[0, 2\pi]$,$\cos x < 0$ for $x \in (\frac{\pi}{2}, \frac{3\pi}{2})$.
Thus,$F$ is increasing in $(0, \frac{\pi}{2})$ and $(\frac{3\pi}{2}, 2\pi)$ and decreasing in $(\frac{\pi}{2}, \frac{3\pi}{2})$.

(D) Given,$f(x) f^{\prime}(x) < 0$. This implies that $f(x)$ and $f^{\prime}(x)$ have opposite signs for all $x \in R$.
Consider the function $g(x) = |f(x)|$.
Then $g(x)^2 = |f(x)|^2 = f(x)^2$.
Differentiating both sides with respect to $x$,we get $2g(x) g^{\prime}(x) = 2f(x) f^{\prime}(x)$.
Thus,$g^{\prime}(x) = \frac{f(x) f^{\prime}(x)}{g(x)} = \frac{f(x) f^{\prime}(x)}{|f(x)|}$.
Since $f(x) f^{\prime}(x) < 0$ and $|f(x)| > 0$ for all $x$ where $f(x) \neq 0$,it follows that $g^{\prime}(x) < 0$.
Therefore,$|f(x)|$ is a decreasing function.

(B) The given function is $f(x) = \begin{cases} 0, & x = 0 \\ x - 3, & x > 0 \end{cases}$.
First,let us check the continuity at $x = 0$.
$f(0) = 0$.
$\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} (x - 3) = -3$.
Since $f(0) \neq \lim_{x \to 0^+} f(x)$,the function is discontinuous at $x = 0$.
Now,consider the interval $x > 0$. For any $x_1, x_2$ such that $0 < x_1 < x_2$,we have $f(x_1) = x_1 - 3$ and $f(x_2) = x_2 - 3$. Since $x_1 < x_2$,it follows that $x_1 - 3 < x_2 - 3$,which means $f(x_1) < f(x_2)$. Therefore,the function is strictly increasing for $x > 0$.
However,for the function to be increasing on the interval $[0, \infty)$,it must be continuous on $[0, \infty)$. Since it is discontinuous at $x = 0$,it is not increasing on $[0, \infty)$.
Thus,the function is strictly increasing when $x > 0$.

(A) function $f(x)$ is strictly increasing if its derivative $f'(x) > 0$ for all $x$ in its domain.
Given the function $f(x) = ax + b$.
Taking the derivative with respect to $x$,we get $f'(x) = \frac{d}{dx}(ax + b) = a$.
For the function to be strictly increasing,we must have $f'(x) > 0$.
Substituting the derivative,we get $a > 0$.
Therefore,the function is strictly increasing if $a > 0$.

(B) To find the intervals where the function $f(x) = x^4 - 4x^3 + 4x^2 + 40$ is monotonically decreasing,we first find its derivative $f'(x)$.
$f'(x) = \frac{d}{dx}(x^4 - 4x^3 + 4x^2 + 40) = 4x^3 - 12x^2 + 8x$.
Next,we factor the derivative:
$f'(x) = 4x(x^2 - 3x + 2) = 4x(x - 1)(x - 2)$.
$A$ function is monotonically decreasing when $f'(x) < 0$.
We analyze the sign of $f'(x) = 4x(x - 1)(x - 2)$ using the wavy curve method (sign scheme):
- For $x < 0$,$f'(x) < 0$.
- For $0 < x < 1$,$f'(x) > 0$.
- For $1 < x < 2$,$f'(x) < 0$.
- For $x > 2$,$f'(x) > 0$.
Thus,the function is monotonically decreasing for $x \in (-\infty, 0) \cup (1, 2)$.
Comparing this with the given options,the correct interval is $1 < x < 2$.

(C) Given that $p^{\prime}(x) > 0$ for all $x \in \mathbb{R}$,the polynomial $p(x)$ is a strictly increasing function on the entire real line.
Since $p(x)$ is strictly increasing,it can have at most one real root.
We are given $p(0) = 1$.
Since $p(x)$ is strictly increasing and $p(0) = 1 > 0$,for any $x > 0$,$p(x) > p(0) = 1$,so $p(x)$ cannot have any positive real roots.
As $x \to -\infty$,$p(x) \to -\infty$ (assuming the degree is odd) or $p(x) \to c$ (if degree is even,but $p^{\prime}(x) > 0$ implies the degree must be odd for the limit to be $-\infty$).
Since $p(0) = 1$ and the function is continuous and strictly increasing,by the Intermediate Value Theorem,there must exist some $x_0 < 0$ such that $p(x_0) = 0$.
Thus,$p(x)$ has exactly one negative real root.

(A) Given $g(x) = f((\tan x - 1)^{2} + a - 1)$.
Taking the derivative,$g^{\prime}(x) = f^{\prime}((\tan x - 1)^{2} + a - 1) \cdot 2(\tan x - 1) \cdot \sec^{2}x$.
Since $f^{\prime\prime}(x) > 0$,$f^{\prime}(x)$ is a strictly increasing function.
We are given $f^{\prime}(a-1) = 0$.
If $(\tan x - 1)^{2} > 0$,then $(\tan x - 1)^{2} + a - 1 > a - 1$,which implies $f^{\prime}((\tan x - 1)^{2} + a - 1) > f^{\prime}(a - 1) = 0$.
For $x \in (0, \frac{\pi}{4})$,$\tan x < 1$,so $(\tan x - 1) < 0$. Thus $g^{\prime}(x) = (\text{positive}) \cdot (\text{negative}) \cdot (\text{positive}) < 0$. So $g$ is decreasing in $(0, \frac{\pi}{4})$.
For $x \in (\frac{\pi}{4}, \frac{\pi}{2})$,$\tan x > 1$,so $(\tan x - 1) > 0$. Thus $g^{\prime}(x) = (\text{positive}) \cdot (\text{positive}) \cdot (\text{positive}) > 0$. So $g$ is increasing in $(\frac{\pi}{4}, \frac{\pi}{2})$.
Therefore,neither statement $(I)$ nor $(II)$ is true.

(A) function $f(x)$ is strictly decreasing on an interval if its derivative $f'(x) < 0$ for all $x$ in that interval.
Given $f(x) = \tan x - 4x$.
Taking the derivative with respect to $x$,we get $f'(x) = \sec^2 x - 4$.
For the function to be strictly decreasing,we require $f'(x) < 0$.
$\sec^2 x - 4 < 0$
$\sec^2 x < 4$
Taking the square root on both sides,we get $|\sec x| < 2$.
Since $|\sec x| = 1/|\cos x|$,this implies $1/|\cos x| < 2$,which means $|\cos x| > 1/2$.
This inequality holds when $-\frac{\pi}{3} < x < \frac{\pi}{3}$.

(A) To find the interval where the function $y = x^2 e^{-x}$ is decreasing,we calculate its derivative $\frac{dy}{dx}$.
Using the product rule: $\frac{dy}{dx} = \frac{d}{dx}(x^2) \cdot e^{-x} + x^2 \cdot \frac{d}{dx}(e^{-x}) = 2x e^{-x} - x^2 e^{-x} = x e^{-x} (2 - x)$.
$A$ function is decreasing when $\frac{dy}{dx} < 0$.
Since $e^{-x} > 0$ for all real $x$,the inequality $\frac{dy}{dx} < 0$ simplifies to $x(2 - x) < 0$.
Multiplying by $-1$ reverses the inequality: $x(x - 2) > 0$.
This inequality holds when $x < 0$ or $x > 2$.
Thus,the function is decreasing on the interval $(-\infty, 0) \cup (2, \infty)$.