If $f(x) = kx^3 - 9x^2 + 9x + 3$ is monotonically increasing in each interval,then

Where does $f(x) = x + \sqrt{1 - x}, 0 < x < 1$ decrease?

Let $f : R \rightarrow R$ be a twice differentiable function such that $f^{\prime\prime}(x) > 0$ for all $x \in R$ and $f^{\prime}(a-1) = 0$, where $a$ is a real number. Let $g(x) = f(\tan^{2}x - 2\tan x + a)$, $0 < x < \frac{\pi}{2}$. Consider the following two statements:
$(I)$ $g$ is increasing in $(0, \frac{\pi}{4})$
$(II)$ $g$ is decreasing in $(\frac{\pi}{4}, \frac{\pi}{2})$
Then,

The function $f(x) = x^3 - 3x^2 - 24x + 5$ is an increasing function in the interval given below:

The values of $a$ for which the function $f(x) = (a + 2)x^3 - 3ax^2 + 9ax - 1$ decreases monotonically for all real $x$ are:

If $f(x) = \frac{a \sin x + b \cos x}{c \sin x + d \cos x}$ is decreasing for all $x$,then