Here are some hints.
Note that as $x$ increases, the sum can get no larger - it is decreasing but not strictly so, because it is sometimes constant. Also with $x=1$ the sum is equal to $7$ so we must have $x \gt 1$ for a sum as low as $5$.
Now if $x\gt 1$ we have immediately that $\left[\frac{3}{x}\right]+\left[\frac{4}{x}\right]\le 2+3=5$ so we must have equality with $\left[\frac{3}{x}\right]=2$ and $\left[\frac{4}{x}\right]=3$
You should be able to complete things from there.
I don't understand your description of the second solution of the second question, but your first solution of that question is correct, the range is $[1,\sqrt{2}]$.
The method for solving the first question is to follow definitions and think logically.
You know that $|\sin(x)| \in [0,1]$ and that $|\cos(x)| \in [0,1]$. Therefore $$|\sin(x)| + |\cos(x)| \in [0,2]
$$
It follows that
$$[|\sin(x)| + |\cos(x)|] \in \{0,1,2\}
$$
In other words, the range of your function is a subset of $\{0,1,2\}$.
So now you have three further questions to pursue:
- Does the range contain $0$?
- Does the range contain $1$?
- Does the range contain $2$?
Question 3 is the easiest: $[|\sin(x)| + |\cos(x)|] = 2$ if and only if $|\sin(x)| = 1$ and $|\cos(x)| = 1$, and I'm sure that you can convince yourself that this is impossible, no matter value of $x$ you pick. Therefore $2$ is not in the range of the function.
Question 2 is also easy: I'm sure that you can find a value of $x$ such that one of $|\sin(x)|$, $|\cos(x)|$ equals $0$ and the other equals $1$, so their sum equals $1$. Therefore $1$ is in the range of the function.
Question 1 is the trickiest. $[|\sin(x)| + |\cos(x)|] = 0$ if and only if $[|\sin(x)| + |\cos(x)|] \in [0,1)$. Can you find a value of $x$ for which this is true? Or perhaps can you prove that this is false for all values of $x$?
Perhaps that's enough for you to proceed.
Best Answer
Let's start with a simple proof showing that when including negative numbers, it's possible for $\bigg[\frac{[x]}{x}\bigg]$ to be any natural number larger than or equal to 1.
Let $x=-\frac{1}{a}$ and $a>1$, then $-1<x<0$. Since $x$ is always between $-1$ and $0$ (in this specific example defining $x$ by $a$), then the greatest integer less than $x$ is always $-1$:
$$\implies[x]=-1$$
So forth, plugging this into the formula
$$\frac{[x]}{x}=\frac{-1}{-\frac{1}{a}}=a$$
Thus, $\forall a>1$ there exists an $x\in(-1,0)$, such that $\frac{[x]}{x}=a$.
So we've shown that $\frac{[x]}{x}$ has range at least $\mathbb{R}_{>1}$, which implies $\bigg[\frac{[x]}{x}\bigg]$, has range of at least $\mathbb{N}_{\geq1}$.
But since $\frac{[x]}{x}\geq0$ $\forall x$ and as you yourself have shown that when $x>0$ then $\bigg[\frac{[x]}{x}\bigg]$ has range $\{0,1\}$ (E.g; $\bigg[\frac{[0.5]}{0.5}\bigg]=0$ and $\bigg[\frac{[1]}{1}\bigg]=1$), then the entire range is $\mathbb{N}_{\geq1}\cup\{0,1\}=\mathbb{N}_0$.
In summary, when $x<0$, we find that $\bigg[\frac{[x]}{x}\bigg]$ can be any natural number greater than or equal to $1$, and when $x>0$ we find that $\bigg[\frac{[x]}{x}\bigg]$ can (and must) be $0$ or $1$, and since $\bigg[\frac{[x]}{x}\bigg]$ as a whole is always greater than or equal to $0$, then the range is all natural numbers including zero.
A good question and a fun solution!