Question

If \(f:(-\infty,0)\to \mathbb{R}\) is defined by \[ f(x)=\frac{[x]}{|x|} \] then \(f(x):x\in(-\infty,0)\) is:

• A. \((-\infty,0)\)
• B. \([-1,0)\)
• C. \((-2,-1]\)
• D. \((-\infty,-1]\)

Hint:

For negative inputs: \[ |x|=-x \] This simplification is very useful in range problems involving modulus.

Answer 1

The Correct Option is D

Step 1: Simplify for negative inputs.
The function is $f(x)=\dfrac{[x]}{|x|}$ on $(-\infty,0)$. For negative $x$ we have $|x|=-x$. So $f(x)=\dfrac{[x]}{-x}$.
Step 2: Look at one block of values.
Take $x$ in $[-n,-n+1)$ where $n$ is a natural number. On this block the greatest integer is $[x]=-n$.
Step 3: Put this in the function.
Then $f(x)=\dfrac{-n}{-x}=\dfrac{n}{x}$. Since $x$ is negative, this value stays negative.
Step 4: Find the value at the left end.
At $x=-n$ we get $f=\dfrac{n}{-n}=-1$. So the block starts at $-1$.
Step 5: See where it goes.
As $x$ moves toward $-n+1$, the value $\dfrac{n}{x}$ drops below $-1$ and keeps going more negative. So each block covers values from $-1$ downward.
Step 6: Join all blocks.
Putting all blocks for $n=1,2,3,\dots$ together, the outputs fill everything from $-1$ down to $-\infty$, with $-1$ included. \[ \boxed{(-\infty,-1]} \]