Let $f(x) = [x]^2 - [x+3] - 3, x \in \mathbb{R}$, where $[\cdot]$ is the greatest integer function. Then

Question

Prepp Team · Accepted Answer

The problem asks for properties of the function $f(x) = [x]^2 - [x+3] - 3$, where $[\cdot]$ denotes the greatest integer function.

Simplifying the Function

Using the property $[x+n] = [x] + n$ for any integer $n$, we can simplify $[x+3]$:

$[x+3] = [x] + 3$

Substitute this back into the definition of $f(x)$:

$f(x) = [x]^2 - ([x] + 3) - 3$

$f(x) = [x]^2 - [x] - 6$

We need to determine the interval where $f(x) < 0$. Substitute the simplified form:

$[x]^2 - [x] - 6 < 0$

Let $y = [x]$. The inequality becomes:

$y^2 - y - 6 < 0$

Factor the quadratic expression:

$(y - 3)(y + 2) < 0$

This inequality holds when $y$ is strictly between the roots $-2$ and $3$.

$-2 < y < 3$

Substitute back $y = [x]$:

$-2 < [x] < 3$

Since $[x]$ must be an integer, the possible values for $[x]$ are $-1, 0, 1, 2$.

We find the corresponding intervals for $x$ for each integer value of $[x]$:

Combining these intervals, we find that $f(x) < 0$ when $x$ is in the range $[-1, 3)$.

This confirms that the statement "$f(x) < 0 \text{ only for } x \in [-1, 3)$" is correct.

Option 1 ($f(x) > 0$): Requires $[x]^2 - [x] - 6 > 0$, meaning $[x] > 3$ or $[x] < -2$. This corresponds to $x \ge 4$ or $x < -3$. The option incorrectly limits it to only $x \in [4, \infty)$.
Option 3 ($\int_{0}^{2} f(x) dx$): For $x \in [0, 2)$, $[x]$ is either $0$ or $1$. In both cases, $f(x) = 0^2 - 0 - 6 = -6$ and $f(x) = 1^2 - 1 - 6 = -6$. So, $\int_{0}^{2} f(x) dx = \int_{0}^{1} (-6) dx + \int_{1}^{2} (-6) dx = -6 + (-6) = -12$. The option states $-6$, which is incorrect.
Option 4 ($f(x) = 0$): Requires $[x]^2 - [x] - 6 = 0$, so $[x] = 3$ or $[x] = -2$. This corresponds to $x \in [-2, -1) \cup [3, 4)$. This is an infinite set of values, not finite.

Therefore, Option B is the correct statement describing the condition for $f(x) < 0$.