Simpler ways of finding solutions to $\int_0^x \lfloor{x\rfloor}^2 dx=2(x-1)$

Question

So I was solving the following question:

Find the number of solutions to $$\int_0^x \lfloor{x\rfloor}^2 dx=2(x-1)$$ for $x<0$.$$$$ Options: $2,3,4,5$

And I managed to show that there's no solution whenever $x$ is an integer, and obtained the following equation for whenever $x$ is not an integer:

$$\lfloor{x\rfloor}^2(\lfloor{x\rfloor}+1-x)-\frac{\lfloor{x\rfloor}(\lfloor{x\rfloor}+1)(2\lfloor{x\rfloor}+1)}{6}=2(1-x)$$

And I managed to solve it, but the method is faaaaar too lengthy(I'd made a silly error before).

Let $n$ denote $\lfloor{x\rfloor}$ and $f$ denote $x-\lfloor{x\rfloor}$.

So, the above equation transforms to: $$f=\frac{-2n^3+3n^2+11n-12}{6(n^2-2)}$$

With some calculus, I was able to show $n=-3$ is the only possible (negative) value such that $0\leq f<1$, giving me the only answer $x=-3+\frac 67=-\frac{15}{7}$

Other methods to solve the original question are welcome.

Edit: The question was on a pen-paper test, so everything has to be done by hand.

---

As the answers show, it either the options or the question that's erroneous.

Answer 1

The problem restricts $x < 0$.

The left side $f(x) = \int \dots$ is just a continuous set of line segments of increasing slope (as x goes more negative), and the right hand side $g(x) = 2(x-1)$ is just a straight line. Since $f(0) < g(0)$, but eventually $f > g$, there is at least 1 intersection. Because $g$ is a straight line and $f$ never curves up, there is at most 1 intersection. So the answer is 1.

The graph here makes it more clear but isn't really necessary for the problem:

https://www.desmos.com/calculator/aprbenpnvd

Answer 2

Let $$f(x)=\int_0^x \lfloor t\rfloor^{2}dt - 2x$$ Clearly as $x \to -\infty$, $f(x) \to -\infty$ while as $x \to \infty$, $f(x) \to \infty$. The relative extrema of $f(x)$ will be where $f'(x) = \lfloor x\rfloor^2-2$ changes sign. This happens at $x = -1, 2$. Then evaluating $f(x)$ at these points finds that $f(-1) = 1$, while $f(2) = -3$. By the mean value theorem, there is a single solution of $f(x) = -2$ between $(-\infty, -1)$, another between $(-1, 2)$, and another between $(2, \infty)$. Therefore, there are $\textbf{3}$ values of $x$ satisfying $$\int_0^x \lfloor x\rfloor^2 dx =2(x-1)$$ for $x \in \mathbb{R}$.

Edit: I realized that the question asked for the number of solutions $x < 0$. If the $x \in (-1, 2)$ that is a solution is negative, then the answer is $2$. Otherwise, the answer is $1$.

To show that there is only one solution less than $0$, you can plug in $x = 0$ and notice that $f(x) = 0> -2$. Therefore, the solution in $(-1, 2)$ must be in $(0, 2)$, and the final answer is actually $\textbf{1}$.