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.
Best Answer
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