Question: Is the following question poorly written? 

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> [![enter image description here](https://i.stack.imgur.com/3uOIp.png)](https://i.stack.imgur.com/3uOIp.png)
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I suppose it is a translation since it comes from a Japanese institution. I am referring to the assumption "... has a solution". Given this information I would assume that the discriminant is equal to $0$ since the equation has "a solution" but the answer involves writing it with a "greater than or equal to" symbol. 

What are your opinions on this? Am I wrong or should the translation specify that it has one or two solutions?

Answer: 
This is almost a comment, rather than an answer, but as the comments above were almost answers, it seems fair. 

To given another example, which might make clear what is going wrong:

Suppose $$ f(x) = \cases{-1, & x<0,\\
 0, & x=0,\\ 
 1, & 0 <x.}$$

Now, except at $x=0$, $f'(x) =0$. (The graph of $f$ consists basically of two lines of slope $0$.) 

Suppose that the question were to evaluate 
$$I= \int\_{-1}^1{d\over dx}f(x) \,dx.$$

The 'FTC method' would give $I= f(1) - f(-1) = 1- (-1) = 2$.

On the other hand, as $f'(x) = 0$ (except at $x=0$!), one should certainly get (geometrically, say, or Riemann sums) that 
$$ \int\_{-1}^1{d\over dx}f(x) \,dx \,= \int\_{-1}^1 0\,dx = 0.$$
This is in fact the 'correct' answer.

This is analogous to the question you were given. 

I think that one can object that the question used the expression, ${d\over dx} f(x)$, which is only defined 'almost everywhere.' In the framework of a calculus class and Riemann integration, $I$ should be treated as an improper integral - and the given answer [although 'correct'!] does not do so... If one treats the integral as an improper integral (i.e., break up the integral around $x=0$, and take limits), and use your 'FTC' method, one gets the correct answer. 

Likewise, the given solution uses
$$\frac{d}{dx}(\tan^{-1}(\frac{1}{x})) = \frac{d}{dx}(\cot^{-1}(x)) = \frac{-1}{1+x^2}.$$
But that equality fails at $x=0$.

FWIW, there is a 'theory of integration' (Lebesgue) that handles 'almost everywhere,' and does not make a distinction between proper and improper integrals.
Here we will address your integral:
\begin{equation}
 \int\_{-1}^{1} \frac{d}{dx} \left[\arctan\left(\frac{1}{x}\right) \right]\:dx
\end{equation}
Now:
\begin{equation}
 \frac{d}{dx} \left[\arctan\left(\frac{1}{x}\right) \right] = \frac{1}{1 + \left(\frac{1}{x}\right)^2} \cdot -\frac{1}{x^2} = -\frac{1}{1 + x^2}
\end{equation}
Thus your integral becomes:
\begin{align}
 \int\_{-1}^{1} \frac{d}{dx} \left[\arctan\left(\frac{1}{x}\right) \right]\:dx &= \int\_{-1}^{1} -\frac{1}{1 + x^2}\:dx = -2\int\_0^1 \frac{1}{1 + x^2}\:dx \nonumber \\
&= -2\left[\arctan(x) \right]\_0^1 = -2 \cdot \frac{\pi}{4} = -\frac{\pi}{2}
\end{align}