TITLE: Is this question wrong? I am getting $\frac{\pi}{4}$
QUESTION [5 upvotes]: Again going through the IIT advanced questions, I came across this one:
Question: Suppose $f(x)=3x^3-13x^2+14x-2$ and $\alpha,\beta,\gamma$ are the roots of $f(x)=0$ such that $\alpha<\beta<\gamma $ then $$\tan^{-1}([\alpha])+\,\tan^{-1}([\beta-1])+\,\tan^{-1}([\gamma-1]) =\,??$$
Note: $[x]$ symbolises Greatest Integer less than or equal to $x$
Options: A: $\frac{3\pi}{4}$     B: ${\pi}$     C: $\frac{\pi}{2}$ D: Can't  be Decided 
I figured out $\alpha$ to be in $(0,1)$,
$\beta$ to be in $(1,2)$ and $\gamma$ to be in $(2,3)$. according to that, the answer should be coming $\frac{\pi}{4}$ but no option given. Some help please?

REPLY [7 votes]: It looks to me like you're right: we have
$$f(0)=-2<0$$
$$f(1)=3-13+14-2=2>0$$
$$f(2)=24-52+28-2=-2<0$$
$$f(3)=81-117+42-2=4>0,$$ so by the Intermediate Value Theorem we have the zeroes arranged as you say. In particular, we have $$[\alpha]=0, [\beta-1]=0, [\gamma-1]=1,$$ so $\tan^{-1}([\alpha])+\tan^{-1}([\beta-1])+\tan^{-1}([\gamma-1])=0+0+{\pi\over 4}={\pi\over 4}.$
Perhaps there is a typo (either in the problem or in how you copied it down)?