If $F(x)$ is onto function such that $F: \mathbb R \to [-1,2]$ , $F(x)= \frac {x^2+9bx+17}{ax^3+x^2+bx+33}$ and $F'(d)=F'(e)=0$ where $d<e$ then The following question appeared in entrance and i din't solve it because i thought it will take lot of time.Is there any "smart" way to solve this?


  If $F(x)$ is onto function such that $F: \mathbb R \to [-1,2]$ , $F(x)= \frac {x^2+9bx+17}{ax^3+x^2+bx+33}$ and $F'(d)=F'(e)=0$ where $d&lt;e$ then 


$(1)a+b+d+e=4$
$(2)a+b+d+e=-4$
$(3) d+e=2$
$(4) d+e=-2$ 

More than one options can also be correct.Any hints/ideas?
Hint: $F'(d) = F'(e) = 0$ gives you two equations. 

Surjectivity allows you to assert that $F(x) = 2$ for some $x$, and $F(x) = -1$ for some $x$, and any other real number between $-1$ and $2$. This gives you some more information. Just calculate the derivative and insert the points. The result should fall out easily.
$a$ must be  zero given that the function is defined for all real.

Moreover subjectivity can be used to determine value of $b$ along with definition of the function.
Then proceed to find $d$ and $e$ with the so formed quadratic expression.