The sum of the values of a for which $$\frac{x^3-6x^2+11x-6}{x^3+x^2-10x+8} + \frac a{30} = 0$$ does not have a real solution is
A. $1$
B. $12$
C. $13$
D. $2$
I tried to factorise the numerator and the denominator and got $$\frac{(x-1)(x-2)(x-3)}{(x-1)(x-2)(x+4)}$$ Then $f(x) = \frac{x-3}{x+4}$, but after this how to get the range of $f(x)$?
Any help will be appreciated ....thanks.
Take $$y=f(x) = \frac{x-3}{x+4}$$ Then, $$\frac{y+1}{y-1}=\frac{2x+1}{-7}$$ or $$x=\frac 1 2 \left(\frac{7y+7}{1-y}-1\right)$$ So, the range is $\mathbb R \backslash \{1\}$ because $x$ is defined iff $y\neq 1$.
Now, check that $y=-\frac{12}{30}$ gives $x=1$, for which the original fraction takes $\frac 0 0$ form. So, $a\neq 12$. Also, we can see that $a\neq -30$ because $y\neq 1$. Again, $a\neq 5$ because $y\neq -\frac 1 6$ (as that would imply $x=2$ which is not possible). Summing these up, we aren't getting $13$, we are getting $-13$. So, I think the option has a sign flip.
Does that help?