Question

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. $-10$
• B. $12$
• C. $5$
• D. $-30$

Answer 1

Answer: (B), (C), (D)

$5$
$-30$
$12$

$\frac{x^3-6x^2+11x-6}{x^3+x^2-10x+8}=\frac{(x-1)(x-2)(x-3)}{(x-1)(x-2)(x+4)}$

$\therefore x\ne 1,2,-4$, then $f\left(x\right)=\frac{x-3}{x+4}$

Range of $f\left(x\right)=R-\{1,-\frac{2}{5},-\frac{1}{6}\}$

So, equation does not have a solution if

$\frac{a}{30}=\{-1,\frac{2}{5},\frac{1}{6}\}$

$\Rightarrow a=\{-30,12,5\}$.