Question

The set of all points of discontinuity of $f\left(x\right)=\frac{x-1}{x^3+6x^2+11x+6}$ is

• A. $\varphi$
• B. $\{-1\}$
• C. $\{-1,-2,-3\}$
• D. $\{1,2,3\}$

Answer 1

The correct option is C $\{-1,-2,-3\}$

The denominator of the equation is given by

$x^3+6x^2+11x+6=x^3+6x^2+9x+2x+6$

$=x(x^2+6x+9)+2(x+3)$

$=x(x+3{)}^2+2(x+3)$

$=(x+3)\left[x\right(x+3)+2]$

$=(x+3)(x^2+3x+2)$

$=(x+3)(x+1)(x+2)$

Thus $f\left(x\right)$ $=(x-1)/(x+3)(x+1)(x+2)$

Hence, $f\left(x\right)$ is discontinuous at $x=-1,-2,-3$