Question

Let $f(x) = \begin{cases} \frac{\left(x^{3}+x^{2}-16 x+20\right)}{(x-2)^{2}}, & \text{if $x \neq 2$ } \\[2ex] k, & \text{if $x=2$ } \end{cases}$
If $f(x)$ is continuous for all $x$, then $k$ =

• A. 2
• B. 3
• C. 6
• D. 7

Answer 1

Answer: (D)

We have $f(x) = \begin{cases} \frac{\left(x^{3}+x^{2}-16 x+20\right)}{(x-2)^{2}}, & \text{if $x \neq 2$ } \\[2ex] k, & \text{if $x=2$ } \end{cases}$
Clearly, $f (x)$ is continuous for all values of x except possibly at $x=2$
It will be continuous at $x=2$ if $\displaystyle\lim _{x \rightarrow 2} f(x)=f(2)$
$\Rightarrow \displaystyle\lim _{x \rightarrow 2} \frac{x^{3}+x^{2}-16 x+20}{(x-2)^{2}}=k$
$\Rightarrow k=\displaystyle\lim _{x \rightarrow 2} \frac{(x-2)^{2}(x+5)}{(x-2)^{2}}$
$=\displaystyle\lim _{x \rightarrow 2}(x+5)=7$