Question

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

• A. 2
• B. 3
• C. 6
• D. 7

Answer 1

Answer: (D)

We have $f(x)=\{\begin{array}{ll}\frac{\left(x^3+x^2-16x+20\right)}{(x-2{)}^2},& \text{if x=2 }\\ k,& \text{if x=2 }\end{array}$
Clearly, $f(x)$ is continuous for all values of x except possibly at $x=2$
It will be continuous at $x=2$ if $\underset{x\to 2}{\lim }f(x)=f(2)$
$\Rightarrow \underset{x\to 2}{\lim }\frac{x^3+x^2-16x+20}{(x-2{)}^2}=k$
$\Rightarrow k=\underset{x\to 2}{\lim }\frac{(x-2{)}^2(x+5)}{(x-2{)}^2}$
$=\underset{x\to 2}{\lim }(x+5)=7$