Question

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

• A. $3$
• B. $5$
• C. $7$
• D. $9$

Answer 1

Answer: (C)

$k=f(2)=\underset{x\to 2}{\lim }f(x)$
$=\underset{x\to 2}{\lim }$ $\frac{x^3+x^2-16x+20}{{\left(x-2\right)}^2}$
Using $L^{\prime }$ Hospital Rule, we get
$\underset{x\to 2}{\lim }$ $\frac{3x^2+2x-16}{2\left(x-2\right)}$ $(\frac{0}{0}$ form$)$
Again using $L^{\prime }$ Hospital Rule, we get
$\underset{x\to 2}{\lim }$ $\frac{6x+2}{2}=7$.
So, $k=7$