Question

$\begin{array}{cc}\text{If}f\left(x\right)& =\frac{x^3+x^2-16x+20}{(x-2{)}^2},\text{if}x\ne 2\\ & =k,\text{if}x=2\end{array}$
is continuous at $x=0$ then

• A. $k=2$
• B. $k=0$
• C. $k=20$
• D. $k=7$

Answer 1

The correct option is D $k=7$

If $f\left(x\right)$ is continuous at $x=a$ then

$l.h.l=r.h.l=f\left(a\right)$

as it is given $f\left(x\right)$ is continuous

$l.h.l=\underset{x\to 2^{-}}{lim}\frac{x^3+x^2-16x+20}{(x-2{)}^2}$

By L. Hopital rule

$\underset{x\to 2^{-}}{lim}\frac{3x^2+2x-16}{2(x-2)}\underset{\underset{h\to 0}{x\to 2^{-}}}{lim}\frac{6x+2}{2}=3(2-h)+1=7$

$\therefore f\left(a\right)=k=l.h.l$

$k=7$