Question

If the function $f\left(x\right)=\{\begin{array}{ccc}\frac{x^2-(k+2)x+2k}{x-2}& for& x\ne 2\\ 2& for& x=2\end{array}$ is continuous at $x=2$, then $k$ is equal to

• A. $-\frac{1}{2}$
• B. $-1$
• C. $0$
• D. $\frac{1}{2}$
• E. $1$

Answer 1

The correct option is C $0$

Given, $f\left(x\right)=\left\{\begin{array}{ccc}\frac{x^2-(k+2)x+2k}{x-2}& for& x\ne 2\\ 2& for& x=3\end{array}\right\}$
Since, $f\left(x\right)$ is continuous at $x=2$.
Therefore, $\underset{x\to 2}{lim}f\left(x\right)=f\left(2\right)$
$\Rightarrow \underset{x\to 2}{lim}\frac{x^2-(k+2)x+2k}{x-2}=2$$Apply{L}^{\prime }-hospitalrule,weget$\displaystyle \lim_{x \rightarrow 2} \dfrac {2x - (k + 2)}{1} = 2\Rightarrow 2(2) - (k + 2) = 2\Rightarrow k + 2 = 2\Rightarrow k = 0$