Question

The value of k that makes function f, defined below, continuous is $f\left(x\right)=\{\begin{array}{cc}\frac{2x^2+5x}{x},& whenx\ne 0\\ 3k-1,& whenx=0\end{array}$

• A. $1/3$
• B. $1$
• C. $2$
• D. $5$
• E. $6$

Answer 1

The correct option is C $2$

Given that $f$ is continuous at $x=0$
$\Rightarrow \underset{x\to 0}{lim}\frac{2x^2+5x}{x}=\underset{x\to 0}{lim}2x+5=5$ , which is equal to $3k-1$
Therefore $3k-1=5$

$\Rightarrow 3k=6$

$\Rightarrow k=2$