Find the value of k making the following function continuous at x = 5
$f(x)= \begin{cases}2 k x-1, & x ≥ 5 \\ 2 x+3, & x<5\end{cases}$
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → $\frac{7}{5}$
$f(5)=10k-1$
$\underset{x→5^-}{\lim}(2x+3)=13$
so for continuity at $x=5$
$12=10k-1⇒k=\frac{7}{5}$