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Find the value of k making the following function continuous at x = 5 $f(x)= \begin{cases}2 k x-1, & x ≥ 5 \\ 2 x+3, & x<5\end{cases}$ |
$\frac{5}{7}$ 0 $\frac{5}{6}$ $\frac{7}{5}$ |
$\frac{7}{5}$ |
The correct answer is Option (4) → $\frac{7}{5}$ $f(5)=10k-1$ $\underset{x→5^-}{\lim}(2x+3)=13$ so for continuity at $x=5$ $12=10k-1⇒k=\frac{7}{5}$ |
