Question:

Step 1: Determine $k$ using continuity at $x=4$. For $f(x)$ to be continuous at $x=4$, the left-hand limit must equal the right-hand limit. $$ \lim_{x \to 4^-} f(x) = \lim_{x \to 4^+} f(x) $$ $$ (4)^2 - 5(4) + 6 = k - \tan\left(\frac{\pi(4)}{4}\right) $$ $$ 16 - 20 + 6 = k - \tan(\pi) $$ $$ 2 = k - 0 $$ $$ k = 2 $$ Step 2: Determine $\alpha$ using continuity at $x=5$. For $f(x)$ to be continuous at $x=5$, the left-hand limit must equal the right-hand limit. $$ \lim_{x \to 5^-} f(x) = \lim_{x \to 5^+} f(x) $$ Substitute $k=2$ into the expression for $f(5^-)$: $$ k - \tan\left(\frac{\pi(5)}{4}\right) = \log_{10}(\alpha(5)) $$ $$ 2 - \tan\left(\frac{5\pi}{4}\right) = \log_{10}(5\alpha) $$ Since $\tan\left(\frac{5\pi}{4}\right) = \tan\left(\pi + \frac{\pi}{4}\right) = \tan\left(\frac{\pi}{4}\right) = 1$: $$ 2 - 1 = \log_{10}(5\alpha) $$ $$ 1 = \log_{10}(5\alpha) $$ Convert the logarithmic equation to an exponential equation: $$ 5\alpha = 10^1 $$ $$ 5\alpha = 10 $$ $$ \alpha = 2 $$