Discussion
Let
\[ f(x) = \begin{cases} x + \alpha , & \text{if } x \lt 0 \\ max(2\cos x, 2\sin x), & \text{if } x \ge 0 \end{cases} \]
If f is continuous at x = 0, then the value of \(\alpha\) is equal to
Answer: A
\(\alpha =\) 2.
For continuity at 0 the left-hand limit must equal \(f(0)\):
\[ \begin{aligned} \lim_{x \to 0^{-}}(x + \alpha) &= \alpha \\ f(0) &= \max(2\cos 0, 2\sin 0) = \max(2, 0) = 2 \end{aligned} \]
so \(\alpha = 2\). Taking \(2\sin 0 = 0\) as the maximum would give 0, and the other values leave a jump at 0.
For \(x \ge 0\) near 0 the maximum is \(2\cos x\), since \(\cos x \gt \sin x\) there, so the right-hand piece starts at 2 and the left-hand piece must meet it.
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