Discussion
Let
\[ f(x) = \begin{cases} \dfrac{\tan\alpha x + (\beta + 1)\tan x}{x}, & \text{for } x \neq 0 \\ 5, & \text{for } x = 0 \end{cases} \]
be continuous at \(x = 0\). Then the value of \(\alpha + \beta\) is equal to
Answer: C
\(\alpha + \beta = \) 4.
Continuity at 0 means the limit of the first branch equals \(f(0) = 5\). Using \(\lim\limits_{x \to 0}\frac{\tan kx}{x} = k\):
\[ \begin{aligned} \lim_{x \to 0}\frac{\tan\alpha x}{x} + (\beta + 1)\frac{\tan x}{x} &= \alpha + \beta + 1 \\ \alpha + \beta + 1 &= 5 \Rightarrow \alpha + \beta = 4 \end{aligned} \]
5 is the answer obtained by forgetting the +1 attached to \(\beta\); 6 adds it on the wrong side.
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