If a function $f(x)=\left\{\begin{array}{cc}\frac{\tan (\alpha+1) x+\tan 2 x}{x} & \text { if } x\gt0 \\…

If a function $f(x)=\left\{\begin{array}{cc}\frac{\tan (\alpha+1) x+\tan 2 x}{x} & \text { if } x\gt0 \\ \beta & \text { at } x=0 \\ \frac{\sin 3 x-\tan 3 x}{x^3} & \text { if } x \lt 0\end{array}\right.$ is continuous at $x=0$ then $|\alpha|+|\beta|=$
  1. $60$
  2. $30$
  3. $15$
  4. $45$

Solution

$f^{\prime}(x)$ is continuous at $x=0$ $\lim _{x \rightarrow 0^{+}} f(x)=f(0)=\beta$ $\lim _{x \rightarrow 0^{+}} \frac{\tan (\alpha+1) x+\tan 2 x}{x}=\alpha+3=\beta \Rightarrow \lim _{x \rightarrow 0^{-}} f(x)=\beta$ $\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{-}} \frac{\sin 3 x-\tan 3 x}{x^3}$ $=\lim _{x \rightarrow 0^{-}} \frac{3 \cos 3 x-3 \sec ^2 3 x}{3 x^2}$ $=\lim _{x \rightarrow 0^{-}} \frac{-9 \sin 3 x-\sec ^2 3 x \tan 3 x \times 18}{6 x}$ $=-\frac{9}{2}-9=-\frac{27}{2}=\beta \Rightarrow \alpha=-\frac{27}{2}-3=-\frac{33}{2}$ $|\alpha|+|\beta|=\frac{60}{2}=30$

Asked in: AP EAMCET 2024 (20 May Shift 1)

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