The values of $a$ and $b$ for which the function $f(x)=\left\{\begin{array}{cc} 1+|\sin x|^{a/|\sin x|}, &…

The values of $a$ and $b$ for which the function $f(x)=\left\{\begin{array}{cc} 1+|\sin x|^{a/|\sin x|}, & \frac{-\pi}{6} \lt x \lt 0 \\ b, & x=0 \\ e^{\tan 2 x / \tan 3 x}, & 0 \lt x \lt \frac{\pi}{6} \end{array}\right.$ is continuous at $x=0$ are
  1. $a=1, b=\frac{3}{2}$
  2. $a=\frac{2}{3} b=e^{2 / 3}$
  3. $a=\frac{2}{3} b=\frac{3}{2}$
  4. $a=-1, b=-e^{2 / 3}$

Solution

$f(x)\left\{\begin{array}{cc}(1+|\sin x|)^{\frac{a}{|\sin x|}} & , \frac{-\pi}{6} \lt x \lt 0 \\ b & x=0 \\ e^{\frac{\tan 2 x}{\tan 3 x}} & , \quad 0 \lt x \lt \frac{\pi}{6}\end{array}\right.$ For $f(x)$ to be continuous at $x=0$ $\begin{aligned} & \lim _{x \rightarrow 0^{-}} f(x)=f(0)=\lim _{x \rightarrow 0^{-}} f(x) \qquad...\mathrm{(i)} \\ & \lim _{x \rightarrow 0^{-}}(1+|\sin x|)^{\frac{a}{|\sin x|}} \\ & =e^{\lim _{x \rightarrow 0}\left(|\sin x| \frac{a}{|\sin x|}\right)}=e^a \qquad \mathrm{(1^{\infty} form)} \end{aligned}$ Now, $\lim _{x \rightarrow 0^{+}} e^{\frac{\tan 2 x}{\tan 3 x}}=\lim _{x \rightarrow 0} e^\frac{\left(\frac{\tan 2 x}{2 x} \cdot 2 x\right)}{\left(\frac{\tan 3 x}{3 x} \cdot 3 x\right)}=e^{\frac{2}{3}}$ $\therefore$ From (i) $e^a=b=e^{\frac{2}{3}} \Rightarrow a=\frac{2}{3}, b=e^{\frac{2}{3}}$

Asked in: AP EAMCET 2024 (20 May Shift 2)

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