Question

If $f\left(x\right)=\{\begin{array}{cc}3(1+|\tan x|{)}^{\frac{\alpha }{|\tan x|}},& -\frac{1}{2}<x<0\\ \beta ,& x=0\\ 3{(1+\left|\frac{\sin x}{3}\right|)}^{\frac{6}{|\sin x|}},& 0<x<\frac{2}{3}\end{array}$ is continuous at $x=0,$ then the ordered pair $(\alpha ,\beta )$ is equal to

• A. $(2,e^2)$
• B. $(2,\frac{2}{e^2})$
• C. $(2,3e^2)$
• D. $(2,\frac{3}{e^2})$

Answer 1

The correct option is C $(2,3e^2)$

$f\left(x\right)=\{\begin{array}{cc}3(1+|\tan x|{)}^{\frac{\alpha }{|\tan x|}},& -\frac{1}{2}<x<0\\ \beta ,& x=0\\ 3{(1+\left|\frac{\sin x}{3}\right|)}^{\frac{6}{|\sin x|}},& 0<x<\frac{2}{3}\end{array}$

$R.H.L.=f\left(0^{+}\right)$
$=\underset{x\to 0^{+}}{lim}3{(1+\left|\frac{\sin x}{3}\right|)}^{\frac{6}{|\sin x|}}$
$=3\times \underset{h\to 0}{lim}{(1+\left|\frac{\sin h}{3}\right|)}^{\frac{6}{|\sin h|}}\left(1^{\infty }\text{form}\right)$
$=3e^2$

and $L.H.L.=f\left(0^{-}\right)$
$=\underset{x\to 0^{-}}{lim}3(1+|\tan x|{)}^{\frac{\alpha }{|\tan x|}}$
$=\underset{h\to 0}{lim}3(1+|\tan h|{)}^{\frac{\alpha }{|\tan h|}}$
$=3e^{\alpha }$
Since $f\left(x\right)$ is continuous at $x=0,$
so $f\left(0^{-}\right)=f\left(0^{+}\right)=f\left(0\right)$
$\Rightarrow 3e^{\alpha }=3e^2=\beta$
$\therefore \alpha =2$ and $\beta =3e^2$