Question

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

• A. $(3,3e^2)$
• B. $(2,e^6)$
• C. $(3,e^6)$
• D. $(2,3e^2)$

Answer 1

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

$f\left(x\right)=\{\begin{array}{ccc}3(1-\tan x{)}^{\frac{\alpha }{-\tan x}}& ;& \frac{-1}{2}<x<0\\ \beta & ;& x=0\\ 3{(1+\frac{\sin x}{3})}^{\frac{6}{\sin x}}& ;& 0<x<\frac{2}{3}\end{array}$
$\because f\left(x\right)$ is continuous at $x=0$
$\therefore f\left(0^{-}\right)=f\left(0\right)$
$\beta =\underset{x\to 0^{-}}{lim}f\left(x\right)$
$=3\underset{x\to 0^{-}}{lim}(1-\tan x{)}^{\frac{\alpha }{-\tan x}}(1^{\infty }$ form$)$
$=3e^{\underset{x\to 0}{lim}\frac{\alpha }{\tan x}\left[\tan x\right]}$
$\Rightarrow \beta =3e^{\alpha }\dots \left(1\right)$
Now, $f\left(0\right)=f\left(0^{+}\right)$
$\Rightarrow \beta =\underset{x\to 0^{+}}{lim}3{(1+\frac{\sin x}{3})}^{\frac{6}{\sin x}}$
$=3\cdot \underset{x\to 0^{+}}{lim}{(1+\frac{\sin x}{3})}^{\frac{6}{\sin x}}(1^{\infty }$ form $)$
$=3e^{\underset{x\to 0^{+}}{lim}\frac{6}{\sin x}\cdot \frac{\sin x}{3}}$
$\Rightarrow \beta =3e^2\dots \left(2\right)$
Now, from $\left(1\right)$ and $\left(2\right)$
$3e^{\alpha }=3e^2=\beta$
$\therefore \alpha =2$ and $\beta =3e^2$