Question

Let $f\left(x\right)=\{\begin{array}{c}\begin{array}{cc}xg\left(x\right),& x\le 0\end{array}\\ \begin{array}{cc}x+a{x}^2-x^3,& x>0\end{array}\end{array}$, where $g\left(t\right)={lim}_{x\to 0}{(1+a\tan x)}^{t/x}$, $a$ is positive constant, then

If $a$ is even prime number, then $g\left(2\right)=$

• A. $e^2$
• B. $e^3$
• C. $e^4$
• D. None of these

Answer 1

Answer: (C)

$e^4$

Given $g\left(t\right)=\underset{x\to 0}{lim}{(1+a\tan x)}^{\frac{t}{x}}$

$=\underset{x\to 0}{lim}{(1+a\tan x)}^{\frac{ta\tan x}{a\tan xx}}$

$=e^{a{lim}_{x\to 0}\left(\frac{\tan x}{x}\right)t}$

$=e^{at};a>0$

$\because a$ is even prime number

$\Rightarrow a=2$

$\therefore g\left(t\right)=e^{2t}$

$\therefore g\left(2\right)=e^4$