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 $f\left(x\right)$ is differentiable at $x=0$ then $a\in$

• A. $(-5,-1)$
• B. $(-10,3)$
• C. $(0,\infty )$
• D. None of these

Answer 1

The correct option is C $(0,\infty )$

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

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

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

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

$L.H.D.$ at $x=0=f^{\prime }\left(0^{-}\right)=\underset{h\to 0^{+}}{lim}\frac{f(-h)-f\left(0\right)}{-h}=\underset{h\to 0^{+}}{lim}\frac{{-he}^{-ah}-0}{-h}=1$

$R.H.D.$ at $x=0=f^{\prime }\left(0^{+}\right)=\underset{h\to 0^{+}}{lim}\frac{f(-h)-f\left(0\right)}{-h}$

$=\underset{h\to 0^{+}}{lim}\frac{h+{ah}^2-h^3}{h}=1$

$\therefore f\left(x\right)$ is differentiable at $x=\forall a>0$

$\therefore a\in (0,\infty ).$