Question

The function
$\left(x\right)=\{\begin{array}{c}\frac{1-a^x+x{a}^x{\log }_{e}a}{a^x{x}^2},x<0\\ \frac{2^x{a}^x-x{\log }_{e}2-x{\log }_{e}a-1}{x^2},x<0\end{array},a>0$
is continuous at $x=0$, then

• A. The value of $a$ is $\frac{1}{\sqrt{2}}$
• B. The value of $a$ is $\frac{1}{2}$
• C. The value of $g\left(0\right)$ is $\frac{{\left({\log }_{e}2\right)}^2}{8}$
• D. The value of $g\left(0\right)$ is $\frac{{\left({\log }_{e}2\right)}^2}{4}$

Answer 1

Answer: (A), (D)

The value of $a$ is $\frac{1}{\sqrt{2}}$
The value of $g\left(0\right)$ is $\frac{{\left({\log }_{e}2\right)}^2}{4}$

$g\left(x\right)=\{\begin{array}{c}\frac{a^{-x}-1+x{\log }_{e}a}{x^2};x<0\\ \frac{{(2\times a)}^x-x{\log }_{e}2a-1}{x^2};x<0\end{array}$

Given it is continuous at $x=0$

$\Rightarrow$LHL $=$RHL

${lim}_{x\to 0^{-}}g\left(x\right)={lim}_{x\to 0^{+}}g\left(x\right)$

$\Rightarrow {lim}_{x\to 0^{-}}\frac{(a^{-x}-1+x{\log }_{e}a)}{x^2}={lim}_{x\to 0^{+}}\frac{{\left(2a\right)}^x-x{\log }_{e}2a-1}{x^2}$

Both are in $\frac{0}{0}$ form

Using L-Hospital Rule

${lim}_{x\to 0^{-}}\frac{(-a^{-x}\left({\log }_{e}a\right)+{\log }_{e}a)}{2x}={lim}_{x\to 0^{+}}\frac{{\left(2a\right)}^x\left({\log }_{e}2a\right)-{\log }_{e}2a}{2x}$

$=\frac{{\log }_{e}a}{2}{lim}_{x\to 0^{-}}\frac{(a^{-x}-1)}{-x}=\frac{{\log }_{e}a}{2}{lim}_{x\to 0^{+}}\frac{{\left(\right(2a)}^x-1)}{x}$

$={\left({\log }_{e}a\right)}^2={\left({\log }_{e}2a\right)}^2$

$\Rightarrow {\log }_{e}a={\log }_{e}2a,{\log }_{e}a+{\log }_{e}2a=0$

$\Rightarrow a=2a$

$\Rightarrow a=0$

a cannot be $0$ since $a>0$ given ${\log }_e{2a}^2=0$

${2a}^2=1$

$\Rightarrow a=+\frac{1}{\sqrt{2}}$

$g\left(0\right)={lim}_{x\to 0^{-}}g\left(x\right)$

$={\left({\log }_{e}a\right)}^2={\left({\log }_e{2}^{-1/2}\right)}^2$

$=\frac{{\left({\log }_{e}2\right)}^2}{4}$