Question

Consider the function $g\left(x\right)=\{\begin{array}{cc}\frac{1+a^x+x{a}^{x}lna}{a^x{x}^2}& x<0\\ \frac{2^x{a}^x-xln2-x\ln a-1}{x^2}& x>0\end{array}$. where $a>0$.
Find the value of $a$ & $g\left(0\right)$ so that the function $g\left(x\right)$ is continuous at $x=0$

• A. a = $\frac{1}{\sqrt{2}}$, g(0) = $\frac{(ln2{)}^2}{8}$
• B. a = $-\frac{1}{\sqrt{2}}$, g(0) = $\frac{(ln2{)}^2}{8}$
• C. a = $-\frac{1}{\sqrt{2}}$, g(0) = $\frac{-}{(ln2{)}^2}8$
• D. a = $\frac{1}{\sqrt{2}}$, g(0) = $-\frac{(ln2{)}^2}{8}$

Answer 1

Answer: (A)

a = $\frac{1}{\sqrt{2}}$, g(0) = $\frac{(ln2{)}^2}{8}$

$g\left(0^{-}\right)=\underset{h\to 0}{lim}\frac{1-a^{-h}+(-h)a^{-h}ln\left(a\right)}{a^{-h}(-h{)}^2}$ = $\underset{h\to 0}{lim}\frac{a^h-1-hlna}{h^2}$

$\underset{h\to 0}{lim}\frac{1+hlna+\frac{h^2}{2!}(lna{)}^2+...-1-hlna}{h^2}=\frac{(lna{)}^2}{2}$

$g\left(0^{+}\right)=\underset{h\to 0}{lim}\frac{2^h{a}^h-hln2-hlna-1}{h^2}$

=$\underset{h\to 0}{lim}\frac{1+hln\left(2a\right)+\frac{h^2}{2!}(ln2a{)}^2+...-hlna-1}{h^2}=\frac{(ln2a{)}^2}{2}$

Now $g\left(x\right)$ is continuous so,
$(lna{)}^2=(ln2a{)}^2$

$\Rightarrow (lna{)}^2=(ln2{)}^2+(lna{)}^2+2ln2lna$
$\Rightarrow lna=\frac{-1}{2}ln2$

$\Rightarrow a=\frac{1}{\sqrt{2}}$
$\therefore g\left(0\right)=\frac{{\left(log\left(\frac{1}{\sqrt{2}}\right)\right)}^2}{2}=\frac{1}{8}(ln2{)}^2$