Question

Let $f\left(x\right)=\{\begin{array}{c}(1+ax{)}^{\frac{1}{x}},\text{if}x<0\\ b,\text{if}x=0\\ \frac{(x+c{)}^{\frac{1}{3}}-1}{(x+1{)}^{\frac{1}{2}}-1},\text{if}x>0\end{array}$ is continuous at $x=0$. Then

• A. $a=\ln \frac{2}{3},b\in R,c=1$
• B. $a=\ln \frac{2}{3},b=\frac{2}{3},c\in R$
• C. $a=\ln \frac{2}{3},b=\frac{2}{3},c=1$
• D. None of these

Answer 1

The correct option is C $a=\ln \frac{2}{3},b=\frac{2}{3},c=1$

Given:$f\left(x\right)=\{\begin{array}{c}{(1+ax)}^{\frac{1}{x}};x<0\\ b;x=0\\ \frac{{(x+c)}^{\frac{1}{3}}-1}{{(x+1)}^{\frac{1}{2}}-1};x>0\end{array}$

If $f\left(x\right)$ is continuous at $x=0$ then

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

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

$\Rightarrow {lim}_{h\to 0}f(-h)=f\left(0\right)$

$\Rightarrow {lim}_{h\to 0}{(1-ah)}^{-\frac{1}{h}}=f\left(0\right)$

$\Rightarrow {lim}_{h\to 0}\left(a\frac{\ln (1-ah)}{-ah}\right)=\ln b$

$\Rightarrow a\times 1=\ln b$ since ${lim}_{x\to 0}\frac{\ln (1+x)}{x}=1$

$\Rightarrow a=\log b$

Also,${lim}_{x\to 0^{+}}f\left(x\right)=f\left(0\right)$

$\Rightarrow {lim}_{h\to 0}f\left(x\right)=f\left(0\right)$

$\Rightarrow {lim}_{h\to 0}\left(\frac{{(h+c)}^{\frac{1}{3}}-1}{{(h+1)}^{\frac{1}{2}}-1}\right)=f\left(0\right)$

$\Rightarrow (\frac{{(h+c)}^{\frac{1}{3}}-1}{{(h+1)}^{\frac{1}{2}}-1}\times \frac{{(h+1)}^{\frac{1}{2}}+1}{{(h+1)}^{\frac{1}{2}}+1})=b$

$\Rightarrow (\frac{{(h+c)}^{\frac{1}{3}}-1}{h}\times {(h+1)}^{\frac{1}{2}}+1)=b$

$\Rightarrow (\frac{{(h+c)}^{\frac{1}{3}}-1}{h}\times 2)=b$

$\Rightarrow \left(\frac{{(h+c)}^{\frac{1}{3}}-1}{h+c-c}\right)=\frac{b}{2}$

$\Rightarrow \frac{c^{\frac{1}{3}-1}}{3}=\frac{b}{2}\because {lim}_{x\to a}\frac{x^n-a^n}{x-a}=n{a}^{n-1},$ where $c=1$

$\Rightarrow \frac{1}{3}=\frac{b}{2}$

$\Rightarrow b=\frac{2}{3}$

$\therefore a=\ln \frac{2}{3},b=\frac{2}{3}$ and $c=1$