Question

The values of the constants a, b and c for which the function
$f\left(x\right)=\left\{\begin{array}{cc}{\left(1+ax\right)}^{1/x},& x<0\\ b,& x=0\\ \frac{{\left(x+c\right)}^{1/3}-1}{{\left(x+1\right)}^{1/2}-1},& x>0\end{array}\right.$may be continuous at x = 0, are

(a) $a={\log }_e\left(\frac{2}{3}\right),b=-\frac{2}{3},c=1$

(b) $a={\log }_e\left(\frac{2}{3}\right),b=\frac{2}{3},c=-1$

(c) $a={\log }_e\left(\frac{2}{3}\right),b=\frac{2}{3},c=1$

(d) none of these

Answer 1

$\left(c\right)a=\log \frac{2}{3},b=\frac{2}{3},c=1$


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

If $f\left(x\right)$ is continuous at $x=0$, then
$\underset{x\to 0^{-}}{\lim }f\left(x\right)=\underset{x\to 0^{+}}{\lim }f\left(x\right)=f\left(0\right)$

$$\begin{gathered} \Rightarrow \underset{x\to 0^{-}}{\lim }f\left(x\right)=f\left(0\right) \\ \Rightarrow \underset{h\to 0}{\lim }f\left(-h\right)=f\left(0\right) \\ \Rightarrow \underset{h\to 0}{\lim }{\left(1-ah\right)}^{\frac{-1}{h}}=f\left(0\right) \\ \Rightarrow \underset{h\to 0}{\lim }\left(a\frac{\log \left(1-ah\right)}{-ah}\right)=\log b \\ \Rightarrow a\times 1=\log b\left[\because \underset{x\to 0}{\lim }\frac{\log \left(1+x\right)}{x}=1\right] \\ \Rightarrow a=\log b \end{gathered}$$

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

$$\begin{gathered} \Rightarrow \underset{x\to 0^{+}}{\lim }f\left(x\right)=f\left(0\right) \\ \Rightarrow \underset{h\to 0}{\lim }f\left(h\right)=f\left(0\right) \\ \Rightarrow \underset{h\to 0}{\lim }\left(\frac{{\left(h+c\right)}^{\frac{1}{3}}-1}{{\left(h+1\right)}^{\frac{1}{2}}-1}\right)=f\left(0\right) \\ \Rightarrow \underset{h\to 0}{\lim }\left(\frac{{\left(h+c\right)}^{\frac{1}{3}}-1}{{\left(h+1\right)}^{\frac{1}{2}}-1}\times \frac{{\left(h+1\right)}^{\frac{1}{2}}+1}{{\left(h+1\right)}^{\frac{1}{2}}+1}\right)=f\left(0\right) \\ \Rightarrow \underset{h\to 0}{\lim }\left(\frac{{\left(h+c\right)}^{\frac{1}{3}}-1}{h}\times \left({\left(h+1\right)}^{\frac{1}{2}}+1\right)\right)=b \\ \Rightarrow \underset{h\to 0}{\lim }\frac{{\left(h+c\right)}^{\frac{1}{3}}-1}{h}\times \underset{h\to 0}{\lim }\left({\left(h+1\right)}^{\frac{1}{2}}+1\right)=b \\ \Rightarrow \underset{h\to 0}{\lim }\left(\frac{{\left(h+c\right)}^{\frac{1}{3}}-1}{h}\right)\times 2=b \\ \Rightarrow \underset{h\to 0}{\lim }\left(\frac{{\left(h+c\right)}^{\frac{1}{3}}-1^{{\displaystyle \frac{1}{3}}}}{\left(h+c\right)-c}\right)=\frac{b}{2} \\ \Rightarrow \frac{c^{\left({\displaystyle \frac{1}{3}}-1\right)}}{3}=\frac{b}{2}\left[\because \underset{x\to a}{\lim }\frac{x^n-a^n}{x-a}=n{a}^{n-1},\mathrm{where}\mathrm{c}=1\right] \\ \Rightarrow \frac{1}{3}=\frac{b}{2} \\ \Rightarrow \frac{2}{3}=b \\ \therefore a=\log \frac{2}{3} \end{gathered}$$