The values of the constants a, b and for which the function f(x)= (1+ax)^ 1 x ,&x>0 ,&x=0 (x+c)^ 1 2 -1 (x+1)^ 1 2 -1 ,&x>0 may be continuous at x = 0, are:

C. a= _ e (2 3 ), b= (2 3 ), c=1

The values of the constants a, b and for which the function f(x)=…

MCQ

The values of the constants $a, b$ and for which the function $\text{f(x)}=\begin{cases}(1+\text{ax})^{\frac{1}{\text{x}}},&\text{x}>0\\\text{b},&\text{x}=0\\\frac{(\text{x}+\text{c})^{\frac{1}{2}}-1}{(\text{x}+1)^{\frac{1}{2}}-1},&\text{x}>0\end{cases}$ may be continuous at $x = 0,$ are:

A
$\text{a}=\log_{\text{e}}\Big(\frac{2}{3}\Big),\text{ b}=-\frac{2}{3},\text{ c}=1$
B
$\text{a}=\log_{\text{e}}\Big(\frac{2}{3}\Big),\text{ b}=\frac{2}{3},\text{ c}=-1$
✓
$\text{a}=\log_{\text{e}}\Big(\frac{2}{3}\Big),\text{ b}=\Big(\frac{2}{3}\Big),\text{ c}=1$
D
none of these

✓

Answer

Correct option: C.

$\text{a}=\log_{\text{e}}\Big(\frac{2}{3}\Big),\text{ b}=\Big(\frac{2}{3}\Big),\text{ c}=1$

$\text{f}(0)=\lim\limits_{\text{x}\rightarrow0}(1+\text{ax})^{\frac{1}{\text{x}}}$
$\text{b}=\lim\limits_{\text{x}\rightarrow{\text{a}}}(1+\text{ax})^{\frac{1}{\text{ax}}\times\text{a}}$
$\text{b}=\text{e}^{\text{a}}$
$\text{a}=\log_{\text{e}}\text{b}$
$\text{f}(0)=\lim\limits_{\text{x}\rightarrow\text{a}^+}\frac{(\text{x}+\text{c})^{\frac{1}{3}}-1}{(\text{x}+1)^{\frac{1}{2}}-1}$
Here, $\text{c}=1$
$\text{x}+1=\text{y}$
$\text{x}\rightarrow0\Rightarrow\text{y}\rightarrow1$
$\text{f}(0)=\lim\limits_{\text{y}\rightarrow1}\frac{\text{y}^{\frac{1}{3}}-1}{\text{y}^{\frac{1}{2}}-1}$
$\text{b}=\lim\limits_{\text{y}\rightarrow1}\frac{\frac{\text{y}^{\frac{1}{3}}-1}{\text{y}-1}}{\frac{\text{y}^{\frac{1}{2}}-1}{\text{y}-1}}=\frac{\frac{1}{3}}{\frac{1}{2}}=\frac{2}{3}$
$\text{a}=\log\text{b}=\log\frac{2}{3}$