The values of the constants a, b and c for which the function


The values of the constants ab and c for which the function

$f(x)=\left\{\begin{array}{ll}(1+a x)^{1 / x} & , x<0 \\ b & , \quad x=0 \\ \frac{(x+c)^{1 / 3}-1}{(x+1)^{1 / 2}-1}, & x>0\end{array}\right.$ 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


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

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

If $f(x)$ is continuous at $x=0$, then

$\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{+}} f(x)=f(0)$

$\Rightarrow \lim _{x \rightarrow 0^{-}} f(x)=f(0)$

$\Rightarrow \lim _{h \rightarrow 0} f(-h)=f(0)$

$\Rightarrow \lim _{h \rightarrow 0}(1-a h)^{\frac{-1}{h}}=f(0)$

$\Rightarrow \lim _{h \rightarrow 0}\left(a \frac{\log (1-a h)}{-a h}\right)=\log b$

$\Rightarrow a \times 1=\log b \quad\left[\because \lim _{x \rightarrow 0} \frac{\log (1+x)}{x}=1\right]$

$\Rightarrow a=\log b$


$\lim _{x \rightarrow 0^{+}} f(x)=f(0)$

$\Rightarrow \lim _{x \rightarrow 0^{+}} f(x)=f(0)$

$\Rightarrow \lim _{h \rightarrow 0} f(h)=f(0)$

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

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

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

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

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

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

$\Rightarrow \frac{c^{\left(\frac{1}{3}-1\right)}}{3}=\frac{b}{2} \quad\left[\because \lim _{x \rightarrow a} \frac{x^{n}-a^{n}}{x-a}=n a^{n-1}\right.$, where $\left.\mathrm{c}=1\right]$

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

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

$\therefore a=\log \frac{2}{3}$

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