Question

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

 

Options

• \[a = \log_e \left( \frac{2}{3} \right), b = - \frac{2}{3}, c = 1\] 

• \[a = \log_e \left( \frac{2}{3} \right), b = \frac{2}{3}, c = - 1\]

• \[a = \log_e \left( \frac{2}{3} \right), b = \frac{2}{3}, c = 1\]

• none of these

Answer 1

\[ a = \log\frac{2}{3}, b = \frac{2}{3}, c = 1\]

\[\text{ Given }: f\left( x \right) = \begin{cases}\left( 1 + ax \right)^\frac{1}{x} , x < 0 \\ b, x = 0 \\ \frac{\left( x + c \right)^\frac{1}{3} - 1}{\left( x + 1 \right)^\frac{1}{2} - 1}, x > 0\end{cases}\]

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

\[\lim_{x \to 0^-} f\left( x \right) = \lim_{x \to 0^+} f\left( x \right) = f\left( 0 \right)\]

\[\Rightarrow \lim_{x \to 0^-} f\left( x \right) = f\left( 0 \right)\]
\[ \Rightarrow \lim_{h \to 0} f\left( - h \right) = f\left( 0 \right)\]
\[ \Rightarrow \lim_{h \to 0} \left( 1 - ah \right)^\frac{- 1}{h} = f\left( 0 \right)\]
\[ \Rightarrow \lim_{h \to 0} \left( a\frac{\log \left( 1 - ah \right)}{- ah} \right) = \log b\]
\[ \Rightarrow a \times 1 = \log b \left[ \because \lim_{x \to 0} \frac{\log \left( 1 + x \right)}{x} = 1 \right]\]
\[ \Rightarrow a = \log b\]

Also,

\[\lim_{x \to 0^+} f\left( x \right) = f\left( 0 \right)\]

\[\Rightarrow \lim_{x \to 0^+} f\left( x \right) = f\left( 0 \right)\]
\[ \Rightarrow \lim_{h \to 0} f\left( h \right) = f\left( 0 \right)\]
\[ \Rightarrow \lim_{h \to 0} \left( \frac{\left( h + c \right)^\frac{1}{3} - 1}{\left( h + 1 \right)^\frac{1}{2} - 1} \right) = f\left( 0 \right)\]
\[ \Rightarrow \lim_{h \to 0} \left( \frac{\left( h + c \right)^\frac{1}{3} - 1}{\left( h + 1 \right)^\frac{1}{2} - 1} \times \frac{\left( h + 1 \right)^\frac{1}{2} + 1}{\left( h + 1 \right)^\frac{1}{2} + 1} \right) = f\left( 0 \right)\]
\[ \Rightarrow \lim_{h \to 0} \left( \frac{\left( h + c \right)^\frac{1}{3} - 1}{h} \times \left( \left( h + 1 \right)^\frac{1}{2} + 1 \right) \right) = b\]
\[ \Rightarrow \lim_{h \to 0} \frac{\left( h + c \right)^\frac{1}{3} - 1}{h} \times \lim_{h \to 0} \left( \left( h + 1 \right)^\frac{1}{2} + 1 \right) = b\]
\[ \Rightarrow \lim_{h \to 0} \left( \frac{\left( h + c \right)^\frac{1}{3} - 1}{h} \right) \times 2 = b\]
\[ \Rightarrow \lim_{h \to 0} \left( \frac{\left( h + c \right)^\frac{1}{3} - 1^\frac{1}{3}}{\left( h + c \right) - c} \right) = \frac{b}{2}\]
\[ \Rightarrow \frac{c^\left( \frac{1}{3} - 1 \right)}{3} = \frac{b}{2} \left[ \because \lim_{x \to a} \frac{x^n - a^n}{x - a} = n a^{n - 1} , \text{ where }c = 1 \right]\]
\[ \Rightarrow \frac{1}{3} = \frac{b}{2}\]
\[ \Rightarrow \frac{2}{3} = b\]
\[ \therefore a = \log\frac{2}{3}\]