Question

The value of a for which the function \[f\left( x \right) = \begin{cases}\frac{\left( 4^x - 1 \right)^3}{\sin\left( x/a \right) \log \left\{ \left( 1 + x^2 /3 \right) \right\}}, & x \neq 0 \\ 12 \left( \log 4 \right)^3 , & x = 0\end{cases}\]may be continuous at x = 0 is

 

विकल्प

• 1

• 2

• 3

• none of these

Answer 1

none of these 

For f(x) to be continuous at  \[x = 0\] , we must have 

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

\[\lim_{x \to 0} \left[ \frac{\left( 4^x - 1 \right)^3}{\sin\frac{x}{a} \log\left( 1 + \frac{x^2}{3} \right)} \right] = 12 \left( \log 4 \right)^3\]

 \[\Rightarrow \lim_{x \to 0} \left[ \frac{\frac{\left( 4^x - 1 \right)^3}{x^3}}{\frac{\sin\frac{x}{a}\log\left( 1 + \frac{x^2}{3} \right)}{x^3}} \right] = 12 \left( \log 4 \right)^3 \]
\[ \Rightarrow \lim_{x \to 0} \left[ \frac{a \left( \frac{4^x - 1}{x} \right)^3}{\left( \frac{\sin\frac{x}{a}}{\frac{x}{a}} \right)\frac{\log\left( 1 + \frac{x^2}{3} \right)}{x^2}} \right] = 12 \left( \log 4 \right)^3 \]
\[ \Rightarrow 3a \lim_{x \to 0} \left[ \frac{\left( \frac{4^x - 1}{x} \right)^3}{\left( \frac{\sin\frac{x}{a}}{\frac{x}{a}} \right)\frac{\log\left( 1 + \frac{x^2}{3} \right)}{\left( \frac{x^2}{3} \right)}} \right] = 12 \left( \log 4 \right)^3 \]
\[ \Rightarrow 3a\left[ \frac{\lim_{x \to 0} \left( \frac{4^x - 1}{x} \right)^3}{\lim_{x \to 0} \left( \frac{\sin\frac{x}{a}}{\frac{x}{a}} \right) \lim_{x \to 0} \frac{\log\left( 1 + \frac{x^2}{3} \right)}{\left( \frac{x^2}{3} \right)}} \right] = 12 \left( \log 4 \right)^3 \]
\[ \Rightarrow 3a \left( \log 4 \right)^3 = 12 \left( \log 4 \right)^3 \left[ \because \lim_{x \to 0} \left( \frac{a^x - 1}{x} \right) = \log a, \lim_{x \to 0} \frac{\log\left( 1 + x \right)}{x} = 1 and \lim_{x \to 0} \frac{\sin x}{x} = 1 \right]\]
\[ \Rightarrow a = 4\]