Question

If  \[f\left( x \right) = \begin{cases}\frac{{36}^x - 9^x - 4^x + 1}{\sqrt{2} - \sqrt{1 + \cos x}}, & x \neq 0 \\ k , & x = 0\end{cases}\]is continuous at x = 0, then k equals

 

Options

• \[16\sqrt{2}\] log 2 log 3

• \[16\sqrt{2}\]

• \[16\sqrt{2}\]  ln 2 ln 3

• none of these

Answer 1

\[16\sqrt{2} \ln2 \ln3\]

Given: 

\[f\left( x \right) = \binom{\frac{{36}^x - 9^x - 4^x + 1}{\sqrt{2} - \sqrt{1 + \ cosx}}, x \neq 0}{k, x = 0}\]

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

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

\[\Rightarrow \lim_{x \to 0} \left( \frac{{36}^x - 9^x - 4^x + 1}{\sqrt{2} - \sqrt{1 + \cos x}} \right) = k\]
\[ \Rightarrow \lim_{x \to 0} \left( \frac{9^x 4^x - 9^x - 4^x + 1}{\sqrt{2} - \sqrt{1 + \cos x}} \right) = k\]
\[ \Rightarrow \lim_{x \to 0} \left( \frac{9^x \left( 4^x - 1 \right) - 1\left( 4^x - 1 \right)}{\sqrt{2} - \sqrt{1 + \cos x}} \right) = k\]
\[ \Rightarrow \lim_{x \to 0} \left( \frac{\left( 9^x - 1 \right)\left( 4^x - 1 \right)}{\sqrt{2} - \sqrt{1 + \cos x}} \right) = k\]
\[ \Rightarrow \lim_{x \to 0} \left( \frac{\left( 9^x - 1 \right)\left( 4^x - 1 \right)}{\sqrt{2} - \sqrt{2}\cos \left( \frac{x}{2} \right)} \right) = k\]
\[ \Rightarrow \lim_{x \to 0} \left( \frac{\left( 9^x - 1 \right)\left( 4^x - 1 \right)}{\sqrt{2}\left[ 1 - \cos \left( \frac{x}{2} \right) \right]} \right) = k\]
\[ \Rightarrow \lim_{x \to 0} \left( \frac{\left( 9^x - 1 \right)\left( 4^x - 1 \right)}{\sqrt{2}\left[ 2 \sin^2 \left( \frac{x}{4} \right) \right]} \right) = k\]
\[ \Rightarrow \lim_{x \to 0} \left( \frac{8\left( 9^x - 1 \right)\left( 4^x - 1 \right)}{16\sqrt{2} x^2 \left[ \frac{\sin^2 \left( \frac{x}{4} \right)}{x^2} \right]} \right) = k\]
\[ \Rightarrow \lim_{x \to 0} \left( \frac{8\left( 9^x - 1 \right)\left( 4^x - 1 \right)}{\sqrt{2} x^2 \left[ \frac{\sin^2 \left( \frac{x}{4} \right)}{\left( \frac{x^2}{16} \right)} \right]} \right) = k\]
\[ \Rightarrow \frac{8}{\sqrt{2}} \lim_{x \to 0} \left( \frac{\left( 9^x - 1 \right)\left( 4^x - 1 \right)}{x^2 \left[ \frac{\sin^2 \left( \frac{x}{4} \right)}{\left( \frac{x}{4} \right)^2} \right]} \right) = k\]
\[ \Rightarrow \frac{8}{\sqrt{2}}\frac{\lim_{x \to 0} \left( \frac{9^x - 1}{x} \right) \lim_{x \to 0} \left( \frac{4^x - 1}{x} \right)}{\lim_{x \to 0} \left[ \frac{\sin \left( \frac{x}{4} \right)}{\left( \frac{x}{4} \right)} \right]^2} = k\]
\[ \Rightarrow \frac{8}{\sqrt{2}} \times \frac{\ln 9 \times \ln 4}{1} = k \left[ \because \lim_{x \to 0} \left( \frac{a^x - 1}{x} \right) = a \right]\]
\[ \Rightarrow \frac{8}{\sqrt{2}} \times \frac{2 \ln 3 \times \left( 2 \ln 2 \right)}{1} = k \]
\[ \]
\[ \Rightarrow \frac{32}{\sqrt{2}} \times \frac{\ln 3 \ln 2}{1} = k\]
\[ \Rightarrow \frac{32\sqrt{2}}{2} \times \frac{\ln 3 \ln 2}{1} = k\]
\[ \Rightarrow k = 16\sqrt{2} \ln 2 \ln 3\]