Question

If \[f\left( x \right) = \begin{cases}\frac{1 - \sin x}{\left( \pi - 2x \right)^2} . \frac{\log \sin x}{\log\left( 1 + \pi^2 - 4\pi x + 4 x^2 \right)}, & x \neq \frac{\pi}{2} \\ k , & x = \frac{\pi}{2}\end{cases}\]is continuous at x = π/2, then k =

 

Options

• \[- \frac{1}{16}\] 

• \[- \frac{1}{32}\] 

• \[- \frac{1}{64}\] 

• \[- \frac{1}{28}\]

Answer 1

\[\frac{- 1}{64}\]

If \[f\left( x \right)\]  is continuous at  \[x = \frac{\pi}{2}\] ,then 

\[\lim_{x \to \frac{\pi}{2}} f\left( x \right) = f\left( \frac{\pi}{2} \right)\]

\[\text{ If }\frac{\pi}{2} - x = t, \text{ then} \]

\[ \Rightarrow \lim_{t \to 0} f\left( \frac{\pi}{2} - t \right) = f\left( \frac{\pi}{2} \right)\]

\[ \Rightarrow \lim_{t \to 0} \left( \frac{1 - \sin \left( \frac{\pi}{2} - t \right)}{4 t^2} \times \frac{\log \sin \left( \frac{\pi}{2} - t \right)}{\log\left( 1 + \pi^2 - 4\pi\left( \frac{\pi}{2} - t \right) + 4 \left( \frac{\pi}{2} - t \right)^2 \right)} \right) = k\]

\[ \Rightarrow \lim_{t \to 0} \left( \frac{\left( 1 - \cos t \right)}{4 t^2} \times \frac{\log \cos t}{\log\left( 1 + \pi^2 - 2 \pi^2 + 4\pi t + 4\left( \frac{\pi^2}{4} + t^2 - \pi t \right) \right)} \right) = k\]

\[ \Rightarrow \lim_{t \to 0} \left( \frac{\left( 1 - \cos t \right)}{4 t^2} \times \frac{\log \cos t}{\log\left( 1 - \pi^2 + 4\pi t + \left( \pi^2 + 4 t^2 - 4\pi t \right) \right)} \right) = k\]

\[ \Rightarrow \lim_{t \to 0} \left( \frac{\left( 1 - \cos t \right)}{4 t^2} \times \frac{\log \cos t}{\log \left( 1 + 4 t^2 \right)} \right) = k\]

\[ \Rightarrow \lim_{t \to 0} \left( \frac{2 \sin^2 \frac{t}{2}}{16 \times \frac{t^2}{4}} \times \frac{\log \cos t}{\log \left( 1 + 4 t^2 \right)} \right) = k\]

\[ \Rightarrow \frac{2}{16} \lim_{t \to 0} \left( \frac{\sin^2 \frac{t}{2}}{\left( \frac{t^2}{4} \right)} \times \frac{\log \cos t}{\left( \frac{4 t^2 \log \left( 1 + 4 t^2 \right)}{4 t^2} \right)} \right) = k\]

\[\]

\[ \Rightarrow \frac{1}{8} \lim_{t \to 0} \left( \frac{\sin^2 \frac{t}{2}}{\left( \frac{t}{2} \right)^2} \times \frac{\left( \frac{\log \cos t}{4 t^2} \right)}{\left( \frac{\log \left( 1 + 4 t^2 \right)}{4 t^2} \right)} \right) = k\]

\[ \Rightarrow \frac{1}{8} \lim_{t \to 0} \left( \frac{\sin^2 \frac{t}{2}}{\left( \frac{t}{2} \right)^2} \times \frac{\left( \frac{\log \sqrt{1 - \sin^2 t}}{4 t^2} \right)}{\left( \frac{\log\left( 1 + 4 t^2 \right)}{4 t^2} \right)} \right) = k\]

\[ \Rightarrow \frac{1}{8} \lim_{t \to 0} \left( \frac{\sin^2 \frac{t}{2}}{\left( \frac{t}{2} \right)^2} \times \frac{\left( \frac{\log\left( 1 - \sin^2 t \right)}{\left( 8 t^2 \right)} \right)}{\left( \frac{\log\left( 1 + 4 t^2 \right)}{4 t^2} \right)} \right) = k\]

\[ \Rightarrow \frac{1}{64} \lim_{t \to 0} \left( \frac{\sin^2 \frac{t}{2}}{\left( \frac{t}{2} \right)^2} \times \frac{\left( \frac{\log\left( 1 - \sin^2 t \right)}{t^2} \right)}{\left( \frac{\log\left( 1 + 4 t^2 \right)}{4 t^2} \right)} \right) = k\]

\[ \Rightarrow \frac{1}{64}\left( \lim_{t \to 0} \left( \frac{\sin\frac{t}{2}}{\left( \frac{t}{2} \right)} \right)^2 \times \frac{\lim_{t \to 0} \left( \frac{\log\left( 1 - \sin^2 t \right)}{t^2} \right)}{\lim_{t \to 0} \left( \frac{\log\left( 1 + 4 t^2 \right)}{4 t^2} \right)} \right) = k\]

\[ \Rightarrow \frac{1}{64}\left( 1 \times \lim_{t \to 0} \frac{\left( - \sin^2 t \right) \log \left( 1 - \sin^2 t \right)}{t^2 \left( - \sin^2 t \right)} \right) = k\]

\[ \Rightarrow \frac{- 1}{64}\left( \lim_{t \to 0} \frac{\left( \sin^2 t \right) \log \left( 1 - \sin^2 t \right)}{t^2 \left( - \sin^2 t \right)} \right) = k\]

\[ \Rightarrow \frac{- 1}{64}\left( \lim_{t \to 0} \left( \frac{\sin t}{t} \right)^2 \lim_{t \to 0} \frac{\log \left( 1 - \sin^2 t \right)}{\left( - \sin^2 t \right)} \right) = k\]

\[\]

\[ \Rightarrow \frac{- 1}{64}\left( \lim_{t \to 0} \left( \frac{\sin t}{t} \right)^2 \lim_{t \to 0} \frac{\log\left( 1 - \sin^2 t \right)}{\left( - \sin^2 t \right)} \right) = k\]

\[ \Rightarrow k = \frac{- 1}{64} \left[ \because \lim_{x \to 0} \frac{\log\left( 1 - x \right)}{x} = 1 \right]\]