Question

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

Options

• 0

• \[\frac{1}{2}\] 

• 1

• −1

Answer 1

Given:  

\[f\left( x \right) = \binom{\frac{\sin\left( \cos x \right) - \cos x}{\left( \pi - 2x \right)^2}, x \neq \frac{\pi}{2}}{k, x = \frac{\pi}{2}}\]

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

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

\[\Rightarrow \lim_{x \to \frac{\pi}{2}} \frac{\sin\left( \cos x \right) - \cos x}{\left( \pi - 2x \right)^2} = k\]

Now,

\[\frac{\pi}{2} - x = y\]

\[\Rightarrow \pi - 2x = 2y\]

Also,

\[x \to \frac{\pi}{2}, y \to 0\]

\[\Rightarrow \lim_{y \to 0} \frac{\sin\left( \cos\left( \frac{\pi}{2} - y \right) \right) - \cos\left( \frac{\pi}{2} - y \right)}{4 y^2} = k\]

\[\Rightarrow \lim_{y \to 0} \frac{\sin\left( \sin y \right) - \sin\left( y \right)}{4 y^2} = k\]

\[\Rightarrow \lim_{y \to 0} \frac{2 \sin\left( \frac{\sin y - y}{2} \right) \cos\left( \frac{\sin y + y}{2} \right)}{4 y^2} = k \left[ \because \sin C - \sin D = 2 sin\left( \frac{C - D}{2} \right) \cos\left( \frac{C + D}{2} \right) \right]\]
\[ \Rightarrow \frac{1}{2} \lim_{y \to 0} \frac{\sin\left( \frac{\sin y - y}{2} \right)}{y}\frac{\cos\left( \frac{\sin y + y}{2} \right)}{y} = k\]
\[ \Rightarrow \frac{1}{2} \lim_{y \to 0} \frac{\left( \frac{\sin y - y}{2} \right) \sin\left( \frac{\sin y - y}{2} \right)}{y\left( \frac{\sin y - y}{2} \right)}\frac{\cos\left( \frac{\sin y + y}{2} \right)}{y} = k\]
\[ \Rightarrow \frac{1}{2} \lim_{y \to 0} \left( \frac{\left( \frac{\sin y - y}{2} \right)}{y} \right)\left( \frac{\sin\left( \frac{\sin y - y}{2} \right)}{\left( \frac{\sin y - y}{2} \right)} \right)\left( \frac{\cos\left( \frac{\sin y + y}{2} \right)}{y} \right) = k\]
\[ \Rightarrow \frac{1}{2} \lim_{y \to 0} \left( \frac{\left( \frac{\sin y - y}{2} \right)}{y} \right) \lim_{y \to 0} \left( \frac{\sin\left( \frac{\sin y - y}{2} \right)}{\left( \frac{\sin y - y}{2} \right)} \right) \lim_{y \to 0} \left( \frac{\cos\left( \frac{\sin y + y}{2} \right)}{y} \right) = k\]
\[ \Rightarrow \frac{1}{4} \lim_{y \to 0} \left( \frac{\sin y}{y} - 1 \right) \lim_{y \to 0} \left( \frac{\sin\left( \frac{\sin y - y}{2} \right)}{\left( \frac{\sin y - y}{2} \right)} \right) \lim_{y \to 0} \left( \frac{\cos\left( \frac{\sin y + y}{2} \right)}{y} \right) = k\]
\[ \Rightarrow \frac{1}{4} \times 0 \times 1 \times \lim_{y \to 0} \left( \frac{\cos\left( \frac{\sin y + y}{2} \right)}{y} \right) = k\]
\[ \Rightarrow 0 = k\]