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प्रश्न
Let \[f\left( x \right) = \begin{cases}\frac{k\cos x}{\pi - 2x}, & where x \neq \frac{\pi}{2} \\ 3, & where x = \frac{\pi}{2}\end{cases}\] and if \[\lim_{x \to \frac{\pi}{2}} f\left( x \right) = f\left( \frac{\pi}{2} \right)\]
उत्तर
We have, \[f\left( x \right) = \begin{cases}\frac{k\cos x}{\pi - 2x}, & where x \neq \frac{\pi}{2} \\ 3, & where x = \frac{\pi}{2}\end{cases}\]
It is given that,
\[\lim_{x \to \frac{\pi}{2}} f\left( x \right) = f\left( \frac{\pi}{2} \right)\]
\[ \Rightarrow \lim_{x \to \frac{\pi}{2}} \frac{k\cos x}{\pi - 2x} = 3\]
\[ \Rightarrow \frac{k}{2} \times \lim_{x - \frac{\pi}{2} \to 0} \frac{\sin\left( \frac{\pi}{2} - x \right)}{\frac{\pi}{2} - x} = 3\]
\[ \Rightarrow \frac{k}{2} \times \lim_{h \to 0} \frac{\sin\left( - h \right)}{- h} = 3 \left( Put x - \frac{\pi}{2} = h \right)\]
\[\lim_{x \to \frac{\pi}{2}} f\left( x \right) = f\left( \frac{\pi}{2} \right)\]
\[ \Rightarrow \lim_{x \to \frac{\pi}{2}} \frac{k\cos x}{\pi - 2x} = 3\]
\[ \Rightarrow \frac{k}{2} \times \lim_{x - \frac{\pi}{2} \to 0} \frac{\sin\left( \frac{\pi}{2} - x \right)}{\frac{\pi}{2} - x} = 3\]
\[ \Rightarrow \frac{k}{2} \times \lim_{h \to 0} \frac{\sin\left( - h \right)}{- h} = 3 \left( Put x - \frac{\pi}{2} = h \right)\]
\[\Rightarrow \frac{k}{2} \times \lim_{h \to 0} \frac{\sinh}{h} = 3 \left[ \sin\left( - \theta \right) = - \sin\theta \right]\]
\[ \Rightarrow \frac{k}{2} = 3 \left( \lim_{x \to 0} \frac{\sin x}{x} = 1 \right)\]
\[ \Rightarrow k = 6\]
Hence, the value of k is 6.
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