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If $f(x)=\frac{1+\sec x}{2(\sec x-1)}$ for $0 < x < \frac{\pi}{2}$ and $f^{\prime}(x)=f(x) \cdot g(x)$, then $g(x)=$
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The correct answer is:
$-2 \operatorname{cosec} x$
$$
\begin{aligned}
& \text { Given, } \mathrm{f}(\mathrm{x})=\frac{1+\sec \mathrm{x}}{2(\sec \mathrm{x}-1)} \\
& \text { now } \mathrm{f}^{\prime}(\mathrm{x})=\frac{1}{2} \frac{(1+\cos \mathrm{x})}{(1-\cos \mathrm{x})} \times \frac{(-\sin \mathrm{x})-\operatorname{Sin}(\mathrm{x})\left(\frac{1-\cos \mathrm{x}}{1+\cos \mathrm{x}}\right)}{(1-\cos \mathrm{x})} \\
& \frac{1}{2} \frac{1+\sec (\mathrm{x})}{(\sec (\mathrm{x})-1)} \cdot-2 \operatorname{cosec}(\mathrm{x}) \\
& \Rightarrow \mathrm{f}^{\prime}(\mathrm{x}) \Rightarrow \mathrm{f}(\mathrm{x}) \cdot \mathrm{g}(\mathrm{x})
\end{aligned}
$$
hence, $g(x)=-2 \operatorname{cosec}(x)$
\begin{aligned}
& \text { Given, } \mathrm{f}(\mathrm{x})=\frac{1+\sec \mathrm{x}}{2(\sec \mathrm{x}-1)} \\
& \text { now } \mathrm{f}^{\prime}(\mathrm{x})=\frac{1}{2} \frac{(1+\cos \mathrm{x})}{(1-\cos \mathrm{x})} \times \frac{(-\sin \mathrm{x})-\operatorname{Sin}(\mathrm{x})\left(\frac{1-\cos \mathrm{x}}{1+\cos \mathrm{x}}\right)}{(1-\cos \mathrm{x})} \\
& \frac{1}{2} \frac{1+\sec (\mathrm{x})}{(\sec (\mathrm{x})-1)} \cdot-2 \operatorname{cosec}(\mathrm{x}) \\
& \Rightarrow \mathrm{f}^{\prime}(\mathrm{x}) \Rightarrow \mathrm{f}(\mathrm{x}) \cdot \mathrm{g}(\mathrm{x})
\end{aligned}
$$
hence, $g(x)=-2 \operatorname{cosec}(x)$
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