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Question
Show that f(x) = cos x is a decreasing function on (0, π), increasing in (−π, 0) and neither increasing nor decreasing in (−π, π).
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Solution
\[Here, \]
\[f\left( x \right) = \cos x\]
\[\text{Domain of cos x is}\left( - \pi, \pi \right).\]
\[ \Rightarrow f'\left( x \right) = - \sin x\]
\[\text{For x} \in \left( - \pi, 0 \right), \sin x < 0 \left[ \because \text{sine function is negative in third and fourth quadrant }\right]\]
\[ \Rightarrow - \sin x > 0\]
\[ \Rightarrow f'\left( x \right) > 0\]
\[So, \text{cos x is increasing in} \left( - \pi, 0 \right) . \]
\[\text{For x} \in \left( 0, \pi \right)),\sin x > 0 \left[ \because \text{sine function is positive in first and second quadrant }\right]\]
\[ \Rightarrow - \sin x < 0\]
\[ \Rightarrow f'\left( x \right) < 0\]
\[\text{So,f(x) is decreasing on}\left( 0, \pi \right).\]
\[\text{Thus,f(x) is neither increasing nor decreasing in}\left( - \pi, \pi \right).\]
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