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