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Solution for Prove that the Function F(X) = Cos X Is: (I) Strictly Decreasing in (0, π) (Ii) Strictly Increasing in (π, 2π) (Iii) Neither Increasing Nor Decreasing in (0, 2π) - CBSE (Science) Class 12 - Mathematics

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Question

Prove that the function f(x) = cos x is:
(i) strictly decreasing in (0, π)
(ii) strictly increasing in (π, 2π)
(iii) neither increasing nor decreasing in (0, 2π).

Solution

\[f\left( x \right) = \cos x\]

\[f'\left( x \right) = - \sin x\]

\[\left( i \right) \] \[\text { Here },\]

\[0 < x < \pi\]

\[ \Rightarrow \sin x > 0 \left[ \because \text { Sine function is positive in first and second quadrant } \right]\]

\[ \Rightarrow - \sin x < 0\]

\[ \Rightarrow f'\left( x \right) < 0, \forall x \in \left( 0, \pi \right)\]

\[\text { So,f(x)is strictly decreasing on } \left( 0, \pi \right) . \]

\[\left( ii \right) \] \[\text { Here, }\]

\[\pi < x < 2\pi\]

\[ \Rightarrow \sin x < 0 \left[ \because \text { Sine function is negative in third and fourth quadrant} \right]\]

\[ \Rightarrow - \sin x > 0\]

\[ \Rightarrow f'\left( x \right) > 0, \forall x \in \left( \pi, 2\pi \right)\]

\[\text { So,f(x)is strictly increasing on } \left( \pi, 2\pi \right) . \]

\[\left( iii \right) \] \[\text { From eqs. (1) and (2), we get }\]

\[f(x)\text { is strictly decreasing on } \left( 0, \pi \right) \text { and is strictly increasing on } \left( \pi, 2\pi \right) . \]

\[\text { So,}f\left( x \right) \text { is neither increasing nor decreasing on}\left( 0, 2\pi \right).\]

  Is there an error in this question or solution?

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Solution for question: Prove that the Function F(X) = Cos X Is: (I) Strictly Decreasing in (0, π) (Ii) Strictly Increasing in (π, 2π) (Iii) Neither Increasing Nor Decreasing in (0, 2π) concept: Increasing and Decreasing Functions. For the courses CBSE (Science), PUC Karnataka Science, CBSE (Arts), CBSE (Commerce)
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