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Solution for Find the Interval in Which F(X) is Increasing Or Decreasing F(X) = Sinx(1 + Cosx), 0 < X < π 2 ? - CBSE (Science) Class 12 - Mathematics

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Question

Find the interval in which f(x) is increasing or decreasing f(x) = sinx(1 + cosx), 0 < x < \[\frac{\pi}{2}\] ?

Solution

\[ f\left( x \right) = \sin x\left( 1 + \cos x \right), 0 < x < \frac{\pi}{2}\]

\[ \Rightarrow f\left( x \right) = \sin x + \sin x\cos x\]

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

\[ \Rightarrow f'\left( x \right) = \cos x - \sin^2 x + \cos^2 x\]

\[ \Rightarrow f'\left( x \right) = \cos x + \cos^2 x - 1 + \cos^2 x\]

\[ \Rightarrow f'\left( x \right) = 2 \cos^2 x + \cos x - 1\]

\[ \Rightarrow f'\left( x \right) = 2 \cos^2 x + 2\cos x - \cos x - 1\]

\[ \Rightarrow f'\left( x \right) = 2\cos x\left( \cos x + 1 \right) - 1\left( \cos x + 1 \right)\]

\[ \Rightarrow f'\left( x \right) = \left( 2\cos x - 1 \right)\left( \cos x + 1 \right)\]

\[\text { For } f\left( x \right) \text { to be increasing, we must have }\]

\[f'\left( x \right) > 0\]

\[ \Rightarrow \left( 2\cos x - 1 \right)\left( \cos x + 1 \right) > 0\]

\[\text { This is only possible when}\]

\[\left( 2\cos x - 1 \right) < 0 \text { and } \left( \cos x + 1 \right) > 0\]

\[ \Rightarrow \left( 2\cos x - 1 \right) < 0 \text { and } \left( \cos x + 1 \right) > 0\]

\[ \Rightarrow \cos x < \frac{1}{2} \text { and} \cos x > - 1\]

\[ \Rightarrow x \in \left( \frac{\pi}{3}, \frac{\pi}{2} \right)\text {  and } x \in \left( 0, \frac{\pi}{2} \right)\]

\[\text { So,} x \in \left( \frac{\pi}{3}, \frac{\pi}{2}\right)\]

\[ \therefore f\left( x \right) \text { is decreasing on } \left( \frac{\pi}{3}, \frac{\pi}{2} \right) .\]

 

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Solution for question: Find the Interval in Which F(X) is Increasing Or Decreasing F(X) = Sinx(1 + Cosx), 0 < X < π 2 ? concept: Increasing and Decreasing Functions. For the courses CBSE (Science), CBSE (Commerce), PUC Karnataka Science, CBSE (Arts)
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