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प्रश्न
Find the interval in which f(x) is increasing or decreasing f(x) = sinx(1 + cosx), 0 < x < \[\frac{\pi}{2}\] ?
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उत्तर
\[\left( iii \right) f\left( x \right) = sin x\left( 1 + cos x \right), 0 < x < \frac{\pi}{2}\]
`f( x) = sin x + sin x\cos x`
` ⇒f(x) = cos x + sin x ( - sin x \right) + cos x\cos x`
`⇒ f(x) = cos x - \sin^2 x + cos^2 x`
`⇒ f(x) = co sx + \cos^2 x - 1 + \cos^2 x`
`⇒ f(x) = 2 \cos^2 x + \cosx - 1`
`⇒ f(x) = 2 \cos^2 x + 2\cosx - \cosx - 1`
`⇒ f(x) = 2\cos x( \cos x + 1 \right) - 1\left( \cos x + 1 )`
`⇒ f(x) = ( 2\cos x - 1 cos x + 1 \right)`
For f(x) to be increasing, we must have
\[f'\left( x \right) > 0\]
`⇒ ( 2 cos x - 1 )( cos x + 1 ) > 0`
This is only possible when
`(2\cos x - 1 ) > 0 and ( cos x + 1 ) > 0`
`⇒ ( 2\cos x - 1 ) > 0 and ( cos x + 1 ) > 0`
`⇒ cos x > 1/2 and cos x > - 1`
`⇒ x ∈ ( 0, π / 3 )and x ∈ ( 0, π / 2 )`
\[So, x \in \left( 0, \frac{\pi}{3} \right)\]
∴ f(x) is increasing on (o, π / 3 )
For f(x) to be decreasing, we must have
\[f'\left( x \right) < 0\]
`⇒ ( 2 cos x - 1 )( cos x + 1 ) < 0`
This is only possible when
`( 2\cosx - 1 ) < 0 and ( cosx + 1 ) > 0`
`⇒ ( 2\cosx - 1 ) < 0 and ( cosx + 1 ) > 0`
`⇒cosx < 1/2 and cosx > - 1`
`⇒ x ∈ (π/3 ,π/2 ) and x ∈ (0 ,π/2 )`
\[So, x \in \left( \frac{\pi}{3}, \frac{\pi}{2} \right)\
` ∴ f (x) \text{ is decreasing on } (π/3 ,π/2 ) .`
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