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प्रश्न
Prove that y = `(4sin theta)/(2 + cos theta) - theta` is an increasing function of θ in `[0, pi/2]`
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उत्तर
Given, `y = (4 sin theta)/(2 + cos theta) - theta` and interval `[0, pi/2]`
`=> "dy"/("d" theta) = ((2 + cos theta) 4 cos theta - 4 sin theta (- sin theta))/((2 + cos theta)^2) - 1`
`= (8 cos theta + 4 cos^2 + 4 sin^2 theta)/((2 + cos theta)^2) - 1`
`= (8 cos theta + 4 (cos^2 theta + sin^2 theta))/((2 + cos theta)^2) - 1`
`= (8 cos theta + 4)/((2 + cos theta)^2) - 1`
`= (8 c0s theta + 4 - (4 + cos^2 theta + 4 cos theta))/((2 + cos theta)^2)`
`= (4 cos theta - cos^2 theta)/((2 + cos theta)^2)`
`= ((4 - cos theta) cos theta)/((2 + cos theta)^2)`
cos θ > 0 in `[0, pi/2] ; 4 - cos theta > 0 [0, pi/2]`
`(∵ -1 <= cos theta <= 1, if theta in [0, pi/2]),`
`(2 + cos theta)^2 > 0 [0, pi/2]` ...(being a perfect square)
= `dy/(d theta) > 0` for all `theta in [0, pi/2]`
= y is strictly increasing function in `[0, pi/2]`
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