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Prove that y = θθθ4sinθ2+cosθ-θ is an increasing function if θθ∈[0,π2]

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प्रश्न

Prove that y = `(4sinθ)/(2 + cosθ) - θ` is an increasing function if `θ ∈[0, pi/2]`

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उत्तर

y = `(4sinθ)/(2 + cosθ) - θ`

∴ `dy/"dθ" = d/"dθ"[(4sinθ)/(2 + cosθ) - θ]`

= `d/"dθ"((4sinθ)/(2 + cosθ)) - d/"dθ"(θ)`

= `((2 + cosθ).d/"dθ"(4sinθ) - 4sinθ.d/"dθ"(2 + cos θ))/((2 + cosθ)^2) - 1`

= `((2 + cosθ)"(4cosθ) - (4sinθ)(0 - sinθ))/((2 + cosθ)^2) - 1`

= `(8cosθ + 4cos^2θ + 4sin^2θ)/(2 + cosθ)^2 - 1`

= `(8cosθ + 4(cos^2θ + sin^2θ))/(2 + cosθ)^2 - 1`

= `(8cosθ + 4)/(2 + cosθ)^2 - 1`

= `((8cos θ + 4) - (2 + cosθ)^2)/(2 + cosθ)^2`

= `(8cosθ + 4 - 4 - 4cosθ - cos^2θ)/(2 + cosθ)^2`

= `(4cosθ - cos^2θ)/(2 + cosθ)^2`

= `(cosθ(4 - cosθ))/(2 + cosθ)^2`

Since, `θ ∈ [0, pi/2], cos θ ≥ 0` Also, cos θ < 4

∴ 4 - cos θ > 0

∴ cos θ (4 - cosθ) ≥ 0

∴ `(cosθ(4 - cosθ))/(2 + cosθ^2)≥ 0`

∴ `dy/"dθ" ≥ 0  "for all"  θ ∈[0, pi/2]`    

Hence, y is an increasing function if `θ ∈[0, pi/2]`.

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अध्याय 2: Applications of Derivatives - Exercise 2.4 [पृष्ठ ९०]

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बालभारती Mathematics and Statistics 2 (Arts and Science) [English] Standard 12 Maharashtra State Board
अध्याय 2 Applications of Derivatives
Exercise 2.4 | Q 24 | पृष्ठ ९०

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