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Ex 6.5

Ex 6.5, 1 (i)
Important

Ex 6.5, 1 (ii)

Ex 6.5, 1 (iii) Important

Ex 6.5, 1 (iv)

Ex 6.5, 2 (i)

Ex 6.5, 2 (ii) Important

Ex 6.5, 2 (iii)

Ex 6.5, 2 (iv) Important

Ex 6.5, 2 (v) Important

Ex 6.5, 3 (i)

Ex 6.5, 3 (ii)

Ex 6.5, 3 (iii)

Ex 6.5, 3 (iv) Important

Ex 6.5, 3 (v)

Ex 6.5, 3 (vi)

Ex 6.5, 3 (vii) Important

Ex 6.5, 3 (viii)

Ex 6.5, 4 (i)

Ex 6.5, 4 (ii) Important

Ex 6.5, 4 (iii)

Ex 6.5, 5 (i)

Ex 6.5, 5 (ii)

Ex 6.5, 5 (iii) Important

Ex 6.5, 5 (iv)

Ex 6.5,6

Ex 6.5,7 Important

Ex 6.5,8

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Ex 6.5,10

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Ex 6.5,13

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Ex 6.5,24 Important

Ex 6.5,25 Important You are here

Ex 6.5, 26 Important

Ex 6.5, 27 (MCQ)

Ex 6.5,28 (MCQ) Important

Ex 6.5,29 (MCQ)

Last updated at April 15, 2021 by Teachoo

Ex 6.5, 25 Show that the semi-vertical angle of the cone of the maximum volume and of given slant height is tan β1 β2Let π be the slant height & ΞΈ be the semi vertical angle of the cone. Now, Height of cone = h = π cos ΞΈ Radius of cone = r = π sin ΞΈ We need to maximize volume of cone V = 1/3 ππ^2 β V = 1/3 ππ^2sin2π l cos π V= 1/3 ππ^3sin2π cos π Differentiating ππ£/ππ = 1/3 ππ^3 (2 sin π cos π . cos π + sin2 π (sin π)) ππ£/ππ = 1/3 ππ^3 [2 sinβ‘π cos^2β‘πβsin^3β‘π ] ππ£/ππ = 1/3 ππ^3 (2 sin π cos2 π sin3 π) ππ£/ππ = 1/3 ππ^3 sin π (2 cos2 π β sin2 π) ππ£/ππ = 1/3 ππ^3 sin π (β2 cosβ‘γπ+sinβ‘γπ)(γ γ β2 cos π β sin π) ππ£/ππ = 1/3 ππ^3 sin π cos π ((β2 cosβ‘γπ+sinβ‘π γ))/cosβ‘π Γ cos π ((β2 cosβ‘γπβsinβ‘π γ))/cosβ‘π ππ£/ππ = 1/3 ππ^3 sin π cos2 π (β2β‘γ+γtan πγβ‘γ )(γ γ β2 β tan π) Putting ππ£/ππ = 0 1/3 ππ^3 sin π cos2 π (β2β‘γ+γtan πγβ‘γ )(γ γ β2 β tan π) = 0 sin π cos2 π (β2β‘γ+γtan πγβ‘γ )(γ γ β2 β tan π) = 0 sin π cos2 π (β2β‘γ+γtan πγβ‘γ )(γ γ β2 β tan π) = 0 sin π½ = 0 π = 0Β° π cannot be 0Β° for cone. cos2 π½ = 0 π = 90Β° π cannot be 90Β° for cone. βπ + tan π½ = 0 tan π = ββ2 For cone, 0Β° < π < 90Β° tan π is (β) ve in II & IV quadrant so tan π = β β2 is not possible βπ β tan π½ = 0 tan π = β2 π = tanβ1 β2 β΄ tan π = β2 is the possible value for cone. The value is either maxima or minima. Finding maxima or minima by first derivative test ππ£/ππ = 1/3 ππ^3 sin π cos2 π (β2 + tanπ) (β2 β tan π) sin π½ 0Β° < π < 90Β° Since π is in 1st quadrant sin π > 0 cos π½ 0Β° < π < 90Β° Since π is in 1st quadrant β΄ cos π > 0 (βπ+πππ π½) 0Β° < ΞΈ < 90Β° Since ΞΈ is in 1st quadrant So, tan ΞΈ > 0 So (β2+π‘ππ ΞΈ) is also (+)ve. Since ππ£/πΞΈ = changes sign from (+) ve to (β) ve π½ = tanβ 1βπ is the maxima. Hence proved