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Ex 6.3, 5 (ii) - Chapter 6 Class 12 Application of Derivatives
Last updated at June 12, 2023 by Teachoo
Ex 6.3
Ex 6.3, 1 (i)
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Ex 6.3, 1 (ii)
Ex 6.3, 1 (iii) Important
Ex 6.3, 1 (iv)
Ex 6.3, 2 (i)
Ex 6.3, 2 (ii) Important
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Ex 6.3, 4 (i)
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Ex 6.3, 5 (i)
Ex 6.3, 5 (ii) You are here
Ex 6.3, 5 (iii) Important
Ex 6.3, 5 (iv)
Ex 6.3,6
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Ex 6.3, 27 (MCQ)
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Ex 6.3,29 (MCQ)
Transcript
Ex 6.3, 5 Find the absolute maximum value and the absolute minimum value of the following functions in the given intervals: (ii) f (π₯) = sinβ‘π₯ + cosβ‘π₯ , π₯ β [0, π ] Finding fβ(π) fβ(π₯)=π(π πππ₯ + πππ π₯)/ππ₯ fβ(π₯)=cosβ‘γπ₯ βsinβ‘π₯ γ Putting fβ(π) cosβ‘γπ₯ βsinβ‘π₯ γ= 0 cosβ‘γπ₯=sinβ‘π₯ γ 1 = sinβ‘π₯/(cosβ‘ π₯) 1 = tan π₯ tan π₯ = 1 We know that know tan ΞΈ = 1 at ΞΈ = π/4 β΄ π₯ = π/4 Since given interval π₯ β [0 , π] Hence calculating f(π₯) at π₯=0 , π/4 ,π Absolute Maximum value of f(π₯) is βπ at π = π /π & Absolute Minimum value of f(π₯) is β1 at π = Ο
