Ex 6.3,9 - Chapter 6 Class 12 Application of Derivatives
Last updated at Dec. 16, 2024 by Teachoo
Ex 6.3
Ex 6.3, 1 (i)
Important
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Transcript
Ex 6.3, 9 What is the maximum value of the function sinβ‘π₯+cosβ‘π₯? Let f(π₯)=sinβ‘π₯+cosβ‘π₯ Consider the interval π₯ β [0 , 2π] Finding fβ(π) fβ(π₯)=π(sinβ‘π₯ + cosβ‘π₯ )/ππ₯ fβ(π₯)=cosβ‘π₯βsinβ‘π₯ Putting fβ(π)=π cosβ‘π₯βsinβ‘π₯=0 cosβ‘π₯=sinβ‘π₯ 1 =sinβ‘π₯/cosβ‘π₯ 1= tanβ‘π₯ tan π₯=1 Since π₯ β [0 , 2π] tan π₯=1 at π₯=π/4 , π₯=5π/4 in the interval [0 , 2π] We have given the interval π₯ β [0 , 2π] Hence Calculating f(π₯) at π₯=0 ,π/4 , 5π/4 & 2π Hence Maximum Value of f(π₯) is βπ at π = π /π
