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Q.

If the function  fx=1+cosxλcosx,0<x<π2μ,x=π2cot 6xcot 4x,π2<x<π     is continuous at
x=π2, then 9λ+6logeμ+μ6e6λ is equal to 

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a

11

b

8

c

2e4+8

d

10

answer is A.

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Detailed Solution

detailed_solution_thumbnail

1+cosxλcosx,0<x<π2μ,x=π2cot6xcot4x,π2<x<πis continuous

limxπ2-f(x)=limxπ2-(1+cosx)λ/cosx=1α form

limxπ2-f(x)=elimxπ2cosxλcosx=eλ
limxπ2+ f(x)=limxπ2+ecot6xcot4x=limxπ2+etan4xtan6x 
=e2/3 (by using L.Hospital) 


f (x) is continuous ax=π2

λ=23,μ=eλ=e23

9λ+6logeμ+μ6e6λ =6+4+e4e4=10

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