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Question

Let f(x)=1+eln(lnx)ln(k2+25) and g(x)=1|x|1. If limx1f(x)g(x)=k(2sin2α+3cosβ+5), k>0 and α,βR, then which of the following statement(s) is (are) CORRECT?

A
k=5
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B
sin10α+cos5βsin2α+cosβ=1
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C
cos2β+sin4α=2
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D
sin2α>cosβ
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Solution

The correct option is C cos2β+sin4α=2
f(x)=1+lnxln(k2+25) and g(x)=1|x|1
limx1f(x)g(x)
=limx1(1+lnxln(k2+25))1|x|1
=limx1(1+lnxln(k2+25))1x1 [1 form]
=exp(limx1lnxln(k2+25)x1)
=eln(k2+25)=k2+25

k2+25=k(2sin2α+3cosβ+5)
2sin2α+3cosβ+5=k+25k (1)
k+25k2k25k
k+25k10
Maximum value of 2sin2α+3cosβ+5 is 10 if α=(2n+1)π2 and β=2nπ
So, for equation (1), L.H.S. 10
So, both L.H.S. and R.H.S. should be equal to 10.
k+25k=10
k210k+25=0
(k5)2=0
k=5

sin2α=1 and cosβ=1
Hence, sin10α+cos5βsin2α+cosβ=1+11+1=1

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