    Question

# Let f(x)=1+eln(lnx)ln(k2+25) and g(x)=1|x|−1. If limx→1+(f(x))g(x)=k(2sin2α+3cosβ+5) for k>0 and α,β∈R, then which of the following is (are) CORRECT?

A
The value of k is equal to 5
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B
The value of sin10α+cos5βsin2α+cosβ is equal to 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+(lnx)ln(k2+25) g(x)=1|x|−1 Now, limx→1+(f(x))g(x) =limx→1+(1+(lnx)ln(k2+25))1/(x−1)=exp(limx→1+(lnx)ln(k2+25)(x−1)) Let x=1+t limx→1+(f(x))g(x)=exp(limt→0+(ln(1+t))ln(k2+25)t)=eln(k2+25)=k2+25 Now, k2+25k=2sin2α+3cosβ+5 ⇒k+25k=2sin2α+3cosβ+5 ⇒(√k−5√k)2+10=2sin2α+3cosβ+5 Above equation is true only when k=5 and sin2α=1 and cosβ=1  Suggest Corrections  1      Similar questions  Explore more