First slide
Introduction to limits
Question

If limx01+xlog1+b21/x=2bsin2θ , b>0 and  θ[π,π] then the value of θ is

Difficult
Solution

limx01+xlog1+b21/x=limx01+xlog1+b21xlog1+b2×log1+b2=elog1+b2=1+b2

thus,

1+b2=2bsin2θ sin2θ=1+b22b1

sin2θ=1sinθ=±1 or θ=±π/2

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