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For \(0\lt\theta\lt\frac{\pi}{4}\), let \(a=\left((\sin \theta)^{\sin \theta}\right)\left(\log _{2} \cos \theta\right), b=\left((\cos \theta)^{\sin \theta}\right)\left(\log _{2} \sin \theta\right),…

Question and options

For 0<θ<π40\lt\theta\lt\frac{\pi}{4}, let a=((sin⁡θ)sin⁡θ)(log⁡2cos⁡θ),b=((cos⁡θ)sin⁡θ)(log⁡2sin⁡θ),c=((sin⁡θ)cos⁡θ)(log⁡2cos⁡θ)a=\left((\sin \theta)^{\sin \theta}\right)\left(\log _{2} \cos \theta\right), b=\left((\cos \theta)^{\sin \theta}\right)\left(\log _{2} \sin \theta\right), c=\left((\sin \theta)^{\cos \theta}\right)\left(\log _{2} \cos \theta\right) and d=((sin⁡θ)sin⁡θ)(log⁡2sin⁡θ)d=\left((\sin \theta)^{\sin \theta}\right)\left(\log _{2} \sin \theta\right). Then, the median value in the sequence a,b,c,da, b, c, d is
  1. a+b2\frac{a+b}{2}
  2. a+d2\frac{a+d}{2}
  3. b+c2\frac{b+c}{2}
  4. c+d2\frac{c+d}{2}

Answer and explanation

Correct answer: B - \(\frac{a+d}{2}\)
For 0<t<pi/4, 0<sin<cos<1 so log2 sin < log2 cos < 0. a vs c: s^s > s^c (base<1), times negative log2 cos gives a<c. b vs d: c^s > s^s, times negative log2 sin gives b<d. d vs a: same positive factor s^s, log2 sin < log2 cos so d<a. Order b<d<a<c, median=(a+d)/2.

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