A) \[cos\left( log\theta \right)>log\left( cos\theta \right)\]
B) \[cos\left( log\theta \right)<log\left( cos\theta \right)\]
C) \[cos\left( log\theta \right)=log\left( cos\theta \right)\]
D) None of these
Correct Answer: A
Solution :
[a] since \[{{e}^{-\pi /2}}<\theta <\pi /2\] \[\Rightarrow log{{e}^{-}}^{\pi /2}<log\theta <log\pi /2\] \[\Rightarrow -\frac{\pi }{2}<log\theta <iog\frac{\pi }{2}<1<\pi /2\] i.e. \[-\frac{\pi }{2}<log\theta <\frac{\pi }{2}\] But \[0<cos\theta <1\] so, \[log\left( cos\theta \right)<log\left( 1 \right)\] \[\Rightarrow log(cos\theta )<0\] Hence \[cos\left( log\theta \right)>log\left( cos\theta \right)\]
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