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Question: If \[{{e}^{\dfrac{-\pi }{2}}}<\theta <\dfrac{\pi }{2}\], then – (a)\[\cos \log \theta <\log \cos ...

If e−π2<θ<π2{{e}^{\dfrac{-\pi }{2}}}<\theta <\dfrac{\pi }{2}, then –
(a)cos⁡log⁡θ<log⁡cos⁡θ\cos \log \theta <\log \cos \theta
(b)cos⁡log⁡θ>log⁡cos⁡θ\cos \log \theta >\log \cos \theta
(c)cos⁡log⁡θ≤log⁡cos⁡θ\cos \log \theta \le \log \cos \theta
(d)cos⁡θ>log⁡θ\cos \theta >\log \theta

Explanation

Solution

Hint: At first write the inequation then apply ln function to all the sides of inequality to get the inequality as −π2<ln⁡θ<ln⁡π2\dfrac{-\pi }{2}<\ln \theta <\ln \dfrac{\pi }{2}. Then apply cos function to write the given inequation as, 0<cos⁡(ln⁡θ)<cos⁡(ln⁡π2)0<\cos \left( \ln \theta \right)<\cos \left( \ln \dfrac{\pi }{2} \right). From here get the inequality cos⁡(ln⁡θ)>0\cos \left( \ln \theta \right)>0. Then use the fact that −1≤cos⁡θ≤1-1\le \cos \theta \le 1. After then apply the in function to get ln⁡(cos⁡θ)≤0\ln \left( \cos \theta \right)\le 0.

Complete step-by-step answer:
In the question we are given as inequality which is e−π2<θ<π2{{e}^{\dfrac{-\pi }{2}}}<\theta <\dfrac{\pi }{2}.
So, we are given that,
e−π2<θ<π2{{e}^{\dfrac{-\pi }{2}}}<\theta <\dfrac{\pi }{2}
Now we will take log⁡e{{\log }_{e}} or ln to all the sides of inequality to write the equation as,
ln⁡e−π2<ln⁡θ<ln⁡π2\ln {{e}^{\dfrac{-\pi }{2}}}<\ln \theta <{{\ln }^{\dfrac{\pi }{2}}}
Here the sides of the inequality do not change because there is a property of in function that is a < b, then, ln⁡a<ln⁡b\ln a<\ln b.
We can write ln⁡e−π2\ln {{e}^{\dfrac{-\pi }{2}}} as −π2\dfrac{-\pi }{2}. So, we can write the inequality as,
−π2<ln⁡θ<ln⁡π2\dfrac{-\pi }{2}<\ln \theta <\ln \dfrac{\pi }{2}
Now, we will take cos to all the sides of inequalities so we get,
cos⁡(−π2)<cos⁡(ln⁡θ)<cos⁡(ln⁡π2)\cos \left( \dfrac{-\pi }{2} \right)<\cos \left( \ln \theta \right)<\cos \left( \ln \dfrac{\pi }{2} \right)
Here cos function will not change the sides of inequality as the values are very less to be considered.
We know that, cos⁡(−θ)=cos⁡θ\cos \left( -\theta \right)=\cos \theta .
So, cos⁡(−π2)\cos \left( \dfrac{-\pi }{2} \right) is equal to cos⁡π2\cos \dfrac{\pi }{2} and as we know value of cos⁡π2\cos \dfrac{\pi }{2} is 0.
So, we can write that, cos⁡(ln⁡θ)\cos \left( \ln \theta \right) is greater than 0.
Now, as know cos⁡θ\cos \theta always less than or equal to 1. So, we say that ln⁡(cos⁡θ)\ln \left( \cos \theta \right) also less than or equal to ln⁡(1)\ln \left( 1 \right) which is equal to ‘0’.
Hence, ln⁡(cos⁡θ)≤0\ln \left( \cos \theta \right)\le 0.
We also found out that, cos⁡(ln⁡θ)>0\cos \left( \ln \theta \right)>0.
Means we can write it as,
ln⁡(cos⁡θ)≤0<cos⁡(ln⁡θ)\ln \left( \cos \theta \right)\le 0<\cos \left( \ln \theta \right)
Or, ln⁡(cos⁡θ)<cos⁡(ln⁡θ)\ln \left( \cos \theta \right)<\cos \left( \ln \theta \right)
We can write it as,
log⁡(cos⁡θ)<cos⁡(log⁡θ)\log \left( \cos \theta \right)<\cos \left( \log \theta \right)
So, the correct option is (b).

Note: While solving inequalities students generally confuse themselves while taking functions to all the sides of the inequality where or where not to change.