Question

If \[{{e}^{\dfrac{-\pi }{2}}}<\theta <\dfrac{\pi }{2}\], then –
(a)\[\cos \log \theta <\log \cos \theta \]
(b)\[\cos \log \theta >\log \cos \theta \]
(c)\[\cos \log \theta \le \log \cos \theta \]
(d)\[\cos \theta >\log \theta \]

Answer 1

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

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