Question

The function \[g\left( x \right) = {e^{{x^2}}}\dfrac{{\log \left( {\pi + x} \right)}}{{\log \left( {e + x} \right)}}\left( {x \geqslant 0} \right)\] is
 $ \left( a \right){\text{ Increasing on [0,}} \propto ) $
 $ \left( b \right){\text{ Decreasing on [0,}} \propto ) $
 $ \left( c \right){\text{ Increasing on [0,}}\pi {\text{/e}}){\text{ and decreasing on [}}\pi {\text{/e,}} \propto {\text{)}} $
 $ \left( d \right){\text{ Decreasing on [0,}}\pi {\text{/e}}){\text{ and increasing on [}}\pi {\text{/e,}} \propto {\text{)}} $

Answer 1

Hint: Here finding whether the function is increasing and decreasing at what interval, we will differentiate the \[f\left( x \right) = \dfrac{{\log \left( {\pi + x} \right)}}{{\log \left( {e + x} \right)}}\] and by solving it we will get the function which will be then compared with the limit. Also, we know that the $ \log $ function is an increasing function. So by all of these, we can tell the condition of the function.

Complete step-by-step answer:
Since, we know that the function $ {e^{{x^2}}} $ increases on $ {\text{[0,}} \propto ) $ so we will consider the function,
 $ \Rightarrow f\left( x \right) = \dfrac{{\log \left( {\pi + x} \right)}}{{\log \left( {e + x} \right)}} $
Now on differentiating this above function with respect to $ x $ , we get the function $ f'\left( x \right) $ as
 $ \Rightarrow f'\left( x \right) = \dfrac{{\log \left( {e + x} \right) \times \dfrac{1}{{\pi + x}} - log\left( {\pi + x} \right) \times \dfrac{1}{{e + x}}}}{{{{\left( {\log \left( {e + x} \right)} \right)}^2}}} $
So on solving the above function, we will get the function as
 $ \Rightarrow f'\left( x \right) = \dfrac{{\log \left( {e + x} \right) \times \left( {e + x} \right) - \left( {\pi + x} \right) \times log\left( {\pi + x} \right)}}{{{{\left( {\log \left( {e + x} \right)} \right)}^2}}} $
Also from the hint, we knew that the $ \log $ function is an increasing function, so to represent it mathematically we can write it as $ e < \pi ,\log \left( {e + x} \right) < \log \left( {\pi + x} \right) $
Therefore from this $ \left( {e + x} \right)\log \left( {e + x} \right) < \left( {e + x} \right)\log \left( {\pi + x} \right) < \left( {\pi + x} \right)\log \left( {\pi + x} \right) $ for all the value of $ x > 0 $ .
Therefore, from the above, we can say that the function $ f'\left( x \right) < 0 $ and will be for all the values of $ x > 0 $ .
So we can say that the $ f\left( x \right) $ decreases in the interval $ \left( {0, \propto } \right) $
Therefore, the option $ \left( b \right) $ will be decreasing in the interval $ {\text{[0,}} \propto ) $ .
So, the correct answer is “Option b”.

Note: When we have the question like increasing or decreasing then the derivative of a function can be used to find it for any intervals in its domain. So if $ f'\left( x \right) > 0 $ in an interval at each of the points then we can say that the function will be said to be increasing. And for any intervals in its domain if $ f'\left( x \right) < 0 $ in an interval at each of the points then we can say that the function will be said to be decreasing. And this info will help you while finding such problems.