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.