Question

Statement-1: $e^{\pi }>{\pi }^e$
Statement-2: ${\pi }^{e\pi }>e^{e\pi }$

• A. Statement-1 is true, Statement-2 is true; Statement-2 is the correct explanation of Statement-1
• B. Statement-1 is true, Statement-2 is true; Statement-2 is not the correct explanation of Statement-1
• C. Statement-1 is true, Statement-2 is false
• D. Statement-1 is false, Statement-2 is true

Answer 1

The correct option is B Statement-1 is true, Statement-2 is true; Statement-2 is not the correct explanation of Statement-1

$e^{\pi }={\left(e^{1/e}\right)}^{e\pi }={\left(x_1^{1/x_1}\right)}^{e\pi }\cdots \left(1\right){\pi }^e={\left({\pi }^{1/\pi }\right)}^{e\pi }={\left(x_2^{1/x_2}\right)}^{e\pi }\cdots \left(2\right)$
Now, let $y=x^{1/x}[x>0]$
Taking ${\log }_e$ both the sides,
${\log }_{e}y=\frac{1}{x}{\log }_{e}x$
Differentiating w.r.t $x$
$\frac{1}{y}\frac{dy}{dx}=\frac{1}{x}.\frac{1}{x}-{\log }_{e}x\times \frac{1}{x^2}$
$\Rightarrow \frac{dy}{dx}=\frac{x^{1/x}}{x^2}(1-{\log }_{e}x)$
For critical point, $\frac{dy}{dx}=0$
$\frac{x^{1/x}}{x^2}(1-{\log }_{e}x)=0\Rightarrow 1-{\log }_{e}x=0(x^{1/x},x^2\text{never be zero})$
$\Rightarrow x=e$
$\frac{dy}{dx}{|}_{x=e^{-}}=+ve,$ $\frac{dy}{dx}{|}_{x=e^{+}}=-ve$
Since, first derivative changes its sign from $+ve$ to $-ve,$ therefore, $e$ is the maximum point.
$\Rightarrow$ $e^{1/e}$ is the maximum value of $x^{1/x}.$
From eqn(1) and (2), we have
${\left(e^{1/e}\right)}^{e\pi }>{\left({\pi }^{1/\pi }\right)}^{e\pi }$
$\Rightarrow e^{\pi }>{\pi }^e$

Now, ${\pi }^{e\pi }>e^{e\pi }[\because \pi >e]$
Both Statement-1 and 2 are true.