Statement-1: For any n ∈ N, we have ∫limits_0^n π≤ft|(sin x)/x| d x ≥…

We have,

$\int\limits_0^{n \pi}\left|\frac{\sin x}{x}\right| d x$

$\Rightarrow I=\sum\limits_{r=1}^n \int\limits_{(r-1) \pi}^{r \pi}\left|\frac{\sin x}{x}\right| d x$

$\Rightarrow I=\sum\limits_{r=1}^n \int\limits_0^\pi\left|\frac{\sin \{(r-1) \pi+u\}}{(r-1) \pi+u}\right| d u$, where $x=(r-1) \pi+u$

$\Rightarrow I=\sum\limits_{r=1}^n \int\limits_0^\pi \frac{\sin u}{(r-1) \pi+u} d u$

$\Rightarrow I \geq \sum\limits_{r=2}^n \int\limits_0^\pi \frac{\sin u}{(r-1) \pi+\pi} d u \left[∵ \frac{\sin u}{(r-1) \pi+u}>\frac{\sin u}{(r-1) \pi+\pi}\right]$

$\Rightarrow I \geq \sum\limits_{r=2}^n \frac{2}{r \pi} \Rightarrow I \geq \frac{2}{\pi}\left(1+\frac{1}{2}+...+\frac{1}{n}\right)$

So, statement -1 is true.

Let $f(x)=\frac{\sin x}{x}$. Then,

$f'(x)=\frac{x \cos x-\sin x}{x^2}=\frac{g(x)}{x^2} \text { where } g(x)=x \cos x-\sin x$

Now,

$g'(x)=-x \text { sin } x<0$ for all $x \in(0, \pi / 2)$

$\Rightarrow g(x)$ is decreasing on $x \in(0, \pi / 2)$

$\Rightarrow g(x)<g(0)$ for all $x \in(0, \pi / 2)$

$\Rightarrow g(x)<0$ for all $x \in(0, \pi / 2)$

∴  $f'(x)<0$ for all $x \in(0, \pi / 2)$

$\Rightarrow f(x)$ is decreasing on $(0, \pi / 2)$

$\Rightarrow f(x)>f(\pi / 2)$ for all $x \in(0, \pi / 2)$

$\Rightarrow \frac{\sin x}{x}>\frac{2}{\pi}$ for all $x \in(0, \pi / 2)$

So, statement-2 is true. But, it is not a correct explanation for statement-1.