Practicing Success
Statement-1: For any $n \in N$, we have $\int\limits_0^{n \pi}\left|\frac{\sin x}{x}\right| d x \geq \frac{2}{\pi}\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{n}\right)$ Statement-2: $\frac{\sin x}{x} \geq \frac{2}{\pi}$ on $\left(0, \frac{\pi}{2}\right)$. |
Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1. Statement-1 is True, Statement-2 is True; Statement-2 is not a correct explanation for Statement-1. Statement-1 is True, Statement-2 is False. Statement-1 is False, Statement-2 is True. |
Statement-1 is True, Statement-2 is True; Statement-2 is not a correct explanation for Statement-1. |
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. |