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TS EAMCET 10-Sep-2020 Shift 2 Solved Paper

Section: Mathematics

Question : 75 of 160

Marks: +1, -0

\underset{n\to\infty}{\text{Lt}}\frac{\pi}{2n}\left[\sin\frac{\pi}{2n}+\sin\frac{2\pi}{2n}+\dots+\sin\frac{\pi}{2}\right]=

Solution:

We have,

\lim\limits_{n\to\infty}\frac{\pi}{2n}\left[\sin\frac{\pi}{2n}+\sin\frac{2\pi}{2n}+\dots+\sin\frac{\pi}{2}\right]

Here,

r

th term

=\sin\left(\frac{r\pi}{2n}\right)

\therefore

The given limit,

\lim\limits_{n\to\infty}\frac{\pi}{2n}\sum\limits_{r=1}^{n}\sin\left(\frac{r\pi}{2n}\right)

Put

\frac{r\pi}{2n}\to x,\;\frac{\pi}{2\pi}\to dx,\;\lim\limits_{n\to\infty}\Sigma\to\int

x_{\min}=\lim\limits_{n\to\infty}\frac{r_{\min}}{\frac{2n}{\pi}}=0

x_{\max}=\lim\limits_{n\to\infty}\frac{r_{\max}}{\frac{2n}{\pi}}=\frac{\pi}{2}=\int\limits_{0}^{\pi/2}\sin x\,dx

=\left(-\cos x\right)_{0}^{\pi/2}=-\left(\cos x\right)_{0}^{\pi 2}

=-\left[\cos\frac{\pi}{2}-\cos0\right]=-[0-1]=1

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