NIMCET 2012 — Mathematics PYQ

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Question

NIMCET 2012 — Mathematics PYQ

NIMCET | Mathematics | 2012

The value of $\lim_{n \to \infty} \frac{\pi}{n} \left[ \sin \frac{\pi}{n} + \sin \frac{2\pi}{n} + \dots + \sin \frac{(n-1)\pi}{n} \right]$ is:

Choose the correct answer:

Explanation

1. Express the expression in summation form:

The given expression is:

L = \lim_{n \to \infty} \frac{\pi}{n} \sum_{r=1}^{n-1} \sin\left(\frac{r\pi}{n}\right)

2. Identify the components for integration:

Let $\frac{r}{n} = x$
Then $\frac{1}{n} = dx$
The lower limit ( $r=1$ ): $\lim_{n \to \infty} \frac{1}{n} = 0$
The upper limit ( $r=n-1$ ): $\lim_{n \to \infty} \frac{n-1}{n} = 1$

3. Convert to a definite integral:

Substituting these into the limit form:

L = \pi \int_{0}^{1} \sin(\pi x) \, dx

4. Evaluate the integral:

The integral of $\sin(\pi x)$ is $-\frac{\cos(\pi x)}{\pi}$ :

L = \pi \left[ \frac{-\cos(\pi x)}{\pi} \right]_{0}^{1}

The $\pi$ outside and inside the bracket cancel out:

L = \left[ -\cos(\pi x) \right]_{0}^{1}

L = -\cos(\pi \cdot 1) - (-\cos(\pi \cdot 0))

L = -\cos(\pi) + \cos(0)

Since $\cos(\pi) = -1$ and $\cos(0) = 1$ :

L = -(-1) + 1

L = 1 + 1

L = 2

Conclusion:

The value of the limit is 2.

Correct Option: (c)