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NDA 2015
Exam
NDA
Year
2015
Subject
Mathematics
Chapter
Definite Integration

NDA 2015 · Mathematics · Definite Integration

Consider the integral I_m=∫_0^π ( 2 m x)/( x) d x , where m is a...

Consider the integral Im=∫0πsin⁡2mxsin⁡xdxI_m=\int_0^\pi \frac{\sin 2 m x}{\sin x} d x, where mm is a positive integer. . Consider the following: 1. Im−Im−1I_m-I_{m-1} is equal to 0. 2. I2m>ImI_{2 m} > I_m Which of the above is/are correct?

Options

  1. A1 only
  2. B2 only
  3. CBoth 1 and 2
  4. DNeither 1 nor 2

Correct answer

A. 1 only

Step-by-step solution

I2m>\mathrm{I}_{2 \mathrm{ m}} >\mathrm{I}m_{\mathrm{m}} is wrong statement Because, Im=\mathrm{I}_{\mathrm{m}}=\mathrm{I}m−1=…..=_{\mathrm{m}-1}=\ldots..=\mathrm{I}n_{\mathrm{n}} Thus, Im−\mathrm{I}_{\mathrm{m}}-\mathrm{I}m−1=0_{\mathrm{m}-1}=0 is the only correct statement. ∴\therefore Option (A) is correct.
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