in Calculus edited by
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27 votes
27 votes

If $f(x) = R \: \sin ( \frac{\pi x}{2}) + S, f’\left(\frac{1}{2}\right) = \sqrt{2}$ and $\int_0^1 f(x) dx = \frac{2R}{\pi}$, then the constants $R$ and $S$ are

  1. $\frac{2}{\pi}$ and $\frac{16}{\pi}$
  2. $\frac{2}{\pi}$ and 0
  3. $\frac{4}{\pi}$ and 0
  4. $\frac{4}{\pi}$ and $\frac{16}{\pi}$
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4 Comments

@97apoorva singh

what is wrong in the question ? 

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@ankitgupta.1729 see the question in 2019 pdf:-

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@97apoorva singh

sorry, I thought you were talking about given question.. I did not know you were talking about pdf.. 

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2 Answers

36 votes
36 votes
Best answer
Correct Option: $C$.

$f(x) = R*\sin(\frac{\pi * x}{2}) + S$

$f'(x) = R*\cos(\frac{\pi*x}{2})*\frac{\pi}{2}$

$f'(\frac{1}{2}) = R*\cos(\frac{\pi}{4})*\frac{\pi}{2} = \sqrt{2}$

$R = \frac{\sqrt{2}*\sqrt{2}*2}{\pi} = \frac{4}{\pi}$

$f(x) = \frac{4}{\pi}*\sin(\frac{\pi*x}{2}) + S$

$\int_{0}^{1} f(x)*dx = \int_{0}^{1}(\frac{4}{\pi}*\sin(\frac{\pi*x}{2}) + S)*dx = \frac{2*R}{\pi} = \frac{8}{\pi^2}$

$\frac{4}{\pi}\int_{0}^{1}\sin(\frac{\pi*x}{2})*dx + \int_{0}^{1}S*dx = \frac{8}{\pi^2}$

$\frac{4}{\pi} \left[-\cos(\frac{\pi*x}{2})*\frac{2}{\pi} \right ]_0^1 + S \left[x \right ]_0^1 = \frac{8}{\pi^2}$

$\frac{8}{\pi^2} \left[-0+1 \right ] + S = \frac{8}{\pi^2}$

$S = 0.$
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1 comment

edited by

here you have calculated " R" first then have put its value in integration to get "S" it's right but you can also calculate "S" first without putting value and after can get " R" both approaches right.

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12
5 votes
5 votes

Option c is right.

2 Comments

why -cos(pie*x/2) / pie/2... why this pie/2 used here..can you help
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https://www.youtube.com/watch?v=SRp51twD9VM 

The video is of changing of limits; nonetheless, it will answer your query. 

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Answer:

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