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JEE Mains · Maths · STD 12 - 7.2 definite integral

ધારો કે ત્રિકોણમિતિય પ્રતિવિધેયોની ફક્ત મુખ્ય કિંમતોનો ઉપયોગ કરતાં \( \lim _{n \rightarrow \infty}\left(\frac{n}{\sqrt{n^4+1}}-\frac{2 n}{\left(n^2+1\right) \sqrt{n^4+1}}+\frac{n}{\sqrt{n^4+16}}-\frac{8 n}{\left(n^2+4\right) \sqrt{n^4+16}}\right. \)  \( \left.+\ldots+\frac{n}{\sqrt{n^4+n^4}}-\frac{2 n \cdot n^2}{\left(n^2+n^2\right) \sqrt{n^4+n^4}}\right)=\frac{\pi}{k}\) છે. તો \(k^2=\) ...........

  1. A \(35\)
  2. B \(36\)
  3. C \(37\)
  4. D \(32\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(32\)

Step-by-step Solution

Detailed explanation

\( \sum_{\mathrm{r}=1}^{\infty} \frac{\mathrm{n}}{\sqrt{\mathrm{n}^4+\mathrm{r}^4}}-\frac{2 \mathrm{nr}^2}{\left(\mathrm{n}^2+\mathrm{r}^2\right) \sqrt{\mathrm{n}^4+\mathrm{r}^4}} \)…
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