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JEE Mains · Maths · STD 12 - 7.1 indefinite integral

माना \(\mathrm{f}(\mathrm{x})=\int \frac{\mathrm{dx}}{\left(3+4 \mathrm{x}^2\right) \sqrt{4-3 \mathrm{x}^2}},|\mathrm{x}|<\frac{2}{\sqrt{3}}\) है। यदि \(\mathrm{f}(0)=0\) तथा \(\mathrm{f}(1)=\frac{1}{\alpha \beta} \tan ^{-1}\left(\frac{\alpha}{\beta}\right), \alpha, \beta>0\) है, तो \(\alpha^2+\beta^2\) बसाबर है :

  1. A \(28\)
  2. B \(26\)
  3. C \(25\)
  4. D \(24\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(28\)

Step-by-step Solution

Detailed explanation

\(f(x)=\int \frac{d x}{\left(3+4 x^2\right) \sqrt{4-3 x^2}}\) \(x=\frac{1}{t}\) \(=\int \frac{\frac{-1}{t^2} d t}{\frac{\left(3 t^2+4\right)}{t^2} \frac{\sqrt{4 t^2-3}}{t}}\) \(=\int \frac{-d t \cdot t}{\left(3 t^2+4\right) \sqrt{4 t^2-3}}: \text { Put } 4 t^2-3=z^2\)…
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