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JEE Mains · Maths · STD 12 - 7.2 definite integral

ધારો કે  \(f(x)=\int_0^x g(t) \log _e\left(\frac{1-\mathrm{t}}{1+\mathrm{t}}\right) \mathrm{dt}\), જ્યાં \(g\) સતત વિષમ વિધેય છે. જો \(\int_{-\pi / 2}^{\pi / 2}\left(f(x)+\frac{x^2 \cos x}{1+\mathrm{e}^x}\right) \mathrm{d} x=\left(\frac{\pi}{\alpha}\right)^2-\alpha\) હોય, તો \(\alpha=\) ...........

  1. A \(0\)
  2. B \(1\)
  3. C \(2\)
  4. D \(3\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(2\)

Step-by-step Solution

Detailed explanation

\(f(x)=\int_0^x g(t) \ln \left(\frac{1-t}{1+t}\right) d t\) \(f(-x)=\int_0^{-x} g(t) \ln \left(\frac{1-t}{1+t}\right) d t\) \(f(-x)=-\int_0^x g(-y) \ln \left(\frac{1+y}{1-y}\right) d y\) \(=-\int_0^x g(y) \ln \left(\frac{1-y}{1+y}\right) d y\) (g is odd)…
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