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JEE MAIN Mathematics QUESTION #967
Question 1
If \(\displaystyle\int e^x\!\left(\dfrac{x\sin^{-1}x}{\sqrt{1-x^2}}+\dfrac{\sin^{-1}x}{(1-x^2)^{3/2}}+\dfrac{x}{1-x^2}\right)dx = g(x)+C\), where \(C\) is the constant of integration, then \(g\!\left(\dfrac{1}{2}\right)\) equals:
  • \(\dfrac{\pi}{4}\sqrt{e^3}\)
  • \(\dfrac{\pi}{6}\sqrt{e^3}\)✔️
  • \(\dfrac{\pi}{4}\sqrt{e^2}\)
  • \(\dfrac{\pi}{6}\sqrt{e^2}\)
Correct Answer Explanation
The integrand can be written as \(\frac{d}{dx}\!\left[e^x \cdot \frac{x\sin^{-1}x}{\sqrt{1-x^2}}\right]\). So \(g(x)=e^x\cdot\frac{x\sin^{-1}x}{\sqrt{1-x^2}}\). At \(x=\frac{1}{2}\): \(g\!\left(\frac{1}{2}\right)=e^{1/2}\cdot\frac{\frac{1}{2}\cdot\frac{\pi}{6}}{\frac{\sqrt{3}}{2}}=\frac{\pi}{6}\sqrt{e^3}\) after simplification.