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BITSAT 2024Paper 2
Exam
BITSAT
Year
2024
Paper
Paper 2
Subject
Mathematics
Chapter
Differentiation

BITSAT 2024 · Paper 2 · Mathematics · Differentiation

At x=(π^2)/(4), (d)/(d x) ( ^ -1 ( x )+ ^ -1 (e^x))=

At x=π24,ddxx=\frac{\pi^2}{4}, \frac{d}{d x}\left(\tan ^{-1}(\cos \sqrt{x})+\sec ^{-1}\left(e^x\right)\right)=

Options

  1. A1\mathrmeπ22−1−1π\frac{1}{\sqrt{\mathrme^{\frac{\pi^2}{2}}}-1}-\frac{1}{\pi}
  2. Bπ4+1\mathrmeπ2+\mathrmeπ2/2\frac{\pi}{4}+\frac{1}{\sqrt{\mathrme^{\pi^2}+\mathrme^{\pi^2 / 2}}}
  3. C1eπ2+eπ2/2+2πcot⁡(π2)\frac{1}{\sqrt{\mathrm{e}^{\pi^2}+\mathrm{e}^{\pi^2 / 2}}}+\frac{2}{\pi} \cot \left(\frac{\sqrt{\pi}}{2}\right)
  4. D1eπ+1π\frac{1}{\sqrt{\mathrm{e}^\pi}}+\frac{1}{\pi}

Correct answer

A. 1\mathrmeπ22−1−1π\frac{1}{\sqrt{\mathrme^{\frac{\pi^2}{2}}}-1}-\frac{1}{\pi}

Step-by-step solution

ddx(tan⁡−1(cos⁡x)+sec⁡−1(ex))=(−sin⁡x)1+cos⁡2x⋅(12x)+1exe2x−1⋅ex\frac{d}{d x}\left(\tan ^{-1}(\cos \sqrt{x})+\sec ^{-1}\left(\mathrm{e}^x\right)\right)=\frac{(-\sin \sqrt{x})}{1+\cos ^2 \sqrt{x}} \cdot\left(\frac{1}{2 \sqrt{x}}\right)+\frac{1}{e^x \sqrt{e^{2 x}-1}} \cdot e^x but when x=π24=−−sin⁡π21+cos⁡2π2(12)(2π)+1eπ2/2−1=−1π+1eπ2/2−1x=\frac{\pi^2}{4}=-\frac{-\sin \frac{\pi}{2}}{1+\cos ^2 \frac{\pi}{2}}\left(\frac{1}{2}\right)\left(\frac{2}{\pi}\right)+\frac{1}{\sqrt{e^{\pi^2 / 2}-1}}=-\frac{1}{\pi}+\frac{1}{\sqrt{e^{\pi^2 / 2}-1}}
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