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NTA Abhyas 2020
Exam
NTA Abhyas
Year
2020
Subject
Mathematics
Chapter
Indefinite Integration

NTA Abhyas 2020 · Mathematics · Indefinite Integration

Let the integral I = ∫ (2020)x + ^ -1 (2020)x 1 - (2020)2x dx= K^ 2...

Let the integral I=∫(2020)x+sin⁡−1(2020)x1−(2020)2xdx=K2(2020)sin⁡−1(2020)x+λI = ∫ \dfrac{\left(2020\right)x + \sin^{-1} \left(2020\right)x}{1 - \left(2020\right)2x}dx= K^{2}\left(2020\right)\sin^{-1} \left(2020\right)x + \lambda (where, λ\lambda is the constant of integration), then the vaue of 2020K{2020}^{K}& is

Options

  1. A20202020
  2. B20192019
  3. Ce
  4. D1e\dfrac{1}{e}

Correct answer

C. e

Step-by-step solution

Let, sin⁡−1(2020x)=t\sin^{-1} \left({2020}^{x}\right) = t
⇒2020xln⁡(2020)1−(2020)2xdx=dt⇒ \dfrac{{2020}^{x}\ln \left(2020\right)}{\sqrt{1 - {\left(2020\right)}^{2x}}}dx = dt
⇒I=∫(2020)tln⁡(2020)dt⇒ I = ∫ \dfrac{{\left(2020\right)}^{t}}{\ln \left(2020\right)}dt
=(2020)tln⁡(2020)2+λ= \dfrac{\left(2020\right)t}{\ln\left(2020\right)^{2}} + \lambda
=(2020)sin⁡−1(2020)x(ln⁡2020)2+λ= \dfrac{\left(2020\right)\sin^{-1} \left(2020\right)x}{\left(\ln 2020\right)^{2}} + \lambda
∴K=1ln⁡2020=log⁡2020e∴ K = \dfrac{1}{\ln 2020} = \log_{2020}e
∴2020K=2020log⁡2020e=e∴ {2020}^{K} = {2020}^{\log_{2020} e} = e
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