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JEE Main 202331 JanuaryMorning Shift
Exam
JEE Main
Year
2023
Date
31 January
Shift
Morning Shift
Subject
Mathematics
Chapter
Definite Integration

JEE Main 2023 · 31 January · Morning Shift · Mathematics · Definite Integration

Let α ∈ (0, 1) and β = _ e (1 - α) . Let P _ n (x) = x + x ^ 2 2 +...

Let α∈(0,1)\alpha \in \left(0, 1\right) and \beta = \log_{e}\left(1 - \alpha\right).Let. Let{P}_{n}\left(x\right) = x + \dfrac{{x}^{2}}{2} + \dfrac{{x}^{3}}{3} + \ldots . + \dfrac{{x}^{n}}{n}, x \in \left(0, 1\right).Thentheintegral. Then the integral{∫}_{0}^{\alpha}\dfrac{{t}^{50}}{1 - t}dt is equal to

Options

  1. Aβ−P50(α)\beta - {P}_{50}\left(\alpha\right)
  2. B−(β+P50(α))- \left(\beta + {P}_{50}\left(\alpha\right)\right)
  3. CP50(α)−β{P}_{50}\left(\alpha\right) - \beta
  4. Dβ+P50(α)\beta + {P}_{50}\left(\alpha\right)

Correct answer

B. −(β+P50(α))- \left(\beta + {P}_{50}\left(\alpha\right)\right)

Step-by-step solution

Let

I=∫0αt501−tdtI = {∫}_{0}^{\alpha}\dfrac{{t}^{50}}{1 - t}dt

⇒I=−∫0α(1−t50−11−t)dt⇒ I = - {∫}_{0}^{\alpha}\left(\dfrac{1 - {t}^{50} - 1}{1 - t}\right)dt

⇒I=−∫0α(1−t501−t)dt+∫0α(11−t)dt⇒ I = - {∫}_{0}^{\alpha}\left(\dfrac{1 - {t}^{50}}{1 - t}\right)dt + {∫}_{0}^{\alpha}\left(\dfrac{1}{1 - t}\right)dt

⇒I=−∫0α(1+t+t2+...+t49)dt−[ln⁡(1−t)]0α⇒ I = - {∫}_{0}^{\alpha}\left(1 + t + {t}^{2} + . . . + {t}^{49}\right)dt - {\left[\ln \left(1 - t\right)\right]}_{0}^{\alpha}

⇒I=−[t+t22+t33+...+t5050]0α−ln⁡(1−α)⇒ I = - {\left[t + \dfrac{{t}^{2}}{2} + \dfrac{{t}^{3}}{3} + . . . + \dfrac{{t}^{50}}{50}\right]}_{0}^{\alpha} - \ln \left(1 - \alpha\right)

⇒I=−[α+α22+α33+...+α5050]−ln⁡(1−α)⇒ I = - \left[\alpha + \dfrac{{\alpha}^{2}}{2} + \dfrac{{\alpha}^{3}}{3} + . . . + \dfrac{{\alpha}^{50}}{50}\right] - \ln \left(1 - \alpha\right)

⇒I=−(β+P50(α))⇒ I = - \left(\beta + {P}_{50}\left(\alpha\right)\right)

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