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JEE Advanced 2014Paper 2
Exam
JEE Advanced
Year
2014
Paper
Paper 2
Subject
Mathematics
Chapter
Definite Integration

JEE Advanced 2014 · Paper 2 · Mathematics · Definite Integration

Paragraph: Given that for each a ∈(0,1) , _ h arrow 0^ + ∫_ h ^ 1-h...

Paragraph: Given that for each a∈(0,1)a \in(0,1), lim⁡h→0+∫h1−ht−a(1−t)a−1dt\lim _{h \rightarrow 0^{+}} \int_{h}^{1-h} t^{-a}(1-t)^{a-1} d t exists. Let this limit be g(a)g(a). In addition, it is given that the function g(a)g(a) is differentiable on (0,1)(0,1).
Question: The value of g\left(\frac{1}{2}\right) is

Options

  1. Aπ\pi
  2. B2π2\pi
  3. Cπ2\dfrac{\pi}{2}
  4. Dπ4\dfrac{\pi}{4}

Correct answer

A. π\pi

Step-by-step solution

g(12)=lim⁡h→0+∫h1−ht−12(1−t)−12dtg\left(\dfrac{1}{2}\right) = \lim_{h \to {0}^{+}} \mathop{∫}_{h}^{1 - h}{t}^{- \dfrac{1}{2}}{\left(1 - t\right)}^{- \dfrac{1}{2}}dt
= {\mathop{∫}_{0}^{1}\dfrac{dt}{\sqrt{t - {t}^{2}}} = \mathop{∫}_{0}^{1}\dfrac{dt}{\sqrt{\dfrac{1}{4} - {\left(t - \dfrac{1}{2}\right)}^{2}}} = \sin^{- 1} \left\left(\dfrac{t - \dfrac{1}{2}}{\dfrac{1}{2}}\right)\right|}_{0}^{1}
=sin⁡−11−sin⁡−1(−1)=π= \sin^{- 1} 1 - \sin^{- 1} (-1) = \pi
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