JEE Advanced · 2014 · 25 May 2014 · Paper 2 · Definite Integrals

Given that for each

a \in (0, 1)

h \to 0^{+} lim \int_{h}^{1 - h} t^{- a} (1 - t)^{a - 1} d t

exists. Let this limit be

g (a)

. In addition, it is given that the function

g (a)

is differentiable on (0,1).

The value of

g (\frac{1}{2})

Answer: (A)

Solution

For

a = \frac{1}{2}

g (\frac{1}{2}) = h \to 0^{+} lim \int_{h}^{1 - h} t^{- 1/2} (1 - t)^{- 1/2} d t = h \to 0^{+} lim \int_{h}^{1 - h} \frac{1}{t ( 1 - t )} d t

This can be rewritten as:

h \to 0^{+} lim \int_{h}^{1 - h} \frac{d t}{\frac{1}{4} - ( t - \frac{1}{2} ) ^{2}}

Using the substitution

u = 2 t - 1

= h \to 0^{+} lim [sin^{- 1} (2 t - 1)]_{h}^{1 - h}

= h \to 0^{+} lim [sin^{- 1} (1 - 2 h) - sin^{- 1} (2 h - 1)]

= sin^{- 1} (1) - sin^{- 1} (- 1) = \frac{π}{2} - (- \frac{π}{2}) = π

JEE Advanced · 2014 · 25 May 2014 · Paper 2 · Single correct option