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: (D)

Solution

Using Leibniz rule for differentiation under the integral sign:

g^{'} (a) = h \to 0^{+} lim \int_{h}^{1 - h} t^{- a} (1 - t)^{a - 1} [ln (\frac{1 - t}{t})] d t

a = \frac{1}{2}

g^{'} (\frac{1}{2}) = h \to 0^{+} lim \int_{h}^{1 - h} t^{- 1/2} (1 - t)^{- 1/2} ln (\frac{1 - t}{t}) d t

By the property

\int_{a}^{b} f (x) d x = \int_{a}^{b} f (a + b - x) d x

, the integrand is an odd function about

t = \frac{1}{2}

, so the integral equals zero.

Therefore

g^{'} (\frac{1}{2}) = 0

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