JEE Advanced · 2015 · 24 May 2015 · Paper 2 · Definite Integrals

Let

f : R \to R

be a continuous odd function, which vanishes exactly at one point and

f (1) = \frac{1}{2}

. Suppose that

F (x) = \int_{- 1}^{x} f (t) d t

for all

x \in [- 1, 2]

and

G (x) = \int_{- 1}^{x} t ∣ f (f (t)) ∣ d t

for all

x \in [- 1, 2]

. If

lim_{x \to 1} \frac{F ( x )}{G ( x )} = \frac{1}{14}

, then the value of

f (\frac{1}{2})

Answer: 7

Solution

Both

F

and

G

vanish at

x = 1

F (1) = \int_{- 1}^{1} f = 0

because

f

is odd, and the integrand

t ∣ f (f (t)) ∣

G

is odd (

f \circ f

is odd, its modulus even, times

t

), so

G (1) = 0

. By L'Hôpital,

lim_{x \to 1} \frac{F ( x )}{G ( x )} = \frac{f ( 1 )}{1 \cdot ∣ f ( f ( 1 )) ∣} = \frac{1/2}{∣ f ( 1/2 ) ∣} = \frac{1}{14}

. Hence

∣ f (1/2) ∣ = 7

, so

f (1/2) = 7

JEE Advanced · 2015 · 24 May 2015 · Paper 2 · Integer answer