JEE-Advanced 2015 (27/40)

Calculus Level 3

Let f : R R f:\mathbb R \to \mathbb R be a continuous odd function, which vanishes exactly at one point and f ( 1 ) = 1 2 f(1)=\frac{1}{2} . Suppose that F ( x ) = 1 x f ( t ) d t F(x)=\displaystyle \int_{-1}^x f(t)dt for all x [ 1 , 2 ] x \in [-1,2] and G ( x ) = 1 x t f ( f ( t ) ) d t G(x)=\displaystyle \int_{-1}^x t|f(f(t))|dt for all x [ 1 , 2 ] x \in [-1,2] . Find the value of f ( 1 2 ) f \left(\frac{1}{2} \right) if lim x 1 F ( x ) G ( x ) = 1 14 \displaystyle \lim_{x \to 1} \frac{F(x)}{G(x)}=\frac{1}{14}


The answer is 7.

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2 solutions

Aqid Khatkhatay
Dec 15, 2016

F'(X)=f(x) G'(X)=x|f(f(x))| Now limit after LHL =f(1)/1×|f(f(1))| =(1/2)/f(1/2) therefore f(1/2)=7

Lipsa Kar
Jul 22, 2015

the limits of the integrals can be changed to x and 1 since f is an odd function and the integral with upper limit 1 and lower limit -1 becomes equal to 0. use the l'hospital rule here since both F(x) and G(x) tend to 0 when x approaches 1. then, use the value f(1)

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