Let f : R → R be a continuous odd function, which vanishes exactly at one point and f ( 1 ) = 2 1 . Suppose that F ( x ) = ∫ − 1 x f ( t ) d t for all x ∈ [ − 1 , 2 ] and G ( x ) = ∫ − 1 x t ∣ f ( f ( t ) ) ∣ d t for all x ∈ [ − 1 , 2 ] . Find the value of f ( 2 1 ) if x → 1 lim G ( x ) F ( x ) = 1 4 1
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
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)
Problem Loading...
Note Loading...
Set Loading...
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