Let `f: RvecR` be a continuous odd function, which vanishes exactly at one point and `f(1)=1/2dot` Suppose that `F(x)=int_(-1)^xf(t)dtfora l lx in [-1,2]a n dG(x)=int_(-1)^x t|f(f(t))|dtfora l lx in [-1,2]dotIf(lim)_(xvec1)(F(x))/(G(x))=1/(14),` Then the value of `f(1/2)` is

Answer

7

Explanation

`f(x)` is continuous odd function and vanishes exactly at one point.
`:.f(0)=0`
`F(x)=int_(-1)^(x)f(t)dt=int_(-1)^(1)f(t)dt+int_(1)^(x)f(t)dt`
`=0+int_(1)^(x)f(t)dt` (as `f(t)` is odd function)
`f(t)` is odd function
`:.f(f(t))` is also odd function
`:.|f(f(t))|` is an even function
`:. t|f(f(t))|` is an odd function
`:.G(x)=int_(1)^(x)t+f(f(t))+dt=int_(1)^(x)t|f(f(t))|dt`
Now `lim_(xto1)(f(x))/(x|f(f(x))|)` (Using L'Hospital's Rule)
`=(1/2)/(1|f(1/2)|)=1/14` (Given)
`impliesf(1/2)=7`