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

Here, `underset( x to 1) lim (F(x))/(G(x)) = 1/14 `
` rArr" " underset( x to 1) lim (F'(x))/(G'(x)) = 1/14 ` [using L'Hospital's rule] …(i)
As ` F(x) = int_(-1)^(x) f(t) dt rArr F'(x) = f(x) ` …(ii)
and ` G(x) = int_(-1)^(x) t|f{f(t)}| dt `
` rArr" " G'(x) = x |f{f(x)}|` ....(iii)
` :." " underset( x to 1) lim (F(x))/(G(x)) = underset( x to 1) lim (F'(x))/(G'(x)) = underset( x to 1) lim (f(x))/(x|f{f(x)}|) = (f(1))/(1|f{f(1)}|)= (1//2)/(|f(1//2)|) ` ...(iv)
Given, ` underset( x to 1) lim (F(x))/(G(x)) = 1/14 `
` :." " (1/2)/(|f(1/2)|) = 1/14 rArr |f(1/2)|= 7`