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Let f : R to R be a continuous odd fun...

Let ` f : R to R ` be a continuous odd function, which vanishes exactly at one point and ` f(1) = 1/2`.
Suppose that
`F (x) = int_(-1)^(x) f(t) dt" for all " x in [-1, 2] and G(x)= int_(-1)^(x) t|f{f(t)}|dt" for all " x in [-1, 2]*" If " lim_( x to 1) (F(x))/(G(x)) = 1/14," then the value of " f(1/2)` is

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The correct Answer is:
7
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`
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