Let f : R → R be a continuous odd function, which vanishes exactly at one point and f 1 = 1 2 . 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 .

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Let f : R &#8594; R be a continuous odd function, which vanishes exactly at one point and f 1 = 1 2 . Suppose that F x = &#8747; - 1 x f ( t ) d t for all x &#8712; - 1 , 2 and G x = &#8747; - 1 x t | f f t | d t for all x &#8712; - 1 , 2 . If lim x &#8594; 1 &#8289; F ( x ) G ( x ) = 1 14 , then the value of f 1 2 is

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Correct answer: 0

f 1 = 1 2 ; f - x = - f x ; f - 1 = - 1 2 a d f ( x ) is zero only at one point F x = ∫ - 1 x f t d t = ∫ 1 x f t d t x ∈ [ - 1 , 2 ] as it is an odd function, F &prime; x = f ( x ) G x = ∫ - 1 x t f ( f t ) d t = ∫ 1 x t f ( f t ) d t , as it is an odd function, ⇒ G &prime; x = x f ( f x ) lim x → 1 ⁡ F ( x ) G ( x ) = lim x → 1 ⁡ F &prime; ( x ) G &prime; ( x ) = lim x → 1 ⁡ f ( x ) x f ( f x ) [ As the limit is in the form of 0/0] = 1 2 f 1 2 = 1 14 ⇒ f 1 2 = 7 ⇒ f 1 2 = 7 a s f 1 2 can not be negative

Let f : R → R be a continuous odd function, which vanishes exactly at one point and f 1 = 1 2 . 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 . If lim x → 1 ⁡ F ( x ) G ( x ) = 1 14 , then the value of f 1 2 is

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