Let f:ℝ→ℝ be a continuous odd function, which vanishes exactly at one point and f1=12.Suppose that Fx=∫−1xftdt for all x∈−1,2 and Gx=∫−1xtfftdtfor all x∈−1,2.If limx→1FxGx=114,then the value of f12 is

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answer is 7.

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Detailed Solution

Given limx→1FxGx=114Use LH rule⇒limx→1F'xG'x=114 ⇒limx→1fxxffx=114⇒f1ff1=114⇒12f12=114⇒f12=7