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Application of Integration question

2015 · Shift 2 · Q26
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  5. /2015 · Shift 2 · Q26

Application of Integration question

2015 · Shift 2 · Q26

JEE AdvancedMathematicsApplication of IntegrationNumerical+4 / −1
Let f:R→Rf:R \to Rf:R→R be a continuous odd function, which vanishes exactly at one point and f(1)=12.f\left( 1 \right) = {1 \over {2.}}f(1)=2.1​ Suppose that F(x)=∫−1xf(t)dtF\left( x \right) = \int\limits_{ - 1}^x {f\left( t \right)dt}F(x)=−1∫x​f(t)dt for all x∈  [−1,2]x \in \,\,\left[ { - 1,2} \right]x∈[−1,2] and G(x)=G(x)=G(x)=∫−1xt∣f(f(t))∣dt\int\limits_{ - 1}^x {t\left| {f\left( {f\left( t \right)} \right)} \right|} dt−1∫x​t∣f(f(t))∣dt for all x∈  [−1,2].x \in \,\,\left[ { - 1,2} \right].x∈[−1,2]. If lim⁡x→1F(x)G(x)=114,\mathop {\lim }\limits_{x \to 1} {{F\left( x \right)} \over {G\left( x \right)}} = {1 \over {14}},x→1lim​G(x)F(x)​=141​, then the value of f(12)f\left( {{1 \over 2}} \right)f(21​) is
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