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

Let f :R→R  be a continuous odd function, which vanishes exactly at one point and f1=12. Suppose that Fx= ∫-1xf(t)dt  for all x∈-1, 2 and Gx= ∫-1xt|fft|dt for all x∈-1, 2. If limx→1⁡F(x)G(x)=114, then the value of f12 is

Solution

f1=12;f-x= -fx;f-1= -12 ad f(x) is zero only at one point
Fx= ∫-1xftdt =  ∫1xftdtx∈[-1, 2] as it is an odd function,
F′x=f(x)
Gx= ∫-1xtf(ft) dt = ∫1xtf(ft) dt , as it is an odd function,

⇒  G′x=x f(fx)
limx→1⁡F(x)G(x)= limx→1⁡F′(x)G′(x)= limx→1⁡f(x)xf(fx) [ As the limit is in the form of 0/0]
= 12f12=114
⇒  f12=7
⇒   f12=7  as f12 can not be negative

Asked in: JEE Advanced 2015 (Paper 2)

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