Let f   :   R → R be a differentiable function such that f 0 = 0 ,   f π 2 = 3 and…

Let f : RR be a differentiable function such that f0=0, fπ2=3 and f0=1. If gx=xπ2 ftcosect-cott cosect ftdt , for x0,π2 then limx0gx=

Solution

g(x)= xπ2[f'(t)cosec tf(t)cosec t cot t]dt

=f(t)cosectxπ2

=fπ2cosec π2-fxsinx=3-fxsinx

     limx0gx=3-limx0fxsinx=limx0f'xcosxas f'0=1

    limx0gx=3-1=2

Asked in: JEE Advanced 2017 (Paper 1)

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