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lf f(x) is a function satisfying f1x+x2f(x)=0 for all non-zero x, then  sinθcosecθf(x)dx is equal to

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a
sin⁡θ+cosec⁡θ
b
sin2⁡θ
c
cosec2⁡θ
d
None of these

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detailed solution

Correct option is D

we have, f1x+x2f(x)=0⇒f(x)=−1x2f1xI=∫sin⁡θcosec⁡θ f(x)dx=∫sin⁡θcosec⁡θ −1x2f1xdx=∫cosec⁡θsin⁡θ f(t)dt, where t=1x⇒I=−∫sin⁡θcosec⁡θ f(t)dt=−I⇒2I=0⇒I=0


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