Let f(x)=9x9x+3, then

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f(x)+f(1−x)=1

f(x)+f(1−x)=−1

f11996+f21996+f31996+…+f19951996=998

f11996+f21996+f31996+…+f19951996=99712

answer is A.

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Detailed Solution

f(x)=9x9x+3----1and f(1−x)=91−x91−x+3 ⇒f(1−x)=99x9yx+3=99+3.9x f(1−x)=93(3+9x)----2Adding (1) and (2), we getf(x)+f(1-x)=9x9x+3+93(3+9x) =3⋅9x+93(9x+3)=3(9x+3)3(9x+3)∴ f(x)+f(1−x)=1Now, putting x=11996,21996,31996,…9981996, , in (3), we get f 11996+f19951996=1f11996+f21996+…+f19951996=(1+1+1+…+997)+12=997+12=99712… … … … … … … … ⇒f(9971996)+f(9991996)=1 ⇒f(9981996)+f(9981996)=1orf(9981996)=12Adding all the above expressions, we get f(11996)+f(21996)+…+f(19951996) =(1+1+1+…+997)+12 =997+12=99712

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