InterviewSolution
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InterviewSolution
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Let F(x) be an indefinite integral of `sin^(2)x` Statement I The function F(x) satisfies `F(x+pi)=F(x)` for all real x. Because Statement II `sin^(2)(x+pi)=sin^(2)x,` for all real x.A. Statement - 1 True , Statement -2 is True , Statement -2 is a correct explanation for Statement -1.B. Statement - 1 is True , Statement -2 is True , Statement -2 is a correct explanation for Statement -1.C. Statement - 1 True ,Statement - 2 is False.D. Statement - 1 is False , Statement - 2 is True. |
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Answer» It is given that `F(x)=intsin^(2)x dx=(1)/(2)int(1-cos2x)dx=(1)/(2)(x-(sin2x)/(2))+C` `thereforeF(pi+x)-F(x)=(1)/(2)[{(pi+x)-(1)/(2)sin2(pi+x)}-{x-(1)/(2)sin2x}]` `=(1)/(2)(pi-(1)/(2)sin2x+(1)/(2)sin2x)=(pi)/(2)` for all x thus,F `(x+pi)` = F (x) for all x is not true. So , statement -2 is not true. Clearly , statement -2 is true as `sin^(2)` a is periodic with period `pi`. |
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