If `f(x)=cos[pi^2]x ,`where `[x]`stands for the greatest integer function, then`f(pi/2)=-1`(b) `f(pi)=1``f(-pi)=0`(d) `f(pi/4)=1`A. `f((pi)/(2))=-1`B. `f(pi)=1`C. `f(-pi)=-1`D. `f((pi)/(4))=2`

If `f(x)=cos[pi^2]x ,`where `[x]`stands for the greatest integer funct

1.	If `f(x)=cos[pi^2]x ,`where `[x]`stands for the greatest integer function, then`f(pi/2)=-1`(b) `f(pi)=1``f(-pi)=0`(d) `f(pi/4)=1`A. `f((pi)/(2))=-1`B. `f(pi)=1`C. `f(-pi)=-1`D. `f((pi)/(4))=2`
Answer» Correct Answer - A We have , `f(x) =cos[pi^(2)]x+cos[-pi^(2)]x` `implies f(x)=cos9x + cos (-10x)` `implies f(x)=cos9x+ cos 10 x` `:. f((pi)/(2)) =cos""(9x)/(2)+ cos 5 pi =0+(-1)^(5)=-1` `f(pi) = cos 9 pi + cos 10 pi=(-1)^(9) +(-1)^(10)=-1+1=0` `f(-pi)= cos (-9pi)+ cos (-10pi) = cos 9 pi + cos 10 pi=0` ` f((pi)/(4))= cos (9pi)/(4)+ cos (10 pi)/(4)=(1)/(sqrt(2))+0=(1)/(sqrt(2))` Hence , option (a) is correct.

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If `f(x)=cos[pi^2]x ,`where `[x]`stands for the greatest integer function, then`f(pi/2)=-1`(b) `f(pi)=1``f(-pi)=0`(d) `f(pi/4)=1`A. `f((pi)/(2))=-1`B. `f(pi)=1`C. `f(-pi)=-1`D. `f((pi)/(4))=2`

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