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Find the period of (i) f(x)=sin pi x ...

Find the period of
(i) `f(x)=sin pi x +{x//3}`, where {.} represents the fractional part.
(ii) `f(x)=|sin 7x|-"cos"^(4)(3x)/(4)+"tan"(2x)/(3)`

Text Solution

Verified by Experts

(i) `f(x)=sin pi x +{x//3}`, where {.} represents the fractional part
Period of ` sin pi x " is " (2pi)/(pi)=2`
Period of `{x//3} " is " (1)/(1//3)=3`
Therefore, period of f(x) is L.C.M. of `(2,3)=6`
(ii) `f(x)=|sin 7x|-"cos"^(4)(3x)/(4)+"tan"(2x)/(3)`
Period of `|sin 7x| " is " (pi)/(7)`
Period of `"cos"^(4)(3x)/(4) " is " (pi)/(3//4)=(4pi)/(3)`
Period of `tan(2x)/(3) " is " (pi)/(2//3)=(3pi)/(2)`
Therefore, period of f(x) is L.C.M. of
`((pi)/(7),(4pi)/(3),(3pi)/(2))=pi xx (L.C.M. of (1,4,3))/(H.C.F. of (7,3,2)`
`=12 pi`
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Knowledge Check

  • If f(x) = {x} + sin ax (where {.} denotes the fractional part function) is periodic then

    A
    a' is a rational multiple of `pi`
    B
    a' is a natural number
    C
    a' is any real number
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  • If f(x)={(sin{cosx}/(x-pi/2), x ne pi/2),(1, x=pi/2):} , where {.} denotes the fractional part of x, then f(x) is :

    A
    continuous at `x=pi/2`
    B
    `lim_(xrarrpi/2)f(x)`, but f(x) is not continuous at `x=pi/2`
    C
    `lim_(xrarrpi/2)` does not exists
    D
    `lim_(xrarr(pi^-)/2)f(x)=1`
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