Period of f x - x -2 x -3 x - - nx - n n -1-2 x - n - N and - denotes greatest integer function isA- 1-2B- 1C- nD- Does not exist-

The correct option is B 1f(x) = [x] + [2x] + [3x] + … + [nx] – (x + 2x + 3x + … + nx) = ( { x }+{ 2x }+{ 3x } +{ 4x }+… +{ nx } ) ∴ period of f(x) = LCM ...

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Standard XII

Mathematics

Using Graph to Find Range of a Function

Period of f x...

Question

Period of $f (x) = [x] + [2 x] + [3 x] + . . . . . + [n x] - \frac{n (n + 1)}{2} x, n ϵ N$ and [∙] denotes greatest integer function is

\frac{1}{2}

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n

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Does not exist.

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Solution

The correct option is B $1$
$f (x) = [x] + [2 x] + [3 x] + \dots + [n x] - (x + 2 x + 3 x + \dots + n x)$
$= - ({x} + {2 x} + {3 x} + {4 x} + \dots + {n x})$
$∴$ period of $f (x) = L C M (1, \frac{1}{2}, \frac{1}{3}, \dots \frac{1}{n}) = 1$

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Period of f(x)=[x]+[2x]+[3x]+.....+[nx]−n(n+1)2x,nϵN and [∙] denotes greatest integer function is

The correct option is B 1f(x)=[x]+[2x]+[3x]+…+[nx]–(x+2x+3x+…+nx) =−({x}+{2x}+{3x}+{4x}+…+{nx}) ∴ period of f(x)=LCM(1,12,13,…1n)=1

Period of $f (x) = [x] + [2 x] + [3 x] + . . . . . + [n x] - \frac{n (n + 1)}{2} x, n ϵ N$ and [∙] denotes greatest integer function is

The correct option is B $1$
$f (x) = [x] + [2 x] + [3 x] + \dots + [n x] - (x + 2 x + 3 x + \dots + n x)$
$= - ({x} + {2 x} + {3 x} + {4 x} + \dots + {n x})$
$∴$ period of $f (x) = L C M (1, \frac{1}{2}, \frac{1}{3}, \dots \frac{1}{n}) = 1$