Solved example on fractional part function

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Q 1: If {x} and [x] represent fractional and integral part of x, then find the value of [x] + \sum\limits_{r = 1}^{2000} {\frac{{\{ x + r\} }}{{2000}}}  ?

 Sol: 

 [x] + \sum\limits_{r = 1}^{2000} {\frac{{\{ x + r\} }}{{2000}}} = [x] + \frac{1}{{2000}}\left[ {\{ x + 1\} + \{ x + 2\} ........\{ x + 2000\} } \right] (expand the summation)

 We know that {x + I} = {x},   
[for properties of fractional x {click here}]

 Therefore,

 [x] + \sum\limits_{r = 1}^{2000} {\frac{{\{ x + r\} }}{{2000}}} = [x] + \frac{1}{{2000}}\left[ {\{ x\} + \{ x\} ........2000times} \right]

                                = [x] + \frac{{2000\{ x\} }}{{2000}}
                              
                               = [x] + {x} = x

  ( since, x = integral part + fractional part = [x] + {x} )

Q 2. Find the domain of \frac{{[x]}}{{\{ x\} }}

 Sol: 
since, {\rm{denominator}} \ne {\rm{0}}

 therefore, {\{ x\} \ne 0}

 we know that fractional part of any number is zero when the number would be an integer.

 => x \notin I

 so, Domain = R - I



Q 3: Find the domain of f(x) = \sqrt {\frac{{x - 1}}{{x - 2\{ x\} }}}

Sol: Know about domain [click here]

 Since, f(x) = \sqrt {\frac{{x - 1}}{{x - 2\{ x\} }}}

for f(x) has to be defined:

 1. {\rm{denominator}} \ne {\rm{0}}

 i.e. {x - 2\{ x\} \ne 0}

 or, x - \{ x\} \ne \{ x\}

or, [x] \ne \{ x\}

Integral part is equal to fractional part only when x = 0 

 therefore, x \ne 0 ...................................... (1)

 2. We know that root is only defined for positive values and zero

 therefore, \frac{{x - 1}}{{x - 2\{ x\} }} \ge 0

 or, \frac{{x - 1}}{{[x] - \{ x\} }} \ge 0

 Consider, [x] - {x},

 See the below graph,

 Observation from graph, 
NOTE THIS, 
if x \ge 1,

 [x] > \{ x\}

and if x < 1,

 \{ x\} \ge [x] 


 Now,\frac{{x - 1}}{{[x] - \{ x\} }} \ge 0

 a. when x \ge 1     

therefore, denominator is positive, ( because [x] > \{ x\} )



so numerator must be positive or zero,

 therefore,  x - 1 \ge 0

x \ge 1 ...................................... (2)

b. when x < 1

 denominator is negative or zero (denominator can't be zero)
so, numerator must be negative or zero

 therefore, x - 1 \le 0 \Rightarrow x \le 1

 but we take x < 1

 take the intersection of both, we get

 x < 1

 combined (2) and (3) i.e. for whole real number line the inequality exists for,

 x \in ( - \infty ,1) \cup [1,\infty )

 \Rightarrow x \in R

 But from (1) x can not be zero because denominator can't be zero

 Therefore, 

 Domain = x \in ( - \infty ,0)\bigcup {(0,\infty )}  or R - {0}





Comments

  1. Unknown11 June 2013 at 00:03

    what is the common factor of the - --
    A to the power x plus B to the power y = C to the power z

    ReplyDelete
    Replies
    1. abhijatindia11 June 2013 at 00:35

      A^x + B^y = C^z

      let common prime factor of A, B and C is C

      so, A = Ca , B = Cb and C = C

      further let x = y = p and z = p + 1

      so, ( Ca )^p + ( Cb)^p = ( C )^(p+1)

      or, C^p.a^p + C^p.b^p = C^p. C

      or, a^p + b^p = C

      so C is common co-prime factor which is equal to a^p + b^p where A = Ca , B = Cb

      Delete

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