Solved example on fractional part function
Q 1: If {x} and [x] represent fractional and integral part of x, then find the value of
?
Sol:
(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]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_sWJ1_vgDulWLGHXELOP2zDK8SCRxnSROeKSx9Ozawc7Z3sxfTFRuhzL0yZ4zShGlpkK9bRJjY5niuAg2utwY6r2_lG1lrniDdSUkL79Ws5zgGadAAYUuop0cJipItnc1wndBPYz5hc5crofVIAjlLv5hqn5vmVIkNyONLxi1n1rzgfP4ld1yqaWXSgKUxVXNKAZUXiMmh50z96mtkCCw6CzKn1MJNJL5o-ZdDMilYOC3LGDQVq_zlWpzkNUEpfIU04C22PDc7pzFjojK7eEKYeOcjFygD28gZdxuWPQ0TS9QGEsEPTQE8adV5wtIjVlIHXnDHCTJ-xkr9OclsvbveiICf1lWNfMdzSJsFXDIweXTDMGBmqQhSC7mQvrrxHPQdSdmANctNmCChazRwbyqrcx0HMbw8_5UIH8PhvB6XW66OQM6AYUG51MWWR5u_bPVoEgUG7-FvhGfdhgcfzbmAxl-2y2qUh0g19mBEpEDw=s0-d)
![= [x] + \frac{{2000\{ x\} }}{{2000}}](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_uG6gD1pVc9DR1QLebeMrUq_Y2i32o8OznzdASkUUK2SHlRUeamP8IvhRSfnvKNrD_flVUhsacY16skZ4-AhbjjM5HH_wSSdpBSpjvwlmCD0qrbXjtmR76W8RV-ckDILnBAHALl1ftkdPKbEhIhVSXFEC-JfQoLR3-ZRBJDFN4EEw-xhmdMaBDT4bIFUaiTIpeYRpAuRx61ahgtfu60-_dzIGtWohmRVELoMRdkjSbplMf89FqiXlEdV6LYOr8rQA=s0-d)
= [x] + {x} = x
( since, x = integral part + fractional part = [x] + {x} )
Q 2. Find the domain of![\frac{{[x]}}{{\{ x\} }}](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_sdSIyJJB0ol5H5ihtz5XeZjfLiG4ADeBfXr0BHTXM6QwAoXv9MI6bADbgAYgGqD7etJ5iQ5aSRF9FGACY47XJWTaZ3DBGA2Vdsj-viFZHIb9Svt8y03Iq9SqaAEJoKOBCP3ubJ9l3yoVnNjfHaG0r1cGl7KJaAR9qZusXY6l4l4nmecotVE_IqoskH5dO2X4IMUTLuJVSLRwMtFxhRV8kvrqjdN6dxOeLhqEAyYw=s0-d)
Sol:
since,
therefore,
we know that fractional part of any number is zero when the number would be an integer.
=>
so, Domain = R - I
Q 3: Find the domain of
Sol: Know about domain [click here]
Since,
for f(x) has to be defined:
1.
i.e.
or,
or,![[x] \ne \{ x\}](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_uRpVXw-XxJSuK9JjGhUzCNzWws-l_IGaXXBbnTQVyTqQ-WKZ5KR_cBtTWihQ01h2-OWL8JMdUy-XLkUE5DY109XwruR24gXnJpVefBZ2qmQLHygKbxLR31sOEycK16IYu22j4B6vXxlte0JFOMA5qfOVCltldKys_h94rjFpX_BqSuhfdkn6yAIpUqinaRB3WWPDMm=s0-d)
Integral part is equal to fractional part only when x = 0
therefore,
...................................... (1)
2. We know that root is only defined for positive values and zero
therefore,
or,![\frac{{x - 1}}{{[x] - \{ x\} }} \ge 0](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_up8k54iEDby7PSlTDF2i-Y3bhGNRxLNeoa2uvSYpJ8ZbJI_y7TdK0xiItkdguz_xNMO39CVWddCfbsZe2nmc17CDhgJjZpoX174DCeztgUdsxrgDTbIzbCJfNXp6TUaPS8KKN1NNSfMWjZwYZcyLk-zBkzm2mZRcYhj0ClRe8vLKFPXqEwkKLNS9wLCQLMSHC5Zfn9B9bmi1QP4GnI5_MRA5a5F5ILN4xD4Iwh8J0SA1R7uqeJ483Jo6iOz_A=s0-d)
Consider, [x] - {x},
See the below graph,
Observation from graph,
NOTE THIS,
if
,
![[x] > \{ x\}](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_ua6Jw_ZAfn6U3_kg4AqLL5ESJB6fJs0PIUiI5LmOnOXZmRRfIS8vBO3I1Svg026eWJS0UaG3Iz_zG34TJk4VnufiyAdAkkuNwSxDwWeQ_rQJU3mJtKmQvCTI9hULhOT0euO6vCdB-KL38LPPYYzVQmLC9FqDONMTDQwZSbiabiBMhkq8rwukpzI2Phi0iFxYNw=s0-d)
and if
,
Now,![\frac{{x - 1}}{{[x] - \{ x\} }} \ge 0](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_up8k54iEDby7PSlTDF2i-Y3bhGNRxLNeoa2uvSYpJ8ZbJI_y7TdK0xiItkdguz_xNMO39CVWddCfbsZe2nmc17CDhgJjZpoX174DCeztgUdsxrgDTbIzbCJfNXp6TUaPS8KKN1NNSfMWjZwYZcyLk-zBkzm2mZRcYhj0ClRe8vLKFPXqEwkKLNS9wLCQLMSHC5Zfn9B9bmi1QP4GnI5_MRA5a5F5ILN4xD4Iwh8J0SA1R7uqeJ483Jo6iOz_A=s0-d)
a. when
therefore, denominator is positive, ( because
)
so numerator must be positive or zero,
therefore,
...................................... (2)
b. when x < 1
denominator is negative or zero (denominator can't be zero)
so, numerator must be negative or zero
therefore,
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,


But from (1) x can not be zero because denominator can't be zero
Therefore,
Domain =
or R - {0}
Sol:
We know that {x + I} = {x},
[for properties of fractional x {click here}]
Therefore,
= [x] + {x} = x
( since, x = integral part + fractional part = [x] + {x} )
Q 2. Find the domain of
Sol:
since,
therefore,
we know that fractional part of any number is zero when the number would be an integer.
=>
so, Domain = R - I
Q 3: Find the domain of
Sol: Know about domain [click here]
Since,
for f(x) has to be defined:
1.
i.e.
or,
or,
Integral part is equal to fractional part only when x = 0
therefore,
2. We know that root is only defined for positive values and zero
therefore,
or,
Consider, [x] - {x},
See the below graph,
Observation from graph,
NOTE THIS,
if
and if
Now,
a. when
therefore, denominator is positive, ( because
so numerator must be positive or zero,
therefore,
b. when x < 1
denominator is negative or zero (denominator can't be zero)
so, numerator must be negative or zero
therefore,
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,
But from (1) x can not be zero because denominator can't be zero
Therefore,
Domain =

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