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_tErLz_4kgxjsFoBR08i1rK9kp8GUuR4K63y2FXa2Mv17w8VYmGCi7YnBjY5Lol5fsIUIYGLylVhzBylka8xwRFXUqRpeO5Qj5o6u4d2A3tvyxAu1e0Yr629glgbx6rtRxC1Fpo4uWC16dGIlnn83LVIurN02Yyqzel9iyYyo4QKv7CcJyjv_bASQIBVsruNOtC3cF4L91V7Q98KmuBZe7s1VtgPHALotJKm1CwKuKlhs4tVZenq-vzWfQNi0UtxRoRX-E4D6D_vNTcqKTTv0RlS4lrQ17B_rdkI6IBro2Y321EjLcKdPZHO4jK4cLJ46x98fBXVtGZgb07o_nMNiQ_yiWlP-j8bLqBTKkeQOEj4WiK25N1NLm68fU_mm-gc6C5OEIYbtpO6OBEGKL4HWGKMVrmK5EvK_JF50bnsHVv7nqV1gp6TaoWChIZa8-Y2gfRNJ1AVyijUcUgwqNzPaGyQm7rpSJ-2YxsiPrB1SM=s0-d)
![= [x] + \frac{{2000\{ x\} }}{{2000}}](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_uPhc3YQCS-LEh96lL6QWWKwWMf8_jjHTlTe10doGj5OYmwiL6lmrJrDXbL9Vtb6Aosr_dKZJJ9juTXeWUv5I4a3oka29J4ivEwH-7V6nGZbSSUFyc37LjEFIaROsDmfN0fnxel_uPPrHUdcKlRTBw12p2Q4n6UxZRnm_JabpuXXf0gmzF_ESY7-97S7ozS8Pi-NOEjIBcBh1tGJRy400Jt2W1nWKErq_87-i5IoZy_nfLwyKf5-Xwo30J3AdRyBA=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_sCsXEIz_GxfOhOsWsPAPXTCSugpMuQQGJYhHgNOSr_seohYRZHtYvLYsp1MNAdGUDTYpVljfZIjOLjLn-RgR0WkbKPsI0FS5IdN73MIdZlws6K0IfmLprG9Bsn-tyCKUYsf0U5M7tYlNUswkisCBHr5FWDb9aW35lmGGc5R067SfuGwLPapMVNAPjMK5TplX2HzXh77s4qd5qW7GgpsmL2tByv_NSicfByTJ3wkQ=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_tncxCYlBovMxDgRkdSWPY4-eSLSPfsGlIRA-tR0yZGLm93DN6ggsB_0q1v0DlrRzu-KR9fm6CXnk7qcmRxEMKZuYi-Wcbi_HM4QWCWKR48Z3OQDkBn6Zyo0-OYWYcHsU4cdqkbut8o-S1J5HtkUPIPmBm2H4pnz75LlRoXgkKFjz9H3yg5w1fHKaQvTzjf8EBC1Igd=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_t6BIPKbIpg6ynXmhpasZ8t3G045D75CQ51QZNuRnBQuaTq9VCaaPpzZmLF_ppdINQ7XYYIEJ6SvtOOZnjI3_OQ0EgCPlbZ480nIznT75-tL9ugzhAXR_LBza_m4w7QRyWHMxiEQ5KzK9qyta-TZKmMRq5vK_0pAowT0miZI8wRwJnPgkhn-2wRnbilv5Wng1i44NAwXj34hZCnOBw7uum___5VYk9eBh0tJr52xlbd1H6iH6gmn-UlLk8R_Ck=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_sVIG_S3TIzKjQdip_pzNsFZtksbPepgHlrnYvAVcnbaZ2UIv48qB6_Sux7rku0mObvrBdhjDuYlPW_bLkpV_tBklVCnG2szaRIGo_n6E0UQqIdgXOeZMl70_lgRdoTQGHjGM-hTaNd30KpGFD7eCCp2IIUH73RfQZ0oR4sdHEL7U-yt6drXvXRlyYrl9IzbdWk=s0-d)
and if
,
Now,![\frac{{x - 1}}{{[x] - \{ x\} }} \ge 0](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_t6BIPKbIpg6ynXmhpasZ8t3G045D75CQ51QZNuRnBQuaTq9VCaaPpzZmLF_ppdINQ7XYYIEJ6SvtOOZnjI3_OQ0EgCPlbZ480nIznT75-tL9ugzhAXR_LBza_m4w7QRyWHMxiEQ5KzK9qyta-TZKmMRq5vK_0pAowT0miZI8wRwJnPgkhn-2wRnbilv5Wng1i44NAwXj34hZCnOBw7uum___5VYk9eBh0tJr52xlbd1H6iH6gmn-UlLk8R_Ck=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