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_sMV8cis48vBMl00YYMfv3Y9TFWLaEkymo3tOW1bfsi2EDyJmlRbjlpUwc2PDtsBUEtlKAH8PkK-aeBHbHcI-T-iaJPf8VX2wKyhYzlItwMmH2W0YoL_3D-4h5PWBxgBjOIprpRXS3DsBuI85mXrehuIeWl4KiBKfXRyQ4voa9S657bFFUNQ5j-wi6irzMkiyfOggRmj10Tc8g7WOPDxpRHBzrQLMxbnRx5ntmu08t7sQpun63PlGMioy_3GcKbBOCMhzpygsDADkW7zUCpiiH8N4ApzOU9nu3DiMwRXmIOx5Kk6ndi5Iq1X-8G2a7tYTmTa0vPSUlY2C2NVGVQLKBPnt7Nk8sodBlgU-YhN8Evk-e9Cxi_1C8IAOhjeKuAZpyTLdXZX0IqA0dUZMxRUt-aQC7RTeKBdZK-aiqzeIjIGvxCW6AqOhNlQzYvIsW89hk9SbE87U9nqYUBXJd7ruH6Gqk1v-ClU5p17v15NGI=s0-d)
![= [x] + \frac{{2000\{ x\} }}{{2000}}](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_s9u3tYYevdL_D5oL5dhEhl2YYwP8RFTPxvEAD3ANVg0tVMcFYuF3bIjfYd-Y1STBuBm3k5vj-PqHpMNFzbmSmhutEYdYbJswtdupXRIrxaY2a5DprYJYiWUgu_OJuSdamh3StuYdBfcXxftYTpEpHOSiVFDeot3xXW6J-fNJHbpTTKJHWYXhzhSPugJVotZqke1HeQ4TDb2r6TUPzjxssyGWKyMh3mpsdHORX7eZrONMxvTRsfVYkjRclvazFlAQ=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_tK4yCK85kWNB14ZG05MZKndG5vbgZYD-BwBYkrAlBRVZmEpbgfusNv6VGT9ywDWhHhD7o9QNxHAmqVrt_cjB1HqsMMKvQQ1Wf-yM8Iz-iF20E8BJ4H1rJf2EuQBX-sPEYh67fu55Kh8TVmTlXzvi-lF5UlaEGSTaMRS-of9o21fucqLDZ0DCHW467I3SO6vxAFvPO00XTCAJBugKIunJH3bWqtcIr5kIKUvZUr_w=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_ukI2suuTBEhIifbeZTzFcJ_EO-SQOkVIxWWCbI45YOj8Ja_xSuzanQ8nSOd9BgbOKFQkzjWEPMffPh1jxvpokrn_FeZJzfhP5TLz6s5pJjUtG_u5gL280KpPFPVjqwd2PYLhVkyKJ9-Av-_lKVFB9O1SOCl6QcQpXi5f6qy1wulbibFha_6Fovmn2v0HGlmHYsnN1q=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_tajSfiWS4Y9BEWUHJJDws_7QKkYp0-FK3PjIkceJ7JfmTRAha2_q_CC6u7P36jIgyU6BVOwoKKqzJbKA7OcMnfQdLv-uZ8VVcdS9NDBrDWQLbizS8naUntTq7rlmj2vkZfoPQ9-FlaBY61dkil2HNfwMI16eNvUHJc94ZoSFT_MwWWJCr5NFhrA-f_bLHOP3DbabbfXojHSs3YKlzIX_RMSUC2o04rQewkEvmGWFDX9X21EeiqOHZjeEbXHCM=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_tCXcCu-rptgE7KoEY4wwTnNUMOoEmuSJTk1wsfzb46ucsE9vsRGWKfXFmSeHEeTPQxfdnl-TFQNvovpvgxVMiBJJ6nzIgmvWbyE-Ls7GYQMYM4w1i9MYuqTn4lCpA0tZuhzMKnApS9IKlISXcMipvrYqoql92JKSfNN2Mc1gKuZge-pEEphkiHe7Aip0wEHCyl=s0-d)
and if
,
Now,![\frac{{x - 1}}{{[x] - \{ x\} }} \ge 0](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_tajSfiWS4Y9BEWUHJJDws_7QKkYp0-FK3PjIkceJ7JfmTRAha2_q_CC6u7P36jIgyU6BVOwoKKqzJbKA7OcMnfQdLv-uZ8VVcdS9NDBrDWQLbizS8naUntTq7rlmj2vkZfoPQ9-FlaBY61dkil2HNfwMI16eNvUHJc94ZoSFT_MwWWJCr5NFhrA-f_bLHOP3DbabbfXojHSs3YKlzIX_RMSUC2o04rQewkEvmGWFDX9X21EeiqOHZjeEbXHCM=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