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_sepI7t1gbXftTjdfdXgSfOduBw6jY7XrWC5ki4kMFVH-MXnCpdJKhgjzruUrO_zgt0ARSFhiYSSDThpPAJ0srg4X87RwqEd7pCRWge0sTlVkfe9H-eS84xTLhxYfAzXEqLXFUf5_GX9TW9Fri-G49oDXJTHI5HInrlHP-gYEcqGl0QpJpZcCvFNad2iw7zpr3sGMw687KootRjMVjz0RAQrKBouqCkSZTFwk5jxKPlYrCpY6fxFzk-nAgiY88fuvPraNzjF7-XLlDJlq8B3bpByFoqc6ytXY9wPxby7kfvYrJcDMOrH7mRtZiGOtNGHkSJ30BZuPR3lo_8kla49EC9hWMRAfkaStMvDLMmZyugYQBBuM5NXTBnm2xdbHLamZnTDPF2fAsQAijUpVToNw3rAzgmHMEgTHmoobDRXXSnq9fOuiVK2UaypxR5BlVWZlSjR178agvGP5jqJjR9siORgFLbdEqz0DKnajxynTM=s0-d)
![= [x] + \frac{{2000\{ x\} }}{{2000}}](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_vJwXJ5Qt7QOqWFEnwsUBALdz8Plvy56wWUHOfkZpWpwCfWv7qF3BMl73TURJs0pjoyZaPGoM-uG6Ff_9cgUlG01kVjhuaKCTDGPGt1zpW6KZe2-1FV8CmTWtCQrUB6LKQ1BrslGLnlOPTVoQ5-iFCVAguF7Bjj8ML-6V9jfp4dmGqDUQhTOgoxCwvWOloAtQS_6VbTgMaKftXEN6pierFVbt5YjBQZomdr3ojA5Jakxl9ey8VHL1OCJSwW9ZIJ4w=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_tbYk14Z6xZRyNwAuviSRXrN_9gYXPDtNfMMY8q5OQRTOq-wzXd-ApWz-HJ9SyjF1eENROYaEBVBdXzlBgi4lW4GgwXosiKci5NMB7SabKDL8kFPKTmW0CLWLNNR9s8jQgxp-NnqKwA7EPMAW7KW84XrviLUFaLWnSJvfoiRLrfTeA-FWdsuhLfxJMHtl_Uw0k3di6eIWeHxoqsUZw9jCKU32uTkIT5lJ-lJjlplA=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_u2LqpK7AFKHXt_tEKd-1MiLM-b60EZr6w9oa_VEfmvFoW9xfYu9198xfNGb-KzPcgzRz_lQz5npD5hvuXHPUfyihhcYOtOI12k57lXL-tDZuIF8EOC0D8zb7VfbGlwFta1vjuaMKyrQcjLojkrRR5QbXLn6a6arQET91NsXlEECXeZ7MzrGCRt9403nK4XoluIEY9C=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_vk-PLgZ14BfMgAy4ILJbBEuDI6dTllF8z9GlOFCY8kFrKil6xtb07Hmdme2it9jqyBrpqilHU6yVRx-kuI46HNk2S31-Ww7jWk0rKo0UauPhjAUwwK-ViVQqLRcsjCalvsTi3R63_zYzLZX-uoooH49uKpIlM6xPzGMLN-caxVCCJ4jtdVv_YV8QaVNPoietpaA0dVFfjq-ySBPzxo8Uum576Z8RWIoi2nTCRkm-owAvSk9f3RWIJA6qLa0MQ=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_tE5gnq5JNkkwCUQvFhbBQBHjSuqIf-tn3yv1pl1DvWeVfUzBF5ReY9KfcCeXLxzWmvDLaiQ9kRUICGUvBZlEPnbzdZp1Il3GZyeu-D01tR39r1JDcXqso8MhOVY0CW_HfOW3OlvUB3mvNxm2NmvLcRdtN27ApepJzjLIn75wsLAKV9JBpvXsUe010q3TGab03G=s0-d)
and if
,
Now,![\frac{{x - 1}}{{[x] - \{ x\} }} \ge 0](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_vk-PLgZ14BfMgAy4ILJbBEuDI6dTllF8z9GlOFCY8kFrKil6xtb07Hmdme2it9jqyBrpqilHU6yVRx-kuI46HNk2S31-Ww7jWk0rKo0UauPhjAUwwK-ViVQqLRcsjCalvsTi3R63_zYzLZX-uoooH49uKpIlM6xPzGMLN-caxVCCJ4jtdVv_YV8QaVNPoietpaA0dVFfjq-ySBPzxo8Uum576Z8RWIoi2nTCRkm-owAvSk9f3RWIJA6qLa0MQ=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