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Ask iit jee aieee pet cbse icse state board community Community Discussion Question: solve the following
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shajikoshy (0)

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Q 1 How many different values are taken by the expression [x] + [2x] + [5x/3] + [3x]+ [4x] for real x in the range 0 < x < 100? [ ] denotes the greatest integer function.
 
Q2. Let a and b be positive real numbers for which
60a = 3 and 60b = 5. Without the use of a calculator or
of logarithms, determine the value of
12[(1-a-b)/{2(1-b)}]
 
    
divalli_oct07 (156)

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can u make ur ques a bit clr???

Bad news is that time always flies,
Good news is that u r the pilot.

yesterday is history,
tomorrow is a mystery,
today is a gift and that is why it's called "the present".
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Greatdreams (3083)

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-------------------------------------------------------------------------------------------------------------------
 
Edited......thanks for pointing my mistake but please provide a solution..

__________________________________________________________________________________________________________
From J.R.R. Tolkien's 'The Lord of the Rings':

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Not all who wander are lost
The old that is strong does not wither,
Deep roots are not reached by frost.
From ashes a fire shall be woken
From shadows a light shall spring
Renewed shall be blade that's broken
The crown less again shall be king.
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elastiboysai (2327)

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Great dreams
your working out is clearly wrong!!
chec urself by plugging in values.
[2x]=2[x]????
plug in x= 0.5
its clearly wrong.
simlarly for all others
 
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vivsarda (151)

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yup u r 100 % rite mr elstiboysai
It can be only taken wen x is an integer
even if x is an integer u cnt take (5/3x)=5/3 x
(X) denotes GIF of x
HOPE HELPS
RATE IF USEFUL!!
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elastiboysai (2327)

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heres da soln 4 da 2nd ques
a ln60=ln3
bln60=ln5
a=ln3/ln12+ln5
b=ln5/ln12+ln5
plug in the values in da expression
1-a-b/2(1-b)
[(1-ln15)/(ln12+ln5)]/[2ln12/(ln12+ln5)]
simplif.
ln12-ln3/2ln12
0.5-log123
so 12 power the abuv is sqrt(12)/3
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elastiboysai (2327)

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well
shajikoshy ji,
for the 1st question its not clear if 0 and 100 are included or not. i've done the problem including them.
[x]+[2x]+[5x/3]+[3x]+[4x]
write it as 35x/3 -{x}-{2x}-{5x/3}-{3x}-{4x}
now the fractional part terms become zero at all integral multiples of 1/coeff.of x
i.e
{x} = 0 at integers
{2x}=0 at integral multiples of1/2
{5x/3}=0 at integral multiples of 3/5
{3x}=0 at integral mult of 1/3
{4x}=0 at integr. multiples of 1/4
 
and importantly*** all the fractional parts are zero at
lcm of 3/5,1/2,1,1/3,1/4 wich is 3.
so total different values =
100[2+5/3+1+3+4]=1166
but we are addin integral points 4times so-(100*3)
subtr. integr pts for 5/3.= -33
u can skip 2 as well so-(100*2) [mult of 1/4 will take care of dem]
and - subtr pts were all change simultaneously = -33
well the ans turns abt to be 600
 
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shajikoshy (0)

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Q 1 How many different values are taken by the expression [x] + [2x] + [5x/3] + [3x]+ [4x] for real x in the range 0 < x <100 ? [ ] denotes the greatest integer function.
 
Sorry for typing error.........
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karthik_18049200 (47)

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anmswer for 2nd Q is 2
try by taking logarithms
cheers
 
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