Find the area under the graph
[y]=sin[x] for x ε [-100,100]
1is it then 29 ?
as Î /2 is an integer for all multiples of 7 and there are 29 multiples of 7 in [-100,100] ?
62No varun :)
not again :P
can u guys give more logic to weigh ur answers so that we can then find some flaws.?
1[y] = sin[x] ε [-100,100]
[y] can have values -1,0,1.
[y] = -1 => y [-1,0)
[y] = 0 => y [ 0,1)
[y] = 1 => y [ 1,2)
Now x ε [100,100] in terms of rad ..
therefore x ε [-5727,5727] in terms of degrees.
[y] can have those values only when [x] is a multiple of 90..
therefore the number of multiples of 90 in [-5727,5727] is 127.
x can also vary ~ 1 unit. therefore the area is 127 * 1 * 1 = 127 ..
Lol that's what I did ...
62varun ur guess was very good..
tell me one point (x,y) where
[y]=-1 and it satisfies the above equation [y]=sin[x]
:P
for all the values that u have found
1(-Î /2 , -1) ..
or is it that -Î /2 is not an integer so [x] can't be -Î /2 ?
62lol.. no dude.. not yet :)
may be u could try to draw graph :)
btw the right answer has already been written on this page
33lol.........
what i was trying was just other's mistake of yesterday.....y=[sin[x]]
Actually i thought the same question can't be today also...
So i thought y=[sin[x]]... Next time will read questions carefully.....
but solvingy=[sin[x]] was a gr8 fun and its answer came 98....
too lengthy.....for that interval....[-100,100]
[3][3][3][3][3][3][3][3][3][3][3][3][3][3][3][3][3][3][3][3][3][3][3]
62lol... no worries :)
i only hope that u guys are really benefiting from this section :)
33But that 98 took so much counting......Oh dear where sin[x] is negative was so tedious but this section is very very useful.....
The best section...
62The Solution:
The LHS is [y] so LHS is always an Integer!
RHS takes values sin[x] between x=-100,100, ..
[x] can take values -100,-99, -98..... 0,1,2,..........99
sin[x] can take values sin(-100) , sin(-99), sin(-98)........ sin(0), sin(1).... sin(99)
of these, only sin (0) is an integer!
so for all values of x other than those for which [x]=0 we have no solution at all!!!
Now we have a block of square from (0,0) to (1,1) as the solution...
For the whole detail of the question visit day before yesterday's question of the day!
13th november (Edited)
1we know that the maxm value of sine is 1 which is at 90 and which can be the only possible value of y except zero. so the area must be 2
am i right nishant?
62No.. i dont see any of these as the answers :)
Give me some more values(or methods) :)
62This is an extreme googly...
Lets see how many of u crack it..
if you really understood the "question of the day" yesterday.. this should be much easier! :)
Lol.. priyam.. i still want to know what ":)" means :D
62No varun and priyam.. neither..
tell me ur methods... in a bit more detail.!
