10≤π{x}<π
so possible values of x are
all integers and all odd integers divided by 2
sin π{x}+cos π{x}=1
Where the angle here is in radians....
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14 Answers
11is that correct sir....??
as this situation is only possible when
π{x}=0 or π{x}=π/2
1π{x}=0 or π{x}=π/2
at only these two point the value of the given function will be 1
so π{x} is 0 at all integral points and at origin only
and π{x} is π/2 only when {x}=1/2 and thats only possible when we divide all odd integers by 2
24did something...
it will be either perfect or disaster :P
√2(sin(Ï€4+Ï€{x})=1
=>(sin(Ï€4+Ï€{x})=1/√2
take sin-1
=> π4+π{x}= π4
=>Ï€{x}=0
=>{x}=0
1sir what abt my solution....??????
sir what have i done wrong...????
106sin π{x}+cos π{x}=1
=> √2 cos(Ï€{x} - Ï€/4) = 1
=> π{x} - π/4 = 2nπ ± π/4
=> {x} = 2n or 2n + 1/2
As {x} = [0,1) So, {x} = 0 or 1/2
=> x = K or K+1/2 where K is an integer
24my mistake was that I didnt write general soln [3]
removing hte hide feature now ..[1]
1so my solution was rite too but i just thought about it in a logical way and didnt give any equations..........
1357yes yours was right in this case, but avoid doing that, because there might be some cases where you will lose some roots.
go by solving the equations only