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@Apu
Note for OP
+solved @user1 @user2... to close the thread when your doubt is solved. Mention the users who helped you solve the doubt. This will be added to their stats.,rotate
Bro spawned instantlyðŸ˜
Arey I remember my friends stuck on the same question lol
Oh ;-;
Consider x=k.pi when k is an integer. The function is zero.
and on other values?
Now consider k not being equal to k.pi.
Since [x+I]=[x]+I, the function becomes $\frac{x(sin(x)+tan(x))}{[\frac{x}{\pi}]+\frac{1}{2}}$
SirLancelotDuLac
Yea typing latex takes time lol.
yup i wrote the func like this only simplifying the denom
After this plug in -x. Since $\frac{x}{\pi}$ is surely not an integer we have $[\frac{-x}{\pi}]=-[\frac{x}{\pi}]-1$
After this you can just see by plugging in -x why it's odd.
Ahh i see
SirLancelotDuLac
the gint function bugs me everywhere ;-;
This is a new thing never thought like this
Thanks
+solved @SirLancelotDuLac
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