Daily Graph 23-03-09

bhiya is dis right?

2n\pi <2\sqrt{x}<(2n+1)\pi

n\pi <\sqrt{x}<\frac{(2n+1)\pi }{2}
squaring

n^{2}\pi^{2} <x<\frac{(2n+1)^{2}\pi^{2} }{4}

region
keeps on increasing

will
post graph quickly

sir see post 14

Sir, I totally Agree wid MAK's SOLUTION,

but as a matter of 1st sight idea, the area wer sin n cos both r -ve will b shaded totally!!

Is dat crrct way of thinkin......

and weat MAK told abt other cases is quiote precise

Sir : ISnt this rite :

pura ka pura 3rd QUADRANT excpet Î and 3Î /2....... seeing frm the fact dat sin²x<|sinx|

as |sinx| < 1

ok nishnt bhaiya can u give IIT-jee problems on chemistry , not seen many problems of iit on chem..

not useful : [2]
0.5* sin(2√x)*(1 + sin√x + cos √x + 0.5* sin(2√x)) > 0

easier wud b :

2 cases :

1) both brackets > 0
2) both brackets < 0

as |sinx| > sin²x excpet wen sinx = 1,0,-1

MATLAB pura ka pura 3rd QUADRANT excpet Î and 3Î /2....... as -ve * -ve = +ve

ohh nishant bhaiya remove that graph, its so funny and even wrong, so i had deleted it[17]

OH YAH!!!!!!

I did pretty much this but then cancelled it!!! [2] [17]

Nice one MAK

=> sin√x.cos√x.(1+sin√x + cos√x + sin√x.cos√x) > 0

=> sin2√x.(1+sin√x + cos√x + sin√x.cos√x) > 0 =>

sin2√x.(1+sin√x).(1+cos√x) > 0

=> sin2√x > 0 ....... since the other two are always greater than zero)

solving dis gives nΠ< √x < (2n+1)Π/2 ..... n>0

also d solution set would be excluding d points where sin√x and cos√x = -1...

let me know if i'm wrong anywhere...!!!