Inverse trigonometric function

Question:
If !!x,y,z∈[-1,1]!! such that
!!Sin^(-1)x + Sin^(-1)y + Sin^(-1)z!! !!=!! !!3(pi)/2!! Then
!!(x^2010 + y^2011 + z^2012)!! !!-9/(x^2010 + y^2011 + z^2012)!! !!=!! !!?!!

  • !!-1!!
  • !!1!!
  • !!0!!
  • !!None!!

Solution:

Here it is being given that,
!!-1≤x≤1!! so !!-pi/2≤sin^(-1)x≤pi/2!! Similarly
!!-1≤y≤1!! !!-pi/2≤sin^(-1)y≤pi/2!!

!!-1≤z≤1!! !!-pi/2≤sin^(-1)z≤pi/2!!

Adding all the three equations,we get;
!!-3pi/2≤Sin^(-1)x + Sin^(-1)y + Sin^(-1)z≤3pi/2!!

By comparison,we get;
!!x=y=z=1!! !!(x^2010 + y^2011 + z^2012) -9/(x^2010 + y^2011 + z^2012)!! !!=!! !!(3-9/3)!! !!=!! !!0!! So !!(c)!! is the correct answer.

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