A) \[x-y\ne 0\]
B) \[x=-y\]
C) \[x=y\]
D) \[x\ne 0,y\ne 0\]
Correct Answer: C
Solution :
\[\because \sin \theta \le 1\] Since, \[{{\sin }^{2}}\theta \le 1\] \[\Rightarrow \] \[\frac{4xy}{{{(x+y)}^{2}}}\le 1\] \[\Rightarrow \] \[0\le {{(x+y)}^{2}}-4xy\] \[\Rightarrow \] \[{{x}^{2}}+{{y}^{2}}+2xy-4xy\ge 0\] \[\Rightarrow \] \[{{(x-y)}^{2}}\ge 0\] which is true for all real \[x\] and \[y\] provided \[x+y\ne 0\], otherwise \[\frac{4xy}{{{(x+y)}^{2}}}\] will be meaningless.
You need to login to perform this action.
You will be redirected in
3 sec
