A) \[f'g=g'f\]
B) \[fg=f'g'\]
C) \[f'g'+fg=0\]
D) \[f'g+g'f=0\]
Correct Answer: A
Solution :
According to the figure, \[OA+{O}'A=OO'\] \[\sqrt{{{g}^{2}}+{{f}^{2}}}+\sqrt{f{{'}^{2}}+g{{'}^{2}}}=\sqrt{{{(g'-g)}^{2}}+{{(f'-f)}^{2}}}\] \[\Rightarrow {{g}^{2}}+{{f}^{2}}+f{{'}^{2}}+g{{'}^{2}}+2\sqrt{{{g}^{2}}+{{f}^{2}}}\times \sqrt{f{{'}^{2}}+g{{'}^{2}}}\] \[={{(g'-g)}^{2}}+{{(f'-f)}^{2}}\] \[\Rightarrow 2\sqrt{{{g}^{2}}+{{f}^{2}}}\sqrt{f{{'}^{2}}+g{{'}^{2}}}=-2(gg'+ff')\] \[\Rightarrow {{g}^{2}}f{{'}^{2}}+{{f}^{2}}g{{'}^{2}}=2gg'ff'\]. \[\therefore {{(gf'-fg')}^{2}}=0\Rightarrow gf'=fg'\].You need to login to perform this action.
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