Question

Let f(x) = (1+x^2)^1/2 then f (xy) = f(x) f(y)

Answer 1

$Given\\f(x)=(1+x^2)^{\frac{1}{2}}\\f(xy)=f(x).f(y)\\f(xy)=(1+(xy)^2)^{\frac{1}{2}}\\f(xy)=(1+x^2.y^2)^{\frac{1}{2}}\\f(x)=(1+x^2)^{\frac{1}{2}}\\f(y)=(1+y^2)^{\frac{1}{2}}\\Now\:consider\:\\f(xy)=f(x).f(y)\\(1+x^2.y^2)^{\frac{1}{2}}=f(x)=(1+x^2)^{\frac{1}{2}}.f(y)=(1+y^2)^{\frac{1}{2}}\\Squaring\:on\:both\:sides\:we\:get\\(1+x^2.y^2)=(1+x^2).(1+y^2)\:Since\:(1+x^2)>0\:and\:(1+y^2)>0\\(1+x^2.y^2)=(1+x^2+y^2+x^2.y^2)\\x^2+y^2=0\\OR\\\frac{x^2}{y^2}=-1$

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